diff --git "a/data/gmat_hf_chunks.jsonl" "b/data/gmat_hf_chunks.jsonl" new file mode 100644--- /dev/null +++ "b/data/gmat_hf_chunks.jsonl" @@ -0,0 +1,1169 @@ +{"id": "GMAT Club Math Book 2024 v8_p1_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 1, "page_end": 1, "topic_guess": "general", "text": "GMAT Club \nMath Book \n8th Edition \nMay 2024"} +{"id": "GMAT Club Math Book 2024 v8_p2_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 2, "page_end": 2, "topic_guess": "word_problems", "text": "Table of Contents: \nNumber Theory ………………………………………………… 2 \nRemainders ..…………………………………………………….18 \nAlgebra …………………………………………………………….25 \nSequences and Progressions ……………………………… 33 \nFunctions / Coordinate Geometry ……………………….. 39 \nWord Problems………………………………………….……….59 \nWork Problems………………………………………….……….64 \nDistance Rate Time Problems……………………………….68 \nOverlapping Sets ……………………………………………….77 \nProbability ……………………………………………………….. 83 \nCombinations …………………………………………………… 91 \nStandard Deviation ……………………………………………. 96 \nThank you for using GMAT Club Math Book. This is not a beginner’s guide as \nthis work focuses on more advanced topics. Significant effort was invested by \nthe authors and long-time GMAT Club members into creating a detailed \noverview of some of the challenging and misunderstood topics. I hope this \nbook serves you well and big thanks to Bunuel, Walker, Shrouded1, and \nsriharimurthy! \n-BB, Founder of GMAT Club\nMay 4, 2024\nGMAT Club Math Book\n2"} +{"id": "GMAT Club Math Book 2024 v8_p3_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 3, "page_end": 3, "topic_guess": "number_theory", "text": "GMAT Club Math Book\n NUMBER THEORY\nDefinition\nNumber Theory is concerned with the properties of numbers in general, and in particular integers.\nAs this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics.\nGMAT Number Types\nGMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers.\nINTEGERS\nDefinition\nIntegers are defined as: all negative natural numbers , zero , and positive natural\nnumbers .\nNote that integers do not include decimals or fractions - just whole numbers.\nEven and Odd Numbers\nAn even number is an integer that is \"evenly divisible\" by 2, i.e., divisible by 2 without a remainder.\nAn even number is an integer of the form , where is an integer.\nAn odd number is an integer that is not evenly divisible by 2.\nAn odd number is an integer of the form , where is an integer.\nZero is an even number.\nAddition / Subtraction:\neven +/- even = even;\neven +/- odd = odd;\nodd +/- odd = even.\nMultiplication:\neven * even = even;\neven * odd = even;\nodd * odd = odd.\nDivision of two integers can result into an even/odd integer or a fraction.\nZERO:\n1. 0 is an integer.\n{. . . , −4, −3, −2, −1} {0}\n{1, 2, 3, 4, . . . }\nn = 2 k k\nn = 2 k+ 1 k\nGMAT Club Math Book\n3"} +{"id": "GMAT Club Math Book 2024 v8_p4_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 4, "page_end": 4, "topic_guess": "fractions_decimals_percents", "text": "2. 0 is an even integer. An even number is an integer that is \"evenly divisible\" by 2, i.e., divisible by 2\nwithout a remainder and as zero is evenly divisible by 2 then it must be even.\n3. 0 is neither positive nor negative integer (the only one of this kind).\n4. 0 is divisible by EVERY integer.\nIRRATIONAL NUMBERS\nFractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals\n(such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as non-\nterminating, non-repeating decimals are non-rational, so they are called the \"irrationals\". Examples would be\n (\"the square root of two\") or the number pi ( ~3.14159..., from geometry). The rationals and the\nirrationals are two totally separate number types: there is no overlap.\nPutting these two major classifications, the rationals and the irrationals, together in one set gives you the\n\"real\" numbers.\nPOSITIVE AND NEGATIVE NUMBERS\nA positive number is a real number that is greater than zero.\nA negative number is a real number that is smaller than zero.\nZero is not positive, nor negative.\nMultiplication:\npositive * positive = positive\npositive * negative = negative\nnegative * negative = positive\nDivision:\npositive / positive = positive\npositive / negative = negative\nnegative / negative = positive\n2√ π =\nGMAT Club Math Book\n4"} +{"id": "GMAT Club Math Book 2024 v8_p5_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 5, "page_end": 5, "topic_guess": "number_theory", "text": "Prime Numbers\nA Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise\na number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1.\nA number is prime if it cannot be written as a product of two factors and , both of which are greater\nthan 1: n = ab.\n• The first twenty-six prime numbers are:\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101\n• Note: only positive numbers can be primes.\n• There are infinitely many prime numbers.\n• The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest\nprime.\n• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are\nmultiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of\nthe form or , because all other numbers are divisible by 2 or 3.\n• Any nonzero natural number can be factored into primes, written as a product of primes or powers of\nprimes. Moreover, this factorization is unique except for a possible reordering of the factors.\n• Prime factorization: every positive integer greater than 1 can be written as a product of one or more\nprime integers in a way which is unique. For instance integer with three unique prime factors , , and \ncan be expressed as , where , , and are powers of , , and , respectively and are .\nExample: .\n• Verifying the primality (checking whether the number is a prime) of a given number can be done by\ntrial division, that is to say dividing by all integer numbers smaller than , thereby checking whether is\na multiple of .\nExample: Verifying the primality of : is little less than , from integers from to , is\ndivisible by , hence is not prime.\nNote that, it is only necessary to try dividing by prime numbers up to , since if n has any divisors at all\n(besides 1 and n), then it must have a prime divisor.\n• If is a positive integer greater than 1, then there is always a prime number with .\nn > 1 a b\n6n− 1 6n+ 1\nn\nn a b c\nn = ∗ ∗ap bq cr p q r a b c ≥ 1\n4200 = ∗ 3 ∗ ∗ 723 52\nn\nn n√ n\nm ≤ n√\n161 161−− −√ 13 2 13 161\n7 161\nn√\nn p n < p < 2 n\nGMAT Club Math Book\n5"} +{"id": "GMAT Club Math Book 2024 v8_p6_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 6, "page_end": 6, "topic_guess": "number_theory", "text": "Factors\nA divisor of an integer , also called a factor of , is an integer which evenly divides without leaving a\nremainder. In general, it is said is a factor of , for non-zero integers and , if there exists an integer \nsuch that .\n• 1 (and -1) are divisors of every integer.\n• Every integer is a divisor of itself.\n• Every integer is a divisor of 0, except, by convention, 0 itself.\n• Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd.\n• A positive divisor of n which is different from n is called a proper divisor.\n• An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a\nprime number is one which has exactly two factors: 1 and itself.\n• Any positive divisor of n is a product of prime divisors of n raised to some power.\n• If a number equals the sum of its proper divisors, it is said to be a perfect number.\nExample: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number.\nThere are some elementary rules:\n• If is a factor of and is a factor of , then is a factor of . In fact, is a factor of for\nall integers and .\n• If is a factor of and is a factor of , then is a factor of .\n• If is a factor of and is a factor of , then or .\n• If is a factor of , and , then a is a factor of .\n• If is a prime number and is a factor of then is a factor of or is a factor of .\nFinding the Number of Factors of an Integer\nFirst make prime factorization of an integer , where , , and are prime factors of and ,\n, and are their powers.\nThe number of factors of will be expressed by the formula . NOTE: this will include 1\nand n itself.\nExample: Finding the number of all factors of 450: \nTotal number of factors of 450 including 1 and 450 itself is \nfactors.\nGreatest Common Factor (Divisior) - GCF (GCD)\nThe greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common\nfactor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without\na remainder.\nTo find the GCF, you will need to do prime-factorization. Then, multiply the common factors (pick the lowest\nn n n\nm n m n k\nn = km\na b a c a (b+ c) a (mb+ nc)\nm n\na b b c a c\na b b a a = b a = − b\na bc gcd(a, b) = 1 c\np p ab p a p b\nn = �� ∗ap bq cr a b c n p\nq r\nn (p+ 1)(q+ 1)(r+ 1)\n450 = ∗ ∗21 32 52\n(1 + 1) ∗ (2 + 1) ∗ (2 + 1) = 2 ∗ 3 ∗ 3 = 18\nGMAT Club Math Book\n6"} +{"id": "GMAT Club Math Book 2024 v8_p7_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 7, "page_end": 7, "topic_guess": "number_theory", "text": "power of the common factors).\n• Every common divisor of a and b is a divisor of gcd(a, b).\n• a*b=gcd(a, b)*lcm(a, b)\nLowest Common Multiple - LCM\nThe lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a\nand b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be\ndivided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then\nlcm(a, b) is defined to be zero.\nTo find the LCM, you will need to do prime-factorization. Then multiply all the factors (pick the highest power\nof the common factors).\nPerfect Square\nA perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2,\nis an perfect square.\nThere are some tips about the perfect square:\n• The number of distinct factors of a perfect square is ALWAYS ODD.\n• The sum of distinct factors of a perfect square is ALWAYS ODD.\n• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.\n• Perfect square always has even number of powers of prime factors.\nDivisibility Rules\n2 - If the last digit is even, the number is divisible by 2.\n3 - If the sum of the digits is divisible by 3, the number is also.\n4 - If the last two digits form a number divisible by 4, the number is also.\n5 - If the last digit is a 5 or a 0, the number is divisible by 5.\n6 - If the number is divisible by both 3 and 2, it is also divisible by 6.\n7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7\n(including 0), then the number is divisible by 7.\n8 - If the last three digits of a number are divisible by 8, then so is the whole number.\n9 - If the sum of the digits is divisible by 9, so is the number.\n10 - If the number ends in 0, it is divisible by 10.\n11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by\n11, then the number is divisible by 11.\nExample: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the\nsum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.\n12 - If the number is divisible by both 3 and 4, it is also divisible by 12.\n25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.\nGMAT Club Math Book\n7"} +{"id": "GMAT Club Math Book 2024 v8_p8_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 8, "page_end": 8, "topic_guess": "number_theory", "text": "• Note: 0!=1.\n• Note: factorial of negative numbers is undefined.\nTrailing zeros:\nTrailing z\neros are a sequence of 0's in the decimal representation (or more generally, in any positional\nrepresentation) of a number, after which no other digits follow.\n125000 has 3 trailing zeros;\nThe number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer , can\nbe determined with this formula:\n, where k must be chosen such that .\nIt's easier if you look at an example:\nHow many zeros are in the end (after which no other digits follow) of ?\n (denominator must be less than 32, is less)\nHence, there are 7 zeros in the end of 32!\nThe formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this\nis equivalent to the number of factors 10, each of which gives one more trailing zero.\nFinding the number of powers of a prime number , in the .\nThe formula is:\n ... till \nWhat is the power of 2 in 25!?\nFinding the power of non-prime in n!:\nHow many powers of 900 are in 50!\nMake the prime factorization of the number: , then find the powers of these prime\nnumbers in the n!.\nFind the power of 2:\n= \nFind the power of 3:\n=\nFind the power of 5:\nn\nFactorials\nFactorial of a non-negative integer, denoted by n!, is the product of all positive integers less than or equal \nto n. E.g. .5! = 1 ∗ 2 ∗ 3 ∗ 4 ∗ 5\nn\n+ + +. . . +n\n5\nn\n52\nn\n53\nn\n5k ≤ n5k\n32!\n+ = 6 + 1 = 732\n5\n32\n52 = 2552\np n!\n+ +n\np\nn\np2\nn\np3 ≤ npx\n+ + + = 12 + 6 + 3 + 1 = 2225\n2\n25\n4\n25\n8\n25\n16\n900 = ∗ ∗22 32 52\n+ + + +50\n2\n50\n4\n50\n8\n50\n16\n50\n32 = 25 + 12 + 6 + 3 + 1 = 47\n247\n+ + = 16 + 5 + 1 = 2250\n3\n50\n9\n50\n27\n322\nGMAT Club Math Book\n8"} +{"id": "GMAT Club Math Book 2024 v8_p9_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 9, "page_end": 9, "topic_guess": "number_theory", "text": "=\nWe need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is\n900 in the power of 6 in 50!.\nConsecutive Integers\nConsecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1,\n0, 1, are consecutive integers.\n• Sum of consecutive integers equals the mean multiplied by the number of terms, . Given consecutive\nintegers , , (mean equals to the average of the first and last\nterms), so the sum equals to .\n• If n is odd, the sum of consecutive integers is always divisible by n. Given , we have \nconsecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.\n• If n is even, the sum of consecutive integers is never divisible by n. Given , we have \nconsecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.\n• The product of consecutive integers is always divisible by .\nGiven consecutive integers: . The product of 3*4*5*6 is 360, which is divisible by 4!=24.\nEvenly Spaced Set\nEvenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two\nsuccessive members of the sequence is a constant. The set of integers is an example of evenly\nspaced set. Set of consecutive integers is also an example of evenly spaced set.\n• If the first term is and the common difference of successive members is , then the term of the\nsequence is given by:\n• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated\nby the formula , where is the first term and is the last term. Given the set\n, .\n• The sum of the elements in any evenly spaced set is given by:\n, the mean multiplied by the number of terms. OR, \n• Special cases:\nSum of n first positive integers: \nSum of n first positive odd numbers: , where is the last, \nterm and given by: . Given first odd positive integers, then their sum equals to\n.\nSum of n first positive even numbers: , where is the last,\n term and given by: . Given first positive even integers, then their sum equals to\n.\n• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is\n+ = 10 + 2 = 1250\n5\n50\n25\n512\nn n\n{−3, −2, −1, 0, 1, 2} mean = = −−3+2\n2\n1\n2\n− ∗ 6 = −31\n2\n{9, 10, 11} n = 3\n{9, 10, 11, 12} n = 4\nn n!\nn = 4 {3, 4, 5, 6}\n{9, 13, 17, 21}\na1 d nth\n= + d(n− 1)an a1\nmean = median =\n+a1 an\n2 a1 an\n{7, 11, 15, 19} mean = median = = 137+19\n2\nSum = ∗ n\n+a1 an\n2 Sum = ∗ n\n2 + d(n−1)a1\n2\n1 + 2+. . . + n = ∗ n1+n\n2\n+ +. . . + = 1 + 3+. . . + =a1 a2 an an n2 an nth\n= 2 n− 1an n = 5\n1 + 3 + 5 + 7 + 9 = = 2552\n+ +. . . + = 2 + 4+. . . +a1 a2 an an= n(n+ 1) an\nnth = 2 nan n = 4\n2 + 4 + 6 + 8 = 4(4 + 1) = 20\nGMAT Club Math Book\n9"} +{"id": "GMAT Club Math Book 2024 v8_p10_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 10, "page_end": 10, "topic_guess": "fractions_decimals_percents", "text": "middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is\n13, so the sum is 13*5 =65.\nFRACTIONS\nDefinition\nFractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one\ninteger by another integer. Set of Fraction is a subset of the set of Rational Numbers.\nFraction can be expressed in two forms fractional representation and decimal representation .\nFractional representation\nFractional representation is a way to express numbers that fall in between integers (note that integers can\nalso be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a\nnumerator (the part) and a denominator (the whole).\n• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is\ncalled denominator. In the fraction, , 9 is the numerator and 7 is denominator.\n• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller\nthan the denominator. is a proper fraction.\n• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a\nmixed number. is improper fraction.\n• An integer combined with a proper fraction is called mixed number. is a mixed number. This can also be\nwritten as an improper fraction: \nConverting Improper Fractions\n• Converting Improper Fractions to Mixed Fractions:\n1. Divide the numerator by the denominator\n2. Write down the whole number answer\n3. Then write down any remainder above the denominator\nExample #1: Convert to a mixed fraction.\nSolution: Divide with a remainder of . Write down the and then write down the remainder above\nthe denominator , like this: \n• Converting Mixed Fractions to Improper Fractions:\n1. Multiply the whole number part by the fraction's denominator\n2. Add that to the numerator\n3. Then write the result on top of the denominator\nExample #2: Convert to an improper fraction.\nSolution: Multiply the whole number by the denominator: . Add the numerator to that:\n. Then write that down above the denominator, like this: \n( )m\nn (a. bcd)\n9\n7\n1\n3\n5\n2\n4 3\n5\n23\n5\n11\n4\n= 211\n4 3 2 3\n4 2 3\n4\n3 2\n5\n3 ∗ 5 = 15\n15 + 2 = 17 17\n5\nGMAT Club Math Book\n10"} +{"id": "GMAT Club Math Book 2024 v8_p11_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 11, "page_end": 11, "topic_guess": "fractions_decimals_percents", "text": "Reciprocal\nReciprocal for a number , denoted by or , is a number which when multiplied by yields . The\nreciprocal of a fraction is . To get the reciprocal of a number, divide 1 by the number. For example\nreciprocal of is , reciprocal of is .\nOperation on Fractions\n• Adding/Subtracting fractions:\nTo add/subtract fractions with the same denominator, add the numerators and place that sum over the\ncommon denominator.\nTo add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the\nfractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions\n• Multiplying fractions: To multiply fractions just place the product of the numerators over the product of\nthe denominators.\n• Dividing fractions: Change the divisor into its reciprocal and then multiply.\nExample #1: \nExample #2: Given , take the reciprocal of . The reciprocal is . Now multiply: .\nDecimal Representation\nThe decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating\n(recuring) decimals (such as 0.333333....).\nReduced fraction (meaning that fraction is already reduced to its lowest term) can be expressed as\nterminating decimal if and only (denominator) is of the form , where and are non-negative\nintegers. For example: is a terminating decimal , as (denominator) equals to . Fraction\n is also a terminating decimal, as and denominator .\nConverting Decimals to Fractions\n• To convert a terminating decimal to fraction:\n1. Calculate the total numbers after decimal point\n2. Remove the decimal point from the number\n3. Put 1 under the denominator and annex it with \"0\" as many as the total in step 1\n4. Reduce the fraction to its lowest terms\nExample: Convert to a fraction.\n1: Total number after decimal point is 2.\n2 and 3: .\n4: Reducing it to lowest terms: \n• To convert a recurring decimal to fraction:\n1. Separate the recurring number from the decimal fraction\n2. Annex denominator with \"9\" as many times as the length of the recurring number\n3. Reduce the fraction to its lowest terms\nExample #1: Convert to a fraction.\nx 1\nx x−1 x 1\na\nb\nb\na\n3 1\n3\n5\n6\n6\n5\n+ = + =3\n7\n2\n3\n9\n21\n14\n21\n23\n21\n3\n5\n2 2 1\n2 ∗ =3\n5\n1\n2\n3\n10\na\nb\nb 2n5m m n\n7\n250 0.028 250 2 ∗ 53\n3\n30 =3\n30\n1\n10 10 = 2 ∗ 5\n0.56\n56\n100\n=56\n100\n14\n25\n0.393939...\nGMAT Club Math Book\n11"} +{"id": "GMAT Club Math Book 2024 v8_p12_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 12, "page_end": 12, "topic_guess": "fractions_decimals_percents", "text": "1: The recurring number is .\n2: , the number is of length so we have added two nines.\n3: Reducing it to lowest terms: .\n• To convert a mixed-recurring decimal to fraction:\n1. Write down the number consisting with non-repeating digits and r\nepeating digits.\n2. Subtract non-repeating number from above.\n3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-\nrepeating digit write down a zero after 9's.\nExample #2: Convert to a fraction.\n1. The number consisting with non-repeating digits and repeating digits is 2512;\n2. Subtract 25 (non-repeating number) from above: 2512-25=2487;\n3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25):\n2487/9900=829/3300.\nRounding\nRounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places,\nand if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4\nor smaller, round down (keep the same) the last digit that you keep.\nExample:\n5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.\n5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.\n5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.\nRatios and Proportions\nGiven that , where a, b, c and d are non-zero real numbers, we can deduce other proportions by\nsimple Algebra. These results are often referred to by the names mentioned along each of the properties\nobtained.\n- invertendo\n- alternendo\n- componendo\n- dividendo\n- componendo & dividendo\n39\n39\n99 39 2\n=39\n99\n13\n33\n0.2512(12)\n=a\nb\nc\nd\n=b\na\nd\nc\n=a\nc\nb\nd\n=a+b\nb\nc+d\nd\n=a−b\nb\nc−d\nd\n=a+b\na−b\nc+d\nc−d\nGMAT Club Math Book\n12"} +{"id": "GMAT Club Math Book 2024 v8_p13_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 13, "page_end": 13, "topic_guess": "word_problems", "text": "EXPONENTS\nExponents are a \"shortcut\" method of showing a number that was multiplied by itself several times. For\ninstance, number multiplied times can be written as , where represents the base, the number that is\nmultiplied by itself times and represents the exponent. The exponent indicates how many times to\nmultiple the base, , by itself.\nExponents one and zero:\n Any nonzero number to the power of 0 is 1.\nFor example: and \n• Note: the case of 0^0 is not tested on the GMAT.\n Any number to the power 1 is itself.\nPowers o\nf zero:\nIf the exponent is positive, the power of zero is zero: , where .\nIf the exponent is negative, the power of zero ( , where ) is undefined, because division by zero is\nimplied.\nPowers of one:\n The integer powers of one are one.\nNegative powers:\nPowers of minus one:\nIf n is an even integer, then .\nIf n is an odd integer, then .\nOperations involving the same exponents:\nKeep the exponent, multiply or divide the bases\n and not \nOperations involving the same bases:\nKeep the base, add or subtract the exponent (add for multiplication, subtract for division)\nFraction as power:\nExponential Equations:\nWhen solving equations with even exponents, we must consider both positive and negative possibilities for the\nsolutions.\na n an a\nn n\na\n= 1a0\n= 150 (−3 = 1)0\n= aa1\n= 00n n > 0\n0n n < 0\n= 11n\n=a−n 1\nan\n(−1 = 1)n\n(−1 = −1)n\n∗ = ( aban bn )n\n= (an\nbn\na\nb )n\n( =am)n amn\n=amn\na( )mn\n(am)n\n∗ =an am an+m\n=an\nam an−m\n=a\n1\nn a√n\n=a\nm\nn am−− −√n\nGMAT Club Math Book\n13"} +{"id": "GMAT Club Math Book 2024 v8_p14_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 14, "page_end": 14, "topic_guess": "sequences_patterns", "text": "For instance , the two possible solutions are and .\nWhen solving equations with odd exponents, we'll have only one solution.\nFor instance for , solution is and for , solution is .\nExponents and divisibility:\n is ALWAYS divisible by .\n is divisible by if is even.\n is divisible by if is odd, and not divisible by a+b if n is even.\nLAST DIGIT OF A PRODUCT\nLast digits of a product of integers are last digits of the product of last digits of these integers.\nFor instance last 2 digits of 845*9512*408*613 would be the last 2 digits of\n45*12*8*13=540*104=40*4=160=60\nExample: The last digit of 85945*89*58307=5*9*7=45*7=35=5?\nLAST DIGIT OF A POWER\nDetermining the last digit of :\n1. Last digit of is the same as that of ;\n2. Determine the cyclicity number of ;\n3. Find the remainder when divided by the cyclisity;\n4. When , then last digit of is the same as that of and when , then last digit of is\nthe same as that of , where is the cyclisity number.\n• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.\n• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.\n• Integers ending with 4 (eg. ) have a cyclisity of 2. When n is odd will end with 4 and when n\nis even will end with 6.\n• Integers ending with 9 (eg. ) have a cyclisity of 2. When n is odd will end with 9 and when n\nis even will end with 1.\nExample: What is the last digit of ?\nSolution: Last digit of is the same as that of . Now we should determine the cyclisity of :\n1. 7^1=7 (last digit is 7)\n2. 7^2=9 (last digit is 9)\n3. 7^3=3 (last digit is 3)\n4. 7^4=1 (last digit is 1)\n5. 7^5=7 (last digit is 7 again!)\n...\nSo, the cyclisity of 7 is 4.\nNow divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of is the same as that of the\nlast digit of , is the same as that of the last digit of , which is .\nROOTS\nRoots (or radicals) are the \"opposite\" operation of applying exponents. For instance x^2=16 and square root\nof 16=4.\n= 25a2 5 −5\n= 8a3 a = 2 = −8a3 a = −2\n−an bn a− b\n−an bn a+ b n\n+an bn a+ b n\nn n n\n(xyz)n\n(xyz)n zn\nc z\nr n\nr > 0 (xyz)n zr r = 0 (xyz)n\nzc c\n(xy4)n (xy4)n\n(xy4)n\n(xy9)n (xy9)n\n(xy9)n\n12739\n12739 739 7\n12739\n739 73 3\nGMAT Club Math Book\n14"} +{"id": "GMAT Club Math Book 2024 v8_p15_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 15, "page_end": 15, "topic_guess": "fractions_decimals_percents", "text": "General rules:\n• and .\n• \n• \n• \n• \n• , when , then and when , then \n• When the GMAT provides the square root sign for an even root, such as or , then the only accepted\nanswer is the positive root.\nThat is, , NOT +5 or -5. In contrast, the equation has TWO solutions, +5 and -5. Even\nroots have only a positive value on the GMAT.\n• Odd roots will have the same sign as the base of the root. For example, and .\n• For GMAT it's good to memorize following values:\nPERCENTS\nA percentage is a way of expressing a number as a fraction of 100 (per cent meaning \"per hundred\"). It is\noften denoted using the percent sign, \"%\", or the abbreviation \"pct\". Since a percent is an amount per 100,\npercents can be represented as fractions with a denominator of 100. For example, 25% means 25 per 100,\n25/100 and 350% means 350 per 100, 350/100.\n• A percent can be represented as a decimal. The following relationship characterizes how percents and\ndecimals interact. Percent Form / 100 = Decimal Form\nFor example: What is 2% represented as a decimal?\nPercent Form / 100 = Decimal Form: 2%/100=0.02\n• Percent change\nGeneral formula for percent increase or decrease, (percent change):\nExample: A company received $2 million in royalties on the first $10 million in sales and then $8 million in\nroyalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from\nthe first $10 million in sales to the next $100 million in sales?\nSolution: Percent decrease can be calculated by the formula above:\n=x√ y√ xy− −√ =\nx√\ny√\nx\ny\n− −\n√\n( =x√ )n xn− −√\n=x\n1\nn x√n\n=x\nn\nm xn− −√m\n+ ≠a√ b√ a+ b− − − −√\n= |x|x2− −√ x ≤ 0 = − xx2− −√ x ≥ 0 = xx2− −√\nx√ x√4\n= 525− −√ = 25x2\n= 5125−− −√3 = −4−64− − − −√3\n≈ 1.412√\n≈ 1.733√\n≈ 2.245√\n≈ 2.456√\n≈ 2.657√\n≈ 2.838√\n≈ 3.1610− −√\nPercent = ∗ 100Change\nOriginal\nGMAT Club Math Book\n15"} +{"id": "GMAT Club Math Book 2024 v8_p16_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 16, "page_end": 16, "topic_guess": "word_problems", "text": ", so the royalties decreased by 60%.\n• Simple Interest\nSimple interest = principal * interest rate * time\nExample: If $15,000 is invested at 10% simple annual interest, how much interest is earned after 9 months?\nSolution: $15,000*0.1*9/12 = $1125\n• Compound Interest\n, where C = the number of times compounded annually.\nIf C=1, meaning that interest is compounded once a year, then the formula will be:\n, where time is number of years.\nExample: If $20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2\nyear?\nSolution: \nORDER OF OPERATIONS - PEMDAS\nPerform the operations inside a Parenthesis first (absolute value signs also fall into this category), then\nExponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to\nright - PEMDAS.\nSpecial cases:\n• An exclamation mark indicates that one should compute the factorial of the term immediately to its left,\nbefore computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But \nmeans while ; a factorial in an exponent applies to the exponent, while a factorial not in\nthe exponent applies to the entire power.\n• If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:\n and not"} +{"id": "GMAT Club Math Book 2024 v8_p16_c2", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 16, "page_end": 16, "topic_guess": "word_problems", "text": "n\nthe exponent applies to the entire power.\n• If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:\n and not\n\nPercent = ∗ 100 =Change\nOriginal\n= ∗ 100 = 60\n−2\n10\n8\n100\n2\n10\nBalance(final) = principal∗ (1 + interest\nC )time∗C\nBalance(final) = principal∗ (1 + interest)time\nBalance = 20, 000 ∗ (1 + = 20, 000 ∗ (1.03 = $25, 335.40.12\n4 )2∗4 )8\n!32\n( )! = 9!32 =25! 2120\n=amn\na( )mn\n(am)n\nGMAT Club Math Book\n16"} +{"id": "GMAT Club Math Book 2024 v8_p17_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 17, "page_end": 17, "topic_guess": "number_theory", "text": "PRACTICE QUESTIONS\nEasy:\n1. https://gmatclub.com/forum/if-n-is-a-pr ... 22490.html\n2. https://gmatclub.com/forum/if-x-is-an-i ... 68233.html\n3. https://gmatclub.com/forum/how-many-two ... 91802.html\n4. https://gmatclub.com/forum/if-n-is-an-i ... 96084.html\n5. https://gmatclub.com/forum/root-80-root-139868.html\n6. https://gmatclub.com/forum/if-x-4-and-y ... 28707.html\n7. https://gmatclub.com/forum/4th-root-of- ... 59975.html\n8. https://gmatclub.com/forum/what-is-the- ... 03027.html\n9. https://gmatclub.com/forum/if-n-is-the- ... 28102.html\n10. https://gmatclub.com/forum/if-x-and-y-a ... 41958.html\nMedium:\n1. https://gmatclub.com/forum/how-many-int ... 10744.html\n2. https://gmatclub.com/forum/topic-143744.html\n3. https://gmatclub.com/forum/if-the-sum-o ... 10512.html\n4. https://gmatclub.com/forum/amy-s-grade- ... 98164.html\n5. https://gmatclub.com/forum/if-3-4-2-3-5 ... l#p1526295\n6. https://gmatclub.com/forum/what-is-63908.html\n7. https://gmatclub.com/forum/the-sum-of-t ... 32773.html\n8. https://gmatclub.com/forum/how-many-of- ... 00708.html\n9. https://gmatclub.com/forum/which-of-the ... 25989.html\n10. https://gmatclub.com/forum/what-is-the- ... 07096.html\nHard:\n1. https://gmatclub.com/forum/for-every-po ... 12521.html\n2. https://gmatclub.com/forum/how-many-int ... 99697.html\n3. https://gmatclub.com/forum/when-the-pos ... 72713.html\n4. https://gmatclub.com/forum/if-x-a-and-b ... 01946.html\n5. https://gmatclub.com/forum/the-function ... 03852.html\n6. https://gmatclub.com/forum/for-prime-nu ... 98508.html\n7. https://gmatclub.com/forum/how-many-pos ... 56052.html\n8. https://gmatclub.com/forum/if-each-expr ... 87153.html\n9. https://gmatclub.com/forum/what-is-the- ... 51159.html\n10. https://gmatclub.com/forum/sqrt-7-sqrt- ... 68800.html"} +{"id": "GMAT Club Math Book 2024 v8_p17_c2", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 17, "page_end": 17, "topic_guess": "general", "text": "forum/if-each-expr ... 87153.html\n9. https://gmatclub.com/forum/what-is-the- ... 51159.html\n10. https://gmatclub.com/forum/sqrt-7-sqrt- ... 68800.html\n\nGMAT Club Math Book\n17"} +{"id": "GMAT Club Math Book 2024 v8_p18_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 18, "page_end": 18, "topic_guess": "number_theory", "text": "GMAT Club Forum\nRemainders\nhttps://gmatclub.com/forum/remainders-144665.html\n REMAINDERS\ncreated by: Bunuel\nDefinition\nIf and are positive integers, there exist unique integers and , called the\nquotient and remainder, respectively, such that\n and .\nFor example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since\n.\nNotice that means that remainder is a non-negative integer and always less\nthan divisor.\nThis formula can also be written as .\nProperties\nWhen is divided by the remainder is 0 if is a multiple of .\nFor example, 12 divided by 3 yields the remainder of 0 since 12 is a multiple of 3\nand .\nWhen a smaller integer is divided by a larger integer, the quotient is 0 and the\nremainder is the smaller integer.\nFor example, 7 divided by 11 has the quotient 0 and the remainder 7 since\nThe possible remainders when positive integer is divided by positive integer can\nrange from 0 to .\nFor example, possible remainders when positive integer is divided by 5 can range\nfrom 0 (when y is a multiple of 5) to 4 (when y is one less than a multiple of 5).\nx y q r\ny = divisor∗ quotient+ remainder = xq+ r 0 ≤ r < x\n15 = 6 ∗ 2 + 3\n0 ≤ r < x\n= q+y\nx\nr\nx\ny x y x\n12 = 3 ∗ 4 + 0\n7 = 11 ∗ 0 + 7\ny x\nx− 1\ny\nGMAT Club Math Book\n18"} +{"id": "GMAT Club Math Book 2024 v8_p19_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 19, "page_end": 19, "topic_guess": "number_theory", "text": "If a number is divided by 10, its remainder is the last digit of that number. If it is\ndivided by 100 then the remainder is the last two digits and so on.\nFor example, 123 divided by 10 has the remainder 3 and 123 divided by 100 has\nthe remainder of 23.\nExample #1 (easy)\nIf the remainder is 7 when positive integer n is divided by 18, what is the\nremainder when n is divided by 6?\nA. 0\nB. 1\nC. 2\nD. 3\nE. 4\nWhen positive integer n is dived by 18 the remainder is 7: .\nNow, since the first term (18q) is divisible by 6, then the remainder will only be from the\nsecond term, which is 7. 7 divided by 6 yields the remainder of 1.\nAnswer: B. Discuss this question \nHERE.\nExample #2 (easy)\nIf n is a prime number greater than 3, what is the remainder when n^2 is\ndivided by 12 ?\nA. 0\nB. 1\nC. 2\nD. 3\nE. 5\nThere are several algebraic ways to solve this question, but the easiest way is as follows:\nsince we cannot have two correct answers just pick a prime greater than 3, square it and\nsee what would be the remainder upon division of it by 12.\nIf , then . The remainder upon division 25 by 12 is 1.\nAnswer: B. Discuss this question \nHERE.\nn = 18q + 7\nn = 5 = 25n2\nGMAT Club Math Book\n19"} +{"id": "GMAT Club Math Book 2024 v8_p20_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 20, "page_end": 20, "topic_guess": "statistics", "text": "Example #3 (easy)\nWhat is the tens digit of positive integer x ?\n(1) x divided by 100 has a remainder of 30.\n(2) x divided by 110 has a remainder of 30.\n(1) x divided by 100 has a remainder of 30. We have that : 30, 130, 230,\n... as you can see every such number has 3 as the tens digit. Sufficient.\n(2) x divided by 110 has a remainder of 30. We have that : 30, 140, 250,\n360, ... so, there are more than 1 value of the tens digit possible. Not sufficient.\nAnswer: A. Discuss this question \nHERE.\nExample #4 (easy)\nWhat is the remainder when the positive integer n is divided by 6?\n(1) n is multiple of 5\n(2) n is a multiple of 12\n(1) n is multiple of 5. If n=5, then n yields the remainder of 5 when divided by 6 but if\nn=10, then n yields the remainder of 4 when divided by 6. We already have two different\nanswers, which means that this statement is not sufficient.\n(2) n is a multiple of 12. Every multiple of 12 is also a multiple of 6, thus n divided by 6\nyields the remainder of 0. Sufficient.\nAnswer: B. Discuss this question \nHERE.\nExample #5 (medium)\nIf s and t are positive integers such that s/t = 64.12, which of the following\ncould be the remainder when s is divided by t ?\nA. 2\nB. 4\nC. 8\nD. 20\nE. 45\n divided by yields the remainder of can always be expressed as: \n(which is the same as ), where is the quotient and is the remainder.\nGiven that , so according to the above ,\nwhich means that must be a multiple of 3. Only option E offers answer which is a\nmultiple of 3\nAnswer. E. Discuss this question \nHERE.\nx = 100q + 30\nx = 110p + 30\ns t r = q+s\nt\nr\nt\ns = qt+ r q r\n= 64.12 = 64 = 64 = 64 +s\nt\n12\n100\n3\n25\n3\n25 =r\nt\n3\n25\nr\nGMAT Club Math Book\n20"} +{"id": "GMAT Club Math Book 2024 v8_p21_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 21, "page_end": 21, "topic_guess": "number_theory", "text": "Example #6 (medium)\nPositive integer n leaves a remainder of 4 after division by 6 and a remainder of\n3 after division by 5. If n is greater than 30, what is the remainder that n leaves\nafter division by 30?\nA. 3\nB. 12\nC. 18\nD. 22\nE. 28\nPositive integer n leaves a remainder of 4 after division by 6: . Thus n could\nbe: 4, 10, 16, 22, 28, ...\nPositive integer n leaves a remainder of 3 after division by 5: . Thus n could\nbe: 3, 8, 13, 18, 23, 28, ...\nThere is a way to derive general formula for (of a type , where is\na divisor and is a remainder) based on above two statements:\nDivisor would be the least common multiple of above two divisors 5 and 6, hence\n.\nRemainder would be the first common integer in above two patterns, hence .\nTherefore general formula based on both statements is . Hence the\nremainder when positive integer n is divided by 30 is 28.\nAnswer. E. Discuss this question \nHERE.\nExample #7 (medium)\nIf x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8?\n(1) When 3x is divided by 2, there is a remainder.\n(2) x = 4y + 1, where y is an integer.\n, notice that we have the product of three\nconsecutive integers. Now, notice that if , then and are consecutive\neven integers, thus one of them will also be divisible by 4, which will make\n divisible by 2*4=8 (basically if then will be\ndivisible by 8*3=24).\n(1) When 3x is divided by 2, there is a remainder. This implies that , which\nmeans that . Therefore is divisible by 8. Sufficient.\n(2) x = 4y + 1, where y is an integer. We have that , thus\n is divisible by 8. Sufficient.\nn = 6 p+ 4\nn = 5 q+ 3\nn n = mx+ r x\nr\nx\nx = 30\nr r = 28\nn = 30m + 28\n− x = x( − 1) = ( x− 1)x(x+ 1)x3 x2\nx = odd x− 1 x+ 1\n(x− 1)(x+ 1) x = odd (x− 1)x(x+ 1)\n3x = odd\nx = odd (x− 1)x(x+ 1)\nx = even+ odd = odd\n(x− 1)x(x+ 1)\nGMAT Club Math Book\n21"} +{"id": "GMAT Club Math Book 2024 v8_p22_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 22, "page_end": 22, "topic_guess": "algebra", "text": "Answer: D. Discuss this question HERE.\nExample #8 (hard)\nWhen 51^25 is divided by 13, the remainder obtained is:\nA. 12\nB. 10\nC. 2\nD. 1\nE. 0\n, now if we expand this expression all terms but the last one will have\n in them, thus will leave no remainder upon division by 13, the last term will\nbe . Thus the question becomes: what is the remainder upon division -1 by\n13? The answer to this question is 12: .\nAnswer: A. Discuss this question \nHERE.\nExample #9 (hard)\nWhen positive integer x is divided by 5, the remainder is 3; and when x is\ndivided by 7, the remainder is 4. When positive integer y is divided by 5, the\nremainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of\nthe following must be a factor of x - y?\nA. 12\nB. 15\nC. 20\nD. 28\nE. 35\nWhen the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively:\n (x could be 3, 8, 13, 18, 23, ...) and (x could be 4, 11, 18, 25,\n...).\nWe can derive general formula based on above two statements the same way as\nfor the example above:\nDivisor will be the least common multiple of above two divisors 5 and 7, hence 35.\nRemainder will be the first common integer in above two patterns, hence 18. So, to satisfy\nboth this conditions x must be of a type (18, 53, 88, ...);\nThe same for y (as the same info is given about y): ;\n. Thus must be a multiple of 35.\n= (52 − 15125 )25\n52 = 13 ∗ 4\n(−1 = −1)25\n−1 = 13 ∗ (−1) + 12\nx = 5 q+ 3 x = 7 p+ 4\nx = 35m + 18\ny = 35n + 18\nx− y = (35m + 18) − (35n + 18) = 35( m− n) x− y\nGMAT Club Math Book\n22"} +{"id": "GMAT Club Math Book 2024 v8_p23_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 23, "page_end": 23, "topic_guess": "number_theory", "text": "Answer: E. Discuss this question HERE.\nExample #10 (hard)\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by\n4?\n(1) When p is divided by 8, the remainder is 5\n(2) x – y = 3\n(1) When p is divided by 8, the remainder is 5. This implies that .\nSince given that , then -->\n.\nSo, . Now, if then\n and if then\n, so in any case\n --> --> in order to be multiple of 4 must be\nmultiple of 16 but as we see it's not, so is not multiple of 4. Sufficient.\n(2) x – y = 3 --> --> but not sufficient to say whether it's\nmultiple of 4.\nAnswer: A. Discuss this question \nHERE.\nExample #11 (hard)\n and are positive integers. Is the remainder of bigger than the\nremainder of ?\n(1) .\n(2) The remainder of is 2\nFirst of all any positive integer can yield only three remainders upon division by 3: 0, 1, or\n2.\nSince, the sum of the digits of and is always 1 then the remainders of \nand are only dependent on the value of the number added to and .\nThere are 3 cases:\nIf the number added to them is: 0, 3, 6, 9, ... then the remainder will be 1 (as the sum of\nthe digits of and will be 1 more than a multiple of 3);\nIf the number added to them is: 1, 4, 7, 10, ... then the remainder will be 2 (as the sum\nof the digits of and will be 2 more than a multiple of 3);\nIf the number added to them is: 2, 5, 8, 11, ... then the remainder will be 0 (as the sum\nof the digits of and will be a multiple of 3).\n(1) . Not sufficient.\np = 8 q+ 5 = +x2 y2\ny = odd = 2 k+ 1 8q+ 5 = + (2 k+ 1x2 )2\n= 8 q+ 4 − 4 − 4 k = 4(2 q+ 1 − − k)x2 k2 k2\n= 4(2 q+ 1 − − k)x2 k2 k = odd\n2q+ 1 − − k = even+ odd− odd− odd = oddk2 k = even\n2q+ 1 − − k = even+ odd− even− even = oddk2\n2q+ 1 − − k = oddk2 = 4 ∗ oddx2 x x2\nx\nx− odd = 3 x = even\nm n +n10m\n3\n+m10n\n3\nm > n\nn\n3\n10m 10n +n10m\n3\n+m10n\n3 10m 10n\n10m 10n\n10m 10n\n10m 10n\nm > n\nGMAT Club Math Book\n23"} +{"id": "GMAT Club Math Book 2024 v8_p24_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 24, "page_end": 24, "topic_guess": "number_theory", "text": "(2) The remainder of is --> is: 2, 5, 8, 11, ... so we have the third case. Which\nmeans that the remainder of is 0. Now, the question asks whether the remainder\nof , which is 0, greater than the reminder of , which is 0, 1, or 2.\nObviously it cannot be greater, it can be less than or equal to. So, the answer to the\nquestion is NO. Sufficient.\nAnswer: B. Discuss this question HERE.\nCheck more PS questions on remainders here \nn\n3 2 n\n+n10m\n3\n+n10m\n3\n+m10n\n3\nGMAT Club Math Book\n24"} +{"id": "GMAT Club Math Book 2024 v8_p25_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 25, "page_end": 25, "topic_guess": "algebra", "text": "GMAT Club Math Book\nAlgebra\nhttps://gmatclub.com/forum/algebra-101576.html\n Algebra 101\nScope\nManipulation of various algebraic expressions\nEquations in 1 & more variables\nDealing with non-linear equations\nAlgebraic identities\nNotation & Assumptions\nIn this document, lower case roman alphabets will be used to denote variables such as a,b,c,x,y,z,w\nIn general it is assumed that the GMAT will only deal with real numbers ( ) or subsets of such as Integers ( ),\nrational numbers ( ) etc\nConcept of variables\nA variable is a place holder, which can be used in mathematical expressions. They are most often used for two\npurposes :\n(a) In Algebraic Equations : To represent unknown quantities in known relationships. For eg : \"Mary's age is 10\nmore than twice that of Jim's\", we can represent the unknown \"Mary's age\" by x and \"Jim's age\" by y and then\nthe known relationship is \n(b) In Algebraic Identities : These are generalized relationships such as , which says for any\nnumber, if you square it and take the root, you get the absolute value back. So the variable acts like a true\nplaceholder, which may be replaced by any number.\nBasic rules of manipulation\nA. When switching terms from one side to the other in an algebraic expression + becomes - and vice versa.\nEg. \nB. When switching terms from one side to the other in an algebraic expression * becomes / and vice versa.\nEg. \nC. you can add/subtract/multiply/divide both sides by the same amount. Eg. \nD. you can take to the exponent or bring from the exponent as long as the base is the same.\nEg 1.\nEg 2.\nIt is important to note that all the operations above are possible not just with constants but also with variables\nthemselves. So you can \"add x\" or \"multiply with y\" on both sides while maintaining the expression. But what you\nneed to be very careful about is when dividing both sides by a variable. When you divide both sides by a\nvariable (or do operations like \"canceling x on both sides\") you implicitly assume that the variable\ncannot be equal to 0, as division by 0 is undefined. This is a concept shows up very often on GMAT\nR R Z\nQ\nx = 2 y+ 10\n= |x|x2− −√\nx+ y = 2 z ⇒ x = 2 z− y\n4 ∗ x = ( y+ 1 ⇒ x =)2 (y+1)2\n4\nx+ y = 10z ⇒ =x+y\n43\n10z\n43\n+ 2 = z ⇒ =x2 4 +2x2 4z\n= ⇒ = ⇒ 4 x = 3 y24x 8y 24x 23y\nGMAT Club Math Book\n25"} +{"id": "GMAT Club Math Book 2024 v8_p26_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 26, "page_end": 26, "topic_guess": "algebra", "text": "questions.\nDegree of an expression\nThe degree of an algebraic expression is defined as the highest power of the variables present in the expression.\nDegree 1 : Linear\nDegree 2 : Quadratic\nDegree 3 : Cubic\nDegree 4 : Bi-quadratic\nEgs : the degree is 1\n the degree is 3\n the degree of x is 3, degree of z is 5, degree of the expression is 5\nSolving equations of degree 1 : LINEAR\nDegree 1 equations or linear equations are equations in one or more variable such that degree of each variable is\none. Let us consider some special cases of linear equations :\nOne variable\nSuch equations will always have a solution. General form is and solution is \nOne equation in Two variables\nThis is not enough to determine x and y uniquely. There can be infinitely many solutions.\nTwo equations in Two variables\nIf you have a linear equation in 2 variables, you need at least 2 equations to solve for both variables. The general\nform is :\nIf then there are infinite solutions. Any point satisfying one equation will always satisfy\nthe second\nIf then there is no such x and y which will satisfy both equations. No solution\nIn all other cases, solving the equations is straight forward, multiply eq (2) by a/d and subtract from (1).\nMore than two equations in Two variables\nPick any 2 equations and try to solve them :\nCase 1 : No solution --> Then there is no solution for bigger set\nCase 2 : Unique solution --> Substitute in other equations to see if the solution works for all others\nCase 3 : Infinite solutions --> Out of the 2 equations you picked, replace any one with an un-picked equation and\nrepeat.\nMore than 2 variables\nThis is not a case that will be encountered often on the GMAT. But in general for n variables you will need at least\nn equations to get a unique solution. Sometimes you can assign unique values to a subset of variables using less\nthan n equations using a small trick. For example consider the equations :\nIn this case you can treat as a single variable to get :\nThese can be solved to get x=0 and 2y+5z=20\nThere is a common misconception that you need n equations to solve n variables. This is not true.\nx+ y\n+ x+ 2x3\n+x3 z5\nax = b x = ( b/a)\nax+ by = c\ndx+ ey = f\n(a/d) = ( b/e) = ( c/f)\n(a/d) = ( b/e) ≠ ( c/f)\nx+ 2y+ 5z = 20\nx+ 4y+ 10z = 40\n2y+ z\nx+ (2y+ 5z) = 20\nx+ 2 ∗ (2 y+ 5z) = 40\nGMAT Club Math Book\n26"} +{"id": "GMAT Club Math Book 2024 v8_p27_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 27, "page_end": 27, "topic_guess": "algebra", "text": "Solving equations of degree 2 : QUADRATIC\nThe general form of a quadratic equation is \nThe equation has no solution if \nThe equation has exactly one solution if \nThis equation has 2 solutions given by if \nThe sum of roots is \nThe product of roots is \nIf the roots are and , the equation can be written as \nA quick way to solve a quadratic, without the above formula is to factorize it :\nStep 1> Divide throughout by coeff of x^2 to put it in the form \nStep 2> Sum of roots = -d and Product = e. Search for 2 numbers which satisfy this criteria, let them be f,g\nStep 3> The equation may be re-written as (x-f)(x-g)=0. And the solutions are f,g\nEg. \nThe sum is -11 and the product is 30. So numbers are -5,-6\nSolvingequations with DEGREE>2\nYou will never be asked to solved higher degree equations, except in some cases where using simple tricks these\nequations can either be factorized or be reduced to a lower degree or both. What you need to note is that an\nequation of degree n has at most n unique solutions.\nFactorization\nThis is the easiest approach to solving higher degree equations. Though there is no general rule to do this,\ngenerally a knowledge of algebraic identities helps. The basic idea is that if you can write an equation in the form\n:\nwhere each of A,B,C are algebraic expressions. Once this is done, the solution is obtained by equating each of\nA,B,C to 0 one by one.\nEg. \nSo the solution is x=0,-5,-6\nReducing to lower degree\nThis is useful sometimes when it is easy to see that a simple variable substitution can reduce the degree.\nEg. \nHere let \na + bx+ c = 0x2\n< 4 acb2\n= 4 acb2\n−b± −4acb2− − − − − −√\n2a > 4 acb2\n−b\na\nc\na\nr1 r2 (x− )( x− ) = 0r1 r2\n+ dx+ e = 0x2\n+ 11x + 30 = 0x2\n+ 11x + 30 = + 5 x+ 6x+ 30 = x(x+ 5) + 6( x+ 5) = ( x+ 5)(x+ 6)x2 x2\nA∗ B∗ C = 0\n+ 11 + 30x = 0x3 x2\nx∗ ( + 11x + 30) = 0x2\nx∗ (x+ 5) ∗ ( x+ 6) = 0\n− 3 + 2 = 0x6 x3\ny = x3\n− 3y+ 2 = 0y2\n(y− 2)(y− 1) = 0\nGMAT Club Math Book\n27"} +{"id": "GMAT Club Math Book 2024 v8_p28_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 28, "page_end": 28, "topic_guess": "number_theory", "text": "So the solution is y=1 or 2 or x^3=1 or 2 or x=1 or \nOther tricks\nSometimes we are given conditions such as the variables being integers which make the solutions much easier to\nfind. When we know that the solutions are integral, often times solutions are easy to find using just brute force.\nEg. and we know a,b are integers such that a=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and\nthe most popular approach too (see below).\n3-steps approach:\nGeneral approach to solving equalities and inequalities with absolute value:\n1. Open modulus and set conditions.\nTo solve/open a modulus, you need to consider 2 situations to find all roots:\nPositive (or rather non-negative)\nNegative\n|x|\n|3| = 3 | − 12| = 12 | − 1.3| = 1.3\n|x| ≥ 0\n|0| = 0\n| − x| = |x|\n|x| + |y| ≥ |x + y|\n|x| = x2− −√\nGMAT Club Math Book\n29"} +{"id": "GMAT Club Math Book 2024 v8_p30_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 30, "page_end": 30, "topic_guess": "algebra", "text": "For example, \na) Positive: if , we can rewrite the equation as: \nb) Negative: if , we can rewrite the equation as: \nWe can also think about conditions like graphics. is a key point in which the expression under modulus\nequals zero. All points right are the first condition and all points left are second condition .\n2. Solve new equations:\na) --> x=5\nb) --> x=-3\n3. Check conditions for each solution:\na) has to satisfy initial condition . . It satisfies. Otherwise, we would have to\nreject x=5.\nb) has to satisfy initial condition . . It satisfies. Otherwise, we would have\nto reject x=-3.\n3-steps approach for complex problems\nLet’s consider following examples,\nExample #1\nQ.: . How many solutions does the equation have?\nSolution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:\na) . --> . We reject the solution because our condition is not\nsatisfied (-1 is not less than -8)\nb) . --> . We reject the solution because our condition is\nnot satisfied (-15 is not within (-8,-3) interval.)\nc) . --> . We reject the solution because our condition is not\nsatisfied (9 is not within (-3,4) interval.)\nd) . --> . We reject the solution because our condition is not\nsatisfied (-1 is not more than 4)\n(Optional) The following illustration may help you understand how to open modulus at different conditions.\nAnswer: 0\nExample #2\nQ.: . What is x?\nSolution: There are 2 conditions:\na) --> or . --> . x e { , } and both solutions satisfy the\ncondition.\n|x− 1| = 4\n(x− 1) ≥ 0 x− 1 = 4\n(x− 1) < 0 −(x− 1) = 4\nx = 1\n(x > 1) (x < 1)\nx− 1 = 4\n−x+ 1 = 4\nx = 5 x− 1 >= 0 5 − 1 = 4 > 0\nx = −3 x− 1 < 0 −3 − 1 = −4 < 0\n|x+ 3| − |4 − x| = |8 + x|\nx < −8 −(x+ 3) − (4 − x) = −(8 + x) x = −1\n−8 ≤ x < −3 −(x+ 3) − (4 − x) = (8 + x) x = −15\n−3 ≤ x < 4 (x+ 3) − (4 − x) = (8 + x) x = 9\nx ≥ 4 (x+ 3) + (4 − x) = (8 + x) x = −1\n| − 4| = 1x2\n( − 4) ≥ 0x2 x ≤ −2 x ≥ 2 − 4 = 1x2 = 5x2 − 5√ 5√\nGMAT Club Math Book\n30"} +{"id": "GMAT Club Math Book 2024 v8_p31_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 31, "page_end": 31, "topic_guess": "word_problems", "text": "b) --> . --> . x e { , } and both solutions satisfy the\ncondition.\n(Optional) The following illustration may help you understand how to open modulus at different conditions.\nAnswer: , , , \nTip & Tricks\nThe 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow\nyou to solve absolute value problems in 10-20 sec.\nI. Thinking of inequality with modulus as a segment at the number line.\nFor example,\nProblem: 13 is equal to x e (-inf,-6)&(0,+inf)\nIII. Thinking about absolute values as the distance between points at the number line.\nFor example,\nProblem: A k >= 1 : \nSummation\nThe sum of an infinite AP can never be finite except if & \nThe general sum of a n term AP with common difference d is given by \nThe sum formula may be re-written as\n= + d = + ( n− 1)dan an−1 a1\nai\nd\na1\n− =ai ai−1\n=\n−ai ak\ni−k\n−aj ak\nj−k\n= 0a1 d = 0\n(2a+ (n− 1)d)n\n2\nn∗ Avg( , ) = ∗ ( FirstTerm + LastTerm)a1 an n\n2\nGMAT Club Math Book\n33"} +{"id": "GMAT Club Math Book 2024 v8_p34_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 34, "page_end": 34, "topic_guess": "sequences_patterns", "text": "Examples\n1. All odd positive integers : {1,3,5,7,...} \n2. All positive multiples of 23 : {23,46,69,92,...} \n3. All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...}\nGeometric Progressions\nDefinition\nIt is a special type of sequence in which the ratio of consequetive terms is constant\nGeneral Term\n is the ith term\n is the common ratio\n is the first term\nDefining Properties\nEach of the following is necessary & sufficient for a sequence to be an GP :\n Constant\nIf you pick any 3 consecutive terms, the middle one is the geometric mean of the\nother two\nFor all i,j > k >= 1 : \nSummation\nThe sum of an infinite GP will be finite if absolute value of r < 1\nThe general sum of a n term GP with common ratio r is given by \nIf an infinite GP is summable (|r|<1) then the sum is \nExamples\n1. All positive powers of 2 : {1,2,4,8,...} \n2. All negative powers of 4 : {1/4,1/16,1/64,1/256,...}\nHarmonic Progressions\nDefinition\nIt is a special type of sequence in which if you take the inverse of every term, this new\nsequence forms an HP\n= 1, d = 2a1\n= 23, d = 23a1\n= −0.1, d = −1a1\n= ∗ r = ∗bn bn−1 a1 rn−1\nbi\nr\nb1\n=\nbi\nbi−1\n( = (\nbi\nbk\n)j−k bj\nbk\n)i−k\n∗b1 −1rn\nr−1\nb1\n1−r\n= 1, r = 2b1\n= 1/4, r = 1/4, sum = = (1/3)b1\n1/4\n(1−1/4)\nGMAT Club Math Book\n34"} +{"id": "GMAT Club Math Book 2024 v8_p35_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 35, "page_end": 35, "topic_guess": "sequences_patterns", "text": "Important Properties\nOf any three consecutive terms of a HP, the middle one is always the harmonic mean of\nthe other two, where the harmonic mean (HM) is defined as :\nOr in other words :\nAPs, GPs, HPs : Linkage\nEach progression provides us a definition of \"mean\" :\nArithmetic Mean : OR \nGeometric Mean : OR \nHarmonic Mean : OR \nFor all non-negative real numbers : AM >= GM >= HM\nIn particular for 2 numbers : AM * HM = GM * GM\nExample :\nLet a=50 and b=2,\nthen the AM = (50+2)*0.5 = 26 ;\nthe GM = sqrt(50*2) = 10 ;\nthe HM = (2*50*2)/(52) = 3.85\nAM > GM > HM\nAM*HM = 100 = GM^2\nMisc Notes\nA subsequence (any set of consequutive terms) of an AP is an AP\nA subsequence (any set of consequutive terms) of a GP is a GP\nA subsequence (any set of consequutive terms) of a HP is a HP\nIf given an AP, and I pick out a subsequence from that AP, consisting of the terms\n such that are in AP then the new subsequence will also be an\nAP\nFor Example : Consider the AP with {1,3,5,7,9,11,...}, so a_n=1+2*(n-\n1)=2n-1\nPick out the subsequence of terms \nNew sequence is {9,19,29,...} which is an AP with and \nIf given a GP, and I pick out a subsequence from that GP, consisting of the terms\n such that are in AP then the new subsequence will also be a GP\n∗ ( + ) =1\n2\n1\na\n1\nb\n1\nHM(a,b)\nHM(a, b) = 2ab\na+b\na+b\n2\na1+..+an\nn\nab− −√ (a1∗. . ∗an)\n1n\n2ab\na+b\nn\n+..+1\na1\n1\nan\n, , , . . .ai1 ai2 ai3 i1, i2, i3\n= 1, d = 2a1\n, , , . . .a5 a10 a15\n= 9a1 d = 10\n, , , . . .bi1 bi2 bi3 i1, i2, i3\nGMAT Club Math Book\n35"} +{"id": "GMAT Club Math Book 2024 v8_p36_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 36, "page_end": 36, "topic_guess": "sequences_patterns", "text": "For Example : Consider the GP with {1,2,4,8,16,32,...}, so b_n=2^(n-1)\nPick out the subsequence of terms \nNew sequence is {4,16,64,...} which is a GP with and \nThe special sequence in which each term is the sum of previous two terms is known as\nthe fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1.\n{1,1,2,3,5,8,13,...}\nIn a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n\nis even and the middle term if n is odd. In either case this is also equal to the mean of the\nfirst and last terms\nSome examples\nExample 1\nA coin is tossed repeatedly till the result is a tails, what is the probability that the total\nnumber of tosses is less than or equal to 5 ?\nSolution\nP(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT)\nWe know that P(H)=P(T)=0.5\nSo Probability = 0.5 + 0.5^2 + ... + 0.5^5\nThis is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence :\nProbability = \nExample 2\nIn an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how\nmany terms are greater than 24 ?\n(1) a1 = 8\n(2) a12 = 24\nSolution\n(1) a1=8, does not tell us anything about the common difference, so impossible to say\nhow many terms are greater than 24\n(2) a12=24, and we know common difference is non-zero. So either all the terms below\na12 are greater than 24 and the terms above it less than 24 or the other way around. In\neither case, there are exactly 11 terms either side of a12. Sufficient\nAnswer is B\nExample 3\nFor positive integers a,b (a=GM>=HM, the solution is :\na <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b\n= 1, r = 2b1\n, , , . . .b2 b4 b6\n= 4b1 r = 4\n0.5 ∗ = ∗ =1−0.55\n1−0.5\n1\n2\n31\n32\n1\n2\n31\n32\nGMAT Club Math Book\n36"} +{"id": "GMAT Club Math Book 2024 v8_p37_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 37, "page_end": 37, "topic_guess": "sequences_patterns", "text": "Example 4\nFor every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by\n(-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is\na)greater than 2\nb)between 1 and 2\nc)between 1/2 and 1\nd)between 1/4 and 1/2\ne)less than 1/4.\nSolution\nThe sequence given has first term 1/2 and each subsequent term can be obtained by\nmultiplying with -1/2. So it is a GP. We can use the GP summation formula\n1023/1024 is very close to 1, so this sum is very close to 1/3\nAnswer is d\nExample 5\nThe sum of the fourth and twelfth term of an arithmetic progression is 20. What is the\nsum of the first 15 terms of the arithmetic progression?\nA. 300\nB. 120\nC. 150\nD. 170\nE. 270\nSolution\nNow we need the sum of first 15 terms, which is given by :\nAnswer is (c)\nS = b = ∗ = ∗1−rn\n1−r\n1\n2\n1−(−1/2)10\n1−(−1/2)\n1\n3\n1023\n1024\n+ 2 = 20a4 a1\n= + 3 d, 2 = + 11da4 a1 a1 a1\n2 + 14d = 20a1\n(2 + (15 − 1) d) = ∗ (2 + 14d ) = 15015\n2 a1 15\n2 a1\nGMAT Club Math Book\n37"} +{"id": "GMAT Club Math Book 2024 v8_p38_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 38, "page_end": 38, "topic_guess": "sequences_patterns", "text": "Practice Questions\nEasy:\n1. https://gmatclub.com/forum/the-infinite ... 00090.html\n2. https://gmatclub.com/forum/in-an-increa ... 43749.html\n3. https://gmatclub.com/forum/the-sequence ... 46630.html\n4. https://gmatclub.com/forum/the-sequence ... 34498.html\n5. https://gmatclub.com/forum/in-a-certain ... 44376.html\n6. https://gmatclub.com/forum/what-is-the- ... 44459.html\n7. https://gmatclub.com/forum/in-the-infin ... 93065.html\n8. https://gmatclub.com/forum/the-sequence ... 66830.html\n9. https://gmatclub.com/forum/in-the-seque ... 98247.html\n10. https://gmatclub.com/forum/for-any-sequ ... 28100.html\nMedium:\n1. https://gmatclub.com/forum/in-a-certain ... 19235.html\n2. https://gmatclub.com/forum/in-the-seque ... 68359.html\n3. https://gmatclub.com/forum/what-is-the- ... 30041.html\n4. https://gmatclub.com/forum/in-the-seque ... 98182.html\n5. https://gmatclub.com/forum/a-sequence-i ... 71548.html\n6. https://gmatclub.com/forum/the-sequence ... 60525.html\n7. https://gmatclub.com/forum/if-the-sum-o ... 98380.html\n8. https://gmatclub.com/forum/if-s1-s2-s3- ... 02927.html\n9. https://gmatclub.com/forum/around-the-w ... 16033.html\n10. https://gmatclub.com/forum/gmat-club-wo ... 95168.html\nHard:\n1. https://gmatclub.com/forum/a-sequence-o ... 20319.html\n2. https://gmatclub.com/forum/a-certain-mu ... -3940.html\n3. https://gmatclub.com/forum/the-sequence ... 27175.html\n4. https://gmatclub.com/forum/if-the-sum-o ... 98379.html\n5. https://gmatclub.com/forum/the-sum-of-n ... 00640.html\n6. https://gmatclub.com/forum/x-y-z-is-an- ... 98430.html\n7. https://gmatclub.com/forum/gmat-club-ol ... 67365.html\n8. https://gmatclub.com/forum/if-the-sum-o ... 98381.html\n9. https://gmatclub.com/forum/around-the-w ... 16261.html\n10. https://gmatclub.com/forum/around-the-w ... 16134.html\nGMAT Club Math Book\n38"} +{"id": "GMAT Club Math Book 2024 v8_p39_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 39, "page_end": 39, "topic_guess": "coordinate_geometry", "text": "GMAT Club Math Book\nAlgebra Functions & Coordinate Geometry\nhttps://gmatclub.com/forum/algebra-functions-87652.html\nFUNCTIONS & COORDINATE GEOMETRY\nDefinition\nCoordinate geometry, or Cartesian geometry, is the study of geometry using a coordinate\nsystem and the principles of algebra and analysis.\nThe Coordinate Plane\nIn coordinate geometry, points are placed on the \"coordinate plane\" as shown below. The\ncoordinate plane is a two-dimensional surface on which we can plot points, lines and\ncurves. It has two scales, called the x-axis and y-axis, at right angles to each other. The\nplural of axis is 'axes' (pronounced \"AXE-ease\").\nA point's location on the plane is given by two numbers, one that tells where it is on the\nx-axis and another which tells where it is on the y-axis. Together, they define a single,\nunique position on the plane. So in the diagram above, the point A has an x value of 20\nand a y value of 15. These are the coordinates of the point A, sometimes referred to as its\nGMAT Club Math Book\n39"} +{"id": "GMAT Club Math Book 2024 v8_p40_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 40, "page_end": 40, "topic_guess": "coordinate_geometry", "text": "\"rectangular coordinates\".\nX axis\nThe horizontal scale is called the x-axis and is usually drawn with the zero point in the\nmiddle. Values to the right are positive and those to the left are negative.\nY axis\nThe vertical scale is called the y-axis and is also usually drawn with the zero point in the\nmiddle. Values above the origin are positive and those below are negative.\nOrigin\nThe point where the two axes cross (at zero on both scales) is called the origin.\nQuadrants\nWhen the origin is in the center of the plane, they divide it into four areas called\nquadrants.\nThe first quadrant, by convention, is the top right, and then they go around counter-\nclockwise. In the diagram above they are labeled Quadrant 1, 2 etc. It is conventional to\nlabel them with numerals but we talk about them as \"first, second, third, and fourth\nquadrant\".\nPoint (x,y)\nThe coordinates are written as an \"ordered pair\". The letter P is simply the name of the\npoint and is used to distinguish it from others.\nThe two numbers in parentheses are the x and y coordinate of the point. The first number\n(x) specifies how far along the x (horizontal) axis the point is. The second is the y\ncoordinate and specifies how far up or down the y axis to go. It is called an ordered pair\nbecause the order of the two numbers matters - the first is always the x (horizontal)\ncoordinate.\nThe sign of the coordinate is important. A positive number means to go to the right (x) or\nGMAT Club Math Book\n40"} +{"id": "GMAT Club Math Book 2024 v8_p41_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 41, "page_end": 41, "topic_guess": "coordinate_geometry", "text": "up (y). Negative numbers mean to go left (x) or down (y).\nDistance between two points\nGiven coordinates of two points, distance D between two points is given by:\n (where is the difference between the x-coordinates and is the\ndifference between the y-coordinates of the points)\nAs you can see, the distance formula on the plane is derived from the Pythagorean\ntheorem.\nAbove formula can be written in the following way for given two points and\n:\nVertical and horizontal lines\nIf the line segment is exactly vertical or horizontal, the formula above will still work fine,\nbut there is an easier way. For a horizontal line, its length is the difference between the x-\ncoordinates. For a vertical line its length is the difference between the y-coordinates.\nDistance between the point A (x,y) and the origin\nAs the one point is origin with coordinate O (0,0) the formula can be simplified to:\nExample #1\nQ: Find the distance between the point A (3,-1) and B (-1,2)\nSolution: Substituting values in the equation we'll get\nD = d + dx2 y2− − − − − − − −√ dx dy\n( , )x1 y1\n( , )x2 y2\nD = ( − + ( −x2 x1)2 y2 y1)2− −− − − − − − − − − − − − − − − − −\n√\nD = +x2 y2− − − − − −√\nD = ( − + ( −x2 x1)2 y2 y1)2− −− − − − − − − − − − − − − − − − −√\nGMAT Club Math Book\n41"} +{"id": "GMAT Club Math Book 2024 v8_p42_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 42, "page_end": 42, "topic_guess": "coordinate_geometry", "text": "Midpoint of a Line Segment\nA line segment on the coordinate plane is defined by two endpoints whose coordinates are\nknown. The midpoint of this line is exactly halfway between these endpoints and it's\nlocation can be found using the Midpoint Theorem, which states:\n• The x-coordinate of the midpoint is the average of the x-coordinates of the two\nendpoints.\n• Likewise, the y-coordinate is the average of the y-coordinates of the endpoints.\nCoordinates of the midpoint of the line segment AB, ( and\n) are and \nLines in Coordinate Geometry\nIn Euclidean geometry, a line is a straight curve. In coordinate geometry, lines in a\nCartesian plane can be described algebraically by linear equations and linear functions.\nEvery straight line in the plane can represented by a first degree equation with two\nvariables.\nD = = = 5(−1 − 3 + (2 − (−1))2 )2− −− − − − − − − − − − − − − − − − −\n√ 16 + 9− − − − −√\nM( , )xm ym A( , )x1 y1\nB( , )x2 y2 =xm\n+x1 x2\n2 =ym\n+y1 y2\n2\nGMAT Club Math Book\n42"} +{"id": "GMAT Club Math Book 2024 v8_p43_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 43, "page_end": 43, "topic_guess": "coordinate_geometry", "text": "There are several approaches commonly used in coordinate geometry. It does not matter\nwhether we are talking about a line, ray or line segment. In all cases any of the below\nmethods will provide enough information to define the line exactly.\n1. General form.\nThe general form of the equation of a straight line is\nWhere , and are arbitrary constants. This form includes all other forms as special\ncases. For an equation in this form the slope is and the y intercept is .\n2. Point-intercept form.\nWhere: is the slope of the line; is the y-intercept of the line; is the independent\nvariable of the function .\n3. Using two points\nIn figure below, a line is defined by the two points A and B. By providing the coordinates\nof the two points, we can draw a line. No other line could pass through both these points\nand so the line they define is unique.\nax+ by+ c = 0\na b c\n− a\nb − c\nb\ny = mx+ b\nm b x\ny\nGMAT Club Math Book\n43"} +{"id": "GMAT Club Math Book 2024 v8_p44_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 44, "page_end": 44, "topic_guess": "algebra", "text": "The equation of a straight line passing through points and is:\nExample #1\nQ: Find the equation of a line passing through the points A (17,4) and B (9,9).\nSolution: Substituting the values in equation we'll get: \n --> --> OR if we want to write\nthe equation in the slope-intercept form: \n4. Using one point and the slope\nSometimes on the GMAT you will be given a point on the line and its slope and from this\ninformation you will need to find the equation or check if this line goes through another\npoint. You can think of the slope as the direction of the line. So once you know that a line\ngoes through a certain point, and which direction it is pointing, you have defined one\nunique line.\nIn figure below, we see a line passing through the point A at (14,23). We also see that it's\nslope is +2 (which means it goes 2 up for every one across). With these two facts we can\nestablish a unique line.\n( , )P1 x1 y1 ( , )P2 x2 y2\n=\ny−y1\nx−x1\n−y1 y2\n−x1 x2\n=\ny−y1\nx−x1\n−y1 y2\n−x1 x2 =y−4\nx−17\n4−9\n17−9\n=y−4\nx−17\n−5\n8 8y− 32 = −5 x+ 85 8y+ 5x− 117 = 0\ny = − x+5\n8\n117\n8\nGMAT Club Math Book\n44"} +{"id": "GMAT Club Math Book 2024 v8_p45_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 45, "page_end": 45, "topic_guess": "algebra", "text": "The equation of a straight line that passes through a point with a\nslope m is:\nExample #2\nQ: Find the equation of a line passing through the point A (14,23) and the slope 2.\nSolution: Substituting the values in equation we'll get\n --> \n4. Intercept form.\nThe equation of a straight line whose x and y intercepts are a and b, respectively, is:\nExample #3\nQ: Find the equation of a line whose x intercept is 5 and y intercept is 2.\nSolution: Substituting the values in equation we'll get -->\n OR if we want to write the equation in the slope-intercept form:\nSlope of a Line\nThe slope or gradient of a line describes its steepness, incline, or grade. A higher slope\nvalue indicates a steeper incline.\nThe slope is defined as the ratio of the \"rise\" divided by the \"run\" between two points on a\nline, or in other words, the ratio of the altitude change to the horizontal distance between\nany two points on the line.\n( , )P1 x1 y1\ny− = m(x− )y1 x1\ny− = m(x− )y1 x1\ny− 23 = 2( x− 14) y = 2 x− 5\n+ = 1x\na\ny\nb\n+ = 1x\na\ny\nb + = 1x\n5\ny\n2\n5y+ 2x− 10 = 0\ny = − x+ 22\n5\nGMAT Club Math Book\n45"} +{"id": "GMAT Club Math Book 2024 v8_p46_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 46, "page_end": 46, "topic_guess": "algebra", "text": "Given two points and on a line, the slope of the line is:\nIf the equation of the line is given in the Point-intercept form: , then is\nthe slope. This form of a line's equation is called the slope-intercept form, because can\nbe interpreted as the y-intercept of the line, the y-coordinate where the line intersects the\ny-axis.\nIf the equation of the line is given in the General form: , then the slope\nis and the y intercept is .\nSLOPE DIRECTION\nThe slope of a line can be positive, negative, zero or undefined.\n( , )x1 y1 ( , )x2 y2 m\nm =\n−y2 y1\n−x2 x1\ny = mx+ b m\nb\nax+ by+ c = 0\n− a\nb − c\nb\nGMAT Club Math Book\n46"} +{"id": "GMAT Club Math Book 2024 v8_p47_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 47, "page_end": 47, "topic_guess": "algebra", "text": "Positive slope\nHere, y increases as x increases, so the line slopes upwards to the right. The slope will be\na positive number. The line below has a slope of about +0.3, it goes up about 0.3 for\nevery step of 1 along the x-axis.\nNegative slope\nHere, y decreases as x increases, so the line slopes downwards to the right. The slope will\nbe a negative number. The line below has a slope of about -0.3, it goes down about 0.3\nfor every step of 1 along the x-axis.\nZero slope\nHere, y does not change as x increases, so the line in exactly horizontal. The slope of any\nhorizontal line is always zero. The line below goes neither up nor down as x increases, so\nits slope is zero.\nUndefined slope\nWhen the line is exactly vertical, it does not have a defined slope. The two x coordinates\nare the same, so the difference is zero. The slope calculation is then something like\n When you divide anything by zero the result has no meaning. The line above\nis exactly vertical, so it has no defined slope.\nSLOPE AND QUADRANTS:\n1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and\nY intercepts of the line with negative slope have the same sign. Therefore if X and Y\nintersects are positive, the line intersects quadrant I; if negative, quadrant III.\n2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X\nintercepts of the line with positive slope have opposite signs. Therefore if X intersect is\nnegative, line intersects the quadrant II too, if positive quadrant IV.\n3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y\naxis themselves) crosses three quadrants. Only the line which crosses origin \nOR is parallel to either of axis crosses only two quadrants.\n4. If a line is horizontal it has a slope of , is parallel to X-axis and crosses quadrant I\nand II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative.\nEquation of such line is y=b, where b is y intersect.\n5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses\nquadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is\nnegative. Equation of such line is , where a is x-intercept.\n6. For a line that crosses two points and , slope \nslope = 15\n0\n(0, 0)\n0\nx = a\n( , )x1 y1 ( , )x2 y2 m =\n−y2 y1\n−x2 x1\nGMAT Club Math Book\n47"} +{"id": "GMAT Club Math Book 2024 v8_p48_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 48, "page_end": 48, "topic_guess": "algebra", "text": "7. If the slope is 1 the angle formed by the line is degrees.\n8. Given a point and slope, equation of a line can be found. The equation of a\nstraight line that passes through a point with a slope is: \nVertical and horizontal lines\nA vertical line is parallel to the y-axis of the coordinate plane. All points on the line will\nhave the same x-coordinate.\nA vertical line has no slope. Or put another way, for a vertical line the slope is undefined.\nThe equation of a vertical line is:\nWhere: x is the coordinate of any point on the line; a is where the line crosses the x-axis\n(x intercept). Notice that the equation is independent of y. Any point on the vertical line\nsatisfies the equation.\nA horizontal line is parallel to the x-axis of the coordinate plane. All points on the line\nwill have the same y-coordinate.\n45\n( , )x1 y1 m y− = m(x− )y1 x1\nx = a\nGMAT Club Math Book\n48"} +{"id": "GMAT Club Math Book 2024 v8_p49_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 49, "page_end": 49, "topic_guess": "algebra", "text": "A horizontal line has a slope of zero.\nThe equation of a horizontal line is:\nWhere: y is the coordinate of any point on the line; b is where the line crosses the y-axis\n(y intercept). Notice that the equation is independent of x. Any point on the horizontal line\nsatisfies the equation.\nParallel lines\nParallel lines have the same slope.\nThe slope can be found using any method that is convenient to you:\ny = b\nGMAT Club Math Book\n49"} +{"id": "GMAT Club Math Book 2024 v8_p50_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 50, "page_end": 50, "topic_guess": "algebra", "text": "From two given points on the line.\nFrom the equation of the line in slope-intercept form\nFrom the equation of the line in point-slope form\nThe equation of a line through the point and parallel to line\n is:\nDistance between two parallel lines and can be found by\nthe formula:\nExample #1\nQ:There are two lines. One line is defined by two points at (5,5) and (25,15). The other is\ndefined by an equation in slope-intercept form form y = 0.52x - 2.5. Are two lines\nparallel?\nSolution:\nFor the top line, the slope is found using the coordinates of the two points that define the\nline. \nFor the lower line, the slope is taken directly from the formula. Recall that the slope\nintercept formula is y = mx + b, where m is the slope. So looking at the formula we see\nthat the slope is 0.52.\nSo, the top one has a slope of 0.5, the lower slope is 0.52, which are not equal.\nTherefore, the lines are not parallel.\nExample #2\nQ: Define a line through a point C parallel to a line passes through the points A and B.\n( , )P1 x1 y1\nax+ by+ c = 0\na(x− ) + b(y− ) = 0x1 y1\ny = mx+ b y = mx+ c\nD = |b−c|\n+1m2− − − − −\n√\nSlope = = 0.515−5\n25−5\nGMAT Club Math Book\n50"} +{"id": "GMAT Club Math Book 2024 v8_p51_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 51, "page_end": 51, "topic_guess": "algebra", "text": "Solution: We first find the slope of the line AB using the same method as in the example\nabove.\nFor the line to be parallel to AB it will have the same slope, and will pass through a given\npoint, C(12,10). We therefore have enough information to define the line by it's equation\nin point-slope form form:\n --> \nPerpendicular lines\nFor one line to be perpendicular to another, the relationship between their slopes has to\nbe negative reciprocal . In other words, the two lines are perpendicular if and only\nif the product of their slopes is .\nSlopeAB = = −0.5220−7\n5−30\ny = −0.52( x− 12) + 10 y = −0.52x + 16.24\n− 1\nm\n−1\nGMAT Club Math Book\n51"} +{"id": "GMAT Club Math Book 2024 v8_p52_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 52, "page_end": 52, "topic_guess": "algebra", "text": "The two lines and are perpendicular if\n.\nThe equation of a line passing through the point ) and perpendicular to line\n is:\nExample #1:\nQ: Are the two lines below perpendicular?\nSolution:\nTo answer, we must find the slope of each line and then check to see if one slope is the\nnegative reciprocal of the other or if their product equals to -1.\nx+ y+ = 0a1 b1 c1 x+ y+ = 0a2 b2 c2\n+ = 0a1a2 b1b2\n( ,P1 x1 y1\nax+ by+ c = 0\nb(x− ) − a(y− ) = 0x1 y1\nSlopeAB = = = 0.3585−19\n9−48\n−14\n−39\nGMAT Club Math Book\n52"} +{"id": "GMAT Club Math Book 2024 v8_p53_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 53, "page_end": 53, "topic_guess": "algebra", "text": "If the lines are perpendicular, each will be the negative reciprocal of the other. It doesn't\nmatter which line we start with, so we will pick AB:\nNegative reciprocal of 0.358 is \nSo, the slope of CD is -2.22, and the negative reciprocal of the slope of AB is -2.79. These\nare not the same, so the lines are not perpendicular, even though they may look as\nthough they are. However, if you looked carefully at the diagram, you might have noticed\nthat point C is a little too far to the left for the lines to be perpendicular.\nExample # 2.\nQ: Define a line passing through the point E and perpendicular to a line passing through\nthe points C and D on the graph above.\nSolution: The point E is on the y-axis and so is the y-intercept of the desired line. Once\nwe know the slope of the line, we can express it using its equation in slope-intercept form\ny=mx+b, where m is the slope and b is the y-intercept.\nFirst find the slope of line CD:\nThe line we seek will have a slope which is the negative reciprocal of:\nSince E is on the Y-axis, we know that the intercept is 10. Plugging these values into the\nline equation, the line we need is described by the equation\nThis is one of the ways a line can be defined and so we have solved the problem. If we\nwanted to plot the line, we would find another point on the line using the equation and\nthen draw the line through that point and the intercept.\nIntersection of two straight lines\nThe point of intersection of two non-parallel lines can be found from the equations of the\ntwo lines.\nSlopeCD = = = −2.2224−4\n22−31\n20\n−9\n− = −2.791\n0.358\nSlopeCD = = = −2.2224−4\n22−31\n20\n−9\n− = 0.451\n−2.22\ny = 0.45x + 10\nGMAT Club Math Book\n53"} +{"id": "GMAT Club Math Book 2024 v8_p54_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 54, "page_end": 54, "topic_guess": "algebra", "text": "To find the intersection of two straight lines:\n1. First we need their equations\n2. Then, since at the point of intersection, the two equations will share a point and thus\nhave the same values of x and y, we set the two equations equal to each other. This gives\nan equation that we can solve for x\n3. We substitute the x value in one of the line equations (it doesn't matter which) and\nsolve it for y.\nThis gives us the x and y coordinates of the intersection.\nExample #1\nQ: Find the point of intersection of two lines that have the following equations (in slope-\nintercept form):\nSolution: At the point of intersection they will both have the same y-coordinate value, so\nwe set the equations equal to each other:\nThis gives us an equation in one unknown (x) which we can solve:\nTo find y, simply set x equal to 10 in the equation of either line and solve for y:\nEquation for a line (Either line will do)\nSet x equal to 10: \nWe now have both x and y, so the intersection point is (10, 27)\nExample #2\nQ: Find the point of intersection of two lines that have the following equations (in slope-\ny = 3 x− 3\ny = 2.3 x+ 4\n3x− 3 = 2.3 x+ 4\nx = 10\ny = 3 x− 3\ny = 30 − 3\ny = 27\nGMAT Club Math Book\n54"} +{"id": "GMAT Club Math Book 2024 v8_p55_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 55, "page_end": 55, "topic_guess": "coordinate_geometry", "text": "intercept form): and (A vertical line)\nSolution: When one of the lines is vertical, it has no defined slope. We find the\nintersection slightly differently.\nOn the vertical line, all points on it have an x-coordinate of 12 (the definition of a vertical\nline), so we simply set x equal to 12 in the first equation and solve it for y.\nEquation for a line \nSet x equal to 12 \nSo the intersection point is at (12,33).\nNote: If both lines are vertical or horizontal, they are parallel and have no\nintersection\nDistance from a point to a line\nThe distance from a point to a line is the shortest distance between them - the length of\na perpendicular line segment from the line to the point.\nThe distance from a point to a line is given by the formula:\nWhen the line is horizontal the formula transforms to: \nWhere: is the y-coordinate of the given point P; is the y-coordinate of any\npoint on the given vertical line L. | | the vertical bars mean \"absolute value\" - make\nit positive even if it calculates to a negative.\nWhen the line is vertical the formula transforms to: \nWhere: is the x-coordinate of the given point P; is the x-coordinate of any\npoint on the given vertical line L. | | the vertical bars mean \"absolute value\" - make\nit positive even if it calculates to a negative.\nWhen the given point is origin, then the distance between origin and line\nax+by+c=0 is given by the formula:\nCircle on a plane\nIn an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set\nof all points (x, y) such that:\ny = 3 x− 3 x = 12\ny = 3 x− 3\ny = 36 − 3\ny = 33\n( , )x0 y0 ax+ by+ c = 0\nD =\n|a +b +c|x0 y0\n+a2 b2− − − − − −√\nD = | − |Py Ly\nPy Ly\nD = | − |Px Lx\nPx Lx\nD = |c|\n+a2 b2− − − − − −\n√\n(x− a + (y− b =)2 )2 r2\nGMAT Club Math Book\n55"} +{"id": "GMAT Club Math Book 2024 v8_p56_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 56, "page_end": 56, "topic_guess": "geometry", "text": "This equation of the circle follows from the Pythagorean theorem applied to any point on\nthe circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled\ntriangle whose other sides are of length x-a and y-b.\nIf the circle is centered at the origin (0, 0), then the equation simplifies to:\nNumber line\nA number line is a picture of a straight line on which every point corresponds to a real\nnumber and every real number to a point.\nOn the GMAT we can often see such statement: is halfway between and on the\nnumber line. Remember this statement can ALWAYS be expressed as:\n.\nAlso on the GMAT we can often see another statement: The distance between and on\nthe number line is the same as the distance between and . Remember this statement\ncan ALWAYS be expressed as:\n.\nParabola\nA parabola is the graph associated with a quadratic function, i.e. a function of the form\n+ =x2 y2 r2\nk m n\n= km+n\n2\np m\np n\n|p− m| = |p − n|\nGMAT Club Math Book\n56"} +{"id": "GMAT Club Math Book 2024 v8_p57_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 57, "page_end": 57, "topic_guess": "coordinate_geometry", "text": ".\nThe general or standard form of a quadratic function is , or in function\nform, , where is the independent variable, is the dependent\nvariable, and , , and are constants.\nThe larger the absolute value of , the steeper (or thinner) the parabola is, since\nthe value of y is increased more quickly.\nIf is positive, the parabola opens upward, if negative, the parabola opens\ndownward.\nx-intercepts: The x-intercepts, if any, are also called the roots of the function. The x-\nintercepts are the solutions to the equation and can be calculated by\nthe formula:\n and \nExpression is called discriminant:\nIf discriminant is positive parabola has two intercepts with x-axis;\nIf discriminant is negative parabola has no intercepts with x-axis;\nIf discriminant is zero parabola has one intercept with x-axis (tangent point).\ny-intercept: Given , the y-intercept is , as y intercept means the\nvalue of y when x=0.\nVertex: The vertex represents the maximum (or minimum) value of the function, and is\nvery important in calculus.\ny = a + bx+ cx2\ny = a + bx+ cx2\nf(x) = a + bx+ cx2 x y\na b c\na\na\n0 = a + bx+ cx2\n=x1\n−b− −4acb2− − − − − −√\n2a =x2\n−b+ −4acb2− − − − − −√\n2a\n− 4acb2\ny = a + bx+ cx2 c\nGMAT Club Math Book\n57"} +{"id": "GMAT Club Math Book 2024 v8_p58_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 58, "page_end": 58, "topic_guess": "coordinate_geometry", "text": "The vertex of the parabola is located at point .\nNote: typically just is calculated and plugged in for x to find y. \nPractice Questions\nEasy:\n1. https://gmatclub.com/forum/which-of-the ... 29938.html\n2. https://gmatclub.com/forum/in-the-recta ... 65534.html\n3. https://gmatclub.com/forum/in-a-rectang ... 94392.html\n4. https://gmatclub.com/forum/in-the-xy-pl ... 05023.html\n5. https://gmatclub.com/forum/in-the-figur ... 68654.html\n6. https://gmatclub.com/forum/in-the-xy-pl ... 30899.html\n7. https://gmatclub.com/forum/in-the-xy-pl ... 21712.html\n8. https://gmatclub.com/forum/in-the-xy-pl ... 40727.html\n9. https://gmatclub.com/forum/if-each-of-t ... 20583.html\n10. https://gmatclub.com/forum/in-the-xy-pl ... 93838.html\n11. https://gmatclub.com/forum/in-the-line- ... 31295.html\nMedium:\n1. https://gmatclub.com/forum/which-of-the ... 55961.html\n2. https://gmatclub.com/forum/line-m-lies- ... 98105.html\n3. https://gmatclub.com/forum/in-the-coord ... 67670.html\n4. https://gmatclub.com/forum/in-the-coord ... 44795.html\n5. https://gmatclub.com/forum/in-the-xy-co ... 07226.html\n6. https://gmatclub.com/forum/in-the-recta ... 20818.html\n7. https://gmatclub.com/forum/on-the-graph ... 36560.html\n8. https://gmatclub.com/forum/in-the-xy-pl ... 42863.html\n9. https://gmatclub.com/forum/the-graph-of ... 07199.html\n10. https://gmatclub.com/forum/in-the-xy-co ... 35012.html\nHard:\n1. https://gmatclub.com/forum/right-triang ... 71597.html\n2. https://gmatclub.com/forum/for-every-po ... 91527.html\n3. https://gmatclub.com/forum/point-1-0-is ... 82265.html\n4. https://gmatclub.com/forum/if-equation- ... 01963.html\n5. https://gmatclub.com/forum/in-the-recta ... 44774.html\n6. https://gmatclub.com/forum/in-the-xy-pl ... 26463.html\n7. https://gmatclub.com/forum/in-the-figur ... 39117.html\n8. https://gmatclub.com/forum/the-graph-of ... 07473.html\n9. https://gmatclub.com/forum/the-center-o ... 89027.html\n10. https://gmatclub.com/forum/in-the-xy-pl ... 00739.html\n(− ,b\n2a c− )b2\n4a\n− ,b\n2a\nGMAT Club Math Book\n58"} +{"id": "GMAT Club Math Book 2024 v8_p59_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 59, "page_end": 59, "topic_guess": "word_problems", "text": "GMAT Club Math Book\nWord Problems Made Easy\nhttps://gmatclub.com/forum/word-problems-made-easy-87346.html\n Word Problems Made Easy\ncreated by: sriharimurthy\nedited by: bb, walker, Bunuel\nThis is an introductory post to word problems.\nIt deals primarily with the translation of word problems into equations.\nDiscussions relating to specific types of word problems will be dealt with separately\n(see end of post).\nThe Following Points Outline a General Approach to\nWord Problems:\n1) Read the entire question carefully and get a feel for what is happening. Identify what\nkind of word problem you're up against.\n2) Make a note of exactly what is being asked.\n3) Simplify the problem - this is what is usually meant by 'translating the English to\nMath'. Draw a figure or table. Sometimes a simple illustration makes the problem much\neasier to approach.\n4) It is not always necessary to start from the first line. Invariably, you will find it easier\nto define what you have been asked for and then work backwards to get the\ninformation that is needed to obtain the answer.\n5) Use variables (a, b, x, y, etc.) or numbers (100 in case of percentages, any common\nmultiple in case of fractions, etc.) depending on the situation.\n6) Use SMART values. Think for a moment and choose the best possible value that would\nhelp you reach the solution in the quickest possible time. DO NOT choose values that\nwould serve only to confuse you. Also, remember to make note of what the value you\nselected stands for.\nGMAT Club Math Book\n59"} +{"id": "GMAT Club Math Book 2024 v8_p60_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 60, "page_end": 60, "topic_guess": "word_problems", "text": "7) Once you have the equations written down it's time to do the math! This is usually\nquite simple. Be very careful so as not to make any silly mistakes in calculations.\n8) Lastly, after solving, cross check to see that the answer you have obtained\ncorresponds to what was asked. The makers of these GMAT questions love to trick\nstudents who don’t pay careful attention to what is being asked. For example, if the\nquestion asks you to find ‘what fraction of the remaining...’ you can be pretty sure one of\nthe answer choices will have a value corresponding to ‘what fraction of the total…’\nTranslating Word Problems\nThese are a few common English to Math translations that will help you break down word\nproblems. My recommendation is to refer to them only in the initial phases of study. With\npractice, decoding a word problem should come naturally. If, on test day, you still have to\ntry and remember what the math translations to some English term is, you haven’t\npracticed enough!\nADDITION: increased by ; more than ; combined ; together ; total of ; sum ; added to ;\nand ; plus\nSUBTRACTION: decreased by ; minus ; less ; difference between/of ; less than ; fewer\nthan ; minus ; subtracted from\nMULTIPLICATION: of ; times ; multiplied by ; product of ; increased/decreased by a\nfactor of (this type can involve both addition or subtraction and multiplication!)\nDIVISION: per ; out of ; ratio of ; quotient of ; percent (divide by 100) ; divided by ;\neach\nEQUALS: is ; are ; was ; were ; will be ; gives ; yields ; sold for ; has ; costs ; adds up to\n; the same as ; as much as\nVARIABLE or VALUE: a number ; how much ; how many ; what\nSome Tricky Forms:\n'per' means 'divided by'\nJack drove at a speed of 40 miles per hour OR 40 miles/hour.\n'a' sometimes means 'divided by'\nJack bought twenty-four eggs for $3 a dozen.\n'less than'\nIn English, the ‘less than’ construction is reverse of what it is in math.\nFor example, ‘3 less than x’ means ‘x – 3’ NOT ‘3 – x’\nSimilarly, if the question says ‘Jack’s age is 3 less than that of Jill’, it means that Jacks age\nis ‘Jill’s age – 3’.\nGMAT Club Math Book\n60"} +{"id": "GMAT Club Math Book 2024 v8_p61_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 61, "page_end": 61, "topic_guess": "word_problems", "text": "The ‘how much is left’ construction\nSometimes, the question will give you a total amount that is made up of a number of\nsmaller amounts of unspecified sizes. In this case, just assign a variable to the unknown\namounts and the remaining amount will be what is left after deducting this named amount\nfrom the total.\nConsider the following:\nA hundred-pound order of animal feed was filled by mixing products from Bins A, B and C,\nand that twice as much was added from Bin C as from Bin A.\nLet \"a\" stand for the amount from Bin A. Then the amount from Bin C was \"2a\", and the\namount taken from Bin B was the remaining portion of the hundred pounds: 100 – a – 2a.\nIn the following cases, order is important:\n‘quotient/ratio of’ construction\nIf a problems says ‘the ratio of x and y’, it means ‘x divided by y’ NOT 'y divided by x'\n‘difference between/of’ construction\nIf the problem says ‘the difference of x and y’ it means |x – y|\nNow that we have seen how it is possible, in theory, to break down word\nproblems, lets go through a few simple examples to see how we can apply this\nknowledge.\nExample 1.\nThe length of a rectangular garden is 2 meters more than its width. Express its length in\nterms of its width.\nSolution:\nKey words: more than (implies addition); is (implies equal to)\nThus, the phrase ‘length is 2 more than width’ becomes:\nLength = 2 + width\nExample 2.\nThe length of a rectangular garden is 2 meters less than its width. Express its length in\nterms of its width.\nSolution:\nKey words: less than (implies subtraction but in reverse order); is (implies equal to)\nThus, the phrase ‘length is 2 less than width’ becomes:\nLength = width - 2\nExample 3.\nThe length of a rectangular garden is 2 times its width. Express its length in terms of its\nwidth.\nSolution:\nKey words: times (implies multiplication); is (implies equal to)\nThus, the phrase ‘length is 2 times width’ becomes:\nGMAT Club Math Book\n61"} +{"id": "GMAT Club Math Book 2024 v8_p62_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 62, "page_end": 62, "topic_guess": "fractions_decimals_percents", "text": "Length = 2*width\nExample 4.\nThe ratio of the length of a rectangular garden to its width is 2. Express its length in terms\nof its width.\nSolution:\nKey words: ratio of (implies division); is (implies equal to)\nThus, the phrase ‘ratio of length to width is 2’ becomes:\nLength/width = 2 → Length = 2*width\nExample 5.\nThe length of a rectangular garden surrounded by a walkway is twice its width. If\ndifference between the length and width of just the rectangular garden is 10 meters, what\nwill be the width of the walkway if just the garden has width 6 meters?\nSolution:\nOk this one has more words than the previous examples, but don’t worry, lets break it\ndown and see how simple it becomes.\nKey words: and (implies addition); twice (implies multiplication); difference between\n(implies subtraction where order is important); what (implies variable); is, will be (imply\nequal to)\nSince this is a slightly more complicated problem, let us first define what we want.\n'What will be the width of the walkway' implies that we should assign a variable for\nwidth of the walkway and find its value.\nThus, let width of the walkway be ‘x’.\nNow, in order to find the width of walkway, we need to have some relation between the\ntotal length/width of the rectangular garden + walkway and the length/width of\njust the garden.\nNotice here that if we assign a variables to the width and length of either\ngarden+walkway or just garden, we can express every thing in terms of just these\nvariables.\nSo, let length of the garden+walkway = L\nAnd width of garden+walkway = W\nThus length of just garden = L – 2x\nWidth of just garden = W - 2x\nNote: Remember that the walkway completely surrounds the garden. Thus its width will\nhave to be accounted for twice in both the total length and total width.\nGMAT Club Math Book\n62"} +{"id": "GMAT Club Math Book 2024 v8_p63_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 63, "page_end": 63, "topic_guess": "algebra", "text": "Now let’s see what the question gives us.\n‘Garden with width 6 meters’ translates to:\nWidth of garden = 6\nW – 2x = 6\nThus, if we know W we can find x.\n‘Length of a rectangular garden surrounded by walkway is twice its width’ translates to:\nLength of garden + length of walkway = 2*(width of garden + width of walkway)\nL = 2*W\n‘Difference between the length and width of just the rectangular garden is 10 meters’\ntranslates to:\nLength of garden – width of garden = 10\n(L – 2x) – (W – 2x) = 10\nL – W = 10\nNow, since we have two equations and two variables (L and W), we can find their values.\nSolving them we get: L = 20 and W = 10.\nThus, since we know the value of W, we can calculate ‘x’\n10 – 2x = 6\n2x = 4\nx = 2\nThus, the width of the walkway is 2 meters.\nEasy wasn't it?\nWith practice, writing out word problems in the form of equations will become second\nnature. How much you need to practice depends on your own individual ability. It could be\n10 questions or it could be 100. But once you’re able to effortlessly translate word\nproblems into equations, more than half your battle will already be won. \nGMAT Club Math Book\n63"} +{"id": "GMAT Club Math Book 2024 v8_p64_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 64, "page_end": 64, "topic_guess": "word_problems", "text": "GMAT Club Math Book\nWork Problems\nhttps://gmatclub.com/forum/work-problems-87357.html\n ‘Work’ Word Problems Made Easy\ncreated by: sriharimurthy\nedited by: bb, walker, Bunuel\nNOTE: In case you are not familiar with translating word problems into equations please\ngo through THIS POST FIRST.\nWhat is a ‘Work’ Word Problem?\nIt involves a number of people or machines working together to complete a task.\nWe are usually given individual rates of completion.\nWe are asked to find out how long it would take if they work together.\nSounds simple enough doesn’t it? Well it is!\nThere is just one simple concept you need to understand in order to solve any ‘work’\nrelated word problem.\nThe ‘Work’ Problem Concept\nSTEP 1: Calculate how much work each person/machine does in one unit of time\n(could be days, hours, minutes, etc).\nHow do we do this? Simple. If we are given that A completes a certain amount of work in\nX hours, simply reciprocate the number of hours to get the per hour work. Thus in one\nhour, A would complete of the work. But what is the logic behind this? Let me explain\nwith the help of an example.\nAssume we are given that Jack paints a wall in 5 hours. This means that in every hour, he\ncompletes a fraction of the work so that at the end of 5 hours, the fraction of work he has\ncompleted will become 1 (that means he has completed the task).\nThus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work\n1\nX\nGMAT Club Math Book\n64"} +{"id": "GMAT Club Math Book 2024 v8_p65_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 65, "page_end": 65, "topic_guess": "word_problems", "text": "completed will be (1*1)/5\nSTEP 2: Add up the amount of work done by each person/machine in that one\nunit of time.\nThis would give us the total amount of work completed by both of them in one hour. For\nexample, if A completes of the work in one hour and B completes of the work in\none hour, then TOGETHER, they can complete of the work in one hour.\nSTEP 3: Calculate total amount of time taken for work to be completed when all\npersons/machines are working together.\nThe logic is similar to one we used in STEP 1, the only difference being that we use it in\nreverse order. Suppose . This means that in one hour, A and B working\ntogether will complete of the work. Therefore, working together, they will\ncomplete the work in Z hours.\nAdvice here would be: DON'T go about these problems trying to remember some\nformula. Once you understand the logic underlying the above steps, you will\nhave all the information you need to solve any ‘work’ related word problem.\n(You will see that the formula you might have come across can be very easily\nand logically deduced from this concept).\nNow, lets go through a few problems so that the above-mentioned concept becomes\ncrystal clear. Lets start off with a simple one :\nExample 1.\nJack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it\ntake if they work together?\nSolution:\nThis is a simple straightforward question wherein we must just follow steps 1 to 3 in order\nto obtain the answer.\nSTEP 1: Calculate how much work each person does in one hour.\nJack → (1/3) of the work\nJohn → (1/5) of the work\nSTEP 2: Add up the amount of work done by each person in one hour.\nWork done in one hour when both are working together = \nSTEP 3: Calculate total amount of time taken when both work together.\nIf they complete of the work in 1 hour, then they would complete 1 job in hours.\n1\nX\n1\nY\n+1\nX\n1\nY\n+ =1\nX\n1\nY\n1\nZ\n1\nZ\n+ =1\n3\n1\n5\n8\n15\n8\n15\n15\n8\nGMAT Club Math Book\n65"} +{"id": "GMAT Club Math Book 2024 v8_p66_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 66, "page_end": 66, "topic_guess": "word_problems", "text": "Example 2.\nWorking, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the\nwork. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time\ndoes Y finish his work?\nSolution:\nNow the only reason this is trickier than the first problem is because the sequence of\nevents are slightly more complicated. The concept however is the same. So if our\nunderstanding of the concept is clear, we should have no trouble at all dealing with this.\n‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us\nthat in one hour, X will finish of the work.\n‘He finishes 2/3 of the work’ This tells us that of the work still remains.\n‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete of the\nwork.\n‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be .\nTherefore, rate at which Y works is .\n‘In how much time does Y finish his work?’ If Y completes of the work in 1 hour, then\nhe will complete of the work in 40 hours.\nSo as you can see, even though the question might have been a little difficult to follow at\nfirst reading, the solution was in fact quite simple. We didn’t use any new concepts. All we\ndid was apply our knowledge of the concept we learnt earlier to the information in the\nquestion in order to answer what was being asked.\nExample 3.\nWorking together, printer A and printer B would finish a task in 24 minutes. Printer A\nalone would finish the task in 60 minutes. How many pages does the task contain if\nprinter B prints 5 pages a minute more than printer A?\nSolution:\nThis problem is interesting because it tests not only our knowledge of the concept of word\nproblems, but also our ability to ‘translate English to Math’\n‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us\nthat A and B combined would work at the rate of per minute.\n‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of\n per minute.\nAt this point, it should strike you that with just this much information, it is possible to\n1\n12\n1\n3\n1\n3\n1\n12\n∗ =1\n10\n1\n12\n1\n120\n1\n120\n1\n3\n1\n24\n1\n60\nGMAT Club Math Book\n66"} +{"id": "GMAT Club Math Book 2024 v8_p67_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 67, "page_end": 67, "topic_guess": "word_problems", "text": "calculate the rate at which B works: Rate at which B works = .\n‘B prints 5 pages a minute more than printer A’ This means that the difference between\nthe amount of work B and A complete in one minute corresponds to 5 pages. So, let us\ncalculate that difference. It will be \n‘How many pages does the task contain?’ If of the job consists of 5 pages, then the 1\njob will consist of pages.\nExample 4.\nMachine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten\nhours longer to produce 660 sprockets than machine B. Machine B produces 10% more\nsprockets per hour than machine A. How many sprockets per hour does machine A\nproduce?\nSolution:\nThe rate of A is sprockets per hour;\nThe rate of B is sprockets per hour.\nWe are told that B produces 10% more sprockets per hour than A, thus \n--> --> the rate of A is sprockets per hour.\nAs you can see, the main reason the 'tough' problems are 'tough' is because they test a\nnumber of other concepts apart from just the ‘work’ concept. However, once you manage\nto form the equations, they are really not all that tough.\nAnd as far as the concept of ‘work’ word problems is concerned – it is always the\nsame!"} +{"id": "GMAT Club Math Book 2024 v8_p67_c2", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 67, "page_end": 67, "topic_guess": "algebra", "text": "age\nto form the equations, they are really not all that tough.\nAnd as far as the concept of ‘work’ word problems is concerned – it is always the\nsame!\n\n− =1\n24\n1\n60\n1\n40\n− =1\n40\n1\n60\n1\n120\n1\n120\n= 600(5∗1)\n1\n120\n660\nt+10\n660\nt\n∗ 1.1 =660\nt+10\n660\nt\nt = 100 = 6660\nt+10\nGMAT Club Math Book\n67"} +{"id": "GMAT Club Math Book 2024 v8_p68_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 68, "page_end": 68, "topic_guess": "word_problems", "text": "GMAT Club Math Book\nDistance, Speed Time\nhttps://gmatclub.com/forum/distance-speed-time-87481.html\n 'Distance/Speed/Time' Word Problems Made Easy\ncreated by: sriharimurthy\nedited by: bb, walker, Bunuel\nNOTE: In case you are not familiar with translating word problems into equations please\ngo through this post first : https://gmatclub.com/forum/word-problem ... 87346.html\nWhat is a ‘D/S/T’ Word Problem?\nUsually involve something/someone moving at a constant or average speed.\nOut of the three quantities (speed/distance/time), we are required to find one.\nInformation regarding the other two will be provided in the question stem.\nThe ‘D/S/T’ Formula: Distance = Speed x Time\nI’m sure most of you are already familiar with the above formula (or some variant of it).\nBut how many of you truly understand what it signifies?\nWhen you see a ‘D/S/T’ question, do you blindly start plugging values into the formula\nwithout really understanding the logic behind it? If then answer to that question is yes,\nthen you would probably have noticed that your accuracy isn’t quite where you’d want it\nto be.\nMy advice here, as usual, is to make sure you understand the concept behind the\nformula rather than just using it blindly.\nSo what’s the concept? Lets find out!\nThe Distance = Speed x Time formula is just a way of saying that the\ndistance you travel depends on the speed you go for any length of time.\nIf you travel at 50 mph for one hour, then you would have traveled 50 miles. If you\nGMAT Club Math Book\n68"} +{"id": "GMAT Club Math Book 2024 v8_p69_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 69, "page_end": 69, "topic_guess": "word_problems", "text": "travel for 2 hours at that speed, you would have traveled 100 miles. 3 hours would\nbe 150 miles, etc.\nIf you were to double the speed, then you would have traveled 100 miles in the first\nhour and 200 miles at the end of the second hour.\nWe can figure out any one of the components by knowing the other two.\nFor example, if you have to travel a distance of 100 miles, but can only go at a\nspeed of 50 mph, then you know that it will take you 2 hours to get there. Similarly,\nif a friend visits you from 100 miles away and tells you that it took him 4 hours to\nreach, you will know that he AVERAGED 25 mph. Right?\nAll calculations depend on AVERAGE SPEED.\nSupposing your friend told you that he was stuck in traffic along the way and that\nhe traveled at 50 mph whenever he could move. Therefore, although practically he\nnever really traveled at 25 mph, you can see how the standstills due to traffic\ncaused his average to reduce. Now, if you think about it, from the information\ngiven, you can actually tell how long he was driving and how long he was stuck due\nto traffic (assuming; what is false but what they never worry about in these\nproblems; that he was either traveling at 50 mph or 0 mph). If he was traveling\nconstantly at 50 mph, he should have reached in 2 hours. However, since he took 4\nhours, he must have spent the other 2 hours stuck in traffic!\nNow lets see how we can represent this using the formula.\nWe know that the total distance is 100 miles and that the total time is 4 hours.\nBUT, his rates were different AND they were different at different times. However,\ncan you see that no matter how many different rates he drove for various different\ntime periods, his TOTAL distance depended simply on the SUM of each of the\ndifferent distances he drove during each time period?\nE.g., if you drive a half hour at 60 mph, you will cover 30 miles. Then if you speed\nup to 80 mph for another half hour, you will cover 40 miles, and then if you slow\ndown to 30 mph, you will only cover 15 miles in the next half hour. But if you drove\nlike this, you would have covered a total of 85 miles (30 + 40 + 15). It is fairly easy\nto see this looking at it this way, but it is more difficult to see it if we scramble it up\nand leave out one of the amounts and you have to figure it out going \"backwards\".\nThat is what word problems do.\nFurther, what makes them difficult is that the components they give you, or ask you\nto find can involve variable distances, variable times, variable speeds, or any two or\nthree of these. How you \"reassemble\" all this in order to use the d = s*t formula\ntakes some reflection that is \"outside\" of the formula itself. You have to think\nGMAT Club Math Book\n69"} +{"id": "GMAT Club Math Book 2024 v8_p70_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 70, "page_end": 70, "topic_guess": "word_problems", "text": "about how to use the formula.\nSo the trick is to be able to understand EXACTLY what they are giving you and\nEXACTLY what it is that is missing, but you do that from thinking, not from the\nformula, because the formula only works for the COMPONENTS of any trip\nwhere you are going an average speed for a certain amount of time. ONCE\nthe conditions deal with different speeds or different times, you have to look at each\nof those components and how they go together. And that can be very difficult if you\nare not methodical in how you think about the components and how they go\ntogether. The formula doesn't tell you which components you need to look at and\nhow they go together. For that, you need to think, and the thinking is not always as\neasy or straightforward as it seems like it ought to be.\nIn the case of your friend above, if we call the time he spent driving 50 mph, T1;\nthen the time he spent standing still is (4 - T1) hours, since the whole trip took 4\nhours. So we have \n100 miles = (50 mph x T1) + (0 mph x [4 - T1]) which is\nequivalent then to: 100 miles = 50 mph x T1\nSo, T1 will equal 2 hours. And, since the time he spent going zero is (4 - 2), it also\nturns out to be 2 hours.\nSometimes the right answers will seem counter-intuitive, so it is really\nimportant to think about the components methodically and systematically.\nThere is a famous trick problem: To qualify for a race, you need to average 60 mph\ndriving two laps around a 1 mile long track. You have some sort of engine difficulty\nthe first lap so that you only average 30 mph during that lap; how fast do you have\nto drive the second lap to average 60 for both of them?\nI will go through THIS problem with you because, since it is SO tricky, it will\nillustrate a way of looking at almost all the kinds of things you have to think about\nwhen working any of these kinds of problems FOR THE FIRST TIME (i.e., before you\ncan do them mechanically because you recognize the TYPE of problem it is).\nIntuitively it would seem you need to drive 90, but this turns out to be wrong for\nreasons I will give in a minute.\nThe answer is that NO MATTER HOW FAST you do the second lap, you can't make it.\nAnd this SEEMS really odd and that it can't possibly be right, but it is. The reason is\nthat in order to average at least 60 mph over two one-mile laps, since 60 mph is\none mile per minute, you will need to do the whole two miles in two minutes or less.\nBut if you drove the first mile at only 30, you used up the whole two minutes just\ndoing IT. So you have run out of time to qualify.\nTo see this with the d = s*t formula, you need to look at the overall trip and break\nit into components, and that is the hardest part of doing this (these) problem(s),\nbecause (often) the components are difficult to figure out, and because it is hard to\nsee which ones you need to put together in which way.\nGMAT Club Math Book\n70"} +{"id": "GMAT Club Math Book 2024 v8_p71_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 71, "page_end": 71, "topic_guess": "word_problems", "text": "In the next section we will learn how to do just that.\nResolving the Components\nWhen you first start out with these problems, the best way to approach\nthem is by organizing the data in a tabular form.\nUse a separate column each for distance, speed and time and a separate row for the\ndifferent components involved (2 parts of a journey, different moving objects, etc.).\nThe last row should represent total distance, total time and average speed for these\nvalues (although there might be no need to calculate these values if the question\ndoes not require them).\nAssign a variable for any unknown quantity.\nIf there is more than one unknown quantity, do not blindly assign another variable\nto it. Look for ways in which you can express that quantity in terms of the quantities\nalready present. Assign another variable to it only if this is not possible.\nIn each row, the quantities of distance, speed and time will always satisfy d\n= s*t.\nThe distance and time column can be added to give you the values of total\ndistance and total time but you CANNOT add the speeds.\nThink about it: If you drive 20 mph on one street, and 40 mph on another street,\ndoes that mean you averaged 60 mph?\nOnce the table is ready, form the equations and solve for what has been\nasked!\nGMAT Club Math Book\n71"} +{"id": "GMAT Club Math Book 2024 v8_p72_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 72, "page_end": 72, "topic_guess": "word_problems", "text": "Warning: Make sure that the units for time and distance agree with the units for the rate.\nFor instance, if they give you a rate of feet per second, then your time must be in seconds\nand your distance must be in feet. Sometimes they try to trick you by using the wrong\nunits, and you have to catch this and convert to the correct units.\nA Few More Points to Note\nMotion in Same Direction (Overtaking): The first thing that should strike you\nhere is that at the time of overtaking, the distances traveled by both will be the\nsame.\nMotion in Opposite Direction (Meeting): The first thing that should strike you\nhere is that if they start at the same time (which they usually do), then at the point\nat which they meet, the time will be the same. In addition, the total distance\ntraveled by the two objects under consideration will be equal to the sum of their\nindividual distances traveled.\nRound Trip: The key thing here is that the distance going and coming back is the\nsame.\nNow that we know the concept in theory, let us see how it works practically,\nwith the help of a few examples.\nNote for tables : All values in black have been given in the question stem. All values in\nblue have been calculated.\nExample 1.\nTo qualify for a race, you need to average 60 mph driving two laps around a 1-mile long\ntrack. You have some sort of engine difficulty the first lap so that you only average 30\nmph during that lap; how fast do you have to drive the second lap to average 60 for both\nof them?\nSolution:\nLet us first start with a problem that has already been introduced. You will see that by\nclearly listing out the given data in tabular form, we eliminate any scope for confusion.\nGMAT Club Math Book\n72"} +{"id": "GMAT Club Math Book 2024 v8_p73_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 73, "page_end": 73, "topic_guess": "word_problems", "text": "In the first row, we are given the distance and the speed. Thus it is possible to calculate\nthe time.\nTime(1) = Distance(1)/Speed(1) = 1/30\nIn the second row, we are given just the distance. Since we have to calculate speed, let\nus give it a variable 'x'. Now, by using the 'D/S/T' relationship, time can also be expressed\nin terms of 'x'.\nTime(2) = Distance(2)/Speed(2) = 1/x\nIn the third row, we know that the total distance is 2 miles (by taking the sum of the\ndistances in row 1 and 2) and that the average speed should be 60 mph. Thus we can\ncalculate the total time that the two laps should take.\nTime(3) = Distance(3)/Speed(3) = 2/60 = 1/30\nNow, we know that the total time should be the sum of the times in row 1 and 2. Thus we\ncan form the following equation :\nTime(3) = Time(1) + Time(2) ---> 1/30 = 1/30 + 1/x\nFrom this, it becomes clear that '1/x' must be 0.\nSince 'x' is the reciprocal of 0, which does not exist, there can be no speed for\nwhich the average can be made up in the second lap.\nExample 2.\nAn executive drove from home at an average speed of 30 mph to an airport where a\nhelicopter was waiting. The executive boarded the helicopter and flew to the corporate\noffices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip\ntook three hours. Find the distance from the airport to the corporate offices.\nSolution:\nSince we have been asked to find the distance from the airport to the corporate office\n(that is the distance he spent flying), let us assign that specific value as 'x'.\nGMAT Club Math Book\n73"} +{"id": "GMAT Club Math Book 2024 v8_p74_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 74, "page_end": 74, "topic_guess": "word_problems", "text": "Thus, the distance he spent driving will be '150 - x'\nNow, in the first row, we have the distance in terms of 'x' and we have been given the\nspeed. Thus we can calculate the time he spent driving in terms of 'x'.\nTime(1) = Distance(1)/Speed(1) = (150 - x)/30\nSimilarly, in the second row, we again have the distance in terms of 'x' and we have been\ngiven the speed. Thus we can calculate the time he spent flying in terms of 'x'.\nTime(2) = Distance(2)/Speed(2) = x/60\nNow, notice that we have both the times in terms of 'x'. Also, we know the total time for\nthe trip. Thus, summing the individual times spent driving and flying and equating it to\nthe total time, we can solve for 'x'.\nTime(1) + Time(2) = Time(3) --> (150 - x)/30 + x/60 = 3 --> x = 120 miles\nAnswer : 120 miles\nNote: In this problem, we did not calculate average speed for row 3 since we did not need\nit. Remember not to waste time in useless calculations!\nExample 3.\nA passenger train leaves the train depot 2 hours after a freight train left the same depot.\nThe freight train is traveling 20 mph slower than the passenger train. Find the speed of\nthe passenger train, if it overtakes the freight train in three hours.\nSolution:\nSince this is an 'overtaking' problem, the first thing that should strike us is that the\ndistance traveled by both trains is the same at the time of overtaking.\nNext we see that we have been asked to find the speed of the passenger train at the time\nof overtaking. So let us represent it by 'x'.\nAlso, we are given that the freight train is 20 mph slower than the passenger train. Hence\nits speed in terms of 'x' can be written as 'x - 20'.\nMoving on to the time, we are told that it has taken the passenger train 3 hours to reach\nthe freight train. This means that the passenger train has been traveling for 3 hours.\nWe are also given that the passenger train left 2 hours after the freight train. This means\nthat the freight train has been traveling for 3 + 2 = 5 hours.\nNow that we have all the data in place, we need to form an equation that will help us\nsolve for 'x'. Since we know that the distances are equal, let us see how we can use this\nto our advantage.\nFrom the first row, we can form the following equation :\nDistance(1) = Speed(1) * Time(1) = x*3\nGMAT Club Math Book\n74"} +{"id": "GMAT Club Math Book 2024 v8_p75_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 75, "page_end": 75, "topic_guess": "word_problems", "text": "From the second row, we can form the following equation :\nDistance(2) = Speed(2) * Time(2) = (x - 20)*5\nNow, equating the distances because they are equal we get the following equation :\n3*x = 5*(x - 20) --> x = 50 mph.\nAnswer : 50 mph.\nExample 4.\nTwo cyclists start at the same time from opposite ends of a course that is 45 miles long.\nOne cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after\nthey begin will they meet?\nSolution:\nSince this is a 'meeting' problem, there are two things that should strike you. First, since\nthey are starting at the same time, when they meet, the time for which both will have\nbeen cycling will be the same. Second, the total distance traveled by the will be equal to\nthe sum of their individual distances.\nSince we are asked to find the time, let us assign it as a variable 't'. (which is same for\nboth cyclists)\nIn the first row, we know the speed and we have the time in terms of 't'. Thus we can get\nthe following equation :\nDistance(1) = Speed(1) * Time(1) = 14*t\nIn the second row, we know the speed and again we have the time in terms of 't'. Thus\nwe can get the following equation :\nDistance(2) = Speed(2) * Time(2) = 16*t\nNow we know that the total distance traveled is 45 miles and it is equal to the sum of the\ntwo distances. Thus we get the following equation to solve for 't' :\nDistance(3) = Distance(1) + Distance(2) --> 45 = 14*t + 16*t --> t = 1.5 hours\nAnswer : 1.5 hours.\nExample 5.\nA boat travels for three hours with a current of 3 mph and then returns the same distance\nagainst the current in four hours. What is the boat's speed in calm water?\nSolution:\nSince this is a question on round trip, the first thing that should strike us is that the\ndistance going and coming back will be the same.\nNow, we are required to find out the boats speed in calm water. So let us assume it to be\n'b'. Now if speed of the current is 3 mph, then the speed of the boat while going\ndownstream and upstream will be 'b + 3' and 'b - 3' respectively.\nGMAT Club Math Book\n75"} +{"id": "GMAT Club Math Book 2024 v8_p76_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 76, "page_end": 76, "topic_guess": "word_problems", "text": "In the first row, we have the speed of the boat in terms of 'b' and we are given the time.\nThus we can get the following equation :\nDistance(1) = Speed(1) * Time(1) = (b + 3)*3\nIn the second row, we again have the speed in terms of 'b' and we are given the time.\nThus we can get the following equation :\nDistance(2) = Speed(2) * Time(2) = (b - 3)*4\nSince the two distances are equal, we can equate them and solve for 'b'.\nDistance(1) = Distance(2) --> (b + 3)*3 = (b - 3)*4 --> b = 21 mph.\nAnswer : 21 mph. \nGMAT Club Math Book\n76"} +{"id": "GMAT Club Math Book 2024 v8_p77_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 77, "page_end": 77, "topic_guess": "sets_probability_counting", "text": "GMAT Club Math Book\nOverlapping Sets\nhttps://gmatclub.com/forum/overlapping-sets-144260.html\nADVANCED OVERLAPPING SETS\nSome hard GMAT quantitative questions will require you to know and understand the formulas for set theory, presenting three sets and\nasking various questions about them. There are two main formulas to solve questions involving three overlapping sets. Consider the\ndiagram below:\nFIRST FORMULA\n.\nLet's see how this formula is derived.\nWhen we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are counted twice\nand section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and subtract section g\nTWICE (again to count section g only once).\nIn the formula above, , where AnB means intersection of A and B (sections d,\nand g), AnC means intersection of A and C (sections e, and g), and BnC means intersection of B and C (sections f, and g).\nNow, when we subtract (d, and g), (e, and g), and (f, and g) from , we are subtract sections d, e, and f\nONCE BUT section g THREE TIMES (and we need to subtract section g only twice), therefor we should add only section g, which is\nintersection of A, B and C (AnBnC) again to get .\nSECOND FORMULA\n.\nNotice that EXACTLY (only) 2-group overlaps is not the same as 2-group overlaps:\nElements which are common only for A and B are in section d (so elements which are common ONLY for A and B refer to the elements\nwhich are in A and B but not in C);\nElements which are common only for A and C are in section e;\nElements which are common only for B and C are in section f.\nLet's see how this formula is derived.\nTotal = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither\nsum of 2 −group overlaps = AnB + AnC + BnC\nAnB AnC BnC A + B + C\nTotal = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither\nTotal = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗ (all three) +Neither\nGMAT Club Math Book\n77"} +{"id": "GMAT Club Math Book 2024 v8_p78_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 78, "page_end": 78, "topic_guess": "geometry", "text": "Again: when we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are\ncounted twice and section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and\nsubtract section g TWICE (again to count section g only once).\nWhen we subtract from A+B+C we subtract sections d, e, and f once (fine) and next we\nneed to subtract ONLY section g ( ) twice. That's it.\nNow, how this concept can be represented in GMAT problem?\nExample 1:\nWorkers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30\nare on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers\nare on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all\nthree teams. How many workers are there in total?\nTranslating:\n\"are placed on at least one team\": members of none =0;\n\"20 are on the marketing team\": M=20;\n\"30 are on the Sales team\": S=30;\n\"40 are on the Vision team\": V=40;\n\"5 workers are on both the Marketing and Sales teams\": MnS=5, note here that some from these 5 can be the members of Vision team\nas well, MnS is sections d an g on the diagram (assuming Marketing = A, Sales = B and Vision = C);\n\"6 workers are on both the Sales and Vision teams\": SnV=6 (the same as above sections f an g);\n\"9 workers are on both the Marketing and Vision teams\": MnV=9.\n\"4 workers are on all three teams\": MnSnV=4, section 4.\nQuestion: Total=?\nApplying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members:\n.\nAnswer: 74. Discuss this question \nHERE.\nExample 2:\nEach of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three\nacademic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22\nstudents sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students\nsign up for exactly two clubs, how many students sign up for all three clubs?\nTranslating:\n\"Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic\nclubs\": Total=59, Neither=0 (as members are required to sign up for a minimum of one);\n\"22 students sign up for the poetry club\": P=22;\n\"27 students for the history club\": H=27;\n\"28 students for the writing club\": W=28;\n\"6 students sign up for exactly two clubs\": (sum of EXACTLY 2-group overlaps)=6, so the sum of sections d, e, and f is given to be 6,\n(among these 6 students there are no one who is a member of ALL 3 clubs)\nQuestion:: \"How many students sign up for all three clubs?\" --> \nApply second formula: -->\n --> .\nAnswer: 6. Discuss this question \nHERE.\nsum of EXACTLY  2 −group overlaps\nAnBnC\nTotal = M + S + V − (MnS + MnV + SnV ) +MnSnV + Neither = 20 + 30 + 40 − (5 + 6 + 9) + 4 + 0 = 74\nPnHnW = g =?\nTotal = P + H + W − (sum of EXACTLY  2 −group overlaps) − 2 ∗PnHnW + Neither\n59 = 22 + 27 + 28 − 6 − 2 ∗x + 0 x = 6\nGMAT Club Math Book\n78"} +{"id": "GMAT Club Math Book 2024 v8_p79_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 79, "page_end": 79, "topic_guess": "statistics", "text": "Example 3:\nOf 20 Adults, 5 belong to A, 7 belong to B, and 9 belong to C. If 2 belong to all three organizations and 3 belong to exactly\n2 organizations, how many belong to none of these organizations?\nTranslating:\n\"20 Adults\": Total=20;\n\"5 belong to A, 7 belong to B, and 9 belong to C\": A=5, B=7, and C=9;\n\"2 belong to all three organizations\": AnBnC=g=2;\n\"3 belong to exactly 2 organizations\": (sum of EXACTLY 2-group overlaps)=3, so the sum of sections d, e, and f is given to be 3,\n(among these 3 adults there are no one who is a member of ALL 3 clubs)\nQuestion:: Neither=?\nApply second formula: -->\n --> .\nAnswer: 6. Discuss this question \nHERE.\nExample 4:\nThis semester, each of the 90 students in a certain class took at least one course from A, B, and C. If 60 students took A,\n40 students took B, 20 students took C, and 5 students took all the three, how many students took exactly two courses?\nTranslating:\n\"90 students\": Total=90;\n\"of the 90 students in a certain class took at least one course from A, B, and C\": Neither=0;\n\"60 students took A, 40 students took B, 20 students took C\": A=60, B=40, and C=20;\n\"5 students took all the three courses\": AnBnC=g=5;\nQuestion:: (sum of EXACTLY 2-group overlaps)=?\nApply second formula: -->\n --> .\nAnswer: 20. Discuss this question \nHERE.\nExample 5:\nIn the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 10 owned\nrabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?\nTranslating:\n\"60 people own cats, dogs, or rabbits\": Total=60; and Neither=0;\n\"30 people owned cats, 40 owned dogs, 10 owned rabbits\": A=30, B=40, and C=10;\n\"12 owned exactly two of the three types of pet\": (sum of EXACTLY 2-group overlaps)=12;\nQuestion:: AnBnC=g=?\nApply second formula: -->\n --> .\nAnswer: 4. Discuss this question \nHERE.\nExample 6:\nWhen Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E)\nhad 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she\ncompared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw\nthat 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name\nlisted more than once, how many names will be on the combined roster?\nTranslating:\n\"E had 26 names, M had 28, and S had 18\": E=26, M=28, and S=18;\n\"E and M had 9 names in common, E and S had 7, and M and S had 10\": EnM=9, EnS=7, and MnS=10;\nTotal = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither\n20 = 5 + 7 + 9 − 3 − 2 ∗ 2 +Neither Neither = 6\nTotal = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither\n90 = 60 + 40 + 20 −x − 2 ∗ 5 + 0 x = 20\nTotal = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + Neither\n60 = 30 + 40 + 10 − 12 − 2 ∗x + 0 x = 4\nGMAT Club Math Book\n79"} +{"id": "GMAT Club Math Book 2024 v8_p80_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 80, "page_end": 80, "topic_guess": "fractions_decimals_percents", "text": "\"4 names were on all 3 rosters\": EnMnS=g=4;\nQuestion:: Total=?\nApply first formula: -->\n --> .\nAnswer: 50. Discuss this question HERE.\nExample 7:\nThere are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a\ncourse in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses\nand 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?\nTranslating:\n\"There are 50 employees in the office of ABC Company\": Total=50;\n\"22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course\"; A=22, B=15, and\nC=14;\n\"Nine of the employees have taken exactly two of the courses\": (sum of EXACTLY 2-group overlaps)=9;\n\"1 employee has taken all three of the courses\": AnBnC=g=1;\nQuestion:: None=?\nApply second formula: -->\n --> .\nAnswer: 10. Discuss this question \nHERE.\nExample 8 (hard):\nIn a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked\nproduct 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the\nproducts, what percentage of the survey participants liked more than one of the three products?\nTranslating:\n\"85% of those surveyed liked at least one of three products: 1, 2, and 3\": Total=100%. Also, since 85% of those surveyed liked at\nleast one of three products then 15% liked none of three products, thus None=15%;\n\"5% of the people in the survey liked all three of the products\": AnBnC=g=5%;\nQuestion:: what percentage of the survey participants liked more than one of the three products?\nApply second formula:\nTotal = {liked product 1} + {liked product 2} + {liked product 3} - {liked exactly two products} - 2*{liked exactly three\nproduct} + {liked none of three products}\n --> , so 5% liked exactly two products. More than one product liked those who liked\nexactly two products, (5%) plus those who liked exactly three products (5%), so 5+5=10% liked more than one product.\nAnswer: 10%. Discuss this question \nHERE.\nExample 9 (hard):\nIn a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play\nCricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many\nstudents play exactly two of these sports?\nTranslating:\n\"In a class of 50 students...\": Total=50;\n\"20 play Hockey, 15 play Cricket and 11 play Football\": H=20, C=15, and F=11;\n\"7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football\": HnC=7, CnF=4, and HnF=5. Notice that\n\"7 play both Hockey and Cricket\" does not mean that out of those 7, some does not play Football too. The same for Cricket/Football\nand Hockey/Football;\n\"18 students do not play any of these given sports\": Neither=18.\nTotal = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither\nTotal = 26 + 28 + 18 − (9 + 7 + 10) + 4 + 0Total = 50\nTotal = A + B + C − (sum of EXACTLY  2 −group overlaps) − 2 ∗AnBnC + None\n50 = 22 + 15 + 14 − 9 − 2 ∗ 1 +None None = 10\n100 = 50 + 30 + 20 −x − 2 ∗ 5 + 15 x = 5\nGMAT Club Math Book\n80"} +{"id": "GMAT Club Math Book 2024 v8_p81_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 81, "page_end": 81, "topic_guess": "fractions_decimals_percents", "text": "Question:: how many students play exactly two of these sports?\nApply first formula:\n{Total}={Hockey}+{Cricket}+{Football}-{HC+CH+HF}+{All three}+{Neither}\n50=20+15+11-(7+4+5)+{All three}+18 --> {All three}=2;\nThose who play ONLY Hockey and Cricket are 7-2=5;\nThose who play ONLY Cricket and Football are 4-2=2;\nThose who play ONLY Hockey and Football are 5-2=3;\nHence, 5+2+3=10 students play exactly two of these sports.\nAnswer: 10. Discuss this question \nHERE.\nExample 10 (hard DS question on three overlapping sets):\nA student has decided to take GMAT and TOEFL examinations, for which he has allocated a certain number of days for\npreparation. On any given day, he does not prepare for both GMAT and TOEFL. How many days did he allocate for the\npreparation?\n(1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days.\n(2) He prepared for either GMAT or TOEFL on 14 days\nWe have: {Total} = {GMAT } + {TOEFL} - {Both} + {Neither}. Since we are told that \"on any given day, he does not prepare for\nboth GMAT and TOEFL\", then {Both} = 0, so {Total} = {GMAT } + {TOEFL} + {Neither}. We need to find {Total}\n(1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days --> {Total} - {GMAT } = 10 and {Total} - {TOEFL} =12.\nNot sufficient.\n(2) He prepared for either GMAT or TOEFL on 14 days --> {GMAT } + {TOEFL} = 14. Not sufficient.\n(1)+(2) We have three linear equations ({Total} - {GMAT } = 10, {Total} - {TOEFL} =12 and {GMAT } + {TOEFL} = 14) with\nthree unknowns ({Total}, {GMAT }, and {TOEFL}), so we can solve for all of them. Sufficient.\nJust to illustrate. Solving gives:\n{Total} = 18 - he allocate total of 18 days for the preparation;\n{GMAT } = 8 - he prepared for the GMAT on 8 days;\n{TOEFL} = 6 - he prepared for the TOEFL on 6 days;\n{Neither} = 4 - he prepared for neither of them on 4 days.\nAnswer: C. Discuss this question \nHERE.\nExample 11 (disguised three overlapping sets problem):\nThree people each took 5 tests. If the ranges of their scores in the 5 practice tests were 17, 28 and 35, what is the\nminimum possible range in scores of the three test-takers?\nA. 17\nB. 28\nC. 35\nD. 45\nE. 80\nConsider this problem to be an overlapping sets problem:\n# of people in group A is 17;\n# of people in group B is 28;\n# of people in group C is 35;\nWhat is the minimum # of total people possible in all 3 groups? Clearly if two smaller groups A and B are subsets of bigger group C (so\nif all people who are in A are also in C and all people who are in B are also in C), then total # of people in all 3 groups will be 35.\nMinimum # of total people cannot possibly be less than 35 since there are already 35 people in group C.\nGMAT Club Math Book\n81"} +{"id": "GMAT Club Math Book 2024 v8_p82_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 82, "page_end": 82, "topic_guess": "sets_probability_counting", "text": "Answer: C.\nP.S. Notice that max range for the original question is not limited when the max # of people in all 3 groups for revised question is\n17+28+35 (in case there is 0 overlap between the 3 groups).\nAnswer: C. Discuss this question HERE.\n____________________________________________________________________________________________________________\nFor more questions on overlapping sets check our Question Banks\nProblem Solving Questions on Overlapping Sets \nGMAT Club Math Book\n82"} +{"id": "GMAT Club Math Book 2024 v8_p83_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 83, "page_end": 83, "topic_guess": "sets_probability_counting", "text": "GMAT Club Math Book\nProbability\nhttps://gmatclub.com/forum/probability-87244.html\n PROBABILITY\nDefinition\nA number expressing the probability (p) that a specific event will occur, expressed as the\nratio of the number of actual occurrences (n) to the number of possible occurrences (N).\nA number expressing the probability (q) that a specific event will not occur:\nExamples\nCoin\nThere are two equally possible outcomes when we toss a coin: a head (H) or tail (T).\nTherefore, the probability of getting head is 50% or and the probability of getting tail is\n50% or .\nAll possibilities: {H,T}\nDice\np = n\nN\nq = = 1 − p(N−n)\nN\n1\n2\n1\n2\nGMAT Club Math Book\n83"} +{"id": "GMAT Club Math Book 2024 v8_p84_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 84, "page_end": 84, "topic_guess": "sets_probability_counting", "text": "There are 6 equally possible outcomes when we roll a die. The probability of getting any\nnumber out of 1-6 is .\nAll possibilities: {1,2,3,4,5,6}\nMarbles, Balls, Cards...\nLet's assume we have a jar with 10 green and 90 white marbles. If we randomly choose a\nmarble, what is the probability of getting a green marble?\nThe number of all marbles: N = 10 + 90 =100\nThe number of green marbles: n = 10\nProbability of getting a green marble: \nThere is one important concept in problems with marbles/cards/balls. When the first\nmarble is removed from a jar and not replaced, the probability for the second marble\ndiffers ( vs. ). Whereas in case of a coin or dice the probabilities are always the\nsame ( and ). Usually, a problem explicitly states: it is a problem \nwith replacement or\nwithout replacement.\nIndependent events\nTwo events are independent if occurrence of one event does not influence occurrence of\nother events. For n independent events the probability is the product of all probabilities of\nindependent events:\np = p1 * p2 * ... * pn-1 * pn\nor\nP(A and B) = P(A) * P(B) - A and B denote independent events\n1\n6\np = = =n\nN\n10\n100\n1\n10\n9\n99\n10\n100\n1\n6\n1\n2\nGMAT Club Math Book\n84"} +{"id": "GMAT Club Math Book 2024 v8_p85_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 85, "page_end": 85, "topic_guess": "sets_probability_counting", "text": "Example #1\nQ:There is a coin and a die. After one flip and one toss, what is the probability of getting\nheads and a \"4\"?\nSolution: Tossing a coin and rolling a die are independent events. The probability of\ngetting heads is and probability of getting a \"4\" is . Therefore, the probability of\ngetting heads and a \"4\" is:\nExample #2\nQ: If there is a 20% chance of rain, what is the probability that it will rain on the first day\nbut not on the second?\nSolution: The probability of rain is 0.2; therefore probability of sunshine is q = 1 - 0.2 =\n0.8. This yields that the probability of rain on the first day and sunshine on the second\nday is:\nP = 0.2 * 0.8 = 0.16\nExample #3\nQ:There are two sets of integers: {1,3,6,7,8} and {3,5,2}. If Robert chooses randomly\none integer from the first set and one integer from the second set, what is the probability\nof getting two odd integers?\nSolution: There is a total of 5 integers in the first set and 3 of them are odd: {1, 3, 7}.\nTherefore, the probability of getting odd integer out of first set is . There are 3 integers\nin the second set and 2 of them are odd: {3, 5}. Therefore, the probability of getting an\nodd integer out of second set is . Finally, the probability of of getting two odd integers\nis:\nMutually exclusive events\nShakespeare's phrase \"To be, or not to be: that is the question\" is an example of two\nmutually exclusive events.\nTwo events are mutually exclusive if they cannot occur at the same time. For n mutually\nexclusive events the probability is the sum of all probabilities of events:\np = p1 + p2 + ... + pn-1 + pn\nor\nP(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events\nExample #1\nQ: If Jessica rolls a die, what is the probability of getting at least a \"3\"?\nSolution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The\nprobability of each outcome is 1/6. The probability of getting at least a \"3\" is:\n1\n2\n1\n6\nP = ∗ =1\n2\n1\n6\n1\n12\n3\n5\n2\n3\nP = ∗ =3\n5\n2\n3\n2\n5\nGMAT Club Math Book\n85"} +{"id": "GMAT Club Math Book 2024 v8_p86_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 86, "page_end": 86, "topic_guess": "sets_probability_counting", "text": "Combination of independent and mutually exclusive\nevents\nMany probability problems contain combination of both independent and mutually\nexclusive events. To solve those problems it is important to identify all events and their\ntypes. One of the typical problems can be presented in a following general form:\nQ: If the probability of a certain event is p, what is the probability of it occurring k times\nin n-time sequence?\n(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)\nSolution: All events are independent. So, we can say that:\n (1)\nBut it isn't the right answer. It would be right if we specified exactly each position for\nevents in the sequence. So, we need to take into account that there are more than one\noutcomes. Let's consider our example with a coin where \"H\" stands for Heads and \"T\"\nstands for Tails:\nHHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3\nheads and 5 tails. Therefore, we need to include all combinations of heads and tails. In\nour general question, probability of occurring event k times in n-time sequence could be\nexpressed as:\n (2)\nIn the example with a coin, right answer is \nExample #1\nQ.:If the probability of raining on any given day in Atlanta is 40 percent, what is the\nprobability of raining on exactly 2 days in a 7-day period?\nSolution: We are not interested in the exact sequence of event and thus apply formula\n#2:\nA few ways to approach a probability problem\nThere are a few typical ways that you can use for solving probability questions. Let's\nconsider example, how it is possible to apply different approaches:\nExample #1\nQ: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly\nchosen to form a committee, what is the probability that the committee includes both Bob\nand Rachel?\nSolution:\nP = + + + =1\n6\n1\n6\n1\n6\n1\n6\n2\n3\n= ∗ (1 − pP ′ pk )n−k\nP = ∗ ∗ (1 − pCn\nk pk )n−k\nP = ∗ ∗ = ∗C8\n3 0.53 0.55 C8\n3 0.58\nP = ∗ ∗C7\n2 0.42 0.65\nGMAT Club Math Book\n86"} +{"id": "GMAT Club Math Book 2024 v8_p87_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 87, "page_end": 87, "topic_guess": "sets_probability_counting", "text": "1) combinatorial approach: The total number of possible committees is . The\nnumber of possible committee that includes both Bob and Rachel is .\n2) reversal combinatorial approach: Instead of counting probability of occurrence of\ncertain event, sometimes it is better to calculate the probability of the opposite and then\nuse formula p = 1 - q. The total number of possible committees is . The number\nof possible committee that does not includes both Bob and Rachel is:\n where,\n - the number of committees formed from 6 other people.\n - the number of committees formed from Rob or Rachel and one out of 6 other\npeople.\n3) probability approach: The probability of choosing Bob or Rachel as a first person in\ncommittee is 2/8. The probability of choosing Rachel or Bob as a second person when first\nperson is already chosen is 1/7. The probability that the committee includes both Bob and\nRachel is.\n4) reversal probability approach: We can choose any first person. Then, if we have\nRachel or Bob as first choice, we can choose any other person out of 6 people. If we have\nneither Rachel nor Bob as first choice, we can choose any person out of remaining 7\npeople. The probability that the committee includes both Bob and Rachel is.\nExample #2\nQ: Given that there are 5 married couples. If we select only 3 people out of the 10, what\nis the probability that none of them are married to each other?\nSolution:\n1) combinatorial approach:\n - we choose 3 couples out of 5 couples.\n - we chose one person out of a couple.\n - we have 3 couple and we choose one person out of each couple.\n - the total number of combinations to choose 3 people out of 10 people.\n2) reversal combinatorial approach: In this example reversal approach is a bit shorter\nN = C8\n2\nn = 1\nP = = =n\nN\n1\nC8\n2\n1\n28\nN = C8\n2\nm = + 2 ∗C6\n2 C6\n1\nC6\n2\n2 ∗ C6\n1\nP = 1 − = 1 −m\nN\n+2∗C6\n2 C6\n1\nC8\n2\nP = 1 − = 1 − =15+2∗6\n28\n27\n28\n1\n28\nP = ∗ = =2\n8\n1\n7\n2\n56\n1\n28\nP = 1 − ( ∗ + ∗ 1) = =2\n8\n6\n7\n6\n8\n2\n56\n1\n28\nC5\n3\nC2\n1\n(C2\n1 )3\nC10\n3\np = = =\n∗(C5\n3 C2\n1 )3\nC10\n3\n10∗8\n10∗3∗4\n2\n3\nGMAT Club Math Book\n87"} +{"id": "GMAT Club Math Book 2024 v8_p88_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 88, "page_end": 88, "topic_guess": "sets_probability_counting", "text": "and faster.\n - we choose 1 couple out of 5 couples.\n - we chose one person out of remaining 8 people.\n - the total number of combinations to choose 3 people out of 10 people.\n3) probability approach:\n1st person: - we choose any person out of 10.\n2nd person: - we choose any person out of 8=10-2(one couple from previous choice)\n3rd person: - we choose any person out of 6=10-4(two couples from previous choices).\nProbability tree\nSometimes, at 700+ level you may see complex probability problems that include\nconditions or restrictions. For such problems it could be helpful to draw a probability tree\nthat include all possible outcomes and their probabilities.\nExample #1\nQ: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If\nthe first ball is not green, she takes the second ball (without replacing first) and she wins\nif the two balls are white or if the first ball is gray and the second ball is white. What is the\nprobability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?\nSolution: Let's draw all possible outcomes and calculate all probabilities.\nNow, It is pretty obvious that the probability of Julia's win is:\nC5\n1\nC8\n1\nC10\n3\np = 1 − = 1 − =\n∗C5\n1 C8\n1\nC10\n3\n5∗8\n10∗3∗4\n2\n3\n= 110\n10\n8\n9\n6\n8\np = 1 ∗ ∗ =8\n9\n6\n8\n2\n3\nP = + ∗ + ∗ =4\n7\n2\n7\n1\n6\n1\n7\n2\n6\n2\n3\nGMAT Club Math Book\n88"} +{"id": "GMAT Club Math Book 2024 v8_p89_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 89, "page_end": 89, "topic_guess": "word_problems", "text": "Tips and Tricks: Symmetry\nSymmetry sometimes lets you solve seemingly complex probability problem in a few\nseconds. Let's consider an example:\nExample #1\nQ: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel.\nHow many ways it can be done ?\nSolution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly\n1/2 of all possible ways:\nN = ∗ = 101\n2 P 5\n2\nGMAT Club Math Book\n89"} +{"id": "GMAT Club Math Book 2024 v8_p90_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 90, "page_end": 90, "topic_guess": "number_theory", "text": "Practice Questions\nEasy:\n1. https://gmatclub.com/forum/in-a-box-of- ... 07395.html\n2. https://gmatclub.com/forum/among-a-grou ... 34499.html\n3. https://gmatclub.com/forum/the-probabil ... 44730.html\n4. https://gmatclub.com/forum/xavier-yvonn ... 65822.html\n5. https://gmatclub.com/forum/carol-purcha ... 21728.html\n6. https://gmatclub.com/forum/two-integers ... 43451.html\n7. https://gmatclub.com/forum/raffle-ticke ... 36563.html\n8. https://gmatclub.com/forum/sixty-percen ... 43900.html\n9. https://gmatclub.com/forum/in-a-set-of- ... 68915.html\n10. https://gmatclub.com/forum/a-committee- ... 44443.html\nMedium:\n1. https://gmatclub.com/forum/if-x-is-to-b ... 67089.html\n2. https://gmatclub.com/forum/a-gardener-i ... 99822.html\n3. https://gmatclub.com/forum/a-contest-wi ... 38710.html\n4. https://gmatclub.com/forum/in-a-certain ... 69115.html\n5. https://gmatclub.com/forum/a-committee- ... 81051.html\n6. https://gmatclub.com/forum/two-thirds-o ... 10059.html\n7. https://gmatclub.com/forum/a-deck-of-ca ... 21976.html\n8. https://gmatclub.com/forum/set-s-consis ... 24713.html\n9. https://gmatclub.com/forum/john-throws- ... 43960.html\n10. https://gmatclub.com/forum/a-fair-die-w ... 84968.html\nHard:\n1. https://gmatclub.com/forum/a-couple-dec ... 68730.html\n2. https://gmatclub.com/forum/mary-and-joe ... 86407.html\n3. https://gmatclub.com/forum/a-box-contai ... 27699.html\n4. https://gmatclub.com/forum/two-dice-are ... 26477.html\n5. https://gmatclub.com/forum/the-table-ab ... 05918.html\n6. https://gmatclub.com/forum/minimum-of-h ... 73869.html\n7. https://gmatclub.com/forum/kate-and-dav ... 97177.html\n8. https://gmatclub.com/forum/when-an-unfa ... 98795.html\n9. https://gmatclub.com/forum/artificial-i ... 21978.html\n10. https://gmatclub.com/forum/a-deck-of-ca ... 21977.html\nGMAT Club Math Book\n90"} +{"id": "GMAT Club Math Book 2024 v8_p91_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 91, "page_end": 91, "topic_guess": "sets_probability_counting", "text": "GMAT Club Math Book\nCombinations\nhttps://gmatclub.com/forum/combinations-87345.html\n COMBINATORICS\ncreated by: walker\nedited by: bb, Bunuel\nDefinition\nCombinatorics is the branch of mathematics studying the enumeration, combination, and\npermutation of sets of elements and the mathematical relations that characterize their\nproperties.\nEnumeration\nEnumeration is a method of counting all possible ways to arrange elements. Although it is\nthe simplest method, it is often the fastest method to solve hard GMAT problems and is a\npivotal principle for any other combinatorial method. In fact, combination and permutation\nis shortcuts for enumeration. The main idea of enumeration is writing down all possible\nways and then count them. Let's consider a few examples:\nExample #1\nQ:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible\nto arrange marbles in a row?\nSolution: Let's write out all possible ways:\nAnswer is 6.\nIn general, the number of ways to arrange n different objects in a row\nExample #2\nQ:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible\nGMAT Club Math Book\n91"} +{"id": "GMAT Club Math Book 2024 v8_p92_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 92, "page_end": 92, "topic_guess": "statistics", "text": "to arrange marbles in a row if blue and green marbles have to be next to each other?\nSolution: Let's write out all possible ways to arrange marbles in a row and then find only\narrangements that satisfy question's condition:\nAnswer is 4.\nExample #3\nQ:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible\nto arrange marbles in a row if gray marble have to be left to blue marble?\nSolution: Let's write out all possible ways to arrange marbles in a row and then find only\narrangements that satisfy question's condition:\nAnswer is 3.\nArrangements of n different objects\nEnumeration is a great way to count a small number of arrangements. But when the total\nnumber of arrangements is large, enumeration can't be very useful, especially taking into\naccount GMAT time restriction. Fortunately, there are some methods that can speed up\ncounting of all arrangements.\nThe number of arrangements of n different objects in a row is a typical problem that can\nbe solve this way:\n1. How many objects we can put at 1st place? n.\n2. How many objects we can put at 2nd place? n - 1. We can't put the object that already\nplaced at 1st place.\n.....\nn. How nany objects we can put at n-th place? 1. Only one object remains.\nTherefore, the total number of arrangements of n different objects in a row is\nN = n ∗ ( n − 1) ∗ ( n − 2). . . .2 ∗ 1 = n!\nGMAT Club Math Book\n92"} +{"id": "GMAT Club Math Book 2024 v8_p93_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 93, "page_end": 93, "topic_guess": "sets_probability_counting", "text": "Combination\nA combination is an unordered collection of k objects taken from a set of n distinct\nobjects. The number of ways how we can choose k objects out of n distinct objects is\ndenoted as:\nknowing how to find the number of arrangements of n distinct objects we can easily find\nformula for combination:\n1. The total number of arrangements of n distinct objects is n!\n2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects\n((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter.\nPermutation\nA permutation is an \nordered collection of k objects taken from a set of n distinct objects.\nThe number of ways how we can choose k objects out of n distinct objects is denoted as:\nknowing how to find the number of arrangements of n distinct objects we can easily find\nformula for combination:\n1. The total number of arrangements of n distinct objects is n!\n2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the\norder of remained (n-k) objects doesn't matter.\nIf we exclude order of chosen objects from permutation formula, we will get combination\nformula:\nCn\nk\n=Cn\nk\nn!\nk!(n−k)!\nP n\nk\n=P n\nk\nn!\n(n−k)!\n=\nP n\nk\nk! Cn\nk\nGMAT Club Math Book\n93"} +{"id": "GMAT Club Math Book 2024 v8_p94_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 94, "page_end": 94, "topic_guess": "statistics", "text": "Circular arrangements\nLet's say we have 6 distinct objects, how many relatively different arrangements do we\nhave if those objects should be placed in a circle.\nThe difference between placement in a row and that in a circle is following: if we shift all\nobject by one position, we will get different arrangement in a row but the same relative\narrangement in a circle. So, for the number of circular arrangements of n objects we\nhave:\nTips and Tricks\nAny problem in Combinatorics is a counting problem. Therefore, a key to solution is a way\nhow to count the number of arrangements. It sounds obvious but a lot of people begin\napproaching to a problem with thoughts like \"Should I apply C- or P- formula here?\".\nDon't fall in this trap: define how you are going to count arrangements first, realize that\nyour way is right and you don't miss something important, and only then use C- or P-\nformula if you need them.\nResources\nWalker's post with Combinatorics/probability problems: \nCombinatorics/probability\nProblems\nR = = ( n − 1)!n!\nn\nGMAT Club Math Book\n94"} +{"id": "GMAT Club Math Book 2024 v8_p95_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 95, "page_end": 95, "topic_guess": "general", "text": "Practice Questions\nEasy:\n1. https://gmatclub.com/forum/there-are-8- ... 34582.html\n2. https://gmatclub.com/forum/there-are-5- ... 88100.html\n3. https://gmatclub.com/forum/a-certain-un ... 68638.html\n4. https://gmatclub.com/forum/the-diagram- ... 44271.html\n5. https://gmatclub.com/forum/in-a-tournam ... 55939.html\n6. https://gmatclub.com/forum/how-many-way ... 47677.html\n7. https://gmatclub.com/forum/in-how-many- ... 05506.html\n8. https://gmatclub.com/forum/a-certain-ba ... 98055.html\n9. https://gmatclub.com/forum/how-many-dif ... 54633.html\n10. https://gmatclub.com/forum/if-a-person- ... 24843.html\nMedium:\n1. https://gmatclub.com/forum/the-letters- ... 20320.html\n2. https://gmatclub.com/forum/team-a-and-t ... 41461.html\n3. https://gmatclub.com/forum/the-subsets- ... 08256.html\n4. https://gmatclub.com/forum/each-in-the- ... 95162.html\n5. https://gmatclub.com/forum/a-company-th ... 68427.html\n6. https://gmatclub.com/forum/clarissa-wil ... 42452.html\n7. https://gmatclub.com/forum/in-the-table ... 66572.html\n8. https://gmatclub.com/forum/if-a-code-wo ... 26652.html\n9. https://gmatclub.com/forum/how-many-two ... 28130.html\n10. https://gmatclub.com/forum/john-has-12- ... 07307.html\nHard:\n1. https://gmatclub.com/forum/pat-will-wal ... 68374.html\n2. https://gmatclub.com/forum/five-integer ... 05923.html\n3. https://gmatclub.com/forum/in-how-many- ... 85707.html\n4. https://gmatclub.com/forum/how-many-fiv ... 91597.html\n5. https://gmatclub.com/forum/nine-family- ... 75071.html\n6. https://gmatclub.com/forum/how-many-pos ... 55804.html\n7. https://gmatclub.com/forum/how-many-odd ... 94655.html\n8. https://gmatclub.com/forum/how-many-dif ... 71774.html\n9. https://gmatclub.com/forum/a-committee- ... 04966.html\n10. https://gmatclub.com/forum/in-how-many- ... 56400.html"} +{"id": "GMAT Club Math Book 2024 v8_p95_c2", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 95, "page_end": 95, "topic_guess": "word_problems", "text": "forum/how-many-dif ... 71774.html\n9. https://gmatclub.com/forum/a-committee- ... 04966.html\n10. https://gmatclub.com/forum/in-how-many- ... 56400.html\n\nPage 1 of 1 All times are UTC - 8 hours [ DST ]\nGMAT Club Math Book\n95"} +{"id": "GMAT Club Math Book 2024 v8_p96_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 96, "page_end": 96, "topic_guess": "statistics", "text": "GMAT Club Math Book\nStandard Deviation\nhttps://gmatclub.com/forum/standard-deviation-87905.html\n STANDARD DEVIATION\nDefinition\nStandard Deviation (SD, or STD or ) - a measure of the dispersion or variation in a\ndistribution, equal to the square root of variance or the arithmetic mean (average) of\nsquares of deviations from the arithmetic mean.\nIn simple terms, it shows how much variation there is from the \"average\" (mean). It may\nbe thought of as the average difference from the mean of distribution, how far data points\nare away from the mean. A low standard deviation indicates that data points tend to be\nvery close to the mean, whereas high standard deviation indicates that the data are\nspread out over a large range of values.\nσ\nvariance =\n∑( −xi xav)2\nN\nσ = =variance− − − − − − −√\n∑( −xi xav)2\nN\n− − − − − − − − −\n√\nGMAT Club Math Book\n96"} +{"id": "GMAT Club Math Book 2024 v8_p97_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 97, "page_end": 97, "topic_guess": "statistics", "text": "Properties\n;\n only if all elements in a set is equal;\nLet standard deviation of be and mean of the set be :\nStandard deviation of is . Decrease/increase in all elements of a set by a\nconstant percentage will decrease/increase standard deviation of the set by the same\npercentage.\nStandard deviation of is . Decrease/increase in all elements of a set by a\nconstant value DOES NOT decrease/increase standard deviation of the set.\nσ ≥ 0\nσ = 0\n{ }xi σ m\n{ }\nxi\na =σ\n′ σ\na\n{ + a}xi = σσ\n′\nGMAT Club Math Book\n97"} +{"id": "GMAT Club Math Book 2024 v8_p98_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 98, "page_end": 98, "topic_guess": "statistics", "text": "if a new element is added to set and standard deviation of a new set is\n, then:\n1) if \n2) if \n3) if \n4) is the lowest if \nTips and Tricks\nGMAC in majority of problems doesn't ask you to calculate standard deviation. Instead it\ntests your intuitive understanding of the concept. In 90% cases it is a faster way to use\njust average of instead of true formula for standard deviation, and treat\nstandard deviation as \"\naverage difference between elements and mean\". Therefore,\nbefore trying to calculate standard deviation, maybe you can solve a problem much faster\nby using just your intuition.\nAdvance tip. Not all points contribute equally to standard deviation. Taking into account\nthat standard deviation uses sum of squares of deviations from mean, the most remote\npoints will essentially contribute to standard deviation. For example, we have a set A that\nhas a mean of 5. The point 10 gives in sum of squares but point 6 gives\nonly . 25 times the difference! So, when you need to find what set has the\nlargest standard deviation, always look for set with the largest range because remote\npoints have a very significant contribution to standard deviation.\nExamples\nExample #1\nQ: There is a set . If we create a new set that consists of\nall elements of the initial set but decreased by 17%, what is the change in standard\ndeviation?\nSolution: We don't need to calculate as we know rule that decrease in all elements of a\nset by a constant percentage will decrease standard deviation of the set by the same\npercentage. So, the decrease in standard deviation is 17%.\nExample #2\nQ: There is a set of consecutive even integers. What is the standard deviation of the set?\n(1) There are 39 elements in the set.\n(2) the mean of the set is 382.\nSolution: Before reading Data Sufficiency statements, what can we say about the\nquestion? What should we know to find standard deviation? \"consecutive even integers\"\ny { }xi {{ }, y}xi\nσ\n′\n> σσ\n′\n|y − m| > σ\n= σσ\n′ |y − m| = σ\n< σσ\n′\n|y − m| < σ\nσ\n′\ny = m\n| − |xi xav\n(10 − 5 = 25)2\n(6 − 5 = 1)2\n{67, 32, 76, 35, 101, 45, 24, 37}\nGMAT Club Math Book\n98"} +{"id": "GMAT Club Math Book 2024 v8_p99_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 99, "page_end": 99, "topic_guess": "statistics", "text": "means that all elements strictly related to each other. If we shift the set by adding or\nsubtracting any integer, does it change standard deviation (average deviation of elements\nfrom the mean)? No. One thing we should know is the number of elements in the set,\nbecause the more elements we have the broader they are distributed relative to the\nmean. Now, look at DS statements, all we need it is just first statement. So, A is\nsufficient.\nExample #3\nQ: Standard deviation of set is 18.3. How many elements\nare 1 standard deviation above the mean?\nSolution: Let's find mean: \nNow, we need to count all numbers greater than 42+18.3=60.3. It is one number - 76.\nThe answer is 1.\nExample #4\nQ: There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form\na new set B by adding 2 more elements to the set A. What two elements will decrease the\nstandard deviation the most?\nA) 9 and 3\nB) -3 and 3\nC) 6 and 1\nD) 4 and 5\nE) 5 and 5\nSolution: The closer to the mean, the greater decrease in standard deviation. D has 4\n(equal our mean) and 5 (differs from mean only by 1). All other options have larger\ndeviation from mean.\nNormal distribution\nIt is a more advance concept that you will never see in GMAT but understanding statistic\nproperties of standard deviation can help you to be more confident about simple\nproperties stated above.\nIn probability theory and statistics, the normal distribution or Gaussian distribution is a\ncontinuous probability distribution that describes data that cluster around a mean or\naverage. Majority of statistical data can be characterized by normal distribution.\n{23, 31, 76, 45, 16, 55, 54, 36}\nm = = 4223+31+76+45+16+55+54+36\n8\nGMAT Club Math Book\n99"} +{"id": "GMAT Club Math Book 2024 v8_p100_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 100, "page_end": 100, "topic_guess": "general", "text": "covers 68% of data\n covers 95% of data\n covers 99% of data\nm − σ < x < m + σ\nm − 2σ < x < m + 2σ\nm − 3σ < x < m + 3σ\nGMAT Club Math Book\n100"} +{"id": "GMAT Club Math Book 2024 v8_p101_c1", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 101, "page_end": 101, "topic_guess": "statistics", "text": "Practice Questions\nEasy:\n1. https://gmatclub.com/forum/the-data-set ... 76313.html\n2. https://gmatclub.com/forum/which-of-the ... 31485.html\n3. https://gmatclub.com/forum/a-certain-ch ... 43982.html\n4. https://gmatclub.com/forum/if-d-is-the- ... 93979.html\n5. https://gmatclub.com/forum/for-a-certai ... 28661.html\n6. https://gmatclub.com/forum/a-certain-li ... 59504.html\n7. https://gmatclub.com/forum/the-standard ... 99221.html\n8. https://gmatclub.com/forum/the-resident ... 83362.html\n9. https://gmatclub.com/forum/the-arithmet ... 29117.html\n10. https://gmatclub.com/forum/the-mean-and ... 98248.html\nMedium:\n1. https://gmatclub.com/forum/a-certain-ch ... 43982.html\n2. https://gmatclub.com/forum/set-s-consis ... 51019.html\n3. https://gmatclub.com/forum/a-vending-ma ... 93351.html\n4. https://gmatclub.com/forum/which-of-the ... 62530.html\n5. https://gmatclub.com/forum/set-j-consis ... 06956.html\n6. https://gmatclub.com/forum/the-table-ab ... 20477.html\n7. https://gmatclub.com/forum/if-a-certain ... 30542.html\n8. https://gmatclub.com/forum/a-researcher ... 34893.html\n9. https://gmatclub.com/forum/a-certain-li ... 03989.html\nHard:\n1. https://gmatclub.com/forum/if-m-is-a-ne ... 26819.html\n2. https://gmatclub.com/forum/set-q-consis ... 55781.html\n3. https://gmatclub.com/forum/which-of-the ... 00345.html\n4. https://gmatclub.com/forum/the-list-sho ... 91966.html\n5. https://gmatclub.com/forum/a-researcher ... 59754.html\n6. https://gmatclub.com/forum/e-is-a-colle ... 99774.html\n7. https://gmatclub.com/forum/set-a-consis ... 10876.html\n8. https://gmatclub.com/forum/if-j-k-m-n-a ... 72294.html\n9. https://gmatclub.com/forum/which-of-the ... 90680.html\n10. https://gmatclub.com/forum/the-list-abo ... 93102.html"} +{"id": "GMAT Club Math Book 2024 v8_p101_c2", "source_name": "GMAT Club Math Book 2024 v8", "source_file": "GMAT Club Math Book 2024 v8.pdf", "source_type": "pdf", "page_start": 101, "page_end": 101, "topic_guess": "word_problems", "text": "forum/if-j-k-m-n-a ... 72294.html\n9. https://gmatclub.com/forum/which-of-the ... 90680.html\n10. https://gmatclub.com/forum/the-list-abo ... 93102.html\n\nPage 1 of 1 All times are UTC - 8 hours [ DST ]\nGMAT Club Math Book\n101"} +{"id": "book 2_p3_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 3, "page_end": 3, "topic_guess": "word_problems", "text": "MANHATTAN PREP\nGMAT Advanced Quant\nGMAT STRATEGY GUIDE\nThis supplemental guide provides in-depth and comprehensive\nexplanations of the advanced math skills necessary for the highest-level\nperformance on the GMAT.\nGMAT® is a registered trademark of the Graduate Management\nAdmissions Council™. Manhattan Prep is neither endorsed by nor\naffiliated with GMAC."} +{"id": "book 2_p4_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 4, "page_end": 4, "topic_guess": "general", "text": "Table of Contents"} +{"id": "book 2_p5_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 5, "page_end": 5, "topic_guess": "strategy", "text": "GMAT Advanced Quant\nCover\nTitle Page\nCopyright\nInstructional Guide Series\nLetter\nIntroduction\nIn This Chapter...\nA Qualified Welcome\nWho Should Use This Book\nTry It Yourself\nThe Purpose of This Book\nAn Illustration\nLearning How to Think\nPlan of This Book\nSolutions to Try-It Problems\nPart 1: Problem Solving and Data Sufficiency Strategies\nChapter 1 Problem Solving: Advanced Principles\nIn This Chapter...\nChapter 1 Problem Solving: Advanced Principles\nPrinciple #1: Understand the Basics\nPrinciple #2: Build a Plan\nPrinciple #3: Solve—and Put Pen to Paper\nPrinciple #4: Review Your Work\nProblem Set\nSolutions\nChapter 2: Problem Solving: Strategies & Tactics\nIn This Chapter...\nChapter 2 Problem Solving: Strategies & Tactics\nAdvanced Strategies\nAdvanced Guessing Tactics\nProblem Set\nSolutions\nChapter 3: Data Sufficiency: Principles\nIn This Chapter...\nChapter 3 Data Sufficiency: Principles\nPrinciple #1: Follow a Consistent Process"} +{"id": "book 2_p6_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 6, "page_end": 6, "topic_guess": "algebra", "text": "Principle #2: Never Rephrase Yes/No as Value\nPrinciple #3: Work from Facts to Question\nPrinciple #4: Be a Contrarian\nPrinciple #5: Assume Nothing\nProblem Set\nSolutions\nChapter 4: Data Sufficiency: Strategies & Tactics\nIn This Chapter...\nChapter 4 Data Sufficiency: Strategies & Tactics\nAdvanced Strategies\nAdvanced Guessing Tactics\nSummary\nCommon Wrong Answers\nProblem Set\nSolutions\nPart 2: Strategies for All Problem Types\nChapter 5 Pattern Recognition\nIn This Chapter...\nPattern Recognition Problems\nSequence Problems\nUnits (Ones) Digit Problems\nRemainder Problems\nOther Pattern Problems\nProblem Set\nSolutions\nChapter 6: Common Terms and Quadratic Templates\nIn This Chapter...\nChapter 6 Common Terms and Quadratic Templates\nCommon Terms\nQuadratic Templates\nQuadratic Templates in Disguise\nProblem Set\nSolutions\nChapter 7: Visual Solutions\nIn This Chapter...\nChapter 7 Visual Solutions\nRepresenting Objects with Pictures"} +{"id": "book 2_p7_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 7, "page_end": 7, "topic_guess": "sequences_patterns", "text": "Rubber Band Geometry\nBaseline Calculations for Averages\nNumber Line Techniques for Statistics Problems\nProblem Set\nSolutions\nChapter 8: Hybrid Problems\nIn This Chapter...\nPop Quiz!\nHybrid Problems\nIdentify and Sequence the Parts\nWhere to Start\nMinor Hybrids\nProblem Set\nSolutions\nPart 3: Practice\nChapter 9 Workout Sets\nIn This Chapter...\nWorkout Set 1\nWorkout Set 1 Answer Key\nWorkout Set 1 Solutions\nWorkout Set 2\nWorkout Set 2 Answer Key\nWorkout Set 2 Solutions\nWorkout Set 3\nWorkout Set 3 Answer Key\nWorkout Set 3 Solutions\nWorkout Set 4\nWorkout Set 4 Answer Key\nWorkout Set 4 Solutions\nWorkout Set 5\nWorkout Set 5 Answer Key\nWorkout Set 5 Solutions\nWorkout Set 6\nWorkout Set 6: Answer Key\nWorkout Set 6 Solutions\nWorkout Set 7\nWorkout Set 7 Answer Key"} +{"id": "book 2_p8_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 8, "page_end": 8, "topic_guess": "general", "text": "Workout Set 7 Solutions\nWorkout Set 8\nWorkout Set 8 Answer Key\nWorkout Set 8 Solutions\nWorkout Set 9\nWorkout Set 9: Answer Key\nWorkout Set 9 Solutions\nWorkout Set 10\nWorkout Set 10 Answer Key\nWorkout Set 10 Solutions\nWorkout Set 11\nWorkout Set 11 Answer Key\nWorkout Set 11 Solutions\nWorkout Set 12\nWorkout Set 12 Answer Key\nWorkout Set 12 Solutions\nWorkout Set 13\nWorkout Set 13 Answer Key\nWorkout Set 13 Solutions\nWorkout Set 14\nWorkout Set 14 Answer Key\nWorkout Set 14 Solutions\nWorkout Set 15\nWorkout Set 15 Answer Key\nWorkout Set 15 Solutions\nWorkout Set 16\nWorkout Set 16 Answer Key\nWorkout Set 16: Answers and Explanations\nmba Mission\nmba Mission\nGo Beyond Books. Try A Free Class Now.\nPrep Made Personal"} +{"id": "book 2_p9_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 9, "page_end": 9, "topic_guess": "coordinate_geometry", "text": "Acknowledgements\nA great number of people were involved in the creation of the book you are holding.\nOur Manhattan Prep resources are based on the continuing experiences of our instructors and\nstudents. The overall vision for this edition was developed by Chelsey Cooley, who determined\nwhat new areas to cover and who wrote all of the problems that are new to this edition.\nChelsey served as the primary author of this edition and Emily Meredith Sledge was the primary\neditor; Emily also served as the primary author of the first edition of this guide. Mario Gambino\nmanaged production for the many—and quite complicated—images that appear in this guide.\nMatthew Callan coordinated the production work for this guide. Once the manuscript was done,\nNaomi Beesen and Ben Ku edited and Cheryl Duckler and Stacey Koprince proofread the entire\nguide from start to finish. Carly Schnur designed the covers.\nRetail ISBN: 978-1-5062-4993-3\nRetail eISBN: 978-1-5062-4994-0\nCourse ISBN: 978-1-5062-4995-7\nCourse eISBN: 978-1-5062-4996-4\nCopyright © 2020 Manhattan Prep, Inc.\nALL RIGHTS RESERVED. No part of this work may be reproduced or used in any form or by any\nmeans—graphic, electronic, or mechanical, including photocopying, recording, taping, web\ndistribution—without the prior written permission of the publisher, MG Prep, Inc.\nGMAT® is a registered trademark of the Graduate Management Admission Council. Manhattan Prep\nis neither endorsed by nor affiliated with GMAC."} +{"id": "book 2_p10_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 10, "page_end": 10, "topic_guess": "word_problems", "text": "GMAT® STRATEGY GUIDES\nGMAT All the Quant\nGMAT All the Verbal\nGMAT Integrated Reasoning and Essay\nSTRATEGY GUIDE SUPPLEMENTS\nMath Verbal \nGMAT Foundations of Math\nGMAT Advanced Quant\nGMAT Foundations of Verbal"} +{"id": "book 2_p11_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 11, "page_end": 11, "topic_guess": "word_problems", "text": "January 7, 2020\nDear Student,\nThank you for picking up a copy of Advanced Quant. I hope this book\nprovides just the guidance you need to get the most out of your GMAT\nstudies.\nAt Manhattan Prep, we continually aspire to provide the best instructors\nand resources possible. If you have any questions or feedback, please do\nnot hesitate to contact us.\nEmail our Student Services team at gmat@manhattanprep.com or give us\na shout at 212-721-7400 (or 800-576-4628 in the United States or Canada).\nWe try to keep all our books free of errors, but if you think we’ve goofed,\nplease visit manhattanprep.com/GMAT/errata.\nOur Manhattan Prep Strategy Guides are based on the continuing\nexperiences of both our instructors and our students. The primary author\nof the this edition of the Advanced Quant guide was Chelsey Cooley and\nthe primary editor was Emily Meredith Sledge. Project management and\ndesign were led by Matthew Callan and Mario Gambino. I’d like to send"} +{"id": "book 2_p12_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 12, "page_end": 12, "topic_guess": "word_problems", "text": "particular thanks to instructors Stacey Koprince and Ben Ku for their\ncontent contributions.\nFinally, we are indebted to all of the Manhattan Prep students who have\ngiven us excellent feedback over the years. This book wouldn’t be half of\nwhat it is without their voice.\nAnd now that you are one of our students too, please chime in! I look\nforward to hearing from you. Thanks again and best of luck preparing for\nthe GMAT!\nSincerely,\nChris Ryan\nExecutive Director, Product Strategy\nwww.manhattanprep.com/gmat 138 West 25th Street, 7th Floor, New\nYork, NY 10001 Tel: 212-721-7400 Fax: 646-514-7425"} +{"id": "book 2_p13_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 13, "page_end": 13, "topic_guess": "general", "text": "Introduction"} +{"id": "book 2_p14_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 14, "page_end": 14, "topic_guess": "fractions_decimals_percents", "text": "In This Chapter...\nA Qualified Welcome\nWho Should Use This Book\nTry It Yourself \nThe Purpose of This Book\nAn Illustration\nLearning How to Think\nPlan of This Book\nSolutions to Try-It Problems"} +{"id": "book 2_p15_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 15, "page_end": 15, "topic_guess": "coordinate_geometry", "text": "Introduction\nA Qualified Welcome\nWelcome to GMAT Advanced Quant! In this venue, we decided to be a\nlittle nerdy and call the introduction Chapter 0. A er all, the point (0, 0) in\nthe coordinate plane is called the origin, isn’t it? (That’s the first and last\nmath joke in this book.)\nUnfortunately, we have to qualify our welcome right away, because this\nbook isn’t for everyone. At least, it’s not for everyone right away."} +{"id": "book 2_p16_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 16, "page_end": 16, "topic_guess": "fractions_decimals_percents", "text": "Who Should Use This Book\nYou should use this book if you meet the following conditions:\nIf you match this description, then please turn the page.\nYou have achieved a scaled score of at least 47 (out of 51) on the Quant\nsection of either the Manhattan Prep practice test or the official practice\ncomputer-adaptive test (CAT).\nYou have worked through the Manhattan Prep All the Quant guide,\nwhich covers all of the topics and strategies you need for the Quant\nsection, or you have worked through similar material from another\ncompany. This material should include the following:\nAlgebra\nFractions, Decimals, Percents, and Ratios\nGeometry\nNumber Properties\nWord Problems\nYou are already comfortable with the core principles in these topics.\nYou want to raise your performance to a scaled score of 49 or higher.\nYou want to become a significantly smarter test-taker."} +{"id": "book 2_p17_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 17, "page_end": 17, "topic_guess": "fractions_decimals_percents", "text": "If you don’t match this description, then you will probably find this book\ntoo difficult at this stage of your preparation.\nFor now, you are better off working on topic-focused material, as found in\nthe All the Quant guide, and ensuring that you have mastered that material\nbefore you return to this book."} +{"id": "book 2_p18_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 18, "page_end": 18, "topic_guess": "general", "text": "Try It Yourself\nThroughout the chapters of this guide, you’ll see Try-It problems—\nproblems designed to test your skills on certain aspects of GMAT problems.\nTake a look at the following three Try-It problems, which are very difficult.\nThey are at least as hard as any real GMAT problem—probably even harder.\nGo ahead and give these problems a try. You should not expect to solve any\nof them in two minutes. In fact, you might find yourself completely stuck. If\nthat’s the case, switch gears. Do your best to eliminate some incorrect\nanswer choices and take an educated guess.\nTry-It #0-1\nA jar is filled with red, white, and blue tokens that are equivalent\nexcept for their color. The chance of randomly selecting a red\ntoken, replacing it, then randomly selecting a white token is the\nsame as the chance of randomly selecting a blue token. If the\nnumber of tokens of every color is a multiple of 3, what is the\nsmallest possible total number of tokens in the jar?\n 9(A)\n12(B)\n15(C)\n18(D)"} +{"id": "book 2_p19_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 19, "page_end": 19, "topic_guess": "coordinate_geometry", "text": "Try-It #0-2\nArrow \n , which is a line segment exactly 5 units long with an\narrowhead at A, is constructed in the xy-plane. The x- and y-\ncoordinates of A and B are integers that satisfy the inequalities 0 ≤\nx ≤ 9 and 0 ≤ y ≤ 9. How many different arrows with these -\nproperties can be constructed?\nTry-It #0-3\nIn the diagram to the right, the value\nof x is closest to which of the\nfollowing? \n21(E)\n 50(A)\n168(B)\n200(C)\n368(D)\n536(E)\n(A)\n2(B)\n(C)\n(D)\n1(E)"} +{"id": "book 2_p20_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 20, "page_end": 20, "topic_guess": "general", "text": "(Note: This problem does not require any non-GMAT math, such as\ntrigonometry.)"} +{"id": "book 2_p21_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 21, "page_end": 21, "topic_guess": "number_theory", "text": "The Purpose of This Book\nThis book is designed to prepare you for the most difficult math problems\non the GMAT.\nSo…what is a difficult math problem, from the point of view of the GMAT?\nA difficult math problem is one that most GMAT test-takers get wrong\nunder exam conditions. In fact, this is essentially how the GMAT measures\ndifficulty: by the percent of test-takers who get the problem wrong.\nSo what kinds of math questions do most test-takers get wrong? What\ncharacterizes these problems? There are two kinds of features:\nTopical nuances or obscure principles\nThese topical nuances are largely covered in the Extra sections of\nthe Manhattan Prep All the Quant guide. This book includes many\nproblems that involve topical nuances. However, the exhaustive\ntheory of divisibility and primes, for instance, is not repeated here.\n1.\nConnected to a particular topic\nInherently hard to grasp or unfamiliar\nEasy to mix up"} +{"id": "book 2_p22_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 22, "page_end": 22, "topic_guess": "word_problems", "text": "Complex structures\nComplex structures are essentially disguises for simpler content.\nThese disguises may be difficult to pierce. The path to the answer is\ntwisted or clouded somehow.\nTo solve problems that have simple content but complex structures,\nyou need approaches that are both more general and more\ncreative. This book focuses on these more general and more\ncreative approaches.\nThe three problems on the previous page have complex structures;\nthe solutions are a bit later in this chapter. In the meantime, take a\nlook at another problem.\n2.\nMay use simple principles in ways that aren’t obvious\nMay require multiple steps\nMay make you consider many cases\nMay combine more than one topic\nMay need a flash of real insight to complete\nMay make you change direction or switch strategies along the\nway"} +{"id": "book 2_p23_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 23, "page_end": 23, "topic_guess": "sequences_patterns", "text": "An Illustration\nGive this problem a whirl. Don’t go on until you have spent a few minutes\non it—or until you have figured it out.\nTry-It #0-4\nWhat should the next number in this sequence be?\n1 2 9 64 __\nNote: This problem is not exactly GMAT-like, because there is no\nmathematically definite rule. However, you’ll know when you’ve solved the\nproblem. The answer will be elegant.\nThis problem has very simple content but a complex structure.\nResearchers in cognitive science have used sequence-completion\nproblems such as this one to develop realistic models of human thought.\nHere is one such model, simplified but practical.\nTop-Down Brain and Bottom-Up Brain\nTo solve the sequence-completion problem above, you need two kinds of\nthinking:"} +{"id": "book 2_p24_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 24, "page_end": 24, "topic_guess": "fractions_decimals_percents", "text": "You might even say that you need two types of brain.\nThe top-down brain is your conscious self. If you imagine the contents of\nyour head as a big corporation, then your top-down brain is the CEO,\nresponding to input, making decisions, and issuing orders. In cognitive\nscience, the top-down brain is called the executive function. Top-down\nthinking and planning is indispensable to any problem-solving process.\nBut the corporation in your head is a big place. For one thing, how does\ninformation get to the CEO? And how preprocessed is that information?\nThe bottom-up brain is your preconscious processor. A er raw sensory\ninput arrives, your bottom-up brain processes that input extensively before\nit reaches your top-down brain.\nFor instance, to your optic nerve, every word on this page is just a lot of\nblack squiggles. Your bottom-up brain immediately turns these squiggles\ninto letters, joins the letters into words, summons relevant images and\nconcepts, and finally serves these images and concepts to your top-down\nbrain. This all happens automatically and swi ly. In fact, it takes effort to\ninterrupt this process. Also, unlike your top-down brain, which does things\none at a time, your bottom-up brain can easily do many things at once.\nHow does all this relate to solving the sequence problem above?"} +{"id": "book 2_p25_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 25, "page_end": 25, "topic_guess": "sequences_patterns", "text": "Each of your brains needs the other one to solve difficult problems.\nYour top-down brain needs your bottom-up brain to notice patterns, sniff\nout valuable leads, and make quick, intuitive leaps and connections.\nBut your bottom-up brain is inarticulate and distractible. Only your top-\ndown brain can build plans, pose explicit questions, follow procedures,\nand state findings.\nImagine that you are trying to solve a tough murder case. To find all the\nclues in the woods, you need both a savvy detective and a sharp-nosed\nbloodhound.\nYour top-down brain is the\ndetective.\nBe organized, fast, and flexible to crack\nthe case.\nYour bottom-up brain is the\nbloodhound.\nTo solve difficult GMAT problems, try to harmonize the activity of your two\nbrains by following an organized, fast, and flexible problem-solving\nprocess.\nOrganized\nYou need a general step-by-step approach to guide you. One such\napproach, inspired by expert mathematician George Pólya, is Understand,\nPlan, Solve (UPS):"} +{"id": "book 2_p26_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 26, "page_end": 26, "topic_guess": "word_problems", "text": "You may never have thought explicitly about steps 1 and 2 before. It may\nhave been easy or even automatic for you to Understand easier problems\nand to Plan your approach to them. As a result, you may tend to dive right\ninto the Solve stage. This is a bad strategy. Mathematicians know that the\nreal math on hard problems is not Solve; the real math is Understand and\nPlan.\nFast\nSpeed is important for its own sake on the GMAT, of course. What you may\nnot have thought as much about is that being fast can also lower your\nstress level and promote good process. If you know you can solve quickly,\nthen you can take more time to comprehend the question, consider the\ngiven information, and select a strategy. To this end, make sure that you\ncan complete calculations and manipulations fairly rapidly so that you can\nafford to spend some time on the Understand and Plan stages of your\nproblem-solving process. A little extra time invested up front can pay off\nhandsomely later.\nFlexible\nTo succeed against difficult problems, you sometimes have to “unstick”\nyourself. Expect to run into brick walls and encounter dead ends. Returning\nto first principles and to the general process (e.g., making sure that you\nfully Understand the problem) can help you back up out of the mud.\nLet’s return to the sequence problem and play out a sample interaction\nbetween the two brains. The path is not linear; there are several dead ends,\nas you would expect. This dialog will lead to the answer, so don’t start\nreading until you’ve given the problem a final shot (if you haven’t already\nUnderstand the problem first.1.\nPlan your attack by adapting known techniques in new ways.2.\nSolve by executing your plan.3."} +{"id": "book 2_p27_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 27, "page_end": 27, "topic_guess": "number_theory", "text": "solved it). The top-down brain is labeled TD; the bottom-up brain is labeled\nBU.\n1 2 9 64 __ TD: “Okay, let’s Understand this thing. At a glance, they’ve\ngiven me an increasing list of numbers, and they want me to\nfind the number that “should” go in the blank, whatever\n“should” means. What’s a good Plan? Hmm. No idea. Stare at\nthe numbers given?”\nBU notices that 9 = 32 and 64 = 82. Likes the two squares.\n1 2 32 82 __ TD: “Write in the two squares.”\n1 2 32 82 __ BU notices that 1 is a square, too.\nsq no sq sq __ TD: “Are they all perfect squares? No, since 2 isn’t.”\nBU doesn’t like this break in the pattern.\n1 2 32 26 — TD: “Wait, back up. What about primes, factoring all the way. 8\n= 23, and so 82 = (23)2 = 26.”\nBU notices 6 = 2 × 3, but so what?\n1 2 32 (23)2 __ TD: “Let’s write 26 as (23)2. Anything there?”\nBU notices lots of 2’s and 3’s, but so what?\nTD: “Okay, keep looking at this. Are the 2’s and 3’s stacked\nsomehow?”"} +{"id": "book 2_p28_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 28, "page_end": 28, "topic_guess": "sequences_patterns", "text": "BU notices no real pattern. There’s 2−3−2 twice as you go\nacross, but so what? And the 1 is weird by itself.\n1 2 9 64 __ TD: “No good leads there. Hmm...time to go back to the\noriginal and try taking differences.”\nBU notices no pattern. The numbers look even uglier.\n1 2 9 64 __ TD: “Hmm. No good. Go back to original numbers again. What’s\ngoing on there?”\nBU notices that the numbers are growing quickly, like squares\nor exponentials.\n12 2 32 82 __ TD: “Must have something to do with those squares. I should\nlook at those again.”\nBU notices a gap on the le , among the powers.\n1? 21 32 82 __ TD: “How about looking at 2. Write it with exponents: 2 = 21. -\nActually, 1 doesn’t have to be 12. One can be to any power and\nstill be 1. The power is a question mark.”\nBU notices 21 then 32. Likes the counting numbers.\nBU really wants 1, 2, 3, 4 somehow.\n1? 21 32 4?? __ TD: “Try 4 in that last position. Could the last term be 4\nsomehow?”\nBU likes the look of this. 8 and 4 are related.\n1? 21 32 43 __ TD: “64 is 4 to the what? 42 = 16, times another 4 equals 64, so\n1 7 55"} +{"id": "book 2_p29_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 29, "page_end": 29, "topic_guess": "number_theory", "text": "it’s 4 to the third power. That fits.”\nBU is thrilled: 1, 2, 3, 4 below and 1, 2, 3 up top.\n10 21 32 43 __ TD: “Extend le . It’s 10. Confirmed. The bases are 1, 2, 3, 4, etc.,\nand the powers are 0, 1, 2, 3, etc.”\nBU is content.\n10 21 32 43 54 TD: “So the answer is 54, which is 252, or 625.”\nYour own process was almost certainly different in the details. Also, your\ninternal dialog was very rapid—parts of it probably only took fractions of a\nsecond to transpire. A er all, you think at the speed of thought.\nThe important thing is to recognize how the bottom-up bloodhound and\nthe top-down detective worked together in the case above. The TD\ndetective set the overall agenda and then pointed the BU bloodhound at\nthe clues. The bloodhound did practically all the “noticing,” which in some\nsense is where all the magic happened. But sometimes the bloodhound\ngot stuck, so the detective had to intervene, consciously trying a new path.\nFor instance, 64 reads so strongly as 82 that the detective had to actively\ngive up on that reading.\nThere are so many possible meaningful sequences that it wouldn’t have\nmade sense to apply a strict recipe from the outset: “Try X first, then Y, then\nZ…” Such an algorithm would require hundreds of possibilities. Should\nyou always look for 1, 2, 3, 4? Should you never find differences or prime"} +{"id": "book 2_p30_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 30, "page_end": 30, "topic_guess": "number_theory", "text": "factors because they weren’t that useful here? Of course not! A computer\ncan rapidly and easily apply a complicated algorithm with hundreds of\nsteps, but humans can’t. (If you are an engineer or programmer, maybe\nyou wish you could program your own brain, but so far, that’s not\npossible!)\nWhat humans are good at, though, is noticing patterns. The bottom-up\nbrain is extremely powerful—far more powerful than any computer yet\nbuilt.\nAs you gather problem-solving tools, the task becomes knowing when to\napply which tool. This task becomes harder as problem structures become\nmore complex. But if you deploy your bottom-up bloodhound according to\na general problem-solving process such as Understand, Plan, Solve, then\nyou can count on the bloodhound to notice the relevant aspects of the\nproblem—the aspects that tell you which tool to use.\nYou can break down Understand, Plan, Solve into several discrete steps:\nUnderstand Glance at the problem briefly: \ndoes anything stand out?\nRead the problem.\nJot down any obvious formulas or\nnumbers.\nPlan Reflect on what you were given: \nwhat clues might help tell you \nhow to approach this problem?"} +{"id": "book 2_p31_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 31, "page_end": 31, "topic_guess": "general", "text": "Organize your approach: \nchoose a solution path.\nSolve Work the problem!\nYou’ll get lots of practice using the UPS process throughout this guide."} +{"id": "book 2_p32_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 32, "page_end": 32, "topic_guess": "word_problems", "text": "Learning How to Think\nThis book is intended to make you smarter.\nIt is also intended to make you scrappier.\nThat description encompasses two main ideas: employing GMAT strategies\nas well as textbook solution methods and knowing when to let go.\nIf you have traditionally been good at paper-based standardized tests, then\nyou may be used to solving practically every problem the “textbook” way.\nProblems that forced you to get down and dirty—to work backwards from\nthe choices, to estimate and eliminate—may have annoyed you.\nA major purpose of this book is to help you learn to choose the best GMAT\napproach. On the hardest Quant problems, the textbook approach is o en\nnot the best GMAT approach.\nUnfortunately, advanced test-takers are sometimes very stubborn.\nSometimes they feel they should solve a problem according to some\ntheoretical approach. Or they fail to move to Plan B or C rapidly enough, so\nthey don’t have enough time le to execute that plan. In the end, they\nmight wind up guessing purely at random—and that’s a shame."} +{"id": "book 2_p33_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 33, "page_end": 33, "topic_guess": "word_problems", "text": "GMAT problems o en have back doors—ways to solve that don’t involve\ncrazy computation or genius-level insights. Remember that in theory,\nGMAT problems can all be solved in two minutes. By searching for the back\ndoor, you might avoid all the bear traps that the problem writer set out by\nthe front door!\nIn addition to learning alternative solution methods, you also need to learn\nwhen to let go. As you know, the GMAT is an adaptive test. If you keep\ngetting questions correct, the test will keep getting harder…and harder…\nand harder…\nAt some point, there will appear a monster problem, one that announces “I\nmust break you.” In your battle with this problem, you could actually lose\nthe bigger war—even if you ultimately conquer this particular problem.\nMaybe it takes you eight minutes, or it beats you up so badly that your\nhead starts pounding. This will take its toll on your score.\nThis will happen to everyone, no matter how good you are at the GMAT.\nWhy?\nThe GMAT is not an academic test, though it certainly appears to be.\nBusiness schools are primarily interested in whether you’re going to be an\neffective businessperson. Good businesspeople are able to assess a\nsituation rapidly, manage scarce resources, distinguish between good\nopportunities and bad ones, and make decisions accordingly.\nThe GMAT wants to put you in a situation where the best decision is, in fact,\nto guess and move on, because business schools are interested in learning"} +{"id": "book 2_p34_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 34, "page_end": 34, "topic_guess": "word_problems", "text": "whether you have the presence of mind to recognize a bad opportunity\nand the discipline to let it go.\nShow the GMAT that you know how to manage your scarce resources (time\nand mental energy) and that you can recognize and cut off a bad\nopportunity."} +{"id": "book 2_p35_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 35, "page_end": 35, "topic_guess": "word_problems", "text": "Plan of This Book\nThe rest of this book has three parts:\nPart One: Problem Solving and Data Sufficiency\nStrategies\nChapter 1: Problem Solving: Advanced\nPrinciples\nChapter 2: Problem Solving: Strategies &\nTactics\nChapter 3: Data Sufficiency: Principles\nChapter 4: Data Sufficiency: Strategies &\nTactics\nPart Two: Strategies for All Problem Types Chapter 5: Pattern Recognition\nChapter 6: Common Terms & Quadratic\nTemplates\nChapter 7: Visual Solutions\nChapter 8: Hybrid Problems\nPart Three: Practice Workouts 1−16: Sixteen sets of 10\nproblems each\nThe four chapters in Part I focus on principles, strategies, and tactics\nrelated to the two types of GMAT math problems: Problem Solving (PS) and\nData Sufficiency (DS). The next four chapters, in Part II, focus on techniques"} +{"id": "book 2_p36_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 36, "page_end": 36, "topic_guess": "statistics", "text": "that apply across several topics but are more specific than the approaches\nin Part I.\nEach of the eight chapters in Part I and Part II contains the following:\nMany of these problems will be GMAT-like in format, but many will not.\nPart III contains sets of GMAT-like Workout problems, designed to exercise\nyour skills as if you were taking the GMAT and seeing its hardest problems.\nSeveral of these sets contain clusters of problems relating to the chapters\nin Parts I and II, although the problems within each set do not all resemble\neach other in obvious ways. Other Workout problem sets are mixed by both\napproach and topic.\nNote that these problems are not arranged in order of difficulty. Also, you\nshould know that some of these problems draw on advanced content\ncovered in the Manhattan Prep All the Quant guide.\nTry-It Problems embedded throughout the text\nProblem Sets at the end of the chapter"} +{"id": "book 2_p37_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 37, "page_end": 37, "topic_guess": "general", "text": "Solutions to Try-It Problems\nIf you haven’t tried to solve the first three Try-It problems in the Try It\nYourself section at the beginning of this chapter, then go back and try them\nnow. Think about how to get your top-down brain and your bottom-up\nbrain to work together like a detective and a bloodhound. Come back\nwhen you’ve tackled the problems, even if you don’t get to an answer (in\nthis case, do make a guess).\nIn these solutions, we’ll outline sample dialogs between the top-down\ndetective and the bottom-up bloodhound.\nTry-It #0-1\nA jar is filled with red, white, and blue tokens that are equivalent\nexcept for their color. The chance of randomly selecting a red\ntoken, replacing it, then randomly selecting a white token is the\nsame as the chance of randomly selecting a blue token. If the\nnumber of tokens of every color is a multiple of 3, what is the\nsmallest possible total number of tokens in the jar? \n 9(A)\n12(B)\n15(C)\n18(D)"} +{"id": "book 2_p38_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 38, "page_end": 38, "topic_guess": "sets_probability_counting", "text": "SOLUTION TO TRY-IT #0-1\n… jar is filled with red, white, and blue tokens …\nchance of randomly selecting …\nTD: “I need to Understand this\nproblem first. There’s a jar, and it’s\ngot red, white, and blue tokens in it.”\nBU notices “chance” and\n“randomly.” That’s probability.\nTD: “All right, this is a probability\nproblem. Now, what’s the situation?”\nBU notices that there are two\nsituations.\n… chance of randomly selecting a red token, replacing\nit, then randomly selecting a white token is the same\nas the chance of randomly selecting a blue token.\nTD: “Let’s rephrase. In simpler\nwords, if I pick a red, then a white,\nthat’s the same chance as if I pick a\nblue. Jot that down. Okay, what\nelse?”\n… number of tokens of every color is a multiple of 3 … BU doesn’t want to deal with this\n“multiple of 3” thing yet.\n… smallest possible total number of tokens in the jar? TD: “Okay, what are they asking\nme?”\nBU notices “smallest possible total\nnumber.” Glances at answer choices.\nThey’re small, but not tiny. Hmm.\n21(E)"} +{"id": "book 2_p39_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 39, "page_end": 39, "topic_guess": "algebra", "text": "TD: “Let’s Reflect for a moment to\nfigure out a Plan. How can I\napproach this? How about algebra—\nif I name the number of each color,\nthen I can represent each fact and\nalso what I’m looking for. Okay, I use\nR, W, and B. Make probability\nfractions. Multiply red and white\nfractions. Simplify algebraically.”\nBU is now unsure. No obvious path\nforward.\nThe chance of randomly selecting a red token,\nreplacing it, then randomly selecting a white token is\nthe same as the chance of randomly selecting a blue\ntoken …\nTD: “Let’s start over conceptually.\nReread the problem. Can I learn\nanything interesting?”\nBU notices that blues are different.\nTD: “How are blues different? Hmm.\nPicking a red, then a white is as likely\nas picking a blue. What does that\nmean?”\nBU notices that it’s unlikely to pick a\nblue. So there aren’t many blues\ncompared to reds or whites.\nFewer blues than reds or whites\nB < R and B < W\nTD: “Are there fewer blues? Yes.\nJustify this. Focus on the algebraic\nsetup.”"} +{"id": "book 2_p40_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 40, "page_end": 40, "topic_guess": "algebra", "text": "In the very first equation above, each fraction on the\nle is less than 1, so their product is even smaller.\nThe denominators of the three fractions are all the\nsame.\nSo the numerator of the product (B) must be smaller\nthan either of the other numerators \n(R and W ).\nBU notices fractions less than 1. All\npositive.\nTD: “Two positive fractions less than\n1 multiplied together give an even\nsmaller number.”\nTD: “Yes, there are fewer blues.”\nBU is quiet.\nIf the number of tokens of every color is a multiple of 3,\nwhat is the smallest possible total number of tokens in\nthe jar?\nNeither R nor W can equal 3 (since B is smaller than\neither).\nLet R = W = 6.\n(A) and (B) are out now. The smallest possible total is\nnow 15.\nTD: “Time to go back and reread the\nrest of the problem.”\nBU again notices “multiple of 3,” also\nin answer choices. Small multiples.\nTD: “Change of Plan: Algebra by itself\nisn’t getting me there. What about\nplugging in a number? Try the most\nconstrained variable: B. Since it’s the\nsmallest quantity, but still positive,\npretend B is 3. Execute this\nalgebraically. Divide by RW.”\nBU likes having only two variables.\nTD: “Need to test other numbers.\nApply constraints I know—B is the\nsmallest number. Rule out answer\nchoices as I go.”\nTD: “6 and 6 don’t work, because the\nright side adds up to larger than 1."} +{"id": "book 2_p41_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 41, "page_end": 41, "topic_guess": "sets_probability_counting", "text": "Let R = 6 and W = 9.\n (C) is out too. Try the next\npossibility.”\nBU doesn’t like breaking the\nsymmetry between R and W. They\nseem to be alike.\nTD: “Does it matter whether R = 6\nand W = 9 or the other way around?\nNo, it doesn’t. One is 6, the other is 9.\nPlug in and go.”\nTD: “This works. The answer is 3 + 6\n+ 9 = 18.”\nThe correct answer is (D). Let’s look at another pathway—one that moves\nmore quickly to the back door.\nALTERNATIVE SOLUTION TO TRY-IT #0-1\n… chance of randomly selecting … BU notices “chance.” BU doesn’t like probability.\nTD: “Oh man, probability. Okay, let’s make sense\nof this and see whether there are any back doors.\nThat’s the Plan.”\n… the number of tokens of every color is\na multiple of 3 …\nBU notices that there are only limited possibilities\nfor each number.\nTD: “Okay, every quantity is a multiple of 3. That\nsimplifies things. There are 3, 6, 9, etc., of each\ncolor.”"} +{"id": "book 2_p42_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 42, "page_end": 42, "topic_guess": "sets_probability_counting", "text": "A jar is filled with red, white, and blue\ntokens …\nBU is alert—what about 0?\nTD: “What about 0? Hmm…the wording at the\nbeginning assumes that there actually are tokens\nof each color. So there can’t be 0 tokens of any\nkind.”\n(A) 9 (B) 12 (C) 15 (D) 18 (E) 21 TD: “Now let’s look at the answer choices.”\nBU notices that they’re small.\n(A) 9 TD: “Try plugging in the choices. Let’s start at the\neasy end—in this case, the smallest number.”\nBU notices 9 = 3 + 3 + 3.\nSelect a red:\n\nSelect a white:\n\n, which is not \"select a\nblue\" \nTD: “The only possible way to have 9 total tokens\nis to have 3 reds, 3 whites, and 3 blues. So…does\nthat work? Plug into probability formula.” \n(A) 9 (B) 12 (C) 15 (D) 18 (E) 21\n(B) 12\nSelect a red: \nTD: “No, that doesn’t work. This is good. Knock out\n(A). Let’s keep going. Try (B).”\nBU notices 12 = 3 + 3 + 6.\nTD: “Only way to have 12 total is 3, 3, and 6. Which\none’s which? Picking a red and then a white is the\nsame as picking a blue, so the blue should be one\nof the 3’s. Let’s say red is 3 and white is 6.”"} +{"id": "book 2_p43_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 43, "page_end": 43, "topic_guess": "general", "text": "Select a white: \n  , which is not “select a\nblue”\n(A) 9 (B) 12 (C) 15 (D) 18 (E) 21\n(C) 15\nTD: “That doesn’t work either. Knock out (B). Keep\ngoing.”\nBU notices 15 has a few options.\nSelect a red:\n\nSelect a white: \n, which is not \"select a\nblue\"\n,\nwhich is not \"select a blue\"\n(A) 9 (B) 12 (C) 15 (D) 18 (E) 21\n(D) 18\n, which IS \"select a blue\" \nTD: “I can make 15 by 3, 6, and 6 or by 3, 3, and 9.\nTry 3−6−6; make blue the 3.”\nTD: “Nope. What about 3−3−9.”\nTD: “Not this one either.”\nTD: “Knock out (C). Try (D).”\nTD: “Maybe 3–6–9 first. Make blue the 3.”\nTD: “That’s it! Answer’s (D).”"} +{"id": "book 2_p44_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 44, "page_end": 44, "topic_guess": "coordinate_geometry", "text": "Many people find this second approach less stressful and more efficient\nthan the textbook approach. In fact, there is no way to find the correct\nanswer by pure algebra. Ultimately, you have to test suitable numbers.\nTry-It #0-2\nArrow \n , which is a line segment exactly 5 units long with an\narrowhead at A, is constructed in the xy-plane. The x- and y-\ncoordinates of A and B are integers that satisfy the inequalities 0 ≤\nx ≤ 9 and 0 ≤ y ≤ 9. How many different arrows with these\nproperties can be constructed? \nSOLUTION TO TRY-IT #0-2\nBU notices “xy-plane.”\nTD: “Let’s Understand first. This is a coordinate-\nplane problem. Sketch it out. Put in boundaries\nas necessary.”\nBU wonders where this is going.\nHow many different arrows with these TD: “What is the question asking for again?\n 50(A)\n168(B)\n200(C)\n368(D)\n536(E)"} +{"id": "book 2_p45_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 45, "page_end": 45, "topic_guess": "number_theory", "text": "properties can be constructed? Reread the question.”\nBU wonders which properties.\n… exactly 5 units long with an arrowhead\nat A … the x- and y-coordinates of A and B\nare to be integers that satisfy the\ninequalities 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9.\nTD: “What are the properties of the arrows\nsupposed to be again? Each arrow is 5 units\nlong.”\nBU notices “integers” and “coordinates” and\npictures a pegboard.\nTD: “Reflect. The tip and the end of the arrow\nhave to touch holes in the pegboard exactly.\nOkay. The Plan is to start counting. How to\nOrganize?”\nBU imagines many possible arrows. Brute force\ncan’t be the right way forward. The arrows can\npoint in all sorts of different ways.\nTD: “Let’s simplify the Plan. Let’s focus on just\none orientation of arrows—pointing straight up.\nDraw this situation. How many places can the\narrow be?”\nBU wants to go up & down, then right & le .\nIn one column, there are 5 positions for the\narrowhead: y = 5, 6, 7, 8, or 9. That’s the\nsame as 9 − 5 + 1, by the way.\nTD: “Count the positions in one column, then\nmultiply by the number of columns. Be careful\nto count endpoints.”"} +{"id": "book 2_p46_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 46, "page_end": 46, "topic_guess": "geometry", "text": "There are 10 identical columns: x = 0\nthrough x = 9. 5 × 10 = 50 possible positions\nfor the arrow pointing straight up.\n50 × 2 = 100 possible positions for the\narrow if it points straight up or to the right.\nTD: “Great. I’ve Solved one part. Other\npossibilities?”\nBU notices the square is the same vertically as\nhorizontally. Go right.\nTD: “I get the same result for arrows pointing\nright. 50 more positions. Is that it? Am I done?”\n50 × 4 = 200 possible positions\nBU wonders about “down” and “le .”\nTD: “These arrows can point straight down or\nstraight le , too. Those would have the same\nresult. So there are 50 positions in each of the\nfour directions. Calculate at this point and\nevaluate answers. Eliminate (A) and (B).”\nAnswer seems to be (C). BU is suspicious: somehow too easy.\nTD: “Tentative answer is (C), but I'm not done.”\nBU wonders about still other ways for the arrows\nto point.\nThree up, four across:\n TD: “Could the arrows be at an angle?”\nBU notices that the arrow is 5 units long,\nassociated with 3−4−5 triangles."} +{"id": "book 2_p47_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 47, "page_end": 47, "topic_guess": "geometry", "text": "TD: “3−4−5 triangles. Yes. Put the arrow as the\nhypotenuse of a 3−4−5 triangle. How can this be\ndone? Try to place the arrow. Remember the\nreversal. Looks like there are four ways if I go 3\nup and 4 across: up right, up le , down right,\ndown le .”\nBU is happy. This is the trick.\nFour up, three across:\n TD: “Likewise, there must be four ways if I go 4\nup and 3 across: again, up right, up le , down\nright, and down le . By the way, the answer\nmust be (D) or (E).”\nThree up, four across, pointing up to the\nright:\nThere are 7 positions vertically for the\narrowhead (9 − 3 + 1) and 6 positions\nhorizontally (9 − 4 + 1), for a total of 7 × 6 =\n42 positions.\n8 × 42 = 336 possible positions at an angle.\nIn total, there are 200 + 336 = 536 positions.\nTD: “Now count just one of these ways. Same\nideas as before. Be sure to include endpoints.”\nBU notices the symmetry. The 3 up, 4 across is\nthe same as the 4 up, 3 across, if you turn the\nsquare.\nTD: “Each of these angled ways will be the same.\nThere are 8 ways to point the arrow at an angle.\nFinish the calculation and confirm the answer.”\nThe correct answer is (E). There isn’t much of an alternative to the\napproach above. With counting problems, it can o en be very difficult to\nestimate the answer or work backwards from the answer choices."} +{"id": "book 2_p48_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 48, "page_end": 48, "topic_guess": "geometry", "text": "Try-It #0-3\nIn the diagram to the right, the value of x is closest to which of the\nfollowing?\nSOLUTION TO TRY-IT #0-3\nTD: “Okay, let’s Understand this. Redraw the\nfigure. The problem wants the value of x. Now…\nhow about a Plan?”\nBU notices this is an isosceles triangle, because\nthere are two sides labeled x. How about the\ntwo equal angles?\n180° − 45° = 135°.\nDivide 135° equally across the two missing\nangles. So each angle is 67.5°.\nTD: “Figure out the two missing angles. Use the\n180° rule.”\nBU doesn’t recognize this triangle.\n(A)\n2(B)\n(C)\n(D)\n1(E)"} +{"id": "book 2_p49_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 49, "page_end": 49, "topic_guess": "geometry", "text": "TD: “Hmm...here’s a Plan: add a perpendicular\nline to make right triangles. Drop the line from\nthe top point. I'll label corners while I'm at it.\nNow fill in angles.”\nBU notices 45−45−90 and is happy.\nTD: “Use the 45−45−90 to write expressions for\nits sides. Then \n can be split up into two\npieces, and I can set up the Pythagorean\ntheorem.”\nBU feels that this process is kind of ugly.\nTD: “Let’s push through. Write the Pythagorean\ntheorem for the small triangle on the le , using\nthe \n as the hypotenuse.”\nBU thinks this equation is really ugly.\nTD: “Push through. Expand the quadratic and\nsimplify.”\nBU doesn’t like the square root on the bottom.\nTD: “Multiply by \n to get rid of it on the -\nbottom of the fraction.”"} +{"id": "book 2_p50_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 50, "page_end": 50, "topic_guess": "algebra", "text": "BU has no idea how to take the square root of\nthis.\nTD: “Neither do I. Let’s try estimating. If x2 is\nabout 3.5, then the square root must be a bit\nless than 2 (since the square root of 4 is 2). 182 is\n324 and 192 is 361, so the answer is around 1.8\nor 1.9.”\nTD: “Answer (A) is about 3.5; that matches the\nsquared value, not the square root. The answer\nneeds to be less than 2, so (B) is also wrong. \nis about 1.7. Answers (D) and (E) are too small,\nso the answer is (C).”\nThe correct answer is (C).\nThe method you just saw is algebraically intensive, and so your bottom-up\nbloodhound might have kicked up a fuss along the way. Sometimes, your\ntop-down brain needs to ignore the bottom-up brain. Remember, when\nyou’re actually taking the GMAT, you have to solve problems quickly—and\nyou don’t need to publish your solutions in a mathematics journal. What\nyou want is to get the correct answer as quickly and as easily as possible. In\nthis regard, the solution above works perfectly well.\n2 + \n(A)\n2(B)\n(C)\n(D)\n1(E)"} +{"id": "book 2_p51_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 51, "page_end": 51, "topic_guess": "geometry", "text": "Alternatively, the question stem asks for an approximate answer, so you\ncan also try estimating from the start. Draw the triangle carefully and start\nwith the same perpendicular line as before. This line is a little shorter than\nthe side of length \n (which is about 1.4). Call the two shorter legs 1.2\nand calculate the hypotenuse. It equals 1.2 multiplied by 1.4, or\napproximately 1.7. (Bonus question: How can you estimate that math\nquickly? Answer below.)\nNow, examine the answer choices using 1.4 for \n and 1.7 for \n :\nThey’re all close, but you can pretty confidently eliminate answers (A) and\n(E). Furthermore, the answer needs to be less than 2, so (B) can’t be it.\nAnswer (C) is closer than (D), so (C) is probably it. Unfortunately, you might\nguess wrong at this point. But the odds are much better than they were\nat the outset.\nIt is worthwhile to look for multiple solution paths as you practice. Your\ntop-down brain will become faster, more organized, and more flexible,\nenabling your bottom-up brain to have more flashes of insight.\nThat was a substantial introduction. Now, on to Chapter 1!\n3.4(A)\n2(B)\n1.7(C)\n1.4(D)\n1(E)"} +{"id": "book 2_p52_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 52, "page_end": 52, "topic_guess": "word_problems", "text": "PART ONE\nProblem Solving and Data\nSufficiency Strategies\nIn This Part\nChapter 1: Problem Solving: Advanced Principles\nChapter 2: Problem Solving: Strategies & Tactic\nChapter 3: Data Sufficiency: Principles\nChapter 4: Data Sufficiency: Strategies & Tactics"} +{"id": "book 2_p53_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 53, "page_end": 53, "topic_guess": "strategy", "text": "CHAPTER 1\nProblem Solving: Advanced\nPrinciples"} +{"id": "book 2_p54_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 54, "page_end": 54, "topic_guess": "strategy", "text": "In This Chapter...\nPrinciple #1: Understand the Basics\nPrinciple #2: Build a Plan\nPrinciple #3: Solve—and Put Pen to Paper\nPrinciple #4: Review Your Work"} +{"id": "book 2_p55_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 55, "page_end": 55, "topic_guess": "strategy", "text": "Chapter 1\nProblem Solving: Advanced\nPrinciples\nChapters 1 and 2 of this book focus on the more fundamental of the two\ntypes of GMAT math questions: Problem Solving (PS). Some of the content\napplies to any kind of math problem, including Data Sufficiency (DS).\nHowever, Chapters 3 and 4 deal specifically with DS issues.\nThis chapter outlines broad principles for solving advanced PS problems.\nYou’ve already seen very basic versions of the first three principles in the\nintroduction, in the dialogs between the top-down and the bottom-up\nbrain.\nAs mentioned earlier, these principles draw on the work of George Pólya,\nwho was a brilliant mathematician and teacher of mathematics. Pólya was\nteaching future mathematicians, not GMAT test-takers, but what he said\nstill applies. His little book How to Solve It has been in print since 1945—it’s\nworth getting a copy.\nIn the meantime, keep reading."} +{"id": "book 2_p56_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 56, "page_end": 56, "topic_guess": "word_problems", "text": "Principle #1: Understand the Basics\nTake time to think and plan before you start solving a difficult problem. If\nQuant is your strength, you may want to dive straight into every problem\nas soon as you see it, without pausing to consider all of the angles. There\nare two good reasons to slow down:\nTo remind yourself to slow down and plan, understand each problem by\ntaking three steps: \nGlance at the entire problem: is it PS or DS? If it’s PS, glance at the answer\nchoices. If it’s DS, glance at the statements. Knowing what type of problem\nyou’re dealing with will help you read more effectively. \nPólya recommended that you ask yourself simple questions as you read a\nproblem. Here are some great Pólya-style questions that can help you\nunderstand:\nYou need to manage your time and mental energy across the entire\nGMAT. If you pause briefly to find a more efficient solution, you’ll save\ntime and energy for other problems. \n1.\nIf you start doing math without thinking first, you might have to change\nyour approach later in the problem, which takes time that you don’t\nhave. You also risk falling for traps.  \n2."} +{"id": "book 2_p57_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 57, "page_end": 57, "topic_guess": "word_problems", "text": "You don’t need to meticulously go through every one of these questions\nwhenever you solve a problem. (However, that’s a good thing to do when\nyou review a problem!) They’re here to help you consider how you might\nread more productively. \nAs you read the problem, jot down any given numbers or formulas on your\nscrap paper. That doesn’t mean you should start doing math while you’re\ntrying to read. If you start trying to solve the problem when you haven’t\neven finished reading it, you’re getting ahead of yourself. \nStart the following problem by taking these three steps: Glance, Read, and\nJot. \nWhat exactly is the problem asking for?\nWhat information would I need in order to find the answer?\nWhat information do I already have? \nWhat information don’t I have?\nSometimes you care about something you don’t know. This could be\nan intermediate unknown quantity that you didn’t think of earlier.\nOther times, you don’t know something, and you don’t care. For\ninstance, if a problem includes the quantity 11! (11 factorial), you will\npractically never need to know the exact value of that quantity.\nWhat is this problem testing? In other words, why is this problem on the\nGMAT? What aspect of math are they testing? What kind of reasoning do\nthey want me to demonstrate?\nHave I seen a problem like this before? Have you already solved a similar\nproblem? What approach worked best?"} +{"id": "book 2_p58_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 58, "page_end": 58, "topic_guess": "number_theory", "text": "Try-It #1-1\nx = 910 − 317 and \n is an integer. If n is a positive integer that has\nexactly two factors, how many different values for n are possible? \nGlance. This is a PS problem. The answers are numbers but in written form;\nthis format is reserved for problems that ask for the number of numbers or\nnumber of possibilities for something. The numbers are small.\nRead. Dive into the text. Here are some possible answers to the Pólya\nquestions:\nWhat\nexactly is\nthe\nproblem\nasking\nfor?\nThe number of possible values for n.\nThis means that n might have multiple possible values. In fact, it probably can\ntake on more than one value.\nI may not need these actual values. I just need to count them.\nWhat are\nthe\nquantities\nI care\nabout?\nI’m given x and n as variables. These are the quantities I care about.\nOne(A)\nTwo(B)\nThree(C)\nFour(D)\nFive(E)"} +{"id": "book 2_p59_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 59, "page_end": 59, "topic_guess": "number_theory", "text": "What do I\nknow?\nx = 910 − 317\nThat is, x = a specific large integer, expressed in terms of powers of 9 and 3.\n is an integer.\nThat is, x is divisible by n, or n is a factor of x.\nFinally, n is a positive integer that has exactly two factors.\nPrime numbers have exactly two factors. So I can rephrase the information: n is\nprime. (Primes are always positive.)\nWhat\ndon’t I\nknow?\nHere’s something I don’t know: I don’t know the value of x as a series of digits.\nUsing a calculator or Excel, I could find out that x equals 3,357,644,238. But I don’t\nknow this number at the outset. Moreover, because this calculation is far too\ncumbersome, it must be the case that I don’t need to find this number.\nWhat is\nthis\nproblem\ntesting?\nFrom the foregoing, I can infer that this problem is testing Divisibility & Primes. I\nwill probably also need to manipulate exponents, since I see them in the\nexpression for x.\nYou can ask these questions in whatever order is most helpful for the\nproblem. For instance, you might not look at what the problem is asking\nfor until you’ve understood the given information.\nJot. As you decide that a piece of information is important, jot it down on\nyour scrap paper. At this point, your scrap paper might look something like\nthis:"} +{"id": "book 2_p61_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 61, "page_end": 61, "topic_guess": "statistics", "text": "Principle #2: Build a Plan\nNext, think about how you will solve the problem:\nReflect. Here are some Pólya questions that help you think about what you\nknow and come up with a plan:\nIs a good\napproach\nalready\nobvious?\nFrom your answers above, you may already see a way to reach the answer. If\nyou can envision the rough outlines of the correct path, then go ahead and\nget started.\nIf not, what in\nthe problem\ncan help me\nfigure out a\ngood\napproach?\nIf you are stuck, look for particular clues to tell you what to do next. Revisit\nyour answers to the basic questions. What do those answers mean? Can you\nrephrase or reword them? Can you combine two pieces of information in any\nway, or can you rephrase the question, given everything you know?\nCan I\nremember a\nsimilar\nproblem?\nTry relating the problem to other problems you’ve faced. This can help you\ncategorize the problem or recall a solution process."} +{"id": "book 2_p62_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 62, "page_end": 62, "topic_guess": "number_theory", "text": "Organize. For the Try-It #1-1 problem, some of the information is already\nrephrased (reorganized). Go further now, combining information and\nsimplifying the question:\nGiven: n is a prime number    AND    n is a factor of x.\nCombined: n is a prime factor of x.\nQuestion: How many different values for n are possible?\nCombined: How many different values for n, a prime factor of x, are possible?\nRephrased: How many distinct prime factors does x have?\nYou need the prime factorization of x. Notice that n is not even in the\nquestion anymore. The variable n just gave you a way to ask this\nunderlying question.\nConsider the other given fact: x = 910 − 317. It can be helpful initially to put\ncertain complicated facts to the side. At this stage, however, you know that\nyou need the prime factors of x. So now you have the beginning of a plan:\nfactor this expression into its primes."} +{"id": "book 2_p63_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 63, "page_end": 63, "topic_guess": "number_theory", "text": "Principle #3: Solve—and Put Pen to\nPaper\nThe third step is to do the work: solve.\nYou’ll want to execute that solution in an error-free way—it would be terrible to\nget all the thinking correct, then make a careless computational mistake. That’s\nwhy we say you should put pen to paper.\nIn the expression 910 − 317, the 3 is prime but the 9 is not. Since the problem is\nasking about prime factors, rewrite the equation in terms of prime numbers:\nNext, pull out a common factor from both terms. The largest common factor is\n317:\nNow, you have what you need: the prime factorization of x. The number x has\nthree distinct prime factors: 2, 3, and 13. The correct answer is (C).\nThe idea of putting pen to paper also applies when you get stuck anywhere\nalong the way on a monster problem."} +{"id": "book 2_p64_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 64, "page_end": 64, "topic_guess": "sequences_patterns", "text": "Think back to those killer Try-It problems in the introduction. Those are not the\nkinds of problems you can figure out just by looking at them.\nWhen you get stuck on a tough problem, take action. Do not just stare, hoping\nthat you suddenly get it.\nInstead, ask yourself the Pólya questions again and write down whatever you\ncan:\nThis way, your top-down brain can help your bottom-up brain find the correct\nleads—or help it let go. In particular, it’s almost impossible to abandon an\nunpromising line of thinking without writing something down.\nThink back to the sequence problem in the introduction. You’ll keep seeing 64\nas 82 unless you try writing it in another way.\nDo not try to juggle everything in your head. Your working memory has limited\ncapacity, and your bottom-up brain needs that space to work. A multistep\nproblem cannot be solved in your brain as quickly, easily, and accurately as it\ncan be on paper.\nAs you put pen to paper, there are three themes you'll want to keep in mind.\n1. LOOK FOR PATTERNS\nReinterpretations of given information or of the question\nIntermediate results, whether specific or general\nAvenues or approaches that didn’t work"} +{"id": "book 2_p65_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 65, "page_end": 65, "topic_guess": "sequences_patterns", "text": "Every GMAT Quant problem has a two-minutes-or-faster solution path, which\nmay depend upon a pattern that you’ll need to extrapolate. You’ll know a\npattern is needed when a problem asks something that would be impossible to\ncalculate (without a calculator) in two minutes. When this happens, write out\nthe first five to eight items in the sequence or list in order to try to spot the\npattern.\nTry-It #1-2\n for all integer values of n greater than 1. If S1 = 1,\nwhat is the sum of the first 61 terms in the sequence?\nNobody is going to write out all 61 terms and then add them up in two minutes.\nThere must be a pattern. The recursive definition of Sn doesn’t yield any secrets\nupon first glance. So write out the early cases in the sequence, starting at n = 1\nand looking for a pattern:\n−48(A)\n−31(B)\n−29(C)\n 1(D)\n 30(E)"} +{"id": "book 2_p66_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 66, "page_end": 66, "topic_guess": "sequences_patterns", "text": "etc.\nThe terms of the sequence are \n . . . . Three terms\nrepeat in this cyclical pattern forever; every third term is the same. Note: If you\ndon’t spot a pattern within the first five to eight terms, stop using this approach\nand see whether there’s another way (including guessing!).\nThe problem asks for the sum, so find the sum of each group of three\nconsecutive terms: \n . There are 20 groups in\nthe first 61 terms, and one additional term that hasn’t been counted yet. So the\nsum of the first 61 terms is as follows:"} +{"id": "book 2_p67_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 67, "page_end": 67, "topic_guess": "sequences_patterns", "text": "The correct answer is (C).\nIt is almost impossible to stare at the recursive definition of this sequence and\ndiscern the resulting pattern.\nThe best way to identify the pattern is to calculate a few values of the sequence\nand look for the pattern. You will learn more about pattern recognition in\nChapter 5.\n2. DRAW IT OUT\nSome problems are much easier to solve if you draw out what’s happening in\nthe problem. Whenever a story problem describes something that could\nactually happen in the real world, you could try to draw out the solution. For\ninstance, if a problem involves motion, you can draw snapshots representing\nthe problem at different points in time.\nTry-It #1-3\nTruck A is on a straight highway heading due south at the same time\nTruck B is on a different straight highway heading due east. At 1:00\np.m., Truck A is exactly 14 miles north of Truck B. If both trucks are\ntraveling at a constant speed of 30 miles per hour, at which of the\nfollowing times will they be exactly 10 miles apart?\n1:10 p.m.(A)\n1:12 p.m.(B)\n1:14 p.m.(C)\n1:15 p.m.(D)\n1:20 p.m.(E)"} +{"id": "book 2_p68_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 68, "page_end": 68, "topic_guess": "word_problems", "text": "Represent Truck A and Truck B as of 1:00 p.m. How does the\ndistance between Truck A and Truck B change as time goes\nby?\nTry another point in time. Since the answers are all a\nmatter of minutes a er 1:00 p.m., try a convenient\nincrement of a few minutes. A er 10 minutes, each truck\nwill have traveled 5 miles (30 miles per 60 minutes = 5 miles in 10 minutes). How\nfar apart will the trucks be then? On the diagram to the right, the distance is\nrepresented by x.\nBecause Truck A is traveling due south and Truck B\nis traveling due east, the triangle must be a right\ntriangle.\nTherefore, x2 = 92 + 52.\nAt this point, you could solve the problem in one of\ntwo ways. The first is to notice that once both trucks\ntravel 6 miles, the diagram will contain a 6 : 8 : 10\ntriangle. Therefore, \n of an hour later, at\n1:12 p.m., the trucks will be exactly 10 miles apart.\nAlternatively, you could set up an algebraic equation\nand solve for the unknown number of miles\ntraveled, such that the distance between the trucks\nis 10. Call that distance y:"} +{"id": "book 2_p69_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 69, "page_end": 69, "topic_guess": "algebra", "text": "Therefore, y could equal\n6 or 8 miles. In other\nwords, the trucks will be\nexactly 10 miles apart at\n1:12 p.m. and at 1:16\np.m. Either way, the\ncorrect answer is (B).\nNotice how instrumental\nthese diagrams were for\nthe solution process. You\nmay already accept that Geometry problems require diagrams. However, many\nother kinds of problems can benefit from visual thinking. You will learn more\nabout advanced visualization techniques in Chapter 7.\n3. SOLVE AN EASIER PROBLEM\nA problem may contain large numbers or complicated expressions that actually\ndistract you from the task at hand: finding a solution path.\nWhen this happens, one tactic is to simplify part of the problem and solve that.\nOnce you understand how the math works, return to the more complex problem\nand apply the same solution path."} +{"id": "book 2_p70_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 70, "page_end": 70, "topic_guess": "number_theory", "text": "Try-It #1-4\nIf x and y are positive integers and \n is the square of an odd\ninteger, what is the smallest possible value of xy ?\nAs you read, jot down the given information.\nNote that you might not immediately write down the square of an odd integer\ninfo if you still have to puzzle out what it means:\nWhat does the square of an odd integer look like? List out a few examples, on\npaper or in your head:\n 1(A)\n 8(B)\n10(C)\n15(D)\n28(E)"} +{"id": "book 2_p71_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 71, "page_end": 71, "topic_guess": "algebra", "text": "Are there any patterns or commonalities? All of the numbers are odd. All of the\nnumbers are perfect squares. Therefore, \n is an odd perfect square.\nAdd that to your notes.\nThe question asks for the smallest possible value of xy. What do you need to\nfigure out in order to find that?\nIf the \n expression is distracting you, try figuring out what this would\nmean for a simpler version of the expression.\nSimpler problem: What if \n is an odd perfect square?\nIn order for the number to be odd, you have to get rid of the even\nnumber 20 (because an even number times any number equals an even\nnumber). The only way to get rid of the even part is to divide it out by y. If\ny is 4, then the expression would become \n . As long as x is\nan odd number, 5x will be odd, too.\nInteresting. How can you apply that thinking to the real problem?\nIt’s still true that, in order for \n to be odd, you have to get rid of the\neven factors in the numerator."} +{"id": "book 2_p72_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 72, "page_end": 72, "topic_guess": "number_theory", "text": "In other words, y2 must cancel out all the even factors in\n1,620. The y2 must contain at least two 2’s, so y itself has to\ncontain at least one 2.\nOkay, that takes care of y: at minimum, y must be 2. If so,\nthen the expression becomes \n .\nNow, what about x? If you’re not sure, return to your simpler problem thinking.\nSimpler problem: In the last step, in order to make 5x odd, x has to be\nodd. 5x also has to be a perfect square. If you make x = 5, then 5x = 25, an\nodd perfect square.\nWhy does that work? A perfect square must contain two of each factor: 5\nand 5, for example. The expression 5x, therefore, needs a second 5 to\nmake this a perfect square.\nBack to the real problem. Make sure that 405x has two pairs\nof every factor. 405 contains only one 5, so x must contain\nanother 5. Also contained in 405 is 81, which is 92. That set\nof factors already represents a perfect square, so the\nminimum requirement is that x equals 5.\nIf y must be 2, at minimum, and x must be 5, at minimum, then the smallest -\npossible value of xy is 10.\nYou can generalize this approach. If a problem has many complexities, you can\nattack it by ignoring some of the complexities at first. Solve a simpler problem."} +{"id": "book 2_p73_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 73, "page_end": 73, "topic_guess": "sequences_patterns", "text": "Then, see whether you can adjust the solution to the simpler problem in order\nto solve the original.\nTo recap, put your work on paper. Don’t try to solve hard problems in your head.\nInstead, do the following:\nIn general, jot down intermediate results as you go. You may see them in a new\nlight and consider how they fit into the solution.\nAlso, try to be organized. For instance, make tables to keep track of cases. The\nmore organized you are, the more insights you will have into difficult problems.\nFind a pattern: Write out the first few cases.\nVisualize a scene: Draw it out!\nSolve an easier problem, then apply your method to the harder problem."} +{"id": "book 2_p74_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 74, "page_end": 74, "topic_guess": "word_problems", "text": "Principle #4: Review Your Work\nWhen you are done with a test or practice set, you are not really done.\nWhen you first do a problem under timed conditions, your brain is too busy\nsolving the problem to effectively learn and remember. What you learn\nfrom a new problem comes a er you’ve finished it and picked your answer,\nwhen you look at it with a clear head and no timer. Give yourself twice as\nmuch time to review each problem as you spent doing the problem in the\nfirst place. \nHere are some things you might consider as you review a problem. Most of\nthese questions are useful even if you got the problem correct. Don’t\nrestrict yourself to reviewing problems you got wrong. Review any problem\nyou might learn something from and ask yourself:\nWhat are all of the pathways to the answer? Which is the best? What is\nthe easiest and fastest way to implement it? \nWhat clues in the problem told me to use a certain approach or take a\ncertain step? If I see one of those clues in a different problem, what\nshould I do? \nWhat traps or tricks are built into this problem?\nWhere could I have made a mistake? \nIf I did make a mistake, what went wrong in my problem-solving\nprocess? Do I need to change how I approach similar problems?"} +{"id": "book 2_p75_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 75, "page_end": 75, "topic_guess": "general", "text": "What could I take from this problem to help me solve other problems in\nthe future?"} +{"id": "book 2_p76_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 76, "page_end": 76, "topic_guess": "sets_probability_counting", "text": "When you do the following problem set, apply the first three principles\nfrom this chapter to each problem: Understand, Plan, and Solve. Then,\nreview each problem in depth. As you review, do two things:\nThe solutions include our own responses to these two tasks. Yours might\nlook different, and that's fine.\nProblem Set\nIdentify exactly what the problem is asking for and what that means in\nthe simplest possible terms. \n1.\nNote at least one general takeaway that might be useful on other\nproblems in the future. \n2.\n1. Each factor of 210 is inscribed on its own plastic ball and all of the\nballs are placed in a jar. If a ball is randomly selected from the jar,\nwhat is the probability that the ball is inscribed with a multiple of\n42 ?"} +{"id": "book 2_p77_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 77, "page_end": 77, "topic_guess": "sequences_patterns", "text": "(A)\n(B)\n\n(C)\n(D)\n\n(E)\n2. If x is a positive integer, what is the units digit of (24)5 + 2x(36)6(17)3 ?\n2(A)\n3(B)\n4(C)\n6(D)\n8(E)\n3. A baker makes a combination of chocolate chip cookies and peanut\nbutter cookies for a school bake sale. His recipes only allow him to\nmake chocolate chip cookies in batches of 7 and peanut butter\ncookies in batches of 6. If he makes exactly 95 cookies for the bake\nsale, what is the minimum number of chocolate chip cookies that\nhe could make?"} +{"id": "book 2_p78_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 78, "page_end": 78, "topic_guess": "algebra", "text": "7(A)\n14(B)\n21(C)\n28(D)\n35(E)\n4. A rectangular solid is changed such that the width and length are\neach increased by 1 inch and the height is decreased by 9 inches.\nDespite these changes, the new rectangular solid has the same\nvolume as the original rectangular solid. If the width and length of\nthe original rectangular solid are equal and the height of the new\nrectangular solid is 4 times the width of the original rectangular\nsolid, what is the volume of the rectangular solid?\n  18(A)\n  50(B)\n100(C)\n200(D)\n400(E)\n5. The sum of all distinct solutions for x in the equation x2 − 8x + 21 = |x\n− 4| + 5 is equal to which of the following?"} +{"id": "book 2_p79_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 79, "page_end": 79, "topic_guess": "general", "text": "−7(A)\n  7(B)\n 10(C)\n 12(D)\n 14(E)"} +{"id": "book 2_p80_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 80, "page_end": 80, "topic_guess": "number_theory", "text": "Solutions\nEach solution addresses the two steps from the instructions:\nIdentify exactly what the problem is asking for, and what that means in\nthe simplest possible terms.\n1)\nNote at least one general takeaway that might useful on other problems\nin the future.\n2)\n1. (C) \n1. What it’s asking: The problem is asking for the probability that the\nselected ball is a multiple of 42.\nThe quantities you care about are the factors of 210.\nWhat you\nknow:\nThere are many balls, each with a different factor of 210. Each factor of 210 is\nrepresented. One ball is selected randomly. Some balls have a multiple of 42\n(e.g., 42 itself ); some do not (e.g., 1).\nWhat you\ndon’t\nknow:\nHow many factors of 210 there are\nHow many of these factors are multiples of 42\nWhat the\nproblem\nis testing:\nProbability; Divisibility & Primes"} +{"id": "book 2_p81_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 81, "page_end": 81, "topic_guess": "number_theory", "text": "The real question:\nPlan: 210 to primes → build full list of factors from prime components →\ndistinguish between multiples of 42 and non-multiples → count factors →\ncompute probability.\nAlternatively, you could list all the factors of 210 using factor pairs.\n1 210\n2 105\n3 70\n5 42\n6 35\n7 30\n10 21\n14 15"} +{"id": "book 2_p82_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 82, "page_end": 82, "topic_guess": "number_theory", "text": "There are 16 factors of 210, and two of them (42 and 210) are multiples\nof 42.\nYou can also count the factors using 210’s prime factorization: (2)(3)(5)\n(7) = (21)(31)(51)(71).\nHere’s a shortcut to determine the number of distinct factors of 210. Add\n1 to the power of each prime factor and multiply:\nThere are 16 different factors of 210: 2 x 2 x 2 x 2 = 16.\nHow many of these 16 factors are multiples of 42? 42 itself is a multiple\nof 42, of course. To find any others, divide 210 by 42 to get 5. This\nnumber is a prime, so the only other possible factor is 42 × 5, or 210.\nThere are two multiples of 42 out of a total of 16 factors, so the\nprobability is \n .\nThe correct answer is (C).\n2. At least one takeaway: The problem is straightforward in one sense: it\nsays the word factor explicitly. Listing all the factors is feasible in two\nminutes, but you do need to be going down that solution path fairly\nquickly because it will take some time. It may be slightly faster to use"} +{"id": "book 2_p83_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 83, "page_end": 83, "topic_guess": "sequences_patterns", "text": "the factor-counting shortcut, but only if you do know how to deal with\nthe multiples of 42.\n2. (A) 2:\n1. What it’s asking: The problem asks for the units digit. Because the\nproblem talks about a product, you care only about the units digits, not\nthe overall values. Furthermore, the problem provides crazy numbers;\nyou are absolutely not going to multiply these out. There must be some\nkind of pattern at work. Use the Last Digit Shortcut (discussed in the All\nthe Quant guide).\nWhat jumps out? If x is a positive integer, then 2x must be even and 5 +\n2x must be odd.\nUnits digit of (24)5 + 2x = units digit of (4)odd. The pattern for the units\ndigit of 4integer = [4, 6]. Thus, the units digit is 4.\nUnits digit of (36)6 must be 6, as every power of 6 ends in 6.\nUnits digit of (17)3 = units digit of (7)3. The pattern for the units digit of\n7integer = [7, 9, 3, 1]. Thus, the units digit is 3.\nThe product of the units digits is (4)(6)(3) = 72, which has a units digit of\n2. The correct answer is (A).\n2. At least one takeaway: Patterns were very important on this one! If\nyou forget any of the units digit patterns, start listing out the early cases.\nAt most, you’ll need to list four cases to find the pattern."} +{"id": "book 2_p84_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 84, "page_end": 84, "topic_guess": "number_theory", "text": "3. (E) 35:\n1. What it’s asking: The problem asks for the minimum number of\nchocolate chip cookies.\nGiven: The baker only makes chocolate chip (C) or peanut butter (P) cookies. He can only\nmake chocolate chip cookies in batches of 7 and peanut butter cookies in batches\nof 6. He makes exactly 95 cookies total.\nWhat jumps out? C and P must be integers. Therefore:\nThe answer choices are small multiples of 7, so work backwards from\nthe answers on this problem. Because the problem asks you to minimize\nthe number of chocolate chip cookies, start with the smallest answer\nchoice.\nMake a chart:\n7C 6P = 95 − 7C Is 6P a multiple of 6?\n(i.e., Is P an integer?)\n7 88 N\n14 81 N\n21 74 N"} +{"id": "book 2_p85_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 85, "page_end": 85, "topic_guess": "algebra", "text": "7C 6P = 95 − 7C Is 6P a multiple of 6?\n(i.e., Is P an integer?)\n28 67 N\n35 60 Y\nUse the answer choices to calculate the value of 6P. Cross off an answer\nchoice if 6P is not a multiple of 6. The first answer choice that works is\nthe last one. The correct answer is (E).\n2. At least one takeaway: The two competing constraints made testing\nchoices the most efficient method.\n4. (E) 400:\nThere are three equations and three variables. What is the easiest way to\nsolve for the volume?\nWhat it’s asking: The question asks for the volume of the box. Draw\nout the scenarios:\n1."} +{"id": "book 2_p86_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 86, "page_end": 86, "topic_guess": "algebra", "text": "The width, w, appears in all three constraint equations, so solve for the\nother variables in terms of w and substitute into the longest constraint:\nSubstitute:\nSince w can’t be zero, you can divide it\nout safely.\nSolve for all variables:\nThe correct answer is (E).\nAt least one takeaway: The question is complex enough that you\ncould check your work at the end by also calculating the volume of\nthe new solid (w + 1)(l + 1)(h − 9). As always, you have to decide\nwhether to spend that time here versus elsewhere.\n2."} +{"id": "book 2_p87_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 87, "page_end": 87, "topic_guess": "algebra", "text": "Notice that, though the initial volume formula seemed long and\nannoying, the calculations canceled out nicely in the end. This is\ncommon on the GMAT—common enough, in fact, to suspect that you\nmay be doing something wrong if the algebra becomes very messy.\n5. (D) 12:\nThe question implies that there may be multiple solutions, as does the\nnonlinear given equation. What is the most efficient way to find those\nsolutions?\nThere are actually two good approaches; choose the one that is easier\nfor you.\nApproach #1: Do the algebra. Split it into two equations, the “positive”\nversion and the “negative” version:\nScenario 1\nx − 4 ≥ 0\nScenario 2\nx − 4 ≤ 0\nWhat it’s asking: The question asks for the sum of the distinct\nsolutions. In other words, if the number 2 were to show up twice as a\nsolution, you would count it only once.\n1."} +{"id": "book 2_p88_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 88, "page_end": 88, "topic_guess": "algebra", "text": "Sum of the different solutions: 5 + 4 + 3 = 12. The correct answer is (D).\nApproach #2: simplify the equation and use theory to finish it off. Isolate\nthe absolute value:\nThink it through. You square a number and get the absolute value of\nthat same number (not squared!). Only a few numbers can make that\ntrue: 1 squared equals |1|, 0 squared equals |0|, −1 squared equals |−1|.\nThat’s it!\nSum of the different solutions: 5 + 4 + 3 = 12.\nAt least one takeaway: When a problem asks for distinct solutions,\ncount each unique solution; ignore multiple instances of the same\nvalue. Consider whether you prefer the pure algebraic approach or\nthe theoretical approach. Both are valid solution methods. Which one\ndo you think you will be able to remember and use more easily?\n2."} +{"id": "book 2_p89_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 89, "page_end": 89, "topic_guess": "word_problems", "text": "CHAPTER 2\nProblem Solving: Strategies &\nTactics"} +{"id": "book 2_p90_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 90, "page_end": 90, "topic_guess": "word_problems", "text": "In This Chapter...\nAdvanced Strategies\nAdvanced Guessing Tactics"} +{"id": "book 2_p91_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 91, "page_end": 91, "topic_guess": "word_problems", "text": "Chapter 2\nProblem Solving: Strategies & Tactics\nSometimes you will encounter a Problem Solving (PS) problem that you\ncan’t answer—either because its content is difficult or obscure or because\nyou don’t have enough time to solve completely in two minutes.\nThis chapter describes a series of different methods you might try in these\ncircumstances. Here, we make the distinction between solution strategies\nand guessing tactics.\nSolution strategies are broad: they apply to a wide variety of problems,\nthey provide a complete approach, and they can be used safely in most\ncircumstances.\nIn contrast, guessing tactics can help you eliminate a few answer choices,\nbut o en leave a fair amount of uncertainty. Moreover, a particular tactic\nmay only be useful in special situations or for parts of a problem.\nThe first section of this chapter outlines four Problem Solving strategies:\nPS Strategy 1: Choose Smart Numbers"} +{"id": "book 2_p92_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 92, "page_end": 92, "topic_guess": "strategy", "text": "The rest of the chapter is devoted to four specialized tactics that can knock\nout answer choices or provide clues about how to approach the problem\nmore effectively:\nOne of the most productive strategies on the GMAT is to pick good\nnumbers and plug them into unknowns. Try this when the concepts are\nespecially complex or when conditions are placed on key inputs that are\notherwise unspecified (e.g., n is a prime number).\nPS Strategy 2: Work Backwards\nAnother common approach is to work backwards from the answer\nchoices, testing to see which one fits. Doing so can o en help you avoid\ndemanding calculations or the need to set up and solve complicated\nalgebraic expressions.\nPS Strategy 3: Test Cases\nIn certain circumstances, a problem allows multiple possible scenarios,\nor cases. On PS questions, the problem usually asks you to find\nsomething that must be true or could be true. On these problems, you\ncan test different numbers to eliminate answers until only one remains.\nPS Strategy 4: Avoid Needless Computation\nThe GMAT rarely requires you to carry out intensive calculations to arrive\nat an answer. Look for opportunities to avoid tedious computation by\nfactoring, simplifying, or estimating."} +{"id": "book 2_p93_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 93, "page_end": 93, "topic_guess": "word_problems", "text": "PS Tactic 1: Look for Answer Pairs\nSome PS questions have answer choices that pair with each other in\nsome way. The correct answer may be part of one of these pairs.\nPS Tactic 2: Apply Cutoffs\nSometimes a back-of-the-envelope estimation can help you eliminate\nany answer choice above or below a certain cutoff.\nPS Tactic 3: Look at Positive/Negative\nSome PS questions include both positive and negative answer choices.\nIn such cases, look for clues as to the correct sign of the correct answer.\nPS Tactic 4: Draw to Scale\nMany Geometry problems allow you to eliminate some answer choices\nusing visual estimation, as long as you draw the diagram accurately\nenough."} +{"id": "book 2_p94_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 94, "page_end": 94, "topic_guess": "algebra", "text": "Advanced Strategies\n1. CHOOSE SMART NUMBERS\nSome types of problems allow you to pick real numbers and solve the\nproblem arithmetically rather than algebraically. For instance, almost any\nProblem Solving problem that has variables in the answer choices gives you\nthis opportunity. Likewise, you can o en pick a smart number for a fraction or\npercent problem without specified absolute value amounts.\nOther problem types allow this strategy as well. For instance, a problem may\nput specific conditions on the inputs but not give you exact numbers. In this\ncase, you can go ahead and just pick inputs that fit the conditions. If a\nproblem specifies that “x must be a positive even integer” but does not\nspecify the value of x, picking 2 for x will probably get you to a solution\nquickly and easily.\nThe GMAT Official Guide contains many problems that are difficult to solve\nalgebraically but much easier to solve with real numbers. However, as an\nadvanced test-taker, you might consider it a point of pride not to plug in a\nnumber. You might want to prove a “theoretically correct” answer. Overcome\nyour pride! The GMAT is not a math test; the GMAT tests you on your flexibility\nof thinking and your ability to manage a very limited amount of time. Use the\neasiest and most efficient solution path, not the textbook math solution path!\nTry-It #2-1"} +{"id": "book 2_p95_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 95, "page_end": 95, "topic_guess": "algebra", "text": "Andra, Elif, and Grady each invested in a certain stock. Andra\ninvested q dollars, which was 40% more than Elif invested. If Elif\ninvested 25% less than Grady invested, what was the total amount\ninvested by all three, in terms of q ?\nThis problem can be solved algebraically: write a couple of equations and\nsolve for all three variables, then add them up. A glance at the answers,\nthough, indicates that the algebra is likely to get messy. Instead, choose a real\nnumber and solve the problem arithmetically.\nIf you’ve made it to the GMAT Advanced Quant book, then you have likely\nused this strategy before (or at least learned about it). At times, you may have\nbeen frustrated because this technique didn’t actually seem easier than doing\nthe math algebraically. If so, here’s the missing piece: you need to learn how\nto choose smart numbers in the best possible way.\nMost of the time, you’re going to choose for the variable given (in this case, q).\nIn some cases, though, starting with the given variable doesn’t make your life\nany easier. The problem above actually has three unknowns: one for Andra,\none for Elif, and one for Grady. Take a look at the relationship between those\nunknowns before you decide which one is the best starting point.\n2q(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p96_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 96, "page_end": 96, "topic_guess": "strategy", "text": "Andra invests 40% more than Elif. For just these two, it would be easier to pick\na number for Elif and then calculate Andra’s amount.\nElif invests 25% less than Grady. For these two, it is easier to start with Grady\nand then calculate Elif. As a result, start with Grady, then find Elif, then find\nAndra.\nIf Grady invests $100, then Elif invests 25% less, or $75. Andra invests 40%\nmore than Elif, or $75 + $30 = $105. Make sure to note on your scrap paper\nthat q = 105.\nCollectively, the three invest $100 + $75 + $105 = $280. Find the answer choice\nthat matches $280:\nThe correct answer is (E).\nNote a few important aspects that will help you to choose smart numbers\nefficiently and effectively.\nFirst, if there are multiple unknowns and you have to choose where to start,\npause to think about how to make the math as easy as possible. In the case of\nthe problem above, if you had picked q = 100 for Andra, your next step would\n2q = 2(105) = $210. Eliminate.(A)\n Eliminate.(B)\nEliminate.\n(C)\n. Eliminate.(D)\n Correct!(E)"} +{"id": "book 2_p97_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 97, "page_end": 97, "topic_guess": "word_problems", "text": "have been to figure out Elif’s amount. It is not the case that Elif would be $60,\nor 40% less than Andra.\nRather, Andra is 40% more than Elif: 1.4e = $100, so \n . Elif would\nactually equal approximately $71.42857. Nobody’s going to want to go down\nthat path! At this stage, you have two choices: you can go back and pick for\nsomeone else or you can think about what numbers would make this\nparticular path easier. For example, a value of q = 140 instead of 100 would\nresult in an easier calculation, \n , and a value of 100 for Elif. (In this case,\nthat would still result in a messy number for the next calculation; on another\nproblem that didn’t have so many numbers, though, making a one-number\nadjustment can still leave you well within the two-minute time frame.)\nSecond, you do not need to calculate the value of every answer choice. You\ncan stop and eliminate a choice whenever you can tell that it will not equal\n$280. When you plug in 105 for q, answer (B) isn’t an integer and answer (C)’s\nunits digit isn’t 0, so neither can be the correct answer.\nThird, practice this strategy extensively in order to expose yourself to these\nlittle variations and possible sticking points. As you become proficient with\nthe strategy, you’ll be amazed at how much time and mental effort it can save\nyou on the GMAT.\n2. WORK BACKWARDS\nIn a number of cases, the easiest way to solve a GMAT problem is to start from\nthe answer choices and work backwards. Don’t be too proud to try this\ntechnique either. The GMAT doesn’t reward perfect math technique; it"} +{"id": "book 2_p98_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 98, "page_end": 98, "topic_guess": "number_theory", "text": "rewards finding the correct answer as efficiently as possible or guessing when\nneeded.\nTry-It #2-2\nIf \n , which of these integers could be the value of z ?\nLook at all of those fractions! Solving for z algebraically in this problem would\nnot be easy. Instead, notice two important clues: the problem asks for the\nvalue of a single variable and the answer choices offer nice-and-easy integers.\nWork backwards! People o en start at the beginning, with choice (A), but\nactually start with (B) or (D); you’ll learn why in a moment:\n(B) \n INCORRECT\nThe right side is smaller than the le . The numerators are always 2, so what\nneeds to happen to bring the two halves of the equation together?\nThe le side needs to be made smaller, so the denominator needs to be\nbigger; eliminate answer choice (A) as well. This is why you start with choice\n(B) or (D)—you can eliminate other answer choices that are too small or too\nlarge without having to test them. Try answer (D) next:\n0(A)\n1(B)\n2(C)\n3(D)\n4(E)"} +{"id": "book 2_p99_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 99, "page_end": 99, "topic_guess": "algebra", "text": "(D) \n CORRECT\nYou can stop when you find an answer that works. If you are paying attention\nto how the math works as you solve, you can o en get away with trying just\ntwo or three answer choices on these problems.\nTry-It #2-3\nA certain college party is attended by both male and female students.\nThe ratio of male-to-female students is 3 to 5. If 5 of the male\nstudents were to leave the party, the ratio would change to 1 to 2.\nHow many total students are at the party?\nOf course, you could set up equations for the unknowns in the problem and\nsolve them algebraically. However, the numbers in the answers are pretty\nstraightforward integers. Try Working Backwards.\nAgain, begin with answer (B) or answer (D), whichever number looks easier\nfor the problem:\n24(A)\n30(B)\n48(C)\n80(D)\n90(E)"} +{"id": "book 2_p100_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 100, "page_end": 100, "topic_guess": "fractions_decimals_percents", "text": "What characteristic must be true of the correct answer? It must be a multiple\nof 8, so eliminate answer (E). Try the remaining answers:\nOnce you have found the answer using this technique, you can stop; you\ndon’t need to test any remaining answers. Also, you don’t have to translate\nequations or figure out how to eliminate variables and solve. You just work\neach number through the problem until you’re done.\nAs you work through an answer, think about how the math is playing out.\nAgain, you will usually be able to eliminate some answers without actually\nhaving to try them.\n3. TEST CASES\nIn the original ratio, you can use an unknown multiplier to represent\nthe total number of students: 3x + 5x = 8x. If there are 30 students\ntotal, then the unknown multiplier is \n = 3.75, leaving you with a\nnon-integer number of students. This is impossible, so this answer\nmust be incorrect.\n(B)\nIf there are 24 total students, then the unknown multiplier is 3: there\nare 9 male students and 15 female students at the party. If 5 males\nleave, there will be 4 le , leaving a ratio of 4 : 15. Eliminate answer (A).\nThis ratio is also pretty far away from the correct ratio of 1 : 2,\nso consider trying the larger remaining answer, 80, next.\n(A)\nIf there are 80 students total, then the unknown multiplier is 10, so\nthere are 30 males and 50 females. If 5 males leave, the new ratio is 25\n: 50, or 1 : 2. This is the correct answer!\n(D)"} +{"id": "book 2_p101_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 101, "page_end": 101, "topic_guess": "number_theory", "text": "Some problems allow you to choose real numbers to solve, but you can’t\nchoose just one set of numbers as you do when you choose smart numbers.\nRather, you have to test multiple scenarios to get yourself to the one correct\nanswer. On Problem Solving problems, this tends to occur with must be true\nor could be true problems.\nTry It #2-4\nIf n is a positive integer, what must be true of n3 − n ?\nThis problem is asking about a theoretical concept: what mathematical\ncharacteristic must be true of this expression? In order to solve, you can test\nallowable cases to narrow down the answers until only one remains.\nNote that in this PS problem, one of the answers must be correct. In this case,\nany positive integer n will help you get to the answer, so start with the\nsimplest possible positive integer: 1. In general, when Testing Cases on PS or\nDS, start with the simplest number that fits the problem’s parameters.\nTry n = 1:\n  n n3 − n\nIt is divisible by 4.(A)\nIt is odd.(B)\nIt is a multiple of 6.(C)\nIt is a prime number.(D)\nIt has, at most, two distinct prime factors.(E)"} +{"id": "book 2_p102_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 102, "page_end": 102, "topic_guess": "number_theory", "text": "n n3 − n\nCase 1 1 0\nTry your next case, ignoring answers (B) and (D) from now on:\n  n  n3 − n\nCase 2 2 6\nYou’re down to (C) and (E). Because one of the statements talks about having\nmore than a certain number of prime factors, try a larger number next:\n  n  n3 − n\nYes, 0 is divisible by 4. (0 is divisible by any number except 0.)(A)\nNo, 0 is not odd. Eliminate.(B)\nYes, 0 is a multiple of 6. (0 is a multiple of any number.)(C)\nNo, 0 is not a prime number. Eliminate.(D)\nYes, 0 does not have more than two prime factors.(E)\nNo, 6 is not divisible by 4. Eliminate.(A)\nYes, 6 is a multiple of 6.(C)\nYes, 6 does not have more than two distinct prime factors. (The prime\nfactors are 2 and 3.)\n(E)"} +{"id": "book 2_p103_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 103, "page_end": 103, "topic_guess": "number_theory", "text": "n  n3 − n\nCase 3 5 120\nThe correct answer is (C). You could save yourself some time on this one by\nrecognizing that the expression n3 − n can be rewritten as:\nIn other words, the expression represents three consecutive integers. Test\nsome cases to discover what must be true about the product of three\nconsecutive integers.\nCase 1: If n = 1, then the three consecutive integers are 0, 1, 2. This product is\neven and not prime, so eliminate answers (B) and (D).\nCase 2: If n = 2, then the three consecutive integers are 1, 2, and 3. This\nproduct does not contain two 4’s, so it is not a multiple of 4. Eliminate (A).\nCase 3: If n = 3, then the three consecutive integers are 3, 4, and 5. You could\nmultiply out the consecutive integers, but only go down the computation\nYes, 120 is a multiple of 6.(C)\nNo, 120 does have more than two distinct prime factors. (The prime\nfactors are 2, 3, and 5.) Eliminate.\n(E)"} +{"id": "book 2_p104_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 104, "page_end": 104, "topic_guess": "number_theory", "text": "path if you must. In this case, you’re trying to find factors, and the three\nconsecutive integers tell you that directly; you don’t need to multiply them\nout. (You’ll learn more about avoiding unnecessary computation in the next\nsection.)\nInstead, if possible, try to notice a pattern. In every set of three consecutive\nnumbers, you will always have at least one even number. You will also always\nhave a multiple of 3. (In Case 1, 0 is a multiple of 3.) As a result, the product\nwill always be a multiple of 6.\nYou may also notice that the problem contains no upper limit. By choosing a\nlarge enough number, you’re going to be able to create a number that\ncontains more than two distinct prime numbers.\nYou can test cases directly to eliminate the four incorrect answers, or you can\nuse a few cases to help you figure out the theory underlying the problem,\nwhich will also get you to the correct answer.\n4. AVOID NEEDLESS COMPUTATION\nYou won’t see many GMAT problems that require substantial calculation to\narrive at a precise answer. Rather, correct answers on difficult problems will\ngenerally be relatively easy to compute once the difficult concept or trick in\nthe problem has been correctly identified and addressed.\nOn several types of GMAT problems, a significant amount of computation can\nbe avoided. A simple rule of thumb is this: if it seems that calculating the\nanswer is going to take a lot of work, there’s a good chance that a shortcut\nexists. Look for the back door!"} +{"id": "book 2_p105_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 105, "page_end": 105, "topic_guess": "fractions_decimals_percents", "text": "Estimation\nIntelligent estimation can save you time and effort on many problems. For\nexample, you can round to nearby benchmarks or be ready to switch a\nfraction to a decimal or percent, and vice versa.\nTry-It #2-5\nThe percent change from 29 to 43 is approximately what percent of\nthe percent change from 43 to 57 ?\nIn this case, the question stem straight up tells you that you can estimate. Any\ntime you see the word approximately (or a synonym), definitely do not try to\nsolve for the exact answer.\nThe question stem is pretty complex. Use the percent change formula:\nA direct translation would look like this:\n50%(A)\n66%(B)\n110%(C)\n133%(D)\n150%(E)"} +{"id": "book 2_p106_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 106, "page_end": 106, "topic_guess": "fractions_decimals_percents", "text": "You can simplify things by solving the question in parts. First, it talks about\nthe percent change from 29 to 43.\nThe percent change for the top half of the fraction, is \n , but that number is\ncumbersome. Estimate! This is approximately \n .\nThe percent change from 43 to 57 is \n . This is about \n . The question is\nreally asking:\nThe fraction \n is equivalent to 150%. The correct answer is (E).\nThe wording of the question can sometimes provide a strong clue that\nestimation should be used. Look for phrases such as these:\nThe test doesn’t have to tell you that you can estimate, though. You can\nusually estimate when the answer choices are far apart.\nHeavy Long Division\n is what percent of \n ?\n. . . the number is approximately equal to which of the following?\n. . . this result is closest to which of the following?\nWhich of the following is most nearly equal to . . .?"} +{"id": "book 2_p107_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 107, "page_end": 107, "topic_guess": "number_theory", "text": "Very few problems on the GMAT truly require long division, even though it\nmight appear otherwise. You can almost always approximate the answer or\nreduce the division by taking out common factors.\nTry-It #2-6\n is equivalent to which of the following?\nAt first glance, it appears that precise long division is necessary. The answer\nchoices are very close together, making estimation difficult. However, with\nsome manipulation and factoring, the solution is much more straightforward.\nThe key to factoring this fraction is to move the decimals of both the\nnumerator and denominator three places to the right so that you’re dealing\nwith integers. Then you might notice that 3,507 is divisible by 7 (3,500 and 7\nare both divisible by 7, providing a clue that you may be able to factor out 7).\nMoreover, 10,020 is divisible by 10 and by 2 (10,020 ends in a 0, and 1,002 is\neven):\nThe correct answer is (A).\n0.35(A)\n0.3505(B)\n0.3509(C)\n0.351(D)\n0.3527(E)"} +{"id": "book 2_p108_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 108, "page_end": 108, "topic_guess": "algebra", "text": "Alternatively, you might observe that 10.02 is very slightly larger than 10.\nTherefore, the correct answer will be slightly smaller than\n Guess between choices (A) and (B).\nTry-It #2-7\nWhat is the value of \nSince 102 is extremely small compared to 105, and the choices are somewhat\nspread out, estimate:\nNotice that by ignoring the 102 term, you made the denominator slightly\nlarger than it originally was. Therefore, 0.81918 is slightly smaller than the\ncorrect answer, 0.82. The correct answer is (D).\nQuadratic Expressions in Word Problems\nSome Word Problems result in a quadratic equation. You are probably pretty\ngood at solving quadratic equations, so your natural bias would be to set up\nand solve the equation. However, if the coefficients are huge, the equation\n8.19(A)\n8.02(B)\n0.89(C)\n0.82(D)\n0.81(E)"} +{"id": "book 2_p109_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 109, "page_end": 109, "topic_guess": "algebra", "text": "may be very difficult to solve. In these cases, try testing the answer choices in\nthe original problem (not in the translated and manipulated quadratic),\nespecially when the answer choices contain easier numbers.\nTry-It #2-8\nA shoe cobbler charges n dollars to repair a single pair of loafers.\nTomorrow, he intends to earn $240 repairing loafers. If he were to\nreduce his fee per pair by $20, he would have to repair an additional\npair of the loafers to earn the same amount of revenue. How many\npairs of loafers does he intend to repair tomorrow?\nThe problem may not seem too bad . . . until you try to set it up algebraically.\nAssign x to represent the number of pairs of loafers the cobbler intends to\nrepair tomorrow. Using the equation for revenue gives you the following:\nFurthermore, reducing his fee by $20 would result in the need to repair an\nadditional pair of shoes for the same amount of revenue gives you the\nfollowing:\n1(A)\n2(B)\n3(C)\n4(D)\n5(E)"} +{"id": "book 2_p110_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 110, "page_end": 110, "topic_guess": "algebra", "text": "From here, the algebra gets complicated very quickly. (The algebra is shown\nlater, if you want to see!)\nSwitch to a different approach—Working Backwards from the answer choices.\nStart with (B) or (D).\nThe revenue in answer (B), $300, is too large. Do you need to make x larger or\nsmaller? If you’re not sure, try (D) next, and notice something very important.\nThe revenue in (D), $200, is too small. The correct revenue, therefore, should\nfall between (B) and (D). The answer must be (C)!\n  x n x × n x + 1 n − 20 (x + 1) × (n − 20)\n(B) 2 $120 $240 3 $100 $300\n(D) 4 $60 $240 5 $40 $200\nIf you’re not sure, try that answer:\n  x n x × n x + 1 n − 20 (x + 1) × (n − 20)\n(C) 3 $80 $240 4 $60 $240\nHere’s the algebra in case you want to see it (but it’s not recommended to do\nthis):"} +{"id": "book 2_p112_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 112, "page_end": 112, "topic_guess": "word_problems", "text": "Advanced Guessing Tactics\nTo repeat, the tactics below are less universally useful than the strategies\nwe just covered. However, when all else fails, “break the glass” and try one\nor more of these tactics. They’re almost always better than guessing\ncompletely randomly. If you’re way behind on time and you have to\nsacrifice a couple of problems, though, guess immediately and move on.\nThe examples below will not be too difficult in order to illustrate the tactic\nclearly and not distract you with other issues. Of course, if you can solve\nthe problem directly, do so. But also study the tactic, so you’re ready to use\nit on a harder problem of the same type.\n1. Look for Answer Pairs            Certainty: Moderate\nGMAT answer choices are sometimes paired in a mathematically relevant\nway. Pairs of answers may:\nThe correct answer is sometimes part of such a pair. Why? The GMAT likes\nto put in a final obstacle. Say you do everything correct, except you solve\nAdd up to 1 on a probability or fraction question\nAdd up to 100% on questions involving percents\nAdd up to 0 (be opposites of each other)\nMultiply to 1 (be reciprocals of each other)"} +{"id": "book 2_p113_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 113, "page_end": 113, "topic_guess": "statistics", "text": "for the wrong unknown or forget to subtract from 1. Under the pressure of\nthe exam, people make this sort of penultimate error all the time\n(penultimate means next to last).\nIn order to catch folks in this trap, the GMAT has to make an answer choice\nthat’s paired to the correct answer—it’s correct except for that one last\nstep.\nThis means that you can o en eliminate unpaired answer choices. Also, the\nway in which the answers are paired may provide clues about the correct\nsolution method and/or traps in the problem.\nTry-It #2-9\nAt a certain high school, the junior class is twice the size of the\nsenior class. If \n of the seniors and \n of the juniors study\nJapanese, what fraction of the students in both classes do not study\nJapanese?\nNote that two pairs of answers each add up to 1: \n , and\n. Answer (A) is not likely to be correct, because it is not part of an\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p114_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 114, "page_end": 114, "topic_guess": "fractions_decimals_percents", "text": "answer pair. The fact that these pairs sum to 1 also provides a clue to\ndouble-check the wording of the question: do vs. do not study Japanese\n(the sum of the fractions of the students that do study Japanese and those\nwho do not will equal 1).\nTo solve, you could use a double-set matrix that shows juniors versus\nseniors and Japanese studiers versus non-Japanese studiers. However,\nlet’s go back to the first strategy in this chapter and choose smart numbers.\nPick a smart number that is a multiple of the denominators in the problem:\n3 × 4 = 12.\nIf the junior class has 12 people, then the senior class thus has 6. If \n of\nthe seniors study Japanese, then \n seniors study Japanese. If\n of the juniors study Japanese, then \n juniors study\nJapanese. There are 12 + 6 = 18 students total, and 2 + 3 = 5 of them study\nJapanese. Thus, \n of the students do study Japanese, so the fraction of\nthe students that do not study Japanese is \n . The correct answer is (E).\n2. Apply Cutoffs           Certainty: High\nYou may be able to eliminate answers above or below some easily\ncalculated threshold value. You may have to imagine that the problem is\nslightly different (and easier) to come up with that threshold value, but\nonce you do, you can o en get rid of two or three answer choices."} +{"id": "book 2_p115_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 115, "page_end": 115, "topic_guess": "word_problems", "text": "This strategy can sometimes be used in combination with an answer pairs\nstrategy, as pairs of answers are o en composed of a high and a low value.\nIn the previous problem, for example, \n of the seniors and \n of the\njuniors study Japanese. Therefore, somewhere between \n   and \n of the\nstudents overall, or less than half, must study Japanese. This implies that\nthe fraction of students who do not study Japanese must be more than\nhalf. You could eliminate answer choices (A), (B), and (C) because each of\nthese answer choices is smaller than \n .\nTry-It #2-10\nThe eSoroban device is available in two colors, orange and green.\nIn 2013, 60% of the eSoroban devices sold were purchased by\nwomen, \n of whom purchased the orange device. If an equal\nnumber of orange and green eSoroban devices were sold in 2013,\nwhat fraction of men who purchased an eSoroban in\n2013 purchased the green device?\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p116_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 116, "page_end": 116, "topic_guess": "statistics", "text": "More than half of the people purchasing this device are women, and more\nthan half of them buy the green version. If an equal number of green and\norange devices are sold, then more than half of the men must buy the\norange version. Therefore, less than half of men buy the green version:\neliminate answers (D) and (E).\nFurthermore, women represent 60% of purchases, while men represent\njust 40%. Pretend for a moment that there were equal numbers of men and\nwomen. If that were the case, then \n of women would buy orange and\n would buy green. On the flip side, \n of men would buy green and \nwould buy orange.\nIn fact, there are more women than men, so the proportion of men buying\norange devices has to be higher than \n . Fewer than \n must buy the\ngreen version. Eliminate answer (C).\nFrom here, if you have a strong number sense, you might guess that\nanswer (B) is more likely the correct answer, because a \n split would\nhave made the answer \n . The actual split is \n , so the correct value is\nnot that much lower than \n . But here’s the actual solution, using a\ndouble-set matrix and a smart number of 100 total people:\n  Men Women Total\nOrange  \n 50"} +{"id": "book 2_p117_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 117, "page_end": 117, "topic_guess": "word_problems", "text": "Men Women Total\nGreen 50 − 35 = 15 60 − 25 = 35 50\nTotal 40 60 100\nBecause 15 men purchased a green device out of 40 men total, the\nproportion is \n = \n . The correct answer is (B).\n3. Look at Positive/Negative            Certainty: High\nA special case of the cutoff strategy occurs when some of the answer\nchoices are positive and others are negative. In this case, focus on figuring\nout the sign of the correct answer, then eliminate any answer choices of\nthe opposite sign. (Again, this is something to do when you have run out of\ndirect approaches or you are short on time!)\nTry-It #2-11\nIf x □ y is defined to equal \n for all x and y, then (−1 □ 2) □ 3 is\nequivalent to which of the following?\n\n(A)\n\n(B)\n\n(C)\n−\n(D)"} +{"id": "book 2_p118_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 118, "page_end": 118, "topic_guess": "word_problems", "text": "The negative sign in the term −1 will be extinguished, because the term\nbefore the □ symbol is squared when this function is calculated. Therefore,\nthe correct answer will be positive. Eliminate (D) and (E). By the way, the\ncorrect answer is (C).\n4. Answer Properties                Certainty: Moderate\nYou may not have time, while doing a Problem Solving problem, to figure\nout the correct answer. However, it might take less time to figure out\nsomething about the correct answer. You saw two examples of this in the\nprevious two guessing strategies: you can sometimes figure out whether\nthe correct answer is above or below a certain value, or whether it’s\npositive or negative. Here are some other things you may be able to\ndetermine: \n5. Draw to Scale            Certainty: Moderate\nThe estimation technique can be extended to some Geometry problems.\nYou can approximate the length of a line segment, size of an angle, or area\nof an object by drawing it as accurately as possible on your scrap paper.\n−\n(E)\nIs the correct answer even or odd?\nIs the correct answer divisible by a certain value? \nWhat is the units digit of the correct answer? \nDoes the correct answer include square roots?\nDoes the correct answer include decimals?"} +{"id": "book 2_p119_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 119, "page_end": 119, "topic_guess": "geometry", "text": "When you take the GMAT, the laminated scratch booklet is printed with a\nlight grid. This grid can help you draw very accurate scale pictures, and\no en the picture alone is enough to answer the question.\nTry-It #2-12\nIn the diagram to the right, equilateral\ntriangle ADE is drawn inside square\nBCDE. A circle is then inscribed inside\ntriangle ADE. What is the ratio of the area\nof the circle to the area of the square?\nIf a Problem Solving question does not say that the diagram is not drawn to\nscale, then the diagram is drawn to scale. The circle is about \n the height\nof the square and about \n the width. Therefore, the area of the circle\nshould be approximately \n of the area of the square. Eliminate any answer\nchoices that are far away from that estimate:\nOk\nOn the high side\nToo high\n(A)\n(B)\n(C)\n(D)\n(E)\n(A)\n(B)\n(C)"} +{"id": "book 2_p120_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 120, "page_end": 120, "topic_guess": "geometry", "text": "Too high\nWay too high\nOnly answers (A) and (B) are remotely reasonable, and only (A) is really\nclose to the estimate.\nThe most efficient way to solve this problem fully is to assign a radius of 1\nto the circle. This way, the circle has an area of π. Now work outwards.\nIf the center of the circle is O, then OF = 1.\nBecause ADE is an equilateral triangle, angle ADE\n= 60°. OD bisects angle ADE, so angle ODE = 30°.\nTherefore, triangle OFD is a 30 : 60 : 90 triangle,\nand \n . DE must be twice the length of\nDF, meaning that \n , and the area of\nsquare BCDE = \n .\nSo the ratio of the area of the circle to the area of the square is \n .\nNotice how much easier the draw to scale tactic was! Will it work every\ntime? No. But it works o en enough that you want to think about using it.\n(D)\n(E)"} +{"id": "book 2_p121_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 121, "page_end": 121, "topic_guess": "strategy", "text": "Problem Set\nSolve problems 1–12. In each case, identify whether you could use\nadvanced strategies (Choose Smart Numbers, Work Backwards, Test Cases,\nor Avoid Needless Computation) and guessing tactics (Look for Answer\nPairs, Apply Cutoffs, Look at Positive-Negative, or Draw to Scale) in any\nbeneficial way. Of course, there are textbook ways to solve these problems\ndirectly; instead of trying the textbook method first, focus on applying the\nstrategies and tactics described in this chapter.\n1. A popular yoga studio that is always filled to capacity moves to a\nnew location that is able to serve 45% more students.\nUnfortunately, 20% of the current students will no longer attend\nclasses at the new location. If classes at the new location are also\nfilled to capacity, what fraction of the students at the new location\nwill be new students?"} +{"id": "book 2_p122_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 122, "page_end": 122, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n2. If x < 10, y < 8, and y < x, what must be true?\nxy < 80I.\nII.\nx 2 + y 2 > 1III.\nNone(A)\nII only(B)\nIII only(C)\nI and II only(D)\nII and III only(E)\n3. If \n , which of the following could be the value of x ?"} +{"id": "book 2_p123_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 123, "page_end": 123, "topic_guess": "general", "text": "6(A)\n  8(B)\n  9(C)\n12(D)\n15(E)\n4. If \n , which of the following could be the value of x  ?\n−2(A)\n  0(B)\n  1(C)\n  3(D)\n  5(E)\n5. If 154 is \n of x, approximately what is the value of 2x  ?\n104(A)\n114(B)\n208(C)\n228(D)\n416(E)\n6. \n is equivalent to which of the following?"} +{"id": "book 2_p124_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 124, "page_end": 124, "topic_guess": "general", "text": "0.6(A)\n0.603(B)\n0.606(C)\n0.615(D)\n0.66(E)\n7. In a certain clothing store, the most expensive pair of socks sells for\n$1 less than twice the price of the cheapest pair of socks. A\ncustomer notices that for exactly $18, she can buy three fewer pairs\nof the most expensive socks than the cheapest socks. What could\nbe the number of pairs of the cheapest socks she could have\npurchased?\n  3(A)\n  5(B)\n  6(C)\n12(D)\n36(E)\n8. If \n , then m must equal which of the following?\n−2(A)\n−1(B)\n  0(C)\n  1(D)\n  2(E)"} +{"id": "book 2_p125_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 125, "page_end": 125, "topic_guess": "number_theory", "text": "9. Simplify \n10. If 3x + 3x + 3x = 1, what is x ?\n−1(A)\n−\n(B)\n  0(C)\n\n(D)\n  1(E)\n11. If a and b are integers and a is a factor of b, what must be true?\na < bI.\nThe distinct prime factors of a2 are also factors of b.II.\nIII."} +{"id": "book 2_p126_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 126, "page_end": 126, "topic_guess": "number_theory", "text": "Without solving problems 13–16, which answers could you confidently\neliminate and why?\nNone(A)\nII only(B)\nIII only(C)\nI and II only(D)\nII and III only(E)\n12. The integer k is positive but less than 400. If 21k is a multiple of 180,\nhow many unique prime factors does k have?\nOne(A)\nTwo(B)\nThree(C)\nFour(D)\nFive(E)\n13. \n0.693(A)\n0.7(B)\n0.71(C)\n6.93(D)\n7.1(E)"} +{"id": "book 2_p127_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 127, "page_end": 127, "topic_guess": "fractions_decimals_percents", "text": "14. \n is equivalent to which of the following?\n−1.8(A)\n−0.18(B)\n  0(C)\n  0.18(D)\n  1.8(E)\n15. In the 7-inch square to the right, another\nsquare is inscribed. What fraction of the\nlarger square is shaded?\n(A)\n(B)\n(C)\n(D)\n(E)\n16. If Mason is now twice as old as Gunther was 10 years ago, and G is\nGunther’s current age in years, which of the following represents\nthe sum of Mason’s and Gunther’s ages 4 years from now?"} +{"id": "book 2_p128_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 128, "page_end": 128, "topic_guess": "general", "text": "(A)\n3G + 28(B)\n3G − 12(C)\n8 − G(D)\n(E)"} +{"id": "book 2_p129_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 129, "page_end": 129, "topic_guess": "fractions_decimals_percents", "text": "Solutions\n1. (D) \n : (Choose Smart Numbers) Both the 45% figure and the answer\nchoices look cumbersome, but the problem never provides a real value\nfor the number of students, either before the move or a er. Choose a\nsmart number of 100 for this Percents problem.\nIf the new studio is filled to capacity, at 145 students, but only 80 old\nstudents continue to attend, then the studio will have 65 new students.\nThe fraction of new students at the new location, then, is \n .\nThe correct answer is (D).\nOld capacity: 100 students\nNew capacity: 145 students\nOld students who will stay with the studio: 80\n2. (A) None: (Test Cases) Test real numbers to prove or disprove the\nstatements. Make sure to choose only values that are allowed by the\nproblem: x < 10, y < 8, and y < x.\nStatement I: If x = 9 and y = 7, then xy = 63, which is less than 80. If, on\nthe other hand, x = −10 and y = −20, then xy = 200, which is greater than\n80. Statement I does not have to be true. Eliminate answer (D).\nStatement II: Positive values for both x and y will make \n , but if x\n= 9 and y = −1, then \n , which is not greater than 1. Statement II"} +{"id": "book 2_p130_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 130, "page_end": 130, "topic_guess": "algebra", "text": "does not have to be true. Eliminate answers (B) and (E).\nStatement III: If x = 3 and y = 2, then x2 + y2 is 13, which is greater than 1.\nIf x = −2 and y = −3, then x2 + y2 is still 13, which is still greater than 1.\nDon’t forget about fractions! If x = 0.3 and y = 0.2, then x2 + y2 = 0.09 +\n0.04, which is not greater than 1. Fractions between 0 and 1 get smaller\nwhen squared. Eliminate answers (C) and (E).\nNone of the statements must be true. The correct answer is (A).\n3. (D) 12: (Work Backwards) This problem can be solved algebraically, but\nnot easily. You’d actually need to use the quadratic formula, and the\nequation would turn out to be really nasty. No thank you!\nInstead, notice that the problem asks for a single variable and the\nanswer choices are decently small integers. Start with answer (B) or (D).\nThat math is a pain. Is answer (D) any easier? If so, start there.\nYes! Much easier: 122 = 144, and that goes very nicely with 288.\nThis equation simplifies to 5 + 2 = 7, so the correct answer is (D). Note\nthat if (D) had not been the correct answer, you would have been able to\n(B) \n(D)"} +{"id": "book 2_p131_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 131, "page_end": 131, "topic_guess": "fractions_decimals_percents", "text": "tell whether to try a larger or smaller answer next based on whether (D)\nwas too small or too large.\n4. (D) 3: (Work Backwards) Typically, when Working Backwards, you start\nwith answer (B) or (D). If you can tell that a problem isn’t likely to have a\nconsistent pattern, though, as with this absolute value equation, then\njust start at one end—answer (A) or answer (E)—and stop when you find\nthe correct answer.\n  x x2 − 6 | x2 − 6 |\n(A) −2 (−2)2 − 6 = 4 − 6 = −2 2\n(B) 0 (0)2 − 6 = 0 − 6 = −6 6\n(C) 1 (1)2 − 6 = 1 − 6 = 5 5\n(D) 3 (3)2 − 6 = 9 − 6 = 3 3\nThe correct answer is (D).\n5. (E) 416: (Avoid Needless Computation) Not only do the numbers in the\nproblem look ugly, but the problem asks for an approximate answer.\nEstimate.\nIt would be much easier if 154 were 150. What about that fraction? First\nof all, put it over 100: \n .\nMuch better! This is almost \n , so use that fraction instead."} +{"id": "book 2_p132_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 132, "page_end": 132, "topic_guess": "algebra", "text": "Translate the equation with the estimated values:\nFinally, make sure to answer the right question that was asked! The\nanswer is not (C), 208, because the question asks for the value of 2x.\nThe correct answer is (E).\n6. (A) 0.6: (Avoid Needless Computation)\nAlternatively, note that 2.010 is very slightly larger than 2. Therefore, the\nfraction is very slightly smaller than \n Only (A) is\nsmaller than 0.603. The correct answer is (A).\n7. (D) 12: (Work Backwards) This problem can be solved algebraically, but\nthe math gets pretty challenging (see end of this explanation). It’s easier\nto work from the answer choices. Start with (B) or (D).\nCall the number of cheap pairs c and the number of expensive pairs e.\nCall the cost of a cheap pair $c and the cost of an expensive pair $e. First,\ntry (B).\n# Cheap Pairs\n(c)\n$c Pairs\n # e Pairs (c −\n3)\n$e Pairs\n Match?\n$e = 2($c) − 1?"} +{"id": "book 2_p133_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 133, "page_end": 133, "topic_guess": "algebra", "text": "# Cheap Pairs\n(c)\n$c Pairs\n # e Pairs (c −\n3)\n$e Pairs\n Match?\n$e = 2($c) − 1?\n(B) 5\n 2 9\nNo ✗\nAnswer (B) is incorrect. Note that you don’t actually have to figure out\nthe value of the right-hand side of the equation, as long as you know\nthat it will not equal 9. The two sides of the equation are pretty far apart,\nso (A) and (C) are also probably not correct answers. Try answer (D) next.\n# Cheap Pairs (c) $c Pairs \n # e Pairs (c − 3) $e Pairs \n Match?\n$e = 2($c) − 1?\n(D) 12 $1.50 9 $2 2 = 2(1.5) – 1\nYes ✓\nAnswer (D) makes the final equation work, so it is the correct answer.\nHere’s how the algebra would have to be set up. Let c = the number of\npairs of cheap socks, e = the number of pairs of expensive socks, x = the\ncost for one pair of cheap socks, and y = the cost for one pair of\nexpensive socks.\nFrom sentence 1: y = 2x − 1"} +{"id": "book 2_p134_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 134, "page_end": 134, "topic_guess": "algebra", "text": "From sentence 2: 18 = cx and 18 = ey = (c − 3)y\nFrom here, you would solve the equations from sentence 2 for x and y,\nrespectively, and plug them into the equation from sentence 1:\nPlug in to equation 1: \nWith the correct manipulation (over multiple, complicated steps!), that\nequation would eventually become c2 − 21 + 108 = 0 and you could solve\nfor the two solutions, c = 9 and c = 12. Only 12 is in the answer choices,\nso choice (D) is the correct answer.\n8. (D) 1: (Work Backwards) In order for the le -hand side to equal 1, m has\nto be positive. Try only the three positive answer choices. In this\ncircumstance, start with the middle number of the remaining choices. If\nm = 1, then the le -hand side simplifies to \n .\nThe correct answer is (D).\n9. \n (Avoid Needless Computation)\n10. (A) −1: (Work Backwards) You might solve this one by inspection: three\nidentical “somethings” sum to 1, so one of those “somethings” equals"} +{"id": "book 2_p135_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 135, "page_end": 135, "topic_guess": "number_theory", "text": ", or \n . Therefore, x = −1.\nTesting choices is fast, too. In this case, (B) and (D) are both fractions,\nbut the other answers are integers, so try testing (A), (C), and (E) instead.\nStop when you find the correct answer.\nMany people would try (C) first because 0 is an easy number. Notice that\nit’s too big. Which number should you try next, −1 or 1? Since 0 leads to a\nnumber that’s too big, try −1 next.\n  x 3x + 3x + 3x\n(C) 0 1 + 1 + 1 = 3\n(A) −1\nThe correct answer is (A).\n11. (B) II Only: (Test Cases) First, consider what the given information a is a\nfactor of b tells you:\nNext, test the statements.\nStatement I: a < b. A factor can be smaller than the main number, but a\nfactor can also equal the main number: 6 is a factor of 6. Statement I\n2 is a factor of 6\n6 is not a factor of 2\n1, 2, 3, and 6 are all factors of 6"} +{"id": "book 2_p136_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 136, "page_end": 136, "topic_guess": "number_theory", "text": "does not have to be true. Eliminate answer (D).\nStatement II: The distinct prime factors of a2 are also factors of b. If a = 2,\nthen a2 = 4. There is just one distinct prime factor of a2: 2. In this case, a\nis a factor of b, yes. If a = 6, then a2 = 36. There are two distinct prime\nfactors of a2: 2 and 3. In this case, a is still a factor of b.\nNotice any patterns? The distinct prime factors of a2 are the same as the\ndistinct prime factors of a, since a2 is made up of a multiplied by itself.\nSo, first, this statement is really saying that the distinct prime factors of\na are also factors of b.\nIf a is a factor of b, then by definition all of a’s prime factors also have to\nbe factors of b. No matter what numbers you try for statement II, a will\nbe a factor of b. Statement II must be true, so eliminate answers (A) and\n(C).\nStatement III: \n  If b is 6, then a could be 6, in which case\n. Alternatively, if b is 6, then a could be 2, in which case \n .\nSo far, this statement looks good.\nDon’t forget about negative numbers! The problem doesn’t specify\npositive integers. What if b is −6? In this case, a could still be 2, so\n. Statement III does not have to be true. Eliminate answer (E)\nand choose correct answer (B).\nNote: The GMAT doesn’t o en test the factors of negative numbers, but\nthe definitions provided in the math review of the Official Guide do"} +{"id": "book 2_p137_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 137, "page_end": 137, "topic_guess": "number_theory", "text": "allow for the possibility.\n12. (C) Three: (Choose Smart Numbers) A laborious way to solve this\nproblem would be to determine all the possible values for k and take a\nprime factorization of each value, counting the number of different\nprime factors that each value has. Ugh!\nAn easier technique is to pick a smart number: one value of k that\nsatisfies the constraints. Any value of k that fits the constraints must\nhave the same number of different prime factors as any other legal value\nof k. Otherwise, the problem could not exist as written. There would be\nmore than one correct answer.\nThe problem states that 21k is a multiple of 180, so \n must be\nan integer. In other words, k must be divisible by 60. The easiest number\nto choose is k = 60.\nThe prime factorization of 60 is 2 × 2 × 3 × 5, so 60 has the unique prime\nfactors 2, 3, and 5. Thus, k has three unique prime factors. The correct\nanswer is (C).\n13. (A), (D), and (E) can be eliminated: If you ignore the 103 in the\ndenominator, the division is \n . This is an\napproximation of the answer, not an exact computation of it, so\neliminate (A). You have slightly overstated the denominator, thus slightly\nunderstated the result. Answers (B) and (C) are possibilities, but (D) and\n(E) are both too large. The correct answer turns out to be (B).\n14. (A), (C), (D), and (E) can be eliminated: The numerator is negative, so\neliminate (C), (D), and (E). The numbers in the numerator will add to"} +{"id": "book 2_p138_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 138, "page_end": 138, "topic_guess": "geometry", "text": "more than –25 (i.e., closer to 0), so eliminate (A).\nThe correct answer must be (B).\n15. (A), (C), (D), and (E) can be eliminated: Ignoring the dimensions 3 and 4\nfor a moment, think about the types of squares that might be inscribed\nin the larger square:\nThe shaded area can be at most \n of the larger square, which occurs\nwhen the smallest possible square is inscribed in the larger square. This\ngives you a great cutoff.\nThe larger the inscribed square, the smaller the shaded area and the\nmore the inscribed square must be rotated from the vertical orientation\nof the minimum inscribed square.\nEliminate (D) and (E), since they are larger than \n . Eliminate (C), since\nthe labeled lengths of 3 and 4 are not equal, indicating that the\ninscribed square is rotated from the minimum square position.\nAnswer (B) is paired with (D) to sum to 1. Answer (A) is unpaired, so\nbetween (A) and (B), the more likely answer is (B). Choice (B) is in fact\nthe correct answer."} +{"id": "book 2_p139_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 139, "page_end": 139, "topic_guess": "fractions_decimals_percents", "text": "There is a way to arrive at the exact answer: compute the relevant areas\nand take the ratio. The point of this exercise, though, is to practice\nguessing tactics.\n16. (D) and (E) can be eliminated: Gunther must be at least 10 years old for\nhim to have had a non-negative age “10 years ago” and for Mason to\nhave a non-negative age now. Therefore, eliminate (D) and (E), as both\nchoices will give a negative result when G > 10. The correct answer is in\nfact (C)."} +{"id": "book 2_p140_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 140, "page_end": 140, "topic_guess": "general", "text": "CHAPTER 3\nData Sufficiency: Principles"} +{"id": "book 2_p141_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 141, "page_end": 141, "topic_guess": "general", "text": "In This Chapter...\nPrinciple #1: Follow a Consistent Process\nPrinciple #2: Never Rephrase Yes/No as Value\nPrinciple #3: Work from Facts to Question\nPrinciple #4: Be a Contrarian\nPrinciple #5: Assume Nothing"} +{"id": "book 2_p142_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 142, "page_end": 142, "topic_guess": "word_problems", "text": "Chapter 3\nData Sufficiency: Principles\nThe goal of every Data Sufficiency (DS) problem is the same: determine\nwhat information will let you answer the given question. This gives you a\nsignificant advantage. Once you know whether a piece of information lets\nyou answer the given question, you can stop calculating. You do not have\nto waste time finishing that calculation.\nHowever, this type of problem presents its own challenges. DS answer\nchoices are never numbers, so you can’t plug them back into the question\nto check your work. Also, answer choice (E)—that the two statements\ncombined are NOT sufficient—leaves open the possibility that the\nembedded math question is not solvable even with all the information\ngiven. That is, unlike Problem Solving (PS), DS may contain math problems\nthat cannot be solved! This aspect of DS is unsettling. (Note: If you don’t\nalready have the five DS answer choices memorized, then you are not yet\nready for this chapter. Practice with other material and then return here\nwhen you feel fully comfortable with how DS works.)\nOn DS problems, the issue being tested is “answer-ability” itself—can the\ngiven question be answered, and if so, with what information? So the GMAT\ndisguises “answer-ability” as best it can. The given facts and the question"} +{"id": "book 2_p143_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 143, "page_end": 143, "topic_guess": "word_problems", "text": "itself are generally presented in ways that make this determination\ndifficult.\nFor instance, information that seems to be sufficient may actually be\ninsufficient if it permits an alternative scenario that leads to a different\nanswer. Likewise, information that seems to be insufficient may actually be\nsufficient if all the possible scenarios lead to the same answer for the\nquestion.\nAdvanced DS problems require you to step up your game. You will have to\nget really good at Testing Cases, which is even more important for\nadvanced DS than for advanced PS problems.\nYou will also have to get really good at simplifying the given facts and the\ngiven question. The GMAT increases the trickiness of the phrasing of the\nquestion and/or statements even more than the complexity of the\nunderlying concepts.\nSo how do you approach advanced DS problems? Unfortunately, there is\nno “one-size-fits-all” approach:\nAll that said, there are a few guiding principles you can follow.\nThe best approach may involve precise application of theory, or it may\ninvolve a “quick and dirty” approach.\nThe statements may be easy or difficult to interpret.\nThe question may require no rephrasing or elaborate rephrasing. In fact,\nthe crux of the problem may rest entirely on a careful rephrasing of the\nquestion."} +{"id": "book 2_p144_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 144, "page_end": 144, "topic_guess": "word_problems", "text": "Principle #1: Follow a Consistent\nProcess\nThis is the most important principle. A consistent process will prevent the\nmost common errors. It will focus your efforts at any stage of the problem.\nPerhaps most importantly, it will reduce your stress level, because you will\nhave confidence in the approach that you’ve practiced.\nFollow these rules of the road:\nDo your work on paper, not in your head. This fits with the put pen to\npaper concept discussed in Chapter 1. Writing down each thought as it\noccurs helps you keep track of the work you’ve done. Your mind is also\nfreed up to think ahead. Many DS questions are explicitly designed to\nconfuse you if you do all the work in your head.\nLabel everything and separate everything physically on your paper. If\nyou mix up the elements of the problem, you will o en mess up the\nproblem itself. Keep these four elements straight:\nFacts given in the question stem. You can leave them unlabeled or\nyou can put them with each statement.\nQuestion. Label the question with a question mark. Obvious, right?\nAmazingly, many people fail to take this simple step. Without a\nquestion mark, you might think the question is a fact—and you will"} +{"id": "book 2_p145_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 145, "page_end": 145, "topic_guess": "word_problems", "text": "Here is one sample schema for setting up your work on a Data Sufficiency\nquestion:\nIt may be worth rewriting the facts from the question stem alongside\neach statement. Although this may seem redundant, the time is well\nspent if it prevents you from forgetting to use those facts.\nget the problem wrong. Keep the question mark as you rephrase. It’s\nalso helpful to keep the helping verb in a Yes/No question:\nOriginal question:\nYou write down:\nYou could just write down:\nDo NOT just write down:\nStatement (1). Label this with a (1).\nStatement (2). Label this with a (2).\nRephrase the question and statements whenever possible. Question\nstems and statements are o en more complex than they need to be; if\nyou can simplify the information up front, you will save yourself time\nand effort later in the problem. In particular, try to rephrase the question\nbefore you dive into the statements.\nEvaluate the easier statement first. If the second statement looks much\neasier to you than the first, then start with the second statement.\nPhysically separate the work that you do on the individual statements.\nDoing so can help reduce the risk of statement carryover—\nunintentionally letting one statement influence you as you evaluate the\nother."} +{"id": "book 2_p146_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 146, "page_end": 146, "topic_guess": "fractions_decimals_percents", "text": "Scrapwork for Rephrasing: ~~~~\n\nConstraints: ~~~~\n\nAD\nBCE\n\n(1)\n\nFACT from question stem: ~~~~\nFACT from (1): ~~~~\n. . . and any work you do to combine\nthese facts\n(2)\n\nFACT from question stem: ~~~~\nFACT from (2): ~~~~\n. . . and any work you do to combine\nthese facts\nQUESTION: .\n. . ?\nANSWER ANSWER\nNotice the physical separation between statement (1) and statement (2).\nYou might even consider going so far as to cover up the statement (1) work\nwhen evaluating statement (2). Also observe that this schema explicitly\nparses out the facts given in the question stem and evaluates those facts\nalongside each statement."} +{"id": "book 2_p147_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 147, "page_end": 147, "topic_guess": "number_theory", "text": "Try this example problem, and then take a look at what the scrap paper\nwould look like according to this schema.\nTry-It #3-1\nIf x and y are integers and 4xy = x2y + 4y, what is the value of xy ?\nBecause the question stem contains an equation, simplify it before\nconsidering the statements:\nTherefore, either x = 2 or y = 0, or both. One of the following scenarios must\nbe true:\nx y xy = ?\n2 Not 0 2y\nNot 2 0 0\n2 0 0\ny − x = 0(1)\nx3 < 0(2)"} +{"id": "book 2_p148_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 148, "page_end": 148, "topic_guess": "algebra", "text": "Interesting. It turns out that you only need to know the value of y. If y = 0,\nthen xy = 0. Otherwise, x must equal 2, in which case the value of xy is still\ndetermined by the value of y.\n(1) INSUFFICIENT: Rather than trying to combine this algebraically with the\nequation in the question stem, try a couple of the possible scenarios that\nfit the statement y − x = 2. Construct scenarios using the earlier table as\ninspiration:\nSince there are two possible answers, this statement is not sufficient.\n(2) SUFFICIENT: If x 3 < 0, then x < 0. If x does not equal 2, then y must equal\n0, according to the fact from the question stem. Therefore, xy = 0.\nNotice how valuable it was to evaluate the fact in the question stem first\nand to use it to rephrase the question. Then, when you reach the\nstatements, your work is made much easier. Here is approximately how\nyour paper could look:\nScrapwork for\nRephrasing:\nDoes x = 2 or does y = 0?\nx = 2, y = 0 or both\nIf y = 0, value of xy = 0.\nIf x = 2, value of xy\ndepends on y.\nIf x = 2, then y = 2 + x = 4, so xy = (2)(4) = 8.\nIf y = 0, then x = y − 2 = −2, so xy = (−2)(0) = 0."} +{"id": "book 2_p149_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 149, "page_end": 149, "topic_guess": "number_theory", "text": "Constraints:                            x and y integers\n\n(1)\n\nCase 1: If x = 2, y = 4, so xy = 8.\nCase 2: If y = 0, x = −2, so xy = 0.\n(2)\n\nIf x is negative, y must equal 0.\nQUESTION:\nDoes y = 0 ?\nWhat is y ?\nINSUFFICIENT y = 0 SUFFICIENT\nThe correct answer is (B).\nIf you don’t rephrase the question, it is easy to fall into the trap of thinking\nthat statement (2) alone is not sufficient.\nYou can lay out your paper in many other ways. For instance, you might go\nwith this version:\nStem:                                           Q: . . . ?"} +{"id": "book 2_p150_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 150, "page_end": 150, "topic_guess": "general", "text": "1)\n2)\nIn this second layout, facts go on the le , while the question and any\nrephrasing go on the right. Then the process is always to see whether you\ncan bridge the gap, going from le to right.\nThe important thing is that you develop a consistent layout that you\nalways use. Don’t give away points on Data Sufficiency because your work\nis sloppy or you mix up the logic."} +{"id": "book 2_p151_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 151, "page_end": 151, "topic_guess": "general", "text": "Principle #2: Never Rephrase Yes/No\nas Value\nAll DS questions can be divided into two types: Value questions (such as\n“What is a?”) and Yes/No questions (such as “Is x an integer?”). Value\nquestions and Yes/No questions are fundamentally different: they require\ndifferent levels of information to answer the question. Therefore, never\nrephrase a Yes/No question as a Value question. Value questions usually\nrequire more information than Yes/No questions.\nTry-It #3-2\nIs the integer n odd?\nYou don’t need to know which value n might be, just whether n is odd.\nTherefore, do not rephrase this question to “What is integer n?” Doing so\nunnecessarily increases the amount of information you need to answer the\nquestion. Of course, if you happen to know what n is, then great, you can\nanswer any Yes/No question about n. But you generally don’t need to know\nthe value of n to answer Yes/No questions about n, and the GMAT loves to\nexploit that truth at your expense.\nn2 − 2n is not a multiple of 4.(1)\nn is a multiple of 3.(2)"} +{"id": "book 2_p152_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 152, "page_end": 152, "topic_guess": "general", "text": "(1) SUFFICIENT: n2 − 2n = n(n − 2). If n is even, both terms in this product\nwill be even and the product will be divisible by 4. Since n2 − 2n is not a\nmultiple of 4, integer n cannot be even—it must be odd.\n(2) INSUFFICIENT: Multiples of 3 can be either odd or even.\nThe correct answer is (A).\nRephrasing a Yes/No question into a Value question makes the question\nunnecessarily picky. Yes/No questions can o en be sufficiently answered\ndespite having multiple possible values for the answer. In the last question,\nfor example, n could be any odd integer. If you rephrased this question to\n“What is n?” you would incorrectly conclude that the answer is (E).\nNote that the converse of this principle is not always true. Occasionally, it’s\nokay to rephrase a Value question as a Yes/No question—specifically, when\nit turns out that there are only two possible values.\nTry-It #3-3\nIf x is a positive integer, what is the remainder of \nSome quick analysis will show that x2 − 1 can be factored into (x + 1)(x − 1).\nIf x is odd, then both of these terms will be even and the product will be\ndivisible by 4, yielding a remainder of 0 when divided by 4. If x is even, then\nx2 will be divisible by 4, so the remainder of x2 − 1 will be 3.\nThere are only two possible values of the remainder: 0 and 3. So this Value\nquestion can be rephrased as the Yes/No question, “Is x odd?” or similarly,"} +{"id": "book 2_p153_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 153, "page_end": 153, "topic_guess": "general", "text": "“Is x even?” Although this Value question seemed at first to have several\ndifferent potential outcomes, only two are possible, so you are able to\nchange the question to a Yes/No format by suitable rephrasing."} +{"id": "book 2_p154_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 154, "page_end": 154, "topic_guess": "word_problems", "text": "Principle #3: Work from Facts to\nQuestion\nEspecially for simple Yes/No questions, people o en assume the answer to\nthe question is Yes before looking at the statements. They only test cases\nfor which the answer is Yes, rather than testing all of the possible cases\nallowed by the statements.\nThis line of thinking is backwards—and tempting, because of the order in\nwhich things are presented.\nInstead ask, “If I start with the applicable facts and consider all\npossibilities, do I get a definitive answer to the question?”\nAlways work from the given facts to the question—never the reverse! This\nis why you have to keep the facts separated from the question and why you\nshould always clearly mark the question on your paper.\nTry-It #3-4\nIf x ≠ 0, is x = 1 ?\n(1)\n(2)"} +{"id": "book 2_p155_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 155, "page_end": 155, "topic_guess": "algebra", "text": "If you work from the question to the facts, you would assume a Yes for the\nquestion, then plug this information into the statements. For instance, you\nwould plug x = 1 into each statement. You would see that the value fits the\nequations in both statements, and you would pick (D) incorrectly. That’s\nespecially easy to do in this case because this particular question is much\nsimpler to think about than the statements (which are nasty little\nequations).\nDon’t start from the question! No matter what, when you are judging\nsufficiency, always proceed from the facts to the question. It doesn’t\nmatter how easy or hard the question is at that point. A er you’ve\nrephrased, put the question on hold and work from the statements and\nany other given facts to the question:\n\nAD\nBCE\n\n(1)\n (2)\nQuestion:\nIs x = 1 ?\nMaybe Yes\nThe correct answer is (B).\nIn a Yes/No question, when evaluating the statements, always try to determine whether\nthe question can be answered the same way under any possibility that is consistent with"} +{"id": "book 2_p156_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 156, "page_end": 156, "topic_guess": "statistics", "text": "those facts. A Yes answer means Always Yes, for all allowed scenarios. Likewise, a No\nanswer means Always No."} +{"id": "book 2_p157_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 157, "page_end": 157, "topic_guess": "word_problems", "text": "Principle #4: Be a Contrarian\nTo avoid statement carryover and to gain insight into the nature of a\nproblem, deliberately try to violate one statement as you evaluate the\nother statement. This will make it much harder for you to make a faulty\nassumption that leads to an incorrect answer. Think outside the first\nstatement’s box.\nTry-It #3-5\nIf x ≠ 0, is xy > 0 ?\nThe question is actually asking whether x and y have the same sign.\n(1) INSUFFICIENT: This indicates nothing about the sign of y.\nIn evaluating statement (2), you might be tempted to assume that x must\nbe positive. A er all, you just read information in statement (1) that\nindicates that x is positive. Besides, it is natural to assume that a given\nvariable will have a positive value, because positive numbers are much\nmore intuitive than negative numbers.\n(1)\n(2)"} +{"id": "book 2_p158_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 158, "page_end": 158, "topic_guess": "sequences_patterns", "text": "Instead, follow Principle #4: actively try to violate statement (1), helping\nyou to expose the trick in this question.\n(2) INSUFFICIENT: Consider the possibility that x is negative. In this case, it\nis necessary to flip the sign of the inequality when you cross-multiply. That\nis, if x < 0, then \n means that 1 > xy, and the answer to the question\nis MAYBE.\n(1) & (2) SUFFICIENT: If x is positive, then statement (2) says that 1 < xy (do\nnot flip the sign when cross-multiplying). Thus, xy > 0.\nThe correct answer is (C).\nWhen evaluating individual statements, deliberately trying to violate the other statement\ncan help you see the full pattern or trick in the problem. You will be less likely to fall victim\nto statement carryover."} +{"id": "book 2_p159_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 159, "page_end": 159, "topic_guess": "number_theory", "text": "Principle #5: Assume Nothing\nThis principle is a corollary of the previous principle: avoid assuming\nconstraints that aren’t actually given in the problem—particularly\nassumptions that seem natural to make.\nTry-It #3-6\nIs z an even integer?\nThe wording of this question has a tendency to bias people toward\nintegers. A er all, the “opposite” of even is odd, and odd numbers are\nintegers, too. However, the question does not state that z must be an\ninteger in the first place, so do not assume that it is.\n(1) SUFFICIENT: The fact that \n is an even integer implies that z = 2 × (an\neven integer), so z must be an even integer. (In fact, according to statement\n(1), z must be divisible by 4.)\n(2) INSUFFICIENT: The fact that 3z is an even integer implies that z =\n, so z might not be an integer at all. For example, z could\n is an even integer.(1)\n3z is an even integer.(2)"} +{"id": "book 2_p160_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 160, "page_end": 160, "topic_guess": "number_theory", "text": "equal \n .\nOne way to avoid assuming is to invoke Principle #3: work from facts to\nquestion. Statement (2) indicates that 3z = even integer = −2, 0, 2, 4, 6, 8, 10,\netc. No even integers have been skipped over, nor have you allowed the\nquestion to suggest z values. That is how assumptions sneak in.\nNext, divide the numbers in your list by 3:\n, etc. Is z an even integer?\nMaybe. Note that you could have stopped computing a er the second\nvalue, since you already achieved a Yes and a No.\nThe correct answer is (A).\nIf you had assumed that z must be an integer, you might have evaluated\nstatement (2) with two cases:\nYou would have incorrectly concluded that statement (2) was sufficient and therefore\nincorrectly selected answer (D).\nAnother common assumption is that a variable must be positive. Do not\nassume that any unknown is positive unless it is stated as such in the\ninformation given (or if the unknown counts physical things or measures\nsome other positive-only quantity).\n3 × even = even, so z could be even.\n3 × odd = odd, so z is definitely not odd."} +{"id": "book 2_p161_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 161, "page_end": 161, "topic_guess": "general", "text": "Problem Set\nFor problems 1–3, apply Principle #1 (follow a consistent process) from this\nchapter to arrive at a solution to each problem. Note that the solutions\npresented for problems 1–3 are specific examples. Your process may be\ndifferent. Also note that the Data Sufficiency answer choices are not listed\nbut all DS problems throughout this guide use the standard DS answers in\nthe standard order.\n1. There are 19 batters on a baseball team. Every batter bats either\nright-handed only, le -handed only, or both right-handed and le -\nhanded. How many of the 19 batters bat le -handed?\nSeven of the batters bat right-handed but do not bat le -\nhanded.\n(1)\nFour of the batters bat both right-handed and le -handed.(2)\n2. If a is a positive integer and 81 divided by a results in a remainder of\n1, what is the value of a ?\nThe remainder when a is divided by 40 is 0.(1)\nThe remainder when 40 is divided by a is 40.(2)"} +{"id": "book 2_p162_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 162, "page_end": 162, "topic_guess": "number_theory", "text": "Solve problems 4 and 5. Apply Principle #2 (Never Rephrase Yes/No as\nValue); describe why these Yes/No questions cannot be rephrased as Value\nquestions.\n3. If a, b, c, d, and e are positive integers such that\n is \n an integer?\nd − e ≥ 4(1)\nd − e > 4(2)\n4. If a, b, and c are each integers greater than 1, is the product abc\ndivisible by 6 ?\nThe product ab is even.(1)\nThe product bc is divisible by 3.(2)\n5. If n is a positive integer, is n − 1 divisible by 3 ?\nn2 + n is not divisible by 6.(1)\n3n = k + 3, where k is a positive multiple of 3(2)\n6. Revisit problems #4 and #5 above, this time deliberately violating\nPrinciple #3 (Work from Facts to Question). Determine the incorrect\nanswer you might have selected if you had reversed the process\nand worked from the question to the facts."} +{"id": "book 2_p163_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 163, "page_end": 163, "topic_guess": "algebra", "text": "First, attempt to solve problem 7 by evaluating statement (1), and then\nevaluating statement (2) without violating the information in statement (1).\nThen, re-solve the problem by applying Principle #4 (be a contrarian). Do\nyou get the same answer? Verify that applying Principle #4 leads to the\ncorrect answer, whereas not following the principle could lead to an\nincorrect answer.\nFor problems 8 and 9, apply Principle #5 (assume nothing) by identifying\nthe explicit constraints given in the problem. What values are still\npermissible? Next, solve using these constraints. Verify that different\n(incorrect) answers are attainable if incorrect assumptions about the\nvariables in the problem are made, and identify examples of such incorrect\nassumptions.\n7. Is m ≠ 0, is m3 > m2 ?\nm > 0(1)\nm2 > m(2)\n8. If yz ≠ 0, is 0 < y < 1 ?\ny = z2                                     (1)\n(2)\n9. Is x > y ? \n|x – y| < |x|(1)"} +{"id": "book 2_p164_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 164, "page_end": 164, "topic_guess": "general", "text": "|x| > |y|(2)"} +{"id": "book 2_p165_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 165, "page_end": 165, "topic_guess": "number_theory", "text": "Solutions\n(Note that the solutions presented for problems 1–3 are specific examples.\nYour notes may be different.)\n1. (A):\nRephrasing:\n19 batters\ntotal →\nintegers\nonly!\nSome R\nonly\nSome L\nonly\nSome R & L\n0 “neither”\nHow many\nle -handed\nbatters?\nL Not L Total\nR\nNot R 0\nTotal x 19\nWhat is the integer x?\n(1)"} +{"id": "book 2_p166_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 166, "page_end": 166, "topic_guess": "number_theory", "text": "L Not L Total\nR 7\nNot R 0\nTotal x 7 19\nQuestion:\nInteger x = ?\nCan find x.\n(SUFFICIENT)\n2. (B):\nRephrasing: 81 divided by a → remainder of 1.\n                          a goes evenly into 80, and a ≠ 1.\n                         a is one of the factors of 80 other than 1.\n                         a = 2, 4, 5, 8, 10, 16, 20, 40, or 80.\nWhich of the numbers listed above is the value of a ?\n(1)\n → remainder of 0.\n(2)\n → remainder of 40."} +{"id": "book 2_p167_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 167, "page_end": 167, "topic_guess": "general", "text": "40 goes evenly into a.\na is a multiple of 40.\na must be larger than 40.\nQuestion:\nWhich of the listed numbers is a ?\na could be 40 or 80.\n(INSUFFICIENT)\na must be 80.\n(SUFFICIENT)\n3. (D):\nRephrasing:\nIs \n an integer?\nIs \n an integer?\nIs \nIs \n(1)\n (2)\nQuestion:\nIs d − e ≥ 4 ?\nYes\n(SUFFICIENT)\nYes\n(SUFFICIENT)"} +{"id": "book 2_p168_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 168, "page_end": 168, "topic_guess": "word_problems", "text": "4. (C): Each of the below is an accurate Yes/No rephrase:\nAlternatively, you could ask “Is there an even integer and a multiple of 3\namong a, b, and c ?”\n(1) INSUFFICIENT: ab is divisible by 2, but it’s unclear whether it is\ndivisible by 3 (or whether c is divisible by 3).\n(2) INSUFFICIENT: bc is divisible by 3, but it’s unclear whether it is\ndivisible by 2 (or whether a is divisible by 2).\n(1) AND (2) SUFFICIENT: Statement (1) indicates that a or b is even, and\nstatement (2) indicates that b or c is divisible by 3. Therefore abc is\ndivisible by both 2 and 3.\nThe correct answer is (C).\nIs \n ?\nIs abc = 6, 12, 18, 24, 30, 36, 42, 48, etc.?\nIs abc divisible by 2 and by 3 ?\n5. (A): An accurate Yes/No rephrase is the following:"} +{"id": "book 2_p169_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 169, "page_end": 169, "topic_guess": "sequences_patterns", "text": "This narrows down the values of interest to a certain type of number,\nwhich follows a pattern: 1, 4, 7, 10, etc.\n(1) SUFFICIENT: If n2 + n = n (n + 1) is not divisible by 6, you can rule out\ncertain values for n.\nn n + 1 n (n + 1) not divisible by 6\n1 2 ✓\n2 3 ✗\n3 4 ✗\n4 5 ✓\n5 6 ✗\n6 7 ✗\n7 8 ✓\nThe pattern from the rephrasing is apparent here: n can only be 1, 4, 7,\n10, etc., all integers that are one greater than a multiple of 3."} +{"id": "book 2_p170_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 170, "page_end": 170, "topic_guess": "number_theory", "text": "Alternatively, use theory. The integers n − 1, n , and n + 1 must be\nconsecutive. If n( n + 1) is not divisible by 3, then n − 1 must be divisible\nby 3, since in any set of three consecutive integers, one of the integers\nmust be divisible by 3.\n(2) INSUFFICIENT: If 3n = 3 × pos integer + 3, then n = pos integer + 1 .\nTherefore, n is an integer such that n ≥ 2. This does not resolve whether\nn is definitely one greater than a multiple of 3.\nThe correct answer is (A).\n6. Revisiting #4, working incorrectly from the Question to the Facts:\nManipulating the question to abc = 6 × integer (and losing track of the\nquestion mark), someone might be tempted to check whether it is\npossible for abc to be a multiple of 6, instead of whether abc is definitely\na multiple of 6:\nThe incorrect thinking would lead someone to wrong answer (D). Be\nsure to revisit how to do this problem correctly!\nRevisiting #5, working incorrectly from the Question to the Facts:\nSomeone working incorrectly might try multiples of 3 for n − 1 to see\nwhether they “work” with the statements:\n(1) abc = 6 × integer, so ab is even. ✓\n(2) abc = 6 × integer, so bc is divisible by 3. ✓"} +{"id": "book 2_p171_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 171, "page_end": 171, "topic_guess": "general", "text": "(1) SUFFICIENT:\nn − 1 n n + 1 n2 + n = n(n+ 1) not divis by 6\n3 4 5 20 ✓\n6 7 8 56 ✓\n9 10 11 110 ✓\n(2) SEEMS SUFFICIENT (incorrectly): If 3n = k + 3, then k = 3n − 3 = 3(n −\n1).\nn − 1 k = 3n − 3 = 3(n − 1) pos mult of 3\n3 9 ✓\n6 18 ✓\n9 27 ✓\nThis incorrect work would lead you to wrong answer (D). Be sure to\nrevisit how to do this problem correctly, so you are certain how to do so\nfor the future!\n7. (C):"} +{"id": "book 2_p172_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 172, "page_end": 172, "topic_guess": "fractions_decimals_percents", "text": "Non-Contrarian Approach Contrarian Approach\n(1) INSUFFICIENT:\nm > 0 or m = pos, so “Is pos3 >\npos2?”\n(1) INSUFFICIENT:\nm > 0 or m = pos, so “Is pos3 > pos2?”\n(2) SEEMS SUFFICIENT:\nm2 > m implies that m is not a\nfraction or 1, therefore m3 >\nm2.\nOr, if you assume that m > 0,\ncarrying over from (1), you\nmight do the following:\nm2 > m\nm > 1 (dividing by m)\nTherefore, m3 > m2.\n(2) INSUFFICIENT:\nm2 > m implies that m is not a fraction or 1, so if m > 1\nthen the answer is Yes.\nBUT, contradicting (1), what if m is negative? That is\npossible according to (2), since\nneg2 > neg.\nIs m3 > m2? → Is neg3 > neg2? No!\nYes, if m > 1.\nNo, if m is a proper fraction\nor 1.\n(That is, 0 < m ≤ 1.)\nYes, if m > 1.\nNo, if m is a proper fraction or 1.\n(That is, 0 < m ≤ 1.)"} +{"id": "book 2_p173_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 173, "page_end": 173, "topic_guess": "fractions_decimals_percents", "text": "Non-Contrarian Approach Contrarian Approach\n(1) AND (2) SUFFICIENT:\nCombined, the statements eliminate negative, zero,\npositive proper fractions, and 1 for the value of m. If m > 1,\nthen m3 > m2. Definite Yes answer.\nThe (incorrect) answer is (B). The correct answer is (C).\nNote that both solutions were hampered by inadequate rephrasing.\nIdeally, you would first rephrase as follows:\nTakeaway: With proper rephrasing, other errors are less likely. But even\nwith inadequate rephrasing, taking a contrarian approach can save you\nfrom a wrong answer.\nIs m3 > m2?\nIs m > 1? (It’s okay to divide by m2, which must be positive: a square is\nnever negative and m is also not 0, according to the question stem.)\n8. (C): The explicit constraint is yz  ≠ 0, which indicates that y ≠ 0 and z ≠ 0.\nBoth y and z could be any nonzero value, including positive integers,\nnegative integers, positive fractions, negative fractions, etc.\nMaking faulty assumptions:"} +{"id": "book 2_p174_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 174, "page_end": 174, "topic_guess": "fractions_decimals_percents", "text": "For (1), a faulty assumption could be made by those who plug in values\nfor z.\nFor example, if you plug in z = −2, −1, 1, 2, 3, 4, etc., you would get y = 1,\n4, 9, 16, etc. That would yield a definite No answer to the question, as all\nthe y values are at least as great as 1. The (unverbalized) assumption is\nthat z is an integer, but that’s not necessarily so.\nFor (2), most people would want to cross-multiply, so a potential false\nassumption is that y is positive (and this is reinforced by the fact that, in\nstatement (1), y is in fact positive):\nConclusion: SEEMS SUFFICIENT (incorrect)\nThese faulty assumptions would lead to the incorrect answers (A) or (D).\nCorrect solution:\n(1) INSUFFICIENT: y must be positive, but is it a fraction or an integer?\nIf z = 2, then y = 4, and the answer is No."} +{"id": "book 2_p175_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 175, "page_end": 175, "topic_guess": "general", "text": "(2) INSUFFICIENT:\nSince you’re assuming y < 0,\nthe rephrasing is y < −1.\nIs 0 < y < 1 ? YES NO\n(1) AND (2) SUFFICIENT: Since y is positive, statement (1) indicates that 0\n< y < 1.\nThe correct answer is (C).\nIf z = \n , then y = \n , and the answer is Yes.\n9. (E):"} +{"id": "book 2_p176_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 176, "page_end": 176, "topic_guess": "fractions_decimals_percents", "text": "(1) INSUFFICIENT: Test cases for x and y. Avoid limiting yourself to cases\nin which x is greater than y. Since “Is x > y?” is a question and not a\nstatement, x might be greater than or less than y. Actively seek cases\nthat give a No answer. \nx y |x – y| < |x| Is x > y ?\n3 4 |–1| < |3|    (valid) No\n4 3   |1| < |3|    (valid) Yes\nSince x could be greater than y or less than y, given the information in\nthe statement, this statement is insufficient. \n(2) INSUFFICIENT: The second case tested above also fits this\nstatement and yields an answer of Yes. Find another case in which |x| >\n|y|, but x is not greater than y. \nx  y |x| > |y| Is x > y ?\n4 3   |4| > |3|    (valid) Yes\n–4 3 |–4| > |3|    (valid) No\n(1) AND (2) INSUFFICIENT: A case that fits both statements and gives an\nanswer of Yes has already been found. Try to find another case for which\nboth statements are true, but which gives an answer of No. As you test\ndifferent values for x and y, you may find certain cases for which one or\nboth statements are not true. Cross these off on your paper, and do not\ntake them into consideration when choosing your answer."} +{"id": "book 2_p177_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 177, "page_end": 177, "topic_guess": "general", "text": "x y (1): |x – y| < |x| (2): |x| > |y| Is x > y ?x y (1): |x – y| < |x| (2): |x| > |y| Is x > y ?\n4 3 |1| < |3|    (valid) |4| > |3|    (valid) Yes\n–4 3 |–7| < |–4|: (not valid) n/a n/a\n–4 –3 |–1| < |–4| |–4| > |–3| No\nThe answer to the question can be either Yes (if x = 4 and y = 3) or No (if\nx = –4 and y = –3), so the statements together are insufficient. \nThe correct answer is (E)."} +{"id": "book 2_p178_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 178, "page_end": 178, "topic_guess": "word_problems", "text": "CHAPTER 4\nData Sufficiency: Strategies & Tactics"} +{"id": "book 2_p179_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 179, "page_end": 179, "topic_guess": "word_problems", "text": "In This Chapter...\nAdvanced Strategies\nAdvanced Guessing Tactics\nSummary\nCommon Wrong Answers"} +{"id": "book 2_p180_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 180, "page_end": 180, "topic_guess": "word_problems", "text": "Chapter 4\nData Sufficiency: Strategies & Tactics\nSometimes you will encounter a Data Sufficiency (DS) problem that you\ncan’t answer—either because its content is difficult or obscure, or because\nyou don’t have enough time to solve completely in two minutes.\nLike Chapter 2, this chapter describes a series of different methods you\nmight try in these circumstances. Again, this chapter distinguishes\nbetween solution strategies and guessing tactics.\nStrategies are broad: they apply to a wide variety of problems, they provide\na complete approach, and they can be used safely in most circumstances.\nIn contrast, tactics can help you eliminate a few answer choices, but o en\nleave a fair amount of uncertainty. Moreover, a particular tactic may only\nbe useful in special situations or for parts of a problem.\nThe first section of this chapter outlines six DS strategies:\nDS Strategy 1: Compute to Completion\nFor some problems, you won’t necessarily be able to tell whether the\nanswer can be calculated until you follow through on the calculations all"} +{"id": "book 2_p181_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 181, "page_end": 181, "topic_guess": "word_problems", "text": "the way.\nDS Strategy 2: Extract the Equation\nFor many Word Problems, you need to represent the problem with\nalgebraic equations to avoid embedded tricks that can be difficult to\nspot otherwise.\nDS Strategy 3: Know the Code\nThe most challenging part of some DS problems is figuring out what the\nquestion and statements are actually saying. If the problem is written in\n“GMAT code,” translate it into simple language before you do anything\nelse. \nDS Strategy 4: Use the Constraints\nMany DS problems provide explicit constraints on the variables. In other\nproblems, these constraints will be implicit (e.g., a variable that refers to\na number of people, houses, or airplanes must be both positive and an\ninteger). In either case, these constraints frequently determine the\ncorrect answer, so you must identify and use them.\nDS Strategy 5: Beware of Inequalities\nWhenever a DS problem involves inequality symbols, be especially\ncareful—the GMAT loves to trick people with inequalities.\nDS Strategy 6: Test Cases\nOne of the best ways to show that a statement is insufficient is to test\ndifferent scenarios, or cases, in which that statement is true."} +{"id": "book 2_p182_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 182, "page_end": 182, "topic_guess": "strategy", "text": "The remainder of the chapter is devoted to tactics that can knock out\nanswer choices or provide clues as to how to approach the problem more\neffectively. These tactics are listed later in the chapter. As with Problem\nSolving (PS) tactics, some of these D S tactics work wonders when used\ncorrectly. Others only slightly improve your guessing odds."} +{"id": "book 2_p183_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 183, "page_end": 183, "topic_guess": "algebra", "text": "Advanced Strategies\n1. COMPUTE TO COMPLETION\nA general principle of Data Sufficiency is that once you have determined\nwhether you can answer the question with a given set of information, you\ncan stop calculating. For some problems, however, you cannot determine\nwhether a single answer can be obtained until you’ve calculated the\nproblem all the way through. This is particularly common in the following\nsituations:\nTry-It #4-1\nWhat is the value of ab ?\n(1) INSUFFICIENT: Statement (1) does not answer the question. For\nexample, if a = 2 and b = 1, then ab = 2, and if a = 3 and b = 2, then ab = 6.\nMultiple equations are involved—particularly if they are non-linear.\nA complicated inequality expression is present.\nVariables hidden within a Geometry problem are related.\na = b + 1 (1)\na2 = b + 1 (2)"} +{"id": "book 2_p184_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 184, "page_end": 184, "topic_guess": "general", "text": "(2) INSUFFICIENT: Statement (2) does not answer the question. For\nexample, if a = 1 and b = 0, then ab = 0, and if a = 2 and b = 3, then ab = 6.\n(1) AND (2) SUFFICIENT: Evaluating both statements together is trickier,\nhowever:\nBased on this work, either a = 0 or a = 1. It would be tempting at this stage\nto decide that since a can have two different values, statements (1) and (2)\ntogether are insufficient. However, this is incorrect.\nLook at the values b can hold in these two scenarios:\na b ab\n1 a − 1 = 0 0\n0 a − 1 = −1 0\nWhile it is true that a can take on different values, ab is equal to zero in\neither case. When a = 1, b = 0, and ab = 0. When a = 0, b = −1, and ab = 0."} +{"id": "book 2_p185_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 185, "page_end": 185, "topic_guess": "word_problems", "text": "Therefore, (1) and (2) combined are SUFFICIENT to answer this specific\nquestion: What is the value of ab ?\nThe correct answer is (C).\nIn a multiple-scenario problem, be sure to compute for the specific question asked (in this\ncase, ab) in order to determine whether the end result for each scenario is actually\ndifferent.\n2. EXTRACT THE EQUATION\nFor Word Problems, setting up an algebraic representation of the question\nis essential. It is very easy to get intellectually lazy and miss an embedded\ntrick in the problem. These tricks are usually much easier to spot if you are\nlooking at the underlying algebra behind the problem.\nTry-It #4-2\nA store sells two types of bird feeders: Alphas and Bravos. Alphas\nfeed 1 bird at a time, whereas Bravos feed 2 birds at a time. The\ntotal number of birds that can be fed at one time by bird feeders\nsold last month is 50. What is the total revenue generated by\nbirdfeeders sold last month?\nLast month, the price of each Alpha was $15 and the price of\neach Bravo was $30.\n(1)\n40 Alphas were sold last month.(2)"} +{"id": "book 2_p186_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 186, "page_end": 186, "topic_guess": "algebra", "text": "From the words in the question stem, Extract the Equation. The problem\nindicates that Alphas can feed 1  bird at a time and Bravos 2 birds at a time.\nThe problem also indicates that last month 50 birds could be fed at a time,\nso you have this:\nTo calculate the revenue, it seems you will need the prices of the bird\nfeeders and the number of bird feeders, A and B.\n(1) SUFFICIENT: Again extract the equation from the wording of the\nquestion:\nIt turns out that you don’t need to know A and B individually, since the\nquestion stem equation indicated that A + 2B = 50. Therefore, total revenue\nequals $15(A + 2B) = $15(50) = $750.\nIt’s true that there are many possible values of A and B that satisfy the\ncondition that A + 2B = 50. However, mathematically every possible\ncombination that satisfies this equation would lead to the same revenue of\n$750. The number of each type of bird feeder sold is irrelevant. In this\nsense, Extract the Equation can be similar to the Compute to Completion\nstrategy because once the equation has been extracted, you may find that\nthe multiple possibilities for the variables might converge to a single\nanswer to the specific question that’s been asked.\nTotal number of birds fed = A + 2B = 50, where A and B represent the\nnumber of birdfeeders of each type that have been sold."} +{"id": "book 2_p187_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 187, "page_end": 187, "topic_guess": "number_theory", "text": "(2) INSUFFICIENT: If there were 40 Alphas sold, there were 5 Bravos sold.\nBut you still don’t know the prices, so you can’t compute revenue.\nThe correct answer is (A).\n!\nRelying on intuition, which indicates a need for the prices and number of the two\nbirdfeeder types, someone might ultimately choose (C) incorrectly. Be sure to translate all\nWord Problems into math so they can be properly evaluated. These issues are sensitive to\nthe exact numbers given.\n3. KNOW THE CODE\nMany DS questions and statements are intentionally written in a\ncomplicated way. Take a look at the two statements below:\nThese statements look very different, but from the perspective of someone\nwho’s solving a DS problem, they say exactly the same thing. In other\nwords, the first statement is just saying that a and b are both odd—but it’s\nsaying it in code to keep you from noticing. \nThere are two clues that the statement is really about odd and even\nnumbers. First, it specifies that a and b are integers. Second, if 2 is not a\nfactor of a certain integer, that’s just another way of saying that the integer\nis odd. You can write that out as follows:\na and b are integers such that 2 is not a factor of 7a2b3.1.\na and b are both odd integers.2."} +{"id": "book 2_p188_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 188, "page_end": 188, "topic_guess": "number_theory", "text": "The next thing to notice is that multiplying a number by 7 has no effect on\nwhether it’s odd or even. Multiplying an even number by 7 gives you an\neven number, and multiplying an odd number by 7 gives you an odd\nnumber. So the 7 can be removed entirely: \nThe exponents also don’t matter. Squaring or cubing an odd integer gives\nyou an odd integer, and squaring or cubing an even integer gives you an\neven integer. So saying that a2b3 is odd is no different from saying that\nab is odd. \nFinally, if the product of two integers is odd, both of those integers have to\nbe odd. Multiplying an even integer by another integer always results in an\neven value. Here’s what the statement really says: \nThe two statements were really the same all along. If you see something in\na DS problem that looks like it could be GMAT code, try to crack the code\nbefore you go further. You may find that a complicated statement is\ndisguising something much easier to work with. \n4. USE THE CONSTRAINTS\na and b are integers, and 7a2b3 is odd.\na and b are integers, and a2b3 is odd.\na and b are odd integers."} +{"id": "book 2_p189_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 189, "page_end": 189, "topic_guess": "algebra", "text": "O en, a DS question will provide relevant constraints on the variables in\nthe problem—for example, that the variables must be integers or must be\npositive, or must be between 0 and 1. When this information is given, it is\nusually essential to the problem. If you don’t use the constraints, you could\neasily end up choosing the wrong answer choice.\nTry-It #4-3\nIf 8x > 3x + 4x, what is the value of the integer x ?\nSimplify the question stem:\nOn top of that, there is also another constraint given: x must be an integer.\nThis limits the scope of the potential values of x even further. Make note of\nthis type of constraint in your work on paper. Write “x = int” or something\nsimilar. You could also incorporate this information by rephrasing the\nquestion to include the constraint: “If the integer x is positive, what is the\nvalue of x?”\n(1) SUFFICIENT: Solve this inequality for x:\n\n8x can only be greater than 7x when x is positive.\n6 − 4x > −2(1)\n3 − 2x ≤ 4 − x ≤ 5 − 2x(2)"} +{"id": "book 2_p190_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 190, "page_end": 190, "topic_guess": "word_problems", "text": "Since you know from the question stem that x > 0, you can conclude that 0\n< x < 2. The only integer between 0 and 2 is 1. Therefore, x = 1.\n(2) SUFFICIENT: Manipulate this compound inequality as follows:\nNote that it’s fine to manipulate all the parts of the compound inequality at\nthe same time as long as you perform each manipulation to all three parts\nof the inequality.\nSince the question stem indicates that x > 0, it must be the case that 0 < x ≤\n1. The only integer that fits this criteria is 1. Therefore, x = 1.\nIf you had overlooked the fact that x is an integer, you would have\ndetermined that there are many values between 0 and 1 or between 0 and\n2. You might have chosen answer (E), incorrectly.\nThe correct answer is (D)."} +{"id": "book 2_p191_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 191, "page_end": 191, "topic_guess": "word_problems", "text": "Make a note of any additional information given to you in the question stem (e.g., “x is\npositive” or “x is an integer”). You o en will have to use this information properly to get the\ncorrect answer.\nInteger constraints in particular are very potent: they o en limit the possible solutions for\na problem to a small set. Sometimes this set is so small that it contains only one item.\nConstraints will not always be explicitly given. The ones the GMAT doesn’t\nexplicitly give you can be called hidden constraints. Hidden constraints are\nmost prevalent in Word Problems and Geometry questions. Here are some\nexamples of hidden constraints that you should train yourself to take note\nof:\nThe number of countable items must be a non-negative integer. Note\nthat zero is only a possibility if it is possible for the items not to exist at\nall—if the problem clearly assumes that the items exist, then the\nnumber of items must be positive. Examples:\nNumber of people\nNumber of yachts\nNumber of books\nMany non-countable quantities must be non-negative numbers, though\nnot necessarily integers. Again, zero is only an option if the underlying\nobject might not exist. If the problem clearly assumes the existence and\ntypical definition of an object, then these quantities must be positive.\nExamples:"} +{"id": "book 2_p192_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 192, "page_end": 192, "topic_guess": "geometry", "text": "These sorts of constraints exist in Problem Solving, but they are even more\nimportant and dangerous on Data Sufficiency. If these constraints are\nimportant in a P S problem, then failing to take a constraint into account\nmay make you unable to solve the problem. That will alert you to the\nexistence of the constraints, since every P S problem must be solvable. In\ncontrast, you will get no such signal on a DS problem. A er all, solvability is\nthe very issue that D S tests!\n5. BEWARE OF INEQUALITIES\nWhenever a D S question involves inequality symbols, be especially careful.\nThe GMAT can employ a variety of different inequality-specific tricks. Here\nare six examples:\nThe side of a triangle must have a positive length. (All geometric\nquantities shown in a diagram, such as lengths, areas, volumes, and\nangles, must be positive. The only exception is negative coordinates\nin a coordinate plane problem.)\nThe weight of a shipment of products must be positive in any unit.\nThe height of a person must be positive in any unit.\nMany other non-countable quantities are theoretically allowed to take\non negative values. Examples:\nThe profit of a company (However, if a company made a profit, then\nthat profit is positive!)\nThe growth rate of a population\nThe change in the value of essentially any variable"} +{"id": "book 2_p193_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 193, "page_end": 193, "topic_guess": "word_problems", "text": "One inequality can imply another seemingly unrelated inequality,\ndepending on the situation. For example, if you need to know\nwhether x > 0, then knowing that x > 5 is sufficient. If x is greater\nthan 5, then it must be positive, thus x > 0. However, the opposite is\nnot the case. If you knew that x < 5, then you would not be able to\ndetermine whether x > 0. A er all, x could be positive but less than\n5, or x could be negative.\n1.\nInequalities can combine with integer constraints to produce a\nsingle value. For example, if 0 < x < 2 and x is an integer, then x must\nequal 1.\n2.\nSome Word Problems can create a hidden constraint involving\ninequalities. These inequalities may come into play in determining\nthe correct answer. For example, a problem might read: “The oldest\nstudent in the class . . . the next oldest student in the class . . . the\nyoungest student in the class  . . .” This can be translated to the\nfollowing inequality: youngest < middle < oldest.\n3.\nInequalities involving a variable in a denominator o en involve two\npossibilities: a positive and a negative one. For example, if you\nknow that \n , you might be tempted to multiply by y and\narrive at 1 < xy. However, this may not be correct. It depends on\nwhether y is a positive or negative number. If y > 0, then it is correct\nto infer that 1 < xy. However, if y < 0, then 1 > xy. Therefore, you’ll\nneed to test two cases (positive and negative) in this situation.\n4.\nAt the same time, hidden constraints may allow you to manipulate\ninequalities more easily. For instance, if a quantity must be positive,\nthen you can multiply both sides of an inequality by that quantity\nwithout having to set up two cases.\n5."} +{"id": "book 2_p194_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 194, "page_end": 194, "topic_guess": "number_theory", "text": "Take a look at some examples that illustrate these concepts.\nTry-It #4-4\nIf \n is a prime number, what is the value of x ?\nIf \n is a prime number, then possible values are 2, 3, 5, and so on.\nTherefore, x must be a perfect square of a prime; possible values include 4,\n9, 25, and so on.\n(1) SUFFICIENT: Manipulate the inequality to isolate x:\nMany questions involving inequalities are actually disguised\npositive/negative questions. For example, if you know that xy > 0,\nthe fact that xy is greater than 0 is not in and of itself very\ninteresting. What is interesting is that the product is positive,\nmeaning both x and y are positive or both x and y are negative.\nThus, x and y have the same sign. Here, the inequality symbol is\nused to disguise the fact that x and y have the same sign.\n6.\n−16 < −3x + 5<22(1)\nx2 is a two-digit number(2)"} +{"id": "book 2_p195_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 195, "page_end": 195, "topic_guess": "number_theory", "text": "Since x is the square of a prime, it can’t be negative or zero; it has to be\npositive. The smallest possible square of a prime is 4 and the next smallest\npossible square of a prime is 9. This inequality allows just one possible\nvalue: x = 4.\n(2) INSUFFICIENT: According to this statement, 10 ≤ x2 ≤ 99. In addition,\nfrom the question stem, \n is a prime number. Determine the possible\nvalue(s) for x:\nBecause there are two possible values for x, this statement is not sufficient.\nThe constraints were so specific that statement (1), which looks at a glance\nas though it will allow more than one possible answer, turns out to be\nsufficient.\nThe correct answer is (A).\nTry-It #4-5\nIf mn ≠ 0, is m > n ?\nIf x = 4, then \n is a prime number, 2, and x2 is a two-digit number, 16.\nIf x = 9, then \n is a prime number, 3, and x2 is a two-digit number, 81.\n(1)\nm2 > n2(2)"} +{"id": "book 2_p196_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 196, "page_end": 196, "topic_guess": "algebra", "text": "The constraint in the question stem indicates that neither m nor n equals\nzero.\n(1) INSUFFICIENT: You can solve algebraically/theoretically or you can test\ncases. If you solve algebraically, be careful: you have to account for\nmultiplying the inequality by a negative:\nAlternatively, test cases:\n(2) INSUFFICIENT: This statement indicates nothing about the signs of the\ntwo variables. Either one could be positive or negative.\n(1) AND (2) INSUFFICIENT. If you are solving algebraically, test the scenarios\nthat you devised for statement (1):\nIf m and n are both positive, then m > n.\nIf m and n are both negative, the sign flips twice, so m > n again.\nIf only one is negative, then the sign flips once and m < n. In this case, m\nmust be the negative number, since any positive is greater than any\nnegative.\nIf m = 3 and n = 2, then statement (1) is true and the answer to the\nquestion is Yes, m > n.\nIf m = −3 and n = 2, then statement (1) is true and the answer to the\nquestion is No, m is not greater than n.\nIf m and n are both positive, then m > n and m² > n². Both statements\nallow this scenario.\nIf m and n are both negative, then m > n but m² is not greater than n²."} +{"id": "book 2_p197_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 197, "page_end": 197, "topic_guess": "algebra", "text": "Alternatively, start by testing whether the cases you already tried for\nstatement (1) also apply to statement (2):\nBecause there are scenarios in which m > n and m < n, both statements\ntogether are still insufficient to answer the question. If you forgot to\naccount for the positive and negative cases, you may end up with (A) or (D)\nas your (incorrect) answer.\nThe correct answer is (E).\n6. TEST CASES\nIn a Data Sufficiency problem, a statement is insufficient if, given the\ninformation in the statement, the question still has more than one possible\nanswer. One way to show that this is true is by Testing Cases. Write down\nseveral scenarios, or cases, in which the statement is true. For instance,\nyou might think of different possible values for the variables in the\nstatement. For each of these cases, determine the answer to the question.\nDiscard this scenario, since it makes statement (2) false.\nIf m is negative and n is positive, then m < n. It could also be true that m²\n> n², as long as m’s magnitude is larger than n’s. If you’re not sure, test\ncases (see below).\nIf m = 3 and n = 2, then m > n and m² > n². Both statements allow this\nscenario.\nIf m = −3 and n = 2, then m < n and m² > n². Both statements allow this\nscenario."} +{"id": "book 2_p198_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 198, "page_end": 198, "topic_guess": "word_problems", "text": "If the question has different answers depending on which case you’re\ntesting, the statement is insufficient. If the question always has the same\nanswer in every case you try, the statement is likely sufficient. \nAs an advanced test-taker, you might be biased against case testing or\n“plugging in numbers”: it might somehow seem less advanced than\ntheoretical approaches. However, the theory required to answer a question\nmay be cumbersome to figure out in two minutes. If implemented\ncorrectly, Testing Cases can be fast, easy, and accurate, so it should be part\nof your toolbox. Particularly, testing cases is o en the simplest way to\nshow that a statement is not sufficient. \nPlugging in random numbers as they come to mind is the most common\napproach to Testing Cases, and while this strategy can be successful, it is\ninherently ad hoc and therefore not the most reliable process. It’s easy to\noverlook a salient scenario. The key is to have a systematic approach to\ntesting cases. This relies on three approaches: the standard number set for\ntesting, discrete number listing, and case testing with concepts.\nSometimes testing enough cases will also allow you to notice a pattern to\nhelp you deduce whether the statement is sufficient.\nStandard Number Set for Testing\nThe GMAT o en tests odd/even rules, positive/negative rules,\nfraction/integer rules, proper vs. improper fractions, etc. On any given\nproblem, you may have trouble identifying which rule is relevant, and in\nfact the GMAT may test more than one rule within a given question.\nTherefore, if you must pick and test numbers, consider a set of numbers\nthat covers every possible combination of properties:"} +{"id": "book 2_p199_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 199, "page_end": 199, "topic_guess": "number_theory", "text": "Odd Even Proper Fraction Improper Fraction\nNegative −1 −2 − \n − \nZero   0    \nPositive 1 2\nThis set includes integers, non-integers, positive and negative numbers,\nand numbers greater than and less than 1. Thus, a comprehensive set of\ntest numbers (to memorize and apply) would be as follows:\nRemember this list as “every integer and half-integer between −2 and 2.”\nNot all of these numbers will be relevant or possible on every problem. For\nexample, if the variable has to be positive, five of the nine values presented\nabove can be ignored. The question itself may suggest certain values to\ntest, but always keep in mind the potential need to test a value of each\nrelevant type—and if a problem really might entail testing nine different\ncases, consider whether that problem is really worth your time. Using the\nStandard Set ensures that you don’t get tripped up by forgetting to try a\nparticular type of value that will give you a different answer, thereby\nallowing you to prove definitively that the statement is not sufficient."} +{"id": "book 2_p200_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 200, "page_end": 200, "topic_guess": "number_theory", "text": "Try-It #4-6\nIs a < 0 ?\nThis problem presents inequalities, and the question asks whether a is\nnegative. Therefore, test different values of a to see which values fit the\nstatements.\nThe question stem doesn’t provide any constraints, so begin by testing\nsome easier integers and then move to fractions if needed:\n(1) INSUFFICIENT: Test the possible integers from the standard number set:\na a3 − a2 − 2a < 0 a < 0?\n−2 (−2)3 − (−2)2 − 2(−2) = −8 (valid) Yes\n1 (1)3 − (1)2 − 2(1) = −2 (valid) No\na3 < a2 + 2a(1)\na2 > a3(2)\n(1) \n(2) \n−1 (−1)3 − (−1)2 − 2(−1) = 0 (invalid)  \n0 (0)3 − (0)2 − 2(0) = 0 (invalid)"} +{"id": "book 2_p201_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 201, "page_end": 201, "topic_guess": "fractions_decimals_percents", "text": "Stop when you find two valid cases that return different answers to the\nquestion.\n(2) INSUFFICIENT: Try integers first:\na a2 − a3 > 0 a < 0?\n−2 (−2)2 − (−2)3 = 12 (valid) Yes\n−1 (−1)2 − (−1)3 = 2 (valid) Yes\nBe careful: You’re not done. Try a fraction. Because a positive and a\nnegative number figured into the mix for statement (1), try one positive\nand one negative fraction:\na a2 − a3 > 0 a < 0?\n−2 (−2)2 − (−2)3 = 12 (valid) Yes\n−1 (−1)2 − (−1)3 = 2 (valid) Yes\n0 (0)2 − (0)3 = 0 (invalid)  \n1 (1)2 − (1)3 = 0 (invalid)  \n2 (2)2 − (2)3 = -4 (invalid)  \n0 (0)2 − (0)3 = 0 (invalid)  \n1 (1)2 − (1)3 = 0 (invalid)  \n2 (2)2 − (2)3 = −4 (invalid)"} +{"id": "book 2_p202_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 202, "page_end": 202, "topic_guess": "algebra", "text": "a a2 − a3 > 0 a < 0?\n (valid) Yes\n(valid) No\nStatement (2) is insufficient. When \n , the constraint is fulfilled, but a\nis positive.\nCombining statements (1) and (2) shows that whenever a = −2, −\n , or \n ,\nboth conditions are fulfilled. The variable a could thus be positive or\nnegative. The correct answer is (E).\nNotice that you did not need to test every possible value for a. For example,\nwhen a = −2, both conditions are easily satisfied. That means that testing −\n was unlikely to be necessary, since that value is not much different from\n−2. Furthermore, when you do find a contradictory answer, you can stop\ntesting. For a Yes/No question, all you need to do is find one valid Yes and\none valid No to prove insufficiency.\nThis problem can also be solved algebraically, but it takes more conceptual\nwork. First solve the inequality as though it were an equation, then map\nthe solutions on a number line and test to see in which regions the\ninequality is true."} +{"id": "book 2_p203_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 203, "page_end": 203, "topic_guess": "word_problems", "text": "(1)\nINSUFFICIENT:\nThis number line demonstrates that either a < −1 or 0 < a < 2.\n(2) INSUFFICIENT:\n\nSince a can’t be zero, and a2 must be\npositive, you can divide by a2.\n\n(1) AND (2) INSUFFICIENT: Overlapping the possible ranges, either a < −1 or\n0 < a < 1. This is still not enough information to tell whether a is negative.\nThe correct answer is (E).\nIn some cases, the standard number testing list may not quite suffice; try\nthe partial DS example below.\nTry-It #4-7\nIf x is positive, is x ≤ 1 ?\nx2 ≤ 1.3(1)"} +{"id": "book 2_p204_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 204, "page_end": 204, "topic_guess": "algebra", "text": "If you used the standard number testing list of \n (ignoring\nthe non-positive values in the set), all of the values for x above 1 would fail\nto fit statement (1) and all of the values for x equal to or below 1 would fit\nstatement (1). Therefore, the standard number testing list would indicate\nthat statement (1) is sufficient. However, x could be 1.1, in which case x2 =\n1.21, which is less than 1.3. So statement (1) is actually insufficient.\nYou could figure out that x could be greater than 1 upon a quick inspection\nof this problem. You might think to try a number slightly larger than 1.\nHowever, if the problem were more complicated, it might not be so\nobvious. In cases like these, use the boundary principle: test values that\nare close to boundaries given in the problem. In this case, the boundary\nvalue is 1, so add 0.9 and 1.1 to your list of numbers to test. You might even\ntry 0.99 or 1.01. You should know that −1, 0, and 1 are natural boundaries,\nbecause numbers behave differently on either side of them (that’s why the\nstandard list contains numbers in the ranges defined by −1, 0, and 1).\nDiscrete Number Listing\nIn the previous problems, the variables a and x were not constrained, so\nyou had to test a series of different numbers to solve the problem. By\ncontrast, many questions suggest specific constraints: x must be odd, for\nexample, or x must be a positive integer. In these cases, just list a series of\nconsecutive numbers that fit these criteria and test them all. For example,\nif x must be positive and even, test x = 2, 4, 6, 8, 10.\nDiscrete number listing can be used whenever a problem specifies a\nsequence of discrete (separate) values that a variable or an expression can"} +{"id": "book 2_p205_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 205, "page_end": 205, "topic_guess": "number_theory", "text": "take on:\nBy contrast, some problems describe a smooth range of potential values\nfor a variable or expression (e.g., 0 < x < 1 or x must be negative). In these\ncases, don’t list consecutive values to test because the set of possible\nvalues is not discrete. If the variable can take on any real number in a\nrange, then rely on the standard number testing list, potentially with some\nmodifications, as described in the previous section.\nA key to the discrete number listing process is to test consecutive values\nthat fit the criteria—it would be too easy to leave out the one exception\nthat proves insufficiency. Never skip numbers that fit the constraint. This is\nespecially important if you are listing discrete numbers to equal an\nexpression, not just a variable. By the way, remember to work from the\nfacts to the question, not the other way around! Don’t assume that the\nquestion should be answered Yes and only test values that make it so.\nTry-It #4-8\nIs x a multiple of 12 ?\nIntegers (the classic case): . . . −3, −2, −1, 0, 1, 2, 3 …\nOdd/even integers: . . . −3, −1, 1, 3 … or . . . −4, −2, 0, 2, 4 . . .\nPositive perfect squares: 1, 4, 9, 16, 25 …\nPositive multiples of 5: 5, 10, 15, 20 …\nAny set that is “integer-like,” with well-defined, separated values\n is odd.(1)\nx is a multiple of 3.(2)"} +{"id": "book 2_p206_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 206, "page_end": 206, "topic_guess": "geometry", "text": "Since statement (1) indicates that \n is odd and the square root\nsign implies a positive answer, list 1, 3, 5, 7, 9, etc.\nNotice that you’re picking values for \n , not x. It would be far too\nmuch work to test different values for x to determine which make\n odd, and you could potentially miss some values that fit the\nstatement. Do not plug in numbers for x here! Instead, list consecutive odd\nvalues for \n , a quick and easy process. Then, solve for x in each\ncase.\nFor this problem, your work on paper may look something like this:\n(1) AND (2) SUFFICIENT: Combine these statements by selecting only the\nvalues for x that are in both lists. On your paper, circle the following values:\nx = 12 and x = 84. These are the values calculated in statement (1) that fit\nthe criteria in statement (2). This seems to be SUFFICIENT—the values for x\nthat fit both statements are multiples of 12. At this point, if you wanted to\ncheck another value, you could, or you could go with the trend, which is\nlikely going to be correct a er testing this many cases.\nThe correct answer is (C).\nINSUFFICIENT: \n = odds = 1, 3, 5, 7, 9, etc.\nx − 3 = 1, 9, 25, 49, 81, etc.\nx = 4, 12, 28, 52, 84, etc.\nIs x divisible by 12? Maybe. For example, 12 is, while 28 is not.\n(1)\nINSUFFICIENT: x = multiples of 3 = 3, 6, 9, 12, 15, etc.\nIs x divisible by 12? Maybe. For example, 12 is, while 15 is not.\n(2)"} +{"id": "book 2_p207_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 207, "page_end": 207, "topic_guess": "algebra", "text": "!\nIn retrospect, it may seem obvious that statement (1) indicates that x is a multiple of 4. But\nif you tried to evaluate statement (1) with algebra, you might reason that \n is odd,\nso (x − 3) is odd2, or an odd perfect square. Thus, x is an odd perfect square plus 3. One\nmight conclude that x is even, which is a true but incomplete description! Listing numbers\nis an easy way to see that these numbers are all multiples of 4.\nAs in the previous example, trying to solve statement (1) algebraically is\ntricky. Yes, it’s worth knowing how to do this algebra. The point is that a\ndiscrete number testing process is quick and simple, so it’s also worth\nknowing how to do.\nCase Testing with Concepts\nYou don’t need to use actual numbers to do case testing. In fact, it’s\nsometimes more efficient to case test using concepts, rather than\nnumbers.\n(1) INSUFFICIENT: \n where k is an integer.\n(2) INSUFFICIENT: x is a multiple of 3, so x must be divisible by 3.\n(1) AND (2) SUFFICIENT: x is divisible by 3 and by 4, so x is divisible by 12."} +{"id": "book 2_p208_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 208, "page_end": 208, "topic_guess": "number_theory", "text": "Try-It #4-9\nIf a, b, and c are integers, is abc > 0 ?\nYou could test cases in the traditional way: find specific values for a, b,\nand/or c that fit one statement or the other, then multiply those three\nvalues together and check whether their product was greater than zero.\nHowever, doing that much actual multiplication would be overkill. Instead\nof case testing by checking specific numbers, try case testing with\nconcepts. \n(1) INSUFFICIENT: Instead of choosing appropriate values for a, b, and c,\nthink about whether they could be positive or negative. For instance, one\nscenario that fits the statement is that a, b, and c are all positive. Another\nscenario is that a and b are positive, but c is negative. You can put this\ninformation in a chart, just as you would with ordinary case testing: \na b c ab > 0 abc > 0?\n+ + + (+)(+) = + (valid) Yes\n+ + — (+)(+) = + (valid) No\n(2) INSUFFICIENT:\nab > 0(1)\nbc > 0(2)"} +{"id": "book 2_p209_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 209, "page_end": 209, "topic_guess": "word_problems", "text": "a b c bc > 0 abc > 0?a b c bc > 0 abc > 0?\n+ + + (+)(+) = + (valid) Yes\n— + + (+)(+) = + (valid) No\nNeither statement is sufficient on its own, so try them together. (1) AND (2)\nINSUFFICIENT:\na b c (1): ab > 0 (2): bc > 0 abc > 0?\n+ + + (+)(+) = + (valid) (+)(+) = + (valid) Yes\n— — — (—)(—) = + (valid) (—)(—) = + (valid) No\nThe answer to this problem is (E). It was never necessary to work with\nspecific numbers to solve this problem.\nThis strategy doesn’t only work on positive/negative or odd/even\nproblems. You can use it for other types of Data Sufficiency problems as\nwell. Before reading the explanation, try to figure out how to solve the\nfollowing problem using concepts (not specific numbers) to test cases.\nTry-It #4-10\nAt a certain company, the bonus pool is divided among a group of\nemployees consisting entirely of engineers and managers so that"} +{"id": "book 2_p210_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 210, "page_end": 210, "topic_guess": "statistics", "text": "each engineer receives a bonus of e dollars and each manager\nreceives a bonus of m dollars. Is m at least 20% greater than e ?\nHere’s how you might case test while solving this Data Sufficiency\nproblem: \n(1) INSUFFICIENT: At least 20% of the employees in the group are\nmanagers, but the amount earned by each employee is unknown. There\ncould be a lot of managers who each earn a large bonus or there could be a\nlot of managers who each earn a small bonus. Think about scenarios on\nthe extreme end of the range, ones that could result in a different answer\nto the question.\n# of\nEngineers\n# of\nManagers\ne m At least 20% managers? m >\n1.2e? \nFew Many Small Large Lots of managers, and each manager\nearns a large bonus: (valid)\nYes\nFew Many Large Small Lots of managers, but each manager\nearns almost no bonus: (valid)\nNo\n(2) INSUFFICIENT: This statement tells you about the total amount\nawarded to the managers. However, this total amount could be received by\nonly a single manager or split across a huge number of managers. \nAt least 20% of the employees in the group are managers.(1)\nThe total amount awarded to managers is at least 20% greater\nthan the total amount awarded to engineers.\n(2)"} +{"id": "book 2_p211_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 211, "page_end": 211, "topic_guess": "general", "text": "# of\nEngineers\n# of\nManagers\ne m m total 20% + greater than e total? m >\n1.2e? \nMany One Very\nsmall\nVery\nlarge\nOnly one manager, but that manager\ngets most of the bonus pool: (valid)\nYes\nFew Many Equal Equal Everybody gets the same bonus, but\nthere are significantly more managers,\nso most of the pool goes to managers:\n(valid)\nNo\n(1) AND (2) INSUFFICIENT: Even considering both statements, there are\nmultiple possible situations. For instance, the group could be composed\nmostly of managers. This matches statement (1). Everyone in the group\ncould earn the same bonus, fitting statement (2), since the number of\nmanagers is much greater than the number of engineers. In this case, since\ne = m, the answer to the question is No. \nAnother possibility is that the group contains an equal number of\nmanagers and engineers. This fits statement (1). If the managers earn very\nlarge bonuses, and the engineers earn very small bonuses, then statement\n(2) is also true, since most of the bonus pool will go to managers. Finally,\nsince m is much greater than e, the answer to the question is Yes.\n# of\nEngineers\n# of\nManagers\ne m (1) 20% +\nmanagers?\n(2) m total 20% +\ngreater than e total?\nm >\n1.2e?\nFew Many Equal Equal Yes: (valid) Yes: (valid) No \nEqual Equal Very\nsmall\nVery\nlarge\nYes: (valid) Yes: (valid) Yes"} +{"id": "book 2_p212_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 212, "page_end": 212, "topic_guess": "number_theory", "text": "Without doing any mathematical calculation, or using any specific\nnumbers, you can prove that the answer to this problem is (E). \nTry one more problem, this time Testing Cases based on odd and even\nnumbers. \nTry-It #4-11\nIf a, b, and c are integers, is abc divisible by 4 ?\nEvaluate the possible odd/even combinations of a, b, and c without testing\nspecific numbers. \n(1) INSUFFICIENT: 2c must be even because c is an integer. This statement\nimplies that a + b is even, which occurs when a and b have the same\nodd/even parity. (There is no constraint on c.) \n(2) INSUFFICIENT: 2b must be even because b is an integer. Thus, this\nstatement implies that a + c is odd, which occurs when a and c have\nopposite odd/even parity. (There is no constraint on b.)\na + b + 2c is even.(1)\na + 2b + c is odd.(2)"} +{"id": "book 2_p213_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 213, "page_end": 213, "topic_guess": "general", "text": "(1) AND (2) INSUFFICIENT: From (1) you know that a and b have the same\nodd/even parity, while from (2) you know that a and c have opposite\nodd/even parity.\na b c (1) a + b + 2c = Even (2) a + 2b + c = Odd Is abc divisible \nby 4?\nEven Even Odd ✓ (valid) ✓  Yes\nOdd Odd Even ✓ (valid) ✓ (valid) Maybe\nEven with these constraints, you do not have a definitive answer to the\nquestion. The correct answer is (E).\nNotice that in evaluating statements (1) and (2) together, you would not\nneed to test completely new cases. You can reuse the work from statement\n(1) and statement (2) to determine the answer (e.g., by circling the cases in\none chart that also appear in the other chart). Be careful as you do this!\nKnow what case you’re considering."} +{"id": "book 2_p214_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 214, "page_end": 214, "topic_guess": "strategy", "text": "Advanced Guessing Tactics\nThe rest of this chapter is devoted to scrappy tactics that can raise your\nodds of success. There will be fewer opportunities to apply these tactics\nthan the strategies mentioned previously. However, when all else fails,\nthese tactics may be your only friend. The tactics are listed according to\ntheir reliability: the earlier tactics nearly always work, while the later\ntactics provide only a modest improvement over random guessing.\n1. Spot Identical Statements            Certainty: Very High\nIF the two statements tell you exactly the same\nthing (a er rephrasing) . . .\nTHEN the answer\nis either (D) or (E).\nBy adding 6 to both sides of statement (1) and multiplying statement (2) by\n3, you can see that both statements indicate that 3y = 2x + 6. Depending on\n( . . . )\n(1) 3y – 6 = 2x\n(2)"} +{"id": "book 2_p215_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 215, "page_end": 215, "topic_guess": "algebra", "text": "the question, either each statement will be sufficient or each will not—\nbecause they are identical, there cannot be any benefit from looking at the\nstatements together. The answer must be (D) or (E).\n2. Spot Clear Sufficiency            Certainty: Very High\nIF the two statements are clearly sufficient\ntogether . . .\nTHEN eliminate\n(E).\nTry-It #4-12\nIf \n and mn ≠ 0, what is the value of Z ?\nIt is obvious that you could plug m = 5 into statement (1) to get a value for\nn, then plug values for m and n into the expression for Z. So you can knock\nout (E) for sure.\nNote that this tactic does not imply that you should assume, or even lean\ntoward, choosing answer choice (C). Many of these types of problems are\ntrying to trap you into choosing (C) because it’s so obvious that the two\ncombined statements are sufficient. Quite o en, some algebraic work will\n(1)\nm = 5(2)"} +{"id": "book 2_p216_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 216, "page_end": 216, "topic_guess": "general", "text": "reveal that one or both of the statements will be sufficient on their own.\nIndeed, in the example problem given above, statement (1) alone is\nsufficient to answer the question, so the correct answer is (A).\n3. Spot One Statement Inside the Other            Certainty: Very High\nIF one statement is “contained within”\n(i.e., is a subset of) the other . . .\nTHEN eliminate (C) and\n“broader statement\nonly.”\n(Some question involving x)\nThis trick only shows up occasionally, but when it does, it’s useful. Notice\nthat narrow statement (1) is completely contained within broad statement\n(2). In other words, any value that satisfies statement (1) also satisfies\nstatement (2). Therefore, if statement (2) is sufficient, statement (1) must\nbe sufficient also. However, the reverse is not true. If (2) is insufficient, (1)\ncould still possibly be sufficient on its own.\nEither way, it is impossible for both of the statements to be required\ntogether to answer the question. It is also impossible for statement (2) to\nbe sufficient without statement (1) being sufficient also. So you can\nx > 50!(1)\nx > 10!(2)"} +{"id": "book 2_p217_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 217, "page_end": 217, "topic_guess": "number_theory", "text": "definitively eliminate (B), as it corresponds to the broader statement, as\nwell as (C) (the together option).\nThis situation can occur with inequalities. Whenever one statement defines\na range that is completely encompassed by the other statement’s range,\nyou can eliminate (C), as well as the broader statement alone.\nBe careful with this tactic, though. It can be easy to think—incorrectly—\nthat one statement is a subset of the other.\n(Some question involving x)\nAt first, it might seem that statement (2) is a subset of statement (1), but 2\nis a prime number that is not odd. Almost every prime number is odd, but\nnot all. Therefore, statement (2) is not a subset of statement (1). Even if just\none value escapes, you cannot use this tactic.\n4. Spot One Statement Adding Nothing            Certainty: High\nIF one statement adds no information to the\nother . . .\nTHEN eliminate\n(C).\n(Some question involving x, y, and z)\nx is an odd number.(1)\nx is a prime number.(2)"} +{"id": "book 2_p218_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 218, "page_end": 218, "topic_guess": "algebra", "text": "This tactic may seem identical to the previous one, but it is not. Notice in\nthis example that statement (1) does not determine whether y is positive or\nnegative, and statement (2) does not even include z. Therefore, neither\nstatement is a subset of the other.\nThat said, the fact that y is negative does not change anything in statement\n(1), because regardless of the value of y, z will remain the same if you swap\ny and −y. If y = 4, then you’d get 4x + (−4)x = z. If y = −4, then you’d get (−4)x +\n4x = z. Those equations are the same! The sign of y doesn’t matter because\ny and −y are symmetric. So knowing the sign of y adds no information to\nstatement (1).\nNote that in this example, (A), (B), (D), and (E) are all still possible answers,\ndepending upon the question. Only (C) can be eliminated.\n5. Spot a (C) Trap            Certainty: Moderate\nIF it is very obvious that the combined statements\nwould be sufficient, but you can’t eliminate the\npossibility that one statement alone is sufficient . . .\nTHEN\neliminate\n(C) and\n(E).\nyx + (−y)x = z(1)\ny < 0(2)"} +{"id": "book 2_p219_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 219, "page_end": 219, "topic_guess": "general", "text": "A (C) Trap is a DS problem that tries to trick you into incorrectly choosing\nanswer (C). It does this by giving you two statements that are obviously\nsufficient when you put them together. The test writers hope that you’ll\njump to putting the two statements together, without testing each\nstatement individually first. The trap is that one or both statements are\nactually sufficient alone, so the correct answer is really (A), (B), or (D). \n(C) Traps frequently appear in Combo problems, which require you to solve\nfor a complicated value, rather than a single variable. The following\nproblem is one example. \nTry-It #4-13\nIf \n and ab ≠ 0, what is the value of K ?\nYou can prove that the two statements combined are sufficient without\nactually doing any math. If you wanted to, you could plug the value of a\nfrom statement (1) into statement (2) to solve for the value of b. Then, you\ncould plug those values into the question to solve for K.\nImmediately be skeptical. Answer (C) should seem too easy. A er all, you\ndidn’t do anything to the question stem. Your next thought should be that\none of the statements alone may be sufficient. Make an informed guess\namong (A), (B), and (D). In some problems, the individual statements may\na = 3(1)\nb − 3(5 − a) = 0(2)"} +{"id": "book 2_p220_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 220, "page_end": 220, "topic_guess": "algebra", "text": "appear equally informative; you would favor choice (D) in such a\nsituation. In this problem, statement (2) gives more information than\nstatement (1) does (two variables vs. just one), so you should favor choice\n(B) as a preliminary guess.\nStart by rephrasing the question:\n“What is the value of K?” rephrases to “What is the value of b + 3a?”\nStatement (2) provides the answer:\nThe correct answer is (B). In these cases, the more complicated statement\nmay be enough."} +{"id": "book 2_p221_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 221, "page_end": 221, "topic_guess": "algebra", "text": "This is not a tactic to use at lower levels of the test. If you’re doing easier\nQuant problems, something that looks like a (C) Trap might just be a\nstraightforward problem with a correct answer of (C). This is one of many\nreasons why the material in this book is only appropriate if you have\nachieved a certain level of proficiency on the math side of the GMAT.\n6. Use Basic Algebraic Reasoning            Certainty: Moderate\nIF possible . . .\n rely on basic algebraic reasoning.\nQuite o en, basic intuition about algebra can lead you to the correct\nanswer. For example, you might have the intuition on a problem that “I\nhave two unknowns and only one equation, so I can’t solve” or “This\nstatement doesn’t even mention the variable(s) that I care about.” O en,\nyou will discover that your intuition is correct.\nTry-It #4-14\nA sales manager at an industrial company has an opportunity to\nswitch to a new, higher-paying job in another state. If his current\nannual salary is $50,000 and his current state tax rate is 5%, how\nmuch income after state tax would he make at the new job?\nHis new salary will be 10% higher than his old salary.(1)\nHis annual state taxes will total $2,200 in the new state at his\nnew job.\n(2)"} +{"id": "book 2_p222_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 222, "page_end": 222, "topic_guess": "algebra", "text": "You may reason that in order to answer this question, you need to know\nhow much his new salary will be and how much his taxes will increase by.\nThis reasoning will lead you to conclude that the correct answer is (C). And\nyou’ll be right.\nOf course, you need to be very careful about using this tactic. Many\nproblems on the GMAT are designed to hoodwink your algebraic reasoning\n—usually to make you think that you need to know every value precisely to\nanswer the question.\nFor example, suppose the question in the previous problem were changed\nto “Will the manager make more money at the new job, a er state taxes?”\nThe correct answer in this case is (B). The problem explicitly states that the\nnew job is higher-paying, so you only need to check whether the change in\nstate tax might lead to a lower a er-tax compensation. Statement (2)\nindicates that his new state taxes ($2,200) will be lower than his current\nstate taxes (5% of $50,000 = $2,500), so his a er-state-tax compensation\nwill definitely be higher in the new job. You do not need to find out how\nmuch higher his pre-tax salary would be to answer this question.\nBecause the GMAT frequently uses traps involving basic algebraic\nreasoning, you should only resort to using it if you are truly stuck.\n7. Spot Cross-Multiplied Inequalities            Certainty: Low\nIF a Y/N question involves inequalities with\nvariables to cross-multiply…\nTHEN guess (C) or\n(E) if you must\nguess."} +{"id": "book 2_p223_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 223, "page_end": 223, "topic_guess": "algebra", "text": "When you are presented with an inequality problem in Data Sufficiency in\nwhich one or more variables appears in a denominator, you may need to\nknow the sign of the variable or variables to answer the question. This\nmight be the hidden trick.\nTry-It #4-15\nIs xy < 1 ?\nStatement (1) might imply that \n , but only if x is positive. If it is\nnegative, you would need to flip the sign: \n . You need to know the\nsign of x—information that is provided in statement (2). The correct answer\nis (C).\nThese types of problems may have an (A) or (B) trap, in that you might fail\nto realize that you have to set up negative and positive cases when you\nmultiply or divide an inequality by a variable.\nWatch out! On some GMAT problems, you can assume the sign of the\nvariables because the variables represent countable quantities of physical\n(1)\nx > 0(2)"} +{"id": "book 2_p224_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 224, "page_end": 224, "topic_guess": "algebra", "text": "things. In those cases, you can assume that the variables are positive, and\nyou can cross-multiply inequalities involving those variables at will.\n8. Judge by Appearance            Certainty: Low\nIF a statement is tricky and you don’t\nknow what to do with it . . .\nTHEN take a guess by\n“judging by\nappearance.”\nSometimes a statement will leave you completely bewildered. In that case,\nthe best (and only!) tactic is to guess whether it will be sufficient judging by\nhow it looks:\nThe general rule is this: if the information in a statement has a structure\nand complexity similar to the question, and has the right ingredients\n(variables, coefficients, etc.), it’s more likely to be sufficient than otherwise.\nThis won’t crack every case by any means, but you’d be surprised at how\nmuch mileage you can get from this tactic.\nTry-It #4-16\nDoes it look like it might be sufficient, even if you can’t see how?\nDoes it use the variables you are looking for?\nCould it likely be manipulated into a form similar to that of the\nquestion?"} +{"id": "book 2_p225_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 225, "page_end": 225, "topic_guess": "algebra", "text": "Does 4a = 4−a + b ? \nDepending on your level of comfort with exponent manipulations, you may\nbe able to prove that statement (1) alone is sufficient:\nHowever, what if you find yourself at a temporary loss on the test? Your\nexponent engine may shut down for a problem or two. Then what do you\ndo?\nOne line of reasoning might be as follows: “Statement (2) is clearly not\nsufficient, since it tells me nothing about b. Now, let’s look at statement (1).\nIt’s a very complicated expression, but it seems to have the right\ningredients. It contains a and b and uses a as an exponent. Also, the\nexponential terms in statement (1) are powers of 2 (2, 4, 16), just like the\n(1)\na = 2(2)"} +{"id": "book 2_p226_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 226, "page_end": 226, "topic_guess": "algebra", "text": "exponential terms in a. I’ll bet if I manipulate this equation right, it will\nanswer the question. I’ll guess (A).” And you’d be correct. The GMAT never\nneeds to know that you guessed."} +{"id": "book 2_p227_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 227, "page_end": 227, "topic_guess": "algebra", "text": "Summary\nThat was a lot of information! To summarize, the following advanced\nstrategies and guessing tactics can be used to solve Data Sufficiency\nproblems effectively or to increase your chances of success:\nAdvanced Data Sufficiency Strategies\nCompute to Completion: If you can’t tell for certain whether the answer\ncan be calculated in theory, keep going on the calculations all the way.\n1.\nExtract the Equation: Represent Word Problems with algebraic\nequations to avoid embedded tricks that can be difficult to find\notherwise.\n2.\nKnow the Code: Translate complicated questions and statements into\nsimpler language.\n3.\nUse the Constraints: Bring explicit and implicit constraints to the\nsurface. These constraints will o en be necessary to determine the\ncorrect answer.\n4.\nBeware of Inequalities: Whenever a problem involves inequality\nsymbols, be careful—there are many ways in which inequalities can be\nused to trick you.\n5.\nTest Cases: There may be many possible scenarios in which a particular\nstatement is true. Methodically write down these cases and check to see\nwhether they result in different answers to the question. \n6."} +{"id": "book 2_p228_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 228, "page_end": 228, "topic_guess": "algebra", "text": "Data Sufficiency Guessing Tactics\nSpot Identical Statements (High): If the two statements say the same\nthing (a er rephrasing), then the answer must be (D) or (E).\n1.\nSpot Clear Sufficiency (High): If the two statements together are clearly\nsufficient, then eliminate (E).\n2.\nSpot One Statement Inside the Other (High): If a narrow statement is\ncompletely contained within a broader statement, then eliminate (C)\nand “broader statement only” (either (A) or (B)).\n3.\nSpot One Statement Adding Nothing (High): If one statement adds no\nvalue to the information given in the other statement, then eliminate\n(C).\n4.\nSpot a (C) Trap (Moderate): If the two statements together are very\nclearly sufficient and at least one of the statements is complex enough\nthat it could be sufficient, then (C) could be a trap answer, so eliminate\n(C).\n5.\nUse Basic Algebraic Reasoning (Moderate): Apply basic knowledge of\nalgebra, such as considering the number of unknowns relative to the\nnumber of known equations, to guide your thought process.\n6.\nSpot Cross-Multiplied Inequalities (Low): If a Yes/No question involves\ninequalities with variables in a denominator, then guess (C) or (E) if you\nmust guess.\n7.\nJudge by Appearance (Low): If you’re completely unsure what to do,\nthen make your best guess as to whether it appears to be sufficient.\n8."} +{"id": "book 2_p229_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 229, "page_end": 229, "topic_guess": "fractions_decimals_percents", "text": "Common Wrong Answers\nThere are two ways to get a Data Sufficiency problem wrong:\nLet’s call the first type of error a Type 1 error: you thought that a statement\nwas sufficient, but it wasn’t. In other words, you thought that it gave you\nan exact answer to the question, but it was actually possible to get at least\ntwo different answers to the question. Here’s why that might happen: \nThe second type of DS error is a Type 2 error. If you made this type of error,\nyou thought that a statement wasn’t sufficient, but it actually was. It\nYou thought that a statement or both statements together\nwere sufficient, but they were actually insufficient.\n1.\nYou thought that a statement or both statements together\nwere insufficient, but they were actually sufficient. \n2.\nIf you tested cases, you might have missed a critical case. For instance,\nmaybe the question has a different answer when you plug in a fraction\nor a negative value. \nYou might have used information from the other statement without\nmeaning to. This can make it seem as though you have enough\ninformation to answer the question, when you actually don’t. Keep the\ntwo statements completely separate on your paper and test them\nindividually before you put them together."} +{"id": "book 2_p230_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 230, "page_end": 230, "topic_guess": "word_problems", "text": "looked like you didn’t have enough information to answer the question,\nbut you actually had more information than you thought: \nThe common thread in many of these errors is that they stem from\nassumptions. The easiest way to miss a DS problem is to assume that a\nstatement is or isn’t sufficient, rather than proving it. You won’t always\nhave time on test day to meticulously prove whether a statement is\nsufficient. But neither should you decide whether a statement is sufficient\njust by glancing at it. At the very least, briefly think about how the\nstatement bears on the specific question you’re being asked. And if you\nhave time, prove your suspicions about each statement before moving on. \nYou may have assumed that you needed a specific value to answer a\nYes/No question. The question might have asked something like “is\nx even?” If you treat this question as though you need to know the exact\nvalue of x, you’ll make Type 2 errors.\nYou may have fallen for a (C) Trap. If you assume that you need both\nstatements in order to answer the question, it’s easy to miss the fact that\none statement is actually sufficient on its own as well. \nNot Testing Cases o en leads to this type of error. Many statements\nlook insufficient, because they don’t appear to include a lot of\ninformation. However, testing cases sometimes reveals that the\nquestion only has one answer."} +{"id": "book 2_p231_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 231, "page_end": 231, "topic_guess": "number_theory", "text": "Problem Set\nFor problems 1−3, list five values that satisfy each of the following\nconstraints.\n1. n is a prime number.\n2. x2 > 0\n3. \n , where M and N are positive integers.\nFor problems 4–6, solve the problem by Testing Cases. Note which\nmethod is appropriate for each problem: testing the standard\nnumber set, listing discrete numbers, or case testing with concepts\nrather than numbers.\n4. If a ≠ 0, is a + a−1 > 2 ?\na > 0(1)\na < 1(2)"} +{"id": "book 2_p232_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 232, "page_end": 232, "topic_guess": "number_theory", "text": "5. At a certain florist shop, roses can be purchased either individually\nor as a bouquet of 12 at a discount of p percent. What is the\ngreatest number of roses that can be purchased with $45 ?\nThe greatest number of roses that can be purchased with $30 is\n24.\n(1)\np = 20(2)\n6. Is y3 ≤ |y| ?\nFor problems 7–9, Test cases to show that one or both statements\nare insufficient.\ny < 1(1)\ny < 0(2)\n7. If a, b, and c are positive integers, is abc an even integer?\na + b is even.(1)\nb + c is odd.(2)\n8. Twenty-eight students were each assigned to one of five classes:\nAnthropology, Biology, Geology, Musicology, and Sociology. Was at\nleast one of the classes assigned fewer than 5 students?"} +{"id": "book 2_p233_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 233, "page_end": 233, "topic_guess": "strategy", "text": "At least 6 students were assigned to each of Biology, Geology,\nand Sociology.\n(1)\nGeology and Sociology were not assigned the same number of\nstudents.\n(2)\n9. Set A consists of the four integers x, x2, x3, and x4, and set B consists\nof the four integers x, 2x, 3x, and 4x. What is the probability that a\nrandomly selected integer from 1 to 100, inclusive, is a member of\nneither set A nor set B ?\nFor problems 10–15, solve the problem. Note the strategies used to\nsolve the problem, and also note any guessing tactics that could be\nemployed to help eliminate answer choices. Even if you weren’t\ncertain as to the correct answer, which answer choices could you\neliminate and why? Also, which guessing tactics would not work,\nand why?\nThe following advanced strategies and guessing tactics were\ndiscussed in this chapter:\nAdvanced Strategies Guessing Tactics\nx > 5(1)\nx < 10(2)"} +{"id": "book 2_p234_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 234, "page_end": 234, "topic_guess": "algebra", "text": "Advanced Strategies Guessing Tactics\nCompute to Completion(1)\nExtract the Equation(2)\nKnow the Code(3)\nUse the Constraints(4)\nBeware of Inequalities(5)\nTest Cases(6)\nSpot Identical Statements(1)\nSpot Clear Sufficiency(2)\nSpot One Statement Inside the Other(3)\nSpot One Statement Adding Nothing(4)\nSpot a (C) Trap(5)\nUse Basic Algebraic Reasoning(6)\nSpot Cross-Multiplied Inequalities(7)\nJudge by Appearance(8)\n10. The average (arithmetic mean) of the original six prices for six coats\nat a clothing store was $85. A er two of the six coats were each\ndiscounted by 20%, the average price of the six coats was $76. Was\nthe coat with the lowest original price one of the two coats that\nwere discounted?\nOne of the discounted coats was the one with the highest\noriginal price.  \n(1)\nBefore the discount, none of the coats had a price greater than\n$180.\n(2)\n11. Amanda and Todd purchase candy, popcorn, and pretzels at the\nstadium. If a package of candy costs half as much as a bag of\npopcorn, how much more money did Amanda and Todd spend on\nthe candy than on the popcorn and pretzels combined?"} +{"id": "book 2_p235_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 235, "page_end": 235, "topic_guess": "number_theory", "text": "The cost of a bag of popcorn is equal to the cost of a pretzel.(1)\nAmanda and Todd purchased 24 packages of candy, 6 pretzels,\nand 6 bags of popcorn.\n(2)\n12. What is the value of |x + 4| ?\nx2 + 8x + 12 = 0(1)\nx2 + 6x = 0(2)\n13. If abcd ≠ 0, is abcd < 0 ?\n(1)\n(2)\n14. If m and n are positive integers, is n a multiple of 24 ?\n(1)\nn is a multiple of (m + 4).(2)\n15. What is the value of x(1 – y)(1 + y) ?\nx2 = x(1)"} +{"id": "book 2_p236_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 236, "page_end": 236, "topic_guess": "general", "text": "y2 = x(2)"} +{"id": "book 2_p237_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 237, "page_end": 237, "topic_guess": "number_theory", "text": "Solutions\n1. n = 2, 3, 5, 7, 11, 13, 17, 19, 23, etc.\n2. x = 0.5, 2, 3, 6.7, 0, −1, etc. (Note that the list should include integers and\nnon-integers, positive and negative values—although the items you\nchoose do not need to match these exact values. Anything other than 0\nwill work!)\n3. M = 10, 17, 24, 31, 38, etc., and N = 1, 2, 3, 4, 5, etc.:\nM must have a remainder of 3 when divided by 7:\nNote that successive values of M differ by 7 and the corresponding\nvalues of N are consecutive integers.\nOne way to generate this list is to pick consecutive numbers for N and\nthen calculate values for M:\nM = 7 + 3, 14 + 3, 21 + 3, 28 + 3, 35 + 3, etc.\nM = 10, 17, 24, 31, 38, etc.\nN = 1, 2, 3, 4, 5, etc."} +{"id": "book 2_p238_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 238, "page_end": 238, "topic_guess": "general", "text": "M = 7N + 3\n        N = 1, 2, 3, 4, 5, etc.\n        M = 7(1) + 3 = 10, 7(2) + 3 = 17, 7(3) + 3 = 24, 7(4) + 3 = 31, 7(5) + 3 = 38,\netc.\n4. (C): Because there are no constraints on a, use the standard number set.\nThe inequality can be rephrased as follows:\n(1) INSUFFICIENT:\na\n > 2?\nYes\n1 1 + 1 = 2 No\n(2) INSUFFICIENT:\na\n > 2?\n−1 −1 + (−1) = −2 No\nNo"} +{"id": "book 2_p239_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 239, "page_end": 239, "topic_guess": "statistics", "text": "a\n > 2?\nYes\n(1) AND (2) SUFFICIENT: Combining the two constraints, 0 < a < 1. Within\nthe standard number set, only \n is within this range. In addition, test\nnumbers close to the boundaries of the range.\na\n > 2?\nYes\nYes\nYes\nIt seems that at the extreme edges, the values are greater than 2, so\nstatements (1) and (2) appear to be SUFFICIENT.\nThe correct answer is (C).\n5. (E): The question asks you to maximize the number of roses that can be\npurchased for $45. To do so, first purchase as many 12-rose bouquets as"} +{"id": "book 2_p240_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 240, "page_end": 240, "topic_guess": "general", "text": "possible, then purchase as many individual roses as possible with the\nremaining money.\n(1) INSUFFICIENT: The greatest number of roses that can be purchased\nwith $30 is 24. In other words, two bouquets can be purchased for $30,\nand any le over money is not enough to purchase one individual rose.\nTest cases to determine how many roses could be purchased with $45.\nCase 1: If a bouquet of 12 roses costs exactly $15, then exactly 24 roses\ncan be purchased with $30, and exactly 36 roses can be purchased with\n$45.\nCase 2: Try to find a case in which only 24 roses can be purchased with\n$30, but more than 36 roses can be purchased with $45. This will\nhappen if two bouquets cost less than $30 and the le over money isn’t\nenough to purchase any more individual roses. But the le over money\nfrom $45 when three bouquets are purchased is just enough to purchase\nan additional rose.\nFor instance, suppose that a bouquet costs $14 and an individual rose\ncosts $2.50 before the discount. (Since this is a Data Sufficiency\nproblem, it isn’t necessary to find the exact value of p for which this will\noccur, as long as you’re confident that such a value exists.) In this case,\nonly 24 roses could be purchased with $30, since the $2 change\nwouldn’t be enough to purchase any individual roses. However, 37 roses\ncould be purchased with $45: three bouquets for $42 and one additional\n$2.50 rose with the remaining $3.\nSince either 36 or 37 roses could be purchased for $45, this statement is\ninsufficient."} +{"id": "book 2_p241_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 241, "page_end": 241, "topic_guess": "general", "text": "(2) INSUFFICIENT: The price of an individual rose is unknown. Therefore,\nany number of roses might be purchased for $45.\n(1) AND (2) INSUFFICIENT: First, use the case in which exactly 36 roses\ncan be purchased for $45. Then, look for a case in which more than 36\nroses can be purchased for $45.\nCase 1: If a bouquet of roses costs exactly $15, then exactly 36 roses can\nbe purchased for $45. (For reference, this implies each rose in a bouquet\ncost \n = $1.25, so the individual [not discounted] rose price is\n = $1.5625.)\nCase 2: Determine whether more than 36 roses could be purchased for\n$45. If roses cost only slightly less than they did in Case 1, it might still\nonly be possible to purchase 24 roses for $30, but it might also be\npossible to purchase one extra individual rose with the le over money.\nTry a cost of $1.50 per rose.\nIf a rose costs $1.50 individually, then a bouquet costs 12($1.50)(0.80) =\n$14.40, Since two bouquets cost 2($14.40) = $28.80, there isn’t enough\nmoney le over to purchase another single rose. Therefore, this case fits\nstatement (1).\nNow three bouquets cost 3($14.40) = $43.20, and one more rose can be\npurchased for $1.50, bringing the total purchase to $44.70. In this case,\n37 roses can be purchased with $45."} +{"id": "book 2_p242_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 242, "page_end": 242, "topic_guess": "coordinate_geometry", "text": "Since either 36 or 37 roses could be purchased for $45, the two\nstatements are insufficient together.\nThe correct answer is (E).\n6. (D): Because there are no constraints on y, use the standard number set,\ndisplayed here in number line graphic form:\nStatement (1): y < 1. If y is less than 1, all tested values show y3 ≤ |y|.\nSUFFICIENT.\nStatement (2): y < 0. If y is less than 0, all tested values show y3 ≤ |y|.\nSUFFICIENT.\nNotice that you could use Guessing Tactic 3: spot one statement inside\nthe other. y < 0 is more limiting than y < 1, so if statement (1) is\nsufficient, statement (2) must be sufficient.\nThe correct answer is (D).\n7. (B): In order for abc to be even, what must be true? If just one of the\nnumbers is even, then the product will be even. The only case that will\nproduce an odd is when all three numbers are odd. Keep this in mind\nwhen testing the statements."} +{"id": "book 2_p243_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 243, "page_end": 243, "topic_guess": "algebra", "text": "(1) INSUFFICIENT: If a + b is even, then the two variables are either both\neven or both odd:\na b c abc even?\nEven Even Even Yes\nOdd Odd Odd No\n(2) SUFFICIENT: b + c is odd, so one of b and c is odd and the other is\neven. In other words, it is impossible for all three variables to be odd.\nTherefore, the product abc must be even.\nThe correct answer is (B).\n8. (C): Set up a table and try to find contradictory scenarios.\n(1) INSUFFICIENT:\nA B G M S < 5 in one class?\n5 6 6 5 6 No\n4 6 6 6 6 Yes\n(2) INSUFFICIENT:"} +{"id": "book 2_p244_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 244, "page_end": 244, "topic_guess": "sets_probability_counting", "text": "A B G M S < 5 in one class?A B G M S < 5 in one class?\n6 6 4 6 6 Yes\n6 5 5 6 6 No\n(1) AND (2) SUFFICIENT:\nA B G M S < 5 in one class?\n4 6 6 5 7 Yes\nSince Geology and Sociology each have to have a minimum of 6\nstudents, and they can’t have the same number of students, one of the\ntwo classes has to have at least 7 students. Between Biology, Geology,\nand Sociology, then, the minimum number of students is 6 + 6 + 7 = 19,\nleaving 9 students to be split among the other two classes. In this case,\nit’s impossible to have 5 or more students in both of those two\nremaining classes.\nThe correct answer is (C).\n9. (C): The question asks for the probability that a randomly selected\ninteger, from 1 to 100, will be in neither one of the two sets. To calculate\nthis probability, you need to know the total number of integers that are\nin one or both of the two sets, as well as within the 1 to 100 range."} +{"id": "book 2_p245_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 245, "page_end": 245, "topic_guess": "number_theory", "text": "Jot down the definitions of the two sets. Integer x is contained in both\nsets, so, at most, the two sets will contain seven distinct integers\nbetween them. However, some of these integers might be less than 1 or\ngreater than 100, and there might be more integers that overlap\nbetween the two sets , decreasing the number of possible values that\noverlap with the selected integer to less than 7. A statement is sufficient\nif every possible value of x yields the same number of distinct values\nbetween 1 and 100, inclusive.\n(1) INSUFFICIENT: Test a straightforward case first, then test a case that\nmight yield a different value.\nCase 1: The smallest possible value for x that fits the statement is 6. If x =\n6, then set A consists of four integers: 6, 62, 63, and 64. Two of those\nintegers, 6 and 62 = 36, are smaller than 100. As long as you know that 63\nand 64 are greater than 100, there is no need to calculate their exact\nvalues. Set B also consists of four integers: 6, 12, 18, and 24.\nTherefore, the sets contain a total of five distinct integers between 1 and\n100, inclusive.\nCase 2: Try an extreme case, such as x = 1,000. In this case, all of the\nintegers in both set A and set B are much greater than 100. Therefore,\nthe sets contain no integers between 1 and 100.\n(2) INSUFFICIENT: The first case tested above, with x = 6, also fits this\nstatement. In this case, there are five distinct integers within the range,\nas described above."} +{"id": "book 2_p246_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 246, "page_end": 246, "topic_guess": "number_theory", "text": "Try another extreme case, such as x = –2. In this case, the only positive\nintegers in either set will be (–2)2 = 4 and (–2)4 = 16, which are both\nbetween 1 and 100. Therefore, there are exactly two distinct integers\nwithin the range. Since the set could contain five integers or two integers\nwithin the given range, this statement is also insufficient. \n(1) AND (2) SUFFICIENT: x = 6 yields five distinct integers that are within\nthe range 1 to 100. Check to see whether this is also true for x = 7, 8, and\n9.\nFor x = 7, 8, or 9, set A contains two integers that are within the range 1\nto 100: x and x2. In all three cases, x3 and x4 are outside of the range.\nAlso, set B contains four integers that are within the range, but one of\nthem, x itself, is also contained in set A. So in every case in which x is\nbetween 5 and 10, there are exactly five distinct integers in the range\nthat are contained in one or both sets.\nThe correct answer is (C).\n10. (B): The average price of six coats was originally $85. So the sum of the\nsix prices was originally 6($85) = $510.\nTwo of the prices were discounted by 20%, resulting in a $9 reduction in\nthe average price, or a 6($9) = $54 reduction in the overall price. Since\n20% (or 1/5 of the cost of these two coats came to $54, the original price\nof these two coats must have been five times this, or 5($54) = $270.\nGiven that the sum of the original prices of the two discounted coats was\n$270, did one of these two coats have the lowest original price?"} +{"id": "book 2_p247_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 247, "page_end": 247, "topic_guess": "general", "text": "(1) INSUFFICIENT: From the question stem, the two discounted coats\nhad a total price of $270, and according to this statement, one of those\ntwo coats was the most expensive. Test cases to see whether the other\ncoat could have been the least expensive. Note that the four non-\ndiscounted coats had a total price of $510 – $270 = $240.\nCase 1: If the discounted coats originally cost $1 and $269 and the other\nfour coats each cost \n , then the least expensive coat was\none of the two that was discounted, and the answer is Yes.\nCase 2: If the discounted coats originally cost $130 and $140 and the\nother four coats each cost \n , then the least expensive\ncoat was not one of the two that was discounted, and the answer is No.\n(2) SUFFICIENT: This statement limits the possible original costs of the\ntwo discounted coats. Since none of the coats originally cost more than\n$180, the two discounted coats could not have cost, for example, $50\nand $220, or $70 and $200. Since neither of the coats could have cost\nmore than $180, the less expensive of the two discounted coats must\nhave cost at least $270 – $180 = $90.\nHowever, this coat could not have been the one with the lowest original\nprice, as $90 is greater than the original average price. The smallest\nnumber in any set cannot be greater than the average of that set.\nTherefore, neither of the two discounted coats could have been the one\nwith the lowest price. The answer to the question is definitely No, and\nthis statement is sufficient."} +{"id": "book 2_p248_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 248, "page_end": 248, "topic_guess": "algebra", "text": "The correct answer is (B).\n11. (C): Assign variables to the unknowns in the problem.\nC = packages of candy\nP = bags of popcorn\nR = number of pretzels\nPc = price of a package of candy\nPp = price of a bag of popcorn\nPr = price of a pretzel\nAlgebraically, the question asks: What is Pc × C − (Pp × P + Pr × R) ?\nThe question stem indicates that \n(1) INSUFFICIENT: Pr = Pp. Substitute and rephrase the question:\nThere are four unknown variables, so statement (1) is not sufficient.\n(2) INSUFFICIENT: C = 24, P = 6, and R = 6:\nWhat is Pc × C − (Pp × P + Pr × R) ?\nWhat is \nWhat is \nWhat is Pc × C − (Pp × P + Pr × R) ?"} +{"id": "book 2_p249_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 249, "page_end": 249, "topic_guess": "algebra", "text": "There are two unknown variables, so statement (2) is not sufficient.\n(1) AND (2) SUFFICIENT:\nGiven the relative prices of the candy, popcorn, and pretzels, and the\nquantity of each purchased, the cost of the candy will always equal the\ncombined cost of the popcorn and pretzels, even though it’s impossible\nto calculate the exact prices. Statements (1) and (2) combined are\nSUFFICIENT to answer the question.\nThe correct answer is (C).\nWhat is \nWhat is 12Pp − 6Pp − 6Pr ?\nWhat is 6Pp − 6Pr ?\nFrom statement (2): What is 6Pp − 6Pr ?\nFrom statement (1): What is 6Pp − 6(Pp) ?\nRephrased question: What is 0 ?\n12. (A): It’s tempting to rephrase the question as “What is x?” but that is\ndangerous when the question contains absolute value symbols. It’s\npossible that two different values of x would resolve to the same value\nof |x + 4|. Leave the question as it is.\n(1) SUFFICIENT: x2 + 8x + 12 = 0. Factor the equation:"} +{"id": "book 2_p250_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 250, "page_end": 250, "topic_guess": "algebra", "text": "Plug these answers into the question stem:\nThe two different values resolve to the same final value for |x + 4|, so\nstatement (1) is sufficient.\n(2) INSUFFICIENT: x2 + 6x = 0. Factor the equation:\nPlug both values into the question stem:\nThere are two different values for the expression |x + 4|, so statement (2)\nis not sufficient.\nThe correct answer is (A).\n13. (C): If abcd ≠ 0, then none of the variables equals 0. In order for the\nproduct abcd to be negative, you would need to have an odd number of\nnegatives in the mix:"} +{"id": "book 2_p251_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 251, "page_end": 251, "topic_guess": "general", "text": "Scenario Product abcdScenario Product abcd\nAll 4 positive or all 4 negative +\n3 positive, 1 negative −\n2 positive, 2 negative +\n1 positive, 3 negative −\n(1) INSUFFICIENT: Beware of inequalities.\n(2) INSUFFICIENT:\nIf b and d are both positive or both negative, then ad > bc.\nIf one is positive and one is negative, then ad < bc.\nThis statement indicates nothing about a and c, though.\nIf a and c are both positive or both negative, then bc > ad.\nIf one is positive and one is negative, then bc < ad.\nThis statement indicates nothing about b and d, though."} +{"id": "book 2_p252_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 252, "page_end": 252, "topic_guess": "number_theory", "text": "(1) AND (2) SUFFICIENT: Here’s where the trick comes in. The two\nstatements allow two possible scenarios:\nIf ad > bc, then b and d have the same sign but a and c have opposite\nsigns. In this case, three signs are the same and one is not, so there are\nan odd number of negatives in the mix, and the product of all four\nvariables must be negative.\nIf ad < bc, b and d have opposite signs, but a and c have the same signs.\nIn this case, three signs are the same and one is not; as before, the\nproduct of all four variables must be negative.\nEither way, the two statements together are sufficient to answer the\nquestion.\nThe correct answer is (C).\n14. (A): The question stem asks whether n = 24, 48, 72, …\n(1) SUFFICIENT: Rewrite and simplify:\nIn other words, n is the product of four consecutive integers. In any four\nconsecutive integers, one number must be divisible by 3 and two\nnumbers must be even. Furthermore, of the two even numbers, one\nmust be divisible by 4. As a result, the product of any four consecutive\nintegers is divisible by (3)(2)(4) = 24."} +{"id": "book 2_p253_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 253, "page_end": 253, "topic_guess": "algebra", "text": "Alternatively, you could figure this out by testing a few cases:\n(2) INSUFFICIENT: If m = 1, then n could equal 5, which is not a multiple\nof 24. If m = 20, then n could equal 24, which is a multiple of 24.\nThe correct answer is (A).\nm = 1: (8)(7)(6)(5), which is divisible by (8)(3) = 24\nm = 2: (9)(8)(7)(6), which is divisible by (8)(3) = 24\nm = 3: (10)(9)(8)(7), which is divisible by (8)(3) = 24\n15. (C): This algebra problem includes a special quadratic: (1 – y)(1 + y) = 1 –\n y2. Rephrase the question as “What is the value of x(1 – y2)?”\n(1) INSUFFICIENT: This statement appears to provide little information.\nHowever, it significantly limits the possible values of x.\nIf x = 1, the answer to the question is the value of 1 – y2, which is\nunknown. If x = 0, the answer to the question is 0. \n(2) INSUFFICIENT: If y2 = x, then the question simplifies to “What is the\nvalue of x(1 – x)?” Different values of x produce different answers, so this\nstatement is insufficient."} +{"id": "book 2_p254_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 254, "page_end": 254, "topic_guess": "general", "text": "(1) AND (2) SUFFICIENT: According to statement (1), the value of x is\neither 0 or 1.\nIf the value of x is 0, then, according to statement (2), y2 = x = 0. In this\ncase, the answer is x(1 – y2) = 0(1 – 0) = 0. \nIf the value of x is 1, then, according to statement (2), y2 = x = 1. In this\ncase, the answer is x(1 – y2) = 1(1 – 1) = 0. \nSince the answer to the question is always 0, the two statements\ntogether are sufficient. \nThe correct answer is (C)."} +{"id": "book 2_p255_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 255, "page_end": 255, "topic_guess": "word_problems", "text": "PART TWO\nStrategies for All Problem Types"} +{"id": "book 2_p256_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 256, "page_end": 256, "topic_guess": "sequences_patterns", "text": "CHAPTER 5\nPattern Recognition"} +{"id": "book 2_p257_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 257, "page_end": 257, "topic_guess": "sequences_patterns", "text": "In This Chapter...\nPattern Recognition Problems\nSequence Problems\nUnits (Ones) Digit Problems\nRemainder Problems\nOther Pattern Problems"} +{"id": "book 2_p258_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 258, "page_end": 258, "topic_guess": "sequences_patterns", "text": "Chapter 5\nPattern Recognition  \nPattern Recognition Problems\nIn the context of the GMAT, pattern recognition involves spotting a\nrepeating cycle or other simple relationship underlying a series of\nnumbers. If you can grasp the rule, you can predict numbers that appear\nlater in the series. The series may be part of a defined sequence or it may\narise from a general list of possibilities. Either way, if you can spot the\npattern, you can eliminate a lot of unnecessary calculation. In fact, o en\nthe only feasible way to get the answer in two minutes (e.g., finding the\n100th number in some series) is to recognize the underlying pattern. Here’s\nan example.\nTry-It #5-1\nEach number in a sequence is 3 more than the previous number. If\nthe first number is 4, what is the value of the 1,000th term in the\nsequence?"} +{"id": "book 2_p259_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 259, "page_end": 259, "topic_guess": "sequences_patterns", "text": "Obviously, finding the 1,000th number the long way (by computing every\nintervening number) is impossible in the time allotted on the GMAT. You\ncan solve this problem in several ways, but one powerful way is to compute\nthe first several terms, spot the underlying pattern, figure out the rule, and\nthen apply it.\nThe first five terms of the sequence are 4, 7, 10, 13, 16. Notice that you\nrepeatedly add 3 to get the next value. Repeated addition is just\nmultiplication, so match these numbers to the multiples of 3. The first five\nmultiples of 3 are 3, 6, 9, 12, 15—all 1 less than the numbers in the\nsequence. Thus, the rule for generating the sequence is “take the\ncorresponding multiple of 3, and add 1.” Therefore, the 1,000th term of the\nsequence is 1 more than the 1,000th multiple of 3, which is 3,000 + 1 =\n3,001. Once you spot the pattern, you can skip over vast amounts of\nunnecessary work.\nThe two most basic patterns are these:\nThe counting numbers (1, 2, 3, 4 . . . ), also known as the positive\nintegers. As simple as this pattern may seem, it is the basis for many\nother patterns. For instance, the sequence of the multiples of 7 (7,\n14, 21, 28 . . . ) can be derived from 1, 2, 3, 4 . . . by multiplying by 7.\nYou can write this sequence as Sn = 7n, where n is the basic\nsequence of positive integers.\n1.\nA repeating cycle of numbers. For instance, the sequence 4, 2, 6, 4,\n2, 6, 4, 2, 6 . . . has a repeating cycle of three terms: 4, 2, 6. Repeating\ncycles can be derived in various ways from the counting numbers\n(e.g., when you raise an integer to increasing powers, the units\ndigits of the results repeat themselves). However, it is o en easier\n2."} +{"id": "book 2_p260_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 260, "page_end": 260, "topic_guess": "sequences_patterns", "text": "When you are examining a string of numbers for a pattern, follow these\nsteps:\nto think of repeating cycles on their own terms, separately from the\ncounting numbers and related patterns.\nCompute the first five to eight terms and try to match them to a\npattern that you already know. The most basic pattern is the\ncounting numbers, but you should also have these related patterns\nup your sleeve:\nFor each of these sequences, notice exactly where the counting\nnumbers come into play. For instance, in the squares series, the\ncounting numbers are the bases, but in the powers series, the\ncounting numbers are the exponents.\n1.\nMultiples (e.g., 7 × 1 = 7, 7 × 2 = 14, 7 × 3 = 21, 7 × 4 = 28, 7 × 5 = 35 \n. . . 7n  . . . )\na.\nSquares (12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25  . . .  n2  . . . )b.\nPowers (21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32  . . . 2n  . . . )c.\nLook for repeating cycles. As soon as you generate a repeated term,\nsee whether the sequence will repeat itself from that point onward.\nRepeating cycles on the GMAT typically begin repeating every four\nterms (or fewer), so five to eight terms should be sufficient to\nidentify the pattern. Of course, some cycles repeat every single term\n—that is, it’s the same number over and over again!\n2.\nIf you are stuck, look for patterns within differences between terms\nor sums across terms:\n3."} +{"id": "book 2_p261_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 261, "page_end": 261, "topic_guess": "sequences_patterns", "text": "Look at the difference between consecutive terms. For instance,\nthis process can help you spot linear sequences (sequences of\nmultiples plus a constant):\na.\nAlso look at the cumulative sum of all the terms up to that point.\nThis is helpful if the terms get closer to zero or alternate in sign:\nSequence:\n ...\n ...\nCumulative Sum:\n ...\n ...\nNotice that the cumulative sum for this sequence approaches 1.\nb.\nSome sums involve matching pairs that sum to the same number\n(or even cancel each other out). Be on the lookout for such\nmatching pairs.\nc.\nWhat is the sum of 1, 5, 8, 10, 11, 11, 12, 14, 17, and 21 ?\nYou can, of course, sum these numbers in order, but look to\nmake natural intermediate sums (subtotals) with matching\npairs. In this example, spot the repeated 11’s in the middle\nand sum them to 22. Working outward, 10 and 12 sum to 22 as"} +{"id": "book 2_p262_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 262, "page_end": 262, "topic_guess": "sequences_patterns", "text": "Perhaps the most important principle to apply on Pattern Recognition\nproblems is to put pen to paper, as discussed in Chapter 1. O en, the\npattern will be completely hidden until you actually compute the first\nseveral values of the sequence or other initial results.\nSeveral types of problems frequently involve underlying patterns. When\nyou see these types of problems on the GMAT, be ready to analyze the\npattern so you can find the rule:\nwell. So do 8 and 14, 5 and 17, and 1 and 21. In all, there are\nfive subtotals of 22, for a grand total of 110.\nLook at characteristics of the numbers: positive/negative,\nodd/even, integer/non-integer, etc. Once you have extended the\npattern for several terms, these characteristics will generally repeat\nor alternate in some predictable way.\n4.\nSequence Problems: Nearly all sequence problems involve a\npattern in the elements (or terms) of the sequence. Sequences can\nbe defined either directly (i.e., each value in the series is a function\nof its location in the order of the sequence) or recursively (i.e., each\nvalue is a function of the previous items in the sequence).\n1.\nUnits (Ones) Digit Problems: Questions involving the last digit\n(sometimes called the units or ones digit) of an integer almost\nalways involve some sort of repeating cycle pattern that can be\nexploited.\n2.\nRemainder Problems: Remainders from the division of one integer\ninto another will result in a pattern. For example, when divided by\n5, the counting numbers will exhibit the following repeating\n3."} +{"id": "book 2_p263_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 263, "page_end": 263, "topic_guess": "number_theory", "text": "remainder pattern: 1, 2, 3, 4, 0, 1, 2, 3, 4, 0 . . . The units digit of an\ninteger is a special case of a remainder: it’s the remainder a er\ndivision by 10.\nOther Pattern Problems: Some pattern problems do not involve\ndeciphering a string of numbers and discovering the rule. For\ninstance, you may have to count a set of numbers that all fit some\nconstraint. The point is to discover a simple rule or group of rules\nthat let you account for all the numbers—and therefore count them\n—without having to generate each one. Here are some ideas:\n4.\nBreak the problem into sub-problems. For instance, a sum may\nbe split into several smaller sums. Or you might count a larger\ntotal, then subtract items that do not fit the constraint. You even\nmight multiply a larger total by the proportion of suitable items,\nif that fraction is easy to calculate.\nRecall counting and summing methods from Manhattan Prep’s\nAll the Quant guide:\nNumber of Choices: When you have a series of successive\ndecisions, you multiply the number of choices you have at\neach stage to find the number of total choices you have. For\ninstance, if you can choose 1 appetizer out of 6 possible\nappetizers and 1 main course out of 7 possible main courses,\nthen you could have 6 × 7 = 42 possible meals.\na.\nNumber of Items in a Consecutive Set of Integers: The number\nof integers in a consecutive set of integers equals the largest\nminus the smallest, plus 1.\nb.\nSum of a Consecutive Set of Integers: The sum of a\nconsecutive set of integers equals the number of integers\n(computed above) multiplied by the average, which is the\nc."} +{"id": "book 2_p264_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 264, "page_end": 264, "topic_guess": "statistics", "text": "average of the largest integer and the smallest integer. This is\nalso equal to the median, or “middle,” number in the set.\nAs you go, always check that the extreme cases are still valid.\nTwo or three constraints can interact in surprising ways,\neliminating some of the values that would seem to work\notherwise."} +{"id": "book 2_p265_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 265, "page_end": 265, "topic_guess": "sequences_patterns", "text": "Sequence Problems\nAny question that involves the definition of a sequence (usually involving\nsubscripted variables, such as An and Sn) is very likely to involve patterns.\nThese patterns can range from relatively straightforward linear patterns to\nmuch more complicated ones.\nWhen you are given a sequence definition, list a few terms of the sequence,\nstarting with any particular terms you are given, and look for a pattern.\nDo not be intimidated by a recursive definition for a sequence, in which\neach term is defined using earlier terms. (By contrast, a direct definition\ndefines each term using the position or index of the term.) To illustrate the\ndifference, here are two ways to define the series of positive odd integers\n{1, 3, 5, 7, 9, etc.}:\nRecursive Definition Direct Definition\nAn = An − 1 + 2 where n > 1 and A1 = 1 \nTranslation:\n“This term = the previous term + 2,\nand the first term is 1.”\nAn = 2n − 1, where n ≥ 1\nTranslation:\n“This term = the index number × 2, minus 1. Thus, the\nfirst term is (2)(1) − 1 = 1.”"} +{"id": "book 2_p266_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 266, "page_end": 266, "topic_guess": "sequences_patterns", "text": "Try-It #5-2\nThe sequence Xn is defined as follows: Xn = 2Xn − 1 − 1 whenever n\nis an integer greater than 1. If X1 = 3, what is the value of X20 −\nX19 ?\nThe pattern underlying this sequence is not obvious, so begin computing a\nfew of the terms in the set:\nn Xn\n1 3\n2 2(3) − 1 = 5\n3 2(5) − 1 = 9\n4 2(9) − 1 = 17\n5 2(17) − 1 = 33\n6 2(33) − 1 = 65\n7 2(65) − 1 = 129"} +{"id": "book 2_p267_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 267, "page_end": 267, "topic_guess": "sequences_patterns", "text": "You might notice that there appears to be a repeating pattern among the\nunits digits of the elements of Xn (3, 5, 9, 7, 3, 5, 9 . . . ). However, this does\nnot help to answer the question, which asks about the difference between\ntwo consecutive elements later in the set. Instead, look at the differences\nbetween consecutive elements:\nn Xn Xn − Xn − 1\n1 3 −\n2 2(3) − 1 = 5 5 − 3 = 2\n3 2(5) − 1 = 9 9 − 5 = 4\n4 2(9) − 1 = 17 17 − 9 = 8\n5 2(17) − 1 = 33 33 − 17 = 16\n6 2(33) − 1 = 65 65 − 33 = 32\n7 2(65) − 1 = 129 129 − 65 = 64\nThe pattern quickly emerges: the difference between consecutive terms in\nthe sequence appears to always be a power of 2. Specifically, X2 − X1 = 2 =\n21, X3 − X2 = 4 = 22, X4 − X3 = 8 = 23, etc."} +{"id": "book 2_p268_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 268, "page_end": 268, "topic_guess": "sequences_patterns", "text": "You can determine the pattern for the sequence: Xn − Xn − 1 equals 2n − 1.\nTherefore, X20 − X19 = 219. This is a difference pattern—a pattern or rule that\nexists among the differences between consecutive terms in the sequence.\nBe careful at this last step! When you extrapolate the pattern, you might\naccidentally think that the number you want is 220. Always explicitly match\nto the index, and realize that you might be slightly shi ed. In this case, the\ndifference you want is not 2n. The difference is 2n − 1.\nTry-It #5-3\nIf \n for all positive integers n, what is the sum of\nthe first 100 elements of An ?\nOnce again, compute the first few elements of An. Because you need to\nknow the sum of the first 100 elements, also track the cumulative sum:\nn An Sum through An\n1\n2\n3\n4"} +{"id": "book 2_p269_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 269, "page_end": 269, "topic_guess": "sequences_patterns", "text": "n An Sum through An\n5\nThe sum of the first n terms of An equals \n . Therefore, the sum of the\nfirst 100 terms is \n . This is a summing pattern—a pattern or rule that\nexists among the cumulative sum of the terms in the sequence."} +{"id": "book 2_p270_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 270, "page_end": 270, "topic_guess": "sequences_patterns", "text": "Units (Ones) Digit Problems\nWhen you raise an integer to a power, the units digit always displays some\nkind of pattern as you increase the power.\nTry-It #5-4\nWhat is the units digit of 4674 ?\nObserve what happens to the units digit of the consecutive powers of 4,\nstarting with 41:\n41 = 4               \n       last digit of 4 = 4\n42 = 4(41)        \n       last digit of 4(4) = last digit of 16 = 6\n43 = 4(42)        \n       last digit of 4(6) = last digit of 24 = 4\n44 = 4(43)        \n       last digit of 4(4) = last digit of 16 = 6\nBecause the computations (4 × 4\n= 16) and (4 × 6 = 24) keep\nrepeating, the units digit will\ncontinue to alternate [4, 6].\nThus, 4x will have a units digit of 4 whenever x is odd and a units digit of 6\nwhenever x is even (assuming, of course, that x is positive). The units digit\nof 4674 is therefore 6.\nAlso notice that in determining the value of the units digit of a product, all\nof the other digits besides the units digit are irrelevant. Therefore, 14674"} +{"id": "book 2_p271_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 271, "page_end": 271, "topic_guess": "sequences_patterns", "text": "and 3,184674 will both also have units digits of 6. This is also true for\nmultiplication of any two integers, as well as the addition of any integers:\nAs mentioned earlier, every integer raised to different positive exponents\nhas a units digit pattern. As an exercise, derive several of the patterns\nyourself for the units digits 2, 3, 5, 7, and 8; you can check your work using\nthe table below:\nSeries Consecutive Powers Units Digit Pattern\n1x 1; 1; 1; 1; 1; 1; etc. [1]\n2x 2; 4; 8; 16; 32; 64; etc. [2, 4, 8, 6]\n3x 3; 9; 27; 81; 243; 729; etc. [3, 9, 7, 1]\n4x 4; 16; 64; 256; 1,024; 4,096; etc. [4, 6]\n5x 5; 25; 125; 625; 3,125; 15,625; etc. [5]\n6x 6; 36; 216; 1,296; 7,776; 46,656; etc. [6]"} +{"id": "book 2_p272_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 272, "page_end": 272, "topic_guess": "sequences_patterns", "text": "Series Consecutive Powers Units Digit Pattern\n7x 7; 49; 343; 2,401; 16,807; 117,649; etc. [7, 9, 3, 1]\n8x 8; 64; 512; 4,096; 32,768; 262,144; etc. [8, 4, 2, 6]\n9x 9; 81; 729; 6,561; 59,049; 531,441; etc. [9, 1]\n10x 10; 100; 1,000; 10,000; 100,000; 1,000,000; etc. [0]\nYou can either memorize this chart or know how to regenerate these\npatterns quickly. The units digits 1, 5, 6, and 0 just repeat the same digit\nforever; there’s no real pattern to memorize. That leaves you with six\npossibilities to memorize (or to re-create when you need them). Note that\nno pattern goes beyond four numbers before repeating, so you don’t have\nto check beyond the first four terms of any pattern.\nTry-It #5-5\nWhat is the units digit of 1940 ?\nAs shown in the table above, 91 = 9, 92 = 81, 93 = 729, etc., and 19 will have\nthe same units digit pattern as 9. Therefore, the pattern is a two-term\nrepeating pattern: 9, 1, 9, 1 . . . . This pattern alternates every two items,\njust like odd and even integers. Since 40 is an even number, the units digit\nof 1940 will equal 1. Try another example:"} +{"id": "book 2_p273_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 273, "page_end": 273, "topic_guess": "sequences_patterns", "text": "What is the remainder when 1940 is divided by 10 ?\nThis alternative question is asking the exact same thing as the original\nquestion. The remainder whenever an integer is divided by 10 will always\nbe the same as the units digit of the original number:"} +{"id": "book 2_p274_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 274, "page_end": 274, "topic_guess": "number_theory", "text": "Remainder Problems\nIn general, remainders provide a means by which the GMAT can disguise an\nunderlying pattern. For example, notice that when the positive integer n is\ndivided by 4, the remainders follow a pattern as n increases consecutively:\n1 div by 4 \n remainder 1 \n2 div by 4 \n remainder 2 \n3 div by 4 \n remainder 3\n4 div by 4 \n remainder 0 \n5 div by 4 \n remainder 1\nA repeating cycle of [1, 2, 3, 0] emerges for the remainders\nwhen dividing the counting numbers by 4. The number of\nterms in the repeat equals the divisor in this case.\nConversely, you can calculate all of the numbers that have a certain\nremainder when divided by a certain value, because they appear at regular\nintervals as well. For instance, the numbers that have a remainder of 3\nwhen divided by 4 are 3, 7, 11, 15, 19, 23, etc. Notice that those numbers\nare evenly spaced exactly 4 apart. \nWhen a problem discusses remainders, look for patterns and take\nadvantage of them. \nTry-It #5-6\nIf x and y are positive integers, what is the remainder when 5x is\ndivided by y ?"} +{"id": "book 2_p275_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 275, "page_end": 275, "topic_guess": "sequences_patterns", "text": "(1) INSUFFICIENT: 53 = 125. The problem provides no information about the\nvalue of y, however. For example, if y = 5, then the remainder equals 0. If y =\n6, then the remainder is 5.\n(2) SUFFICIENT: This statement may not initially appear to be sufficient,\nbut test some different values for x:\nx 5x Remainder of \n1 5 1\n2 25 1\n3 125 1\n4 625 1\nThe pattern is clear: no matter what exponent 5 is raised to, the remainder\nwhen divided by 4 will always equal 1—a fact that you probably did not\nexpect before testing the rule for this problem.\nThe correct answer is (B).\nx = 3(1)\ny = 4(2)"} +{"id": "book 2_p276_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 276, "page_end": 276, "topic_guess": "sequences_patterns", "text": "This problem is an example of a (C) Trap. If you used both statements, you\ncould calculate the exact remainder. However, that would be too easy for a\ntough GMAT Quant problem. You don’t actually need both statements to\ndetermine that the answer to the question is 1. Thanks to the remainder\npattern, statement (2) is sufficient on its own."} +{"id": "book 2_p277_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 277, "page_end": 277, "topic_guess": "number_theory", "text": "Other Pattern Problems\nMany questions will not at first glance demonstrate an obvious pattern. For\nexample, a Word Problem involving counting a collection of objects or\nmaximizing some number may hide some sort of regularity. The point is to\ndiscover a simple rule or group of rules that let you account for all the\npossibilities—and therefore count or maximize them��without having to\ngenerate each possibility separately.\nTry-It #5-7\nHow many of the integers between 1 and 400, inclusive, are not\ndivisible by 4 and do not contain any 4’s as a digit?\nThis problem involves a counting pattern. It’s clear that there are 400\nintegers between 1 and 400, inclusive. You’ll need to subtract the integers\nthat are divisible by 4 or contain a 4 as a digit. The tricky part is the overlap:\nsome numbers, such as 64 and 124, violate both constraints.\nIt is easier to determine the number of multiples of 4. Since there are 400\nintegers in the set, and those 400 integers are consecutive, there must be\n integers that are divisible by 4. There are 400 − 100 = 300\nintegers remaining."} +{"id": "book 2_p278_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 278, "page_end": 278, "topic_guess": "number_theory", "text": "Next, consider the remaining numbers that have a 4 among their digits.\nThese numbers have a 4 among their digits but are not themselves\nmultiples of 4.\nOnly one of the integers between 1 and 400, inclusive, has a 4 in the\nhundreds place: 400. You have already eliminated that one from the count\nbecause it is a multiple of 4.\nTo eliminate integers with a 4 in the tens place, of the form x4y, count only\nthose that are not multiples of 4. These are the numbers whose last two\ndigits are 41, 42, 43, 45, 46, 47, or 49. There are seven such numbers in each\nset of “hundreds,” that is, the 300’s, the 200’s, the 100’s, and the no\nhundreds. That is a total of 7 × 4 = 28 terms. There are 300 − 28 = 272\nintegers remaining.\nLast, count and eliminate the integers with 4 in the units digit, of the form\nxy4, that have not already been subtracted. These are the numbers whose\nlast two digits are 14, 34, 54, 74, or 94 (numbers with an even integer in the\ntens place and 4 in the units are all divisible by 4, so they have been\neliminated already). There are 5 such numbers in each set of “hundreds,”\nso that is a total of 5 × 4 = 20 terms. There are 272 − 20 = 252 integers\nremaining.\nThe correct answer is 400 − 100 − 28 − 20 = 252.\nTry-It #5-8\nx = 1010 − z, where z is a two-digit integer. If the sum of the digits\nof x equals 84, how many values for z are possible?"} +{"id": "book 2_p279_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 279, "page_end": 279, "topic_guess": "sequences_patterns", "text": "The first thing to do in solving this problem is to subtract any two-digit\nnumber from 1010 and look for a pattern. Try subtracting 24:\nNotice the pattern: the first eight digits of x are all 9’s, so those digits sum\nto 72. This will be true no matter which two-digit integer you try for z.\nTherefore, the final two digits of x must sum to 84 − 72 = 12.\nWhat possibilities would work for these final two digits of x? 39, 48, 57, 66,\n75, 84, and 93 all add to 12. When subtracted from 100, these numbers will\nhave to produce a two-digit integer. Try subtracting from 100 to find the\npattern:\nHmm. Each successive number is larger, so it will result in a smaller two-\ndigit integer. Do the numbers actually drop to one digit at some point?\nJump to the other end of the scale and try 93:\nIn order for x to end in 93, z would have to be 7, which is not a two-digit\ninteger. According to the GMAT, a two-digit integer must have a nonzero"} +{"id": "book 2_p280_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 280, "page_end": 280, "topic_guess": "sequences_patterns", "text": "tens digit and zeros for all higher places. What about the next number up?\nIf x ends in 84, then it would result in the two-digit integer 100 − 84 = 16. All\nof the other numbers are two-digit integers; only 93 doesn’t work.\nTherefore, the final two digits of x can only be 39, 48, 57, 66, 75, and 84,\nresulting in six possible values for z.\nAs the two previous problems demonstrate, unusual patterns can appear\nin problems on the GMAT. Sometimes you must “think outside the box” to\nidentify the wanted pattern.\nLook for the examples in the following chart:\nPattern\nType\nExample Comments\nRepeats 1, 3, −2, 1, 3,\n−2, 1, etc.\nO en, these repeating patterns can only be identified by listing\nout a few values in the pattern. On the GMAT, the cycle is\nusually 4 numbers or fewer (the cycle in the example shown is\n3).\nConsecutive\nIntegers\n10, 11, 12,\n13, 14, etc.\nCan be defined as follows: An = n + k, where n and k are\nintegers. In this example, An = n + 9, so that the first term is 1 +\n9 = 10. Note that the average term = the median term = \n ×\n(First + Last)."} +{"id": "book 2_p281_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 281, "page_end": 281, "topic_guess": "sequences_patterns", "text": "Pattern\nType\nExample Comments\nConsecutive\nMultiples\n7, 14, 21, 28,\netc.\nConsecutive multiples of 7, for example, can be defined as\nfollows: An = 7n, where n is a set of consecutive integers. The\nevens are just a special case (multiples of 2, or 2n). Note that\nthe average term = the median term = \n × (First + Last).\nEvenly\nSpaced Sets\n9, 16, 23, 30,\netc.\n(Constant\ndifference of\n7 between\nconsecutive\nterms)\nWhen dividing this series by 7, each of the terms leaves a\nremainder of 2.\nCan be defined as a multiple plus/minus a constant: An = 7n +\nk, where n is a set of consecutive integers, in this example. The\nodds are a special case (multiples of 2, plus 1, or 2n + 1).\nNote that the average term = the median term = \n × (First +\nLast), as for consecutive multiples.\nNon-\nUniform\nSpacing\nthat Itself\nFollows a\nPattern\n0, 1, 3, 6, 10,\n15, etc.\n(Spacing\nbetween\nterms\nfollows 1, 2,\n3, 4, 5, etc.\npattern.)\nAnother example of this type is the perfect squares: 0, 1, 4, 9,\n16, 25, etc.\n(Spacing between squares = 1, 3, 5, 7, 9, etc. = the odd\nintegers!)\nAlternating\nSign\n−1, 1, −2, 2,\n−3, 3, etc.\nCan result from a (−1)n term in a direct sequence definition, or\na (−(An − 1)) term in a recursive sequence definition."} +{"id": "book 2_p282_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 282, "page_end": 282, "topic_guess": "sequences_patterns", "text": "In general, listing five to eight examples or terms will usually be sufficient\nto identify a pattern on the GMAT—you can stop with fewer examples if\nyou’ve identified the pattern by that point."} +{"id": "book 2_p283_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 283, "page_end": 283, "topic_guess": "sequences_patterns", "text": "Problem Set\nFor the following problems, use the Pattern Recognition techniques\ndiscussed in this chapter to solve.\n1. In the sequence 4, 9, 14, 19 ..., each term is 5 greater than the\nprevious term. What is the remainder when the 75th term is divided\nby 9 ?\n2. If x and y are integers between 0 and 9, inclusive, and the units digit\nof xy is 5, what are the possible values of x and y  ?\n3. What is the remainder when 1317 + 1713 is divided by 10 ?\n4. If y is a positive integer, what is the units digit of y  ?\nThe units digit of y2 equals 6.(1)\nThe units digit of (y + 1)2 equals 5.(2)\n5. If y is a positive integer, what is the units digit of y  ?"} +{"id": "book 2_p284_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 284, "page_end": 284, "topic_guess": "sequences_patterns", "text": "The units digit of y2 = 1.(1)\nThe units digit of y does not equal 1.(2)\n6. In sequence A, A1 = 1, A2 = 100, and the value of An is strictly\nbetween the values of An – 1 and An – 2 for all n ≥ 3. Which of the\nfollowing must be true? \nA100 < A200 < A300 < A400(A)\nA100 < A300 < A400 < A200(B)\nA200 < A400 < A300 < A100(C)\nA400 < A200 < A300 < A100(D)\nA400 < A300 < A200 < A100(E)\n7. If x is an integer, what is the remainder when x is divided by 5 ?\nx2 has a remainder of 4 when divided by 5.(1)\nx3 has a remainder of 2 when divided by 5.(2)\n8. If x and y are positive integers, what is the remainder when 5x is\ndivided by y  ?\nx is an even integer.(1)\ny = 3(2)"} +{"id": "book 2_p285_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 285, "page_end": 285, "topic_guess": "sequences_patterns", "text": "9. a, b, c, and d are positive integers. If \n has a remainder of 9 and \nhas a remainder of 10, what is the minimum possible value for bd  ?\n10. What is the sum of the numbers in the grid below?\n1 −2 −1 2 3\n3 −4 −2 3 6\n5 −6 −3 4 9\n7 −8 −4 5 12\n9 −10 −5 6 15\n11 −12 −6 7 18\n11. The sequence a1, a2, a3 … an is defined such that an = an−1 + 9 + n for\nall n > 1. If a1 = 10, what is the value of a11  ?\n12. The sequence S is defined as follows for all n ≥ 1:"} +{"id": "book 2_p286_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 286, "page_end": 286, "topic_guess": "sequences_patterns", "text": "The sum of the first 10 terms of S is\nBetween −1 and \n(A)\nBetween \n and 0(B)\nBetween 0 and \n(C)\nBetween \n and 1(D)\nGreater than 1(E)\n13. In sequence Q, the first number is 3, and each subsequent number\nin the sequence is determined by doubling the previous number\nand then adding 2. How many times does the digit 8 appear in the\nunits digit of the first 10 terms of the sequence?\n14. ♦ P♦ is defined as the product of all even integers r such that 0 < r ≤\nP. For example, ♦ 14 ♦ = 2 × 4 × 6 × 8 × 10 × 12 × 14. If ♦ K♦ is divisible\nby 411, what is the smallest possible value for K ?"} +{"id": "book 2_p287_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 287, "page_end": 287, "topic_guess": "fractions_decimals_percents", "text": "22(A)\n24(B)\n28(C)\n32(D)\n44(E)\n15. Mitchell plans to work at a day camp over the summer. Each week,\nhe will be paid according to the following schedule: At the end of\nthe first week, he will receive $1. At the end of each subsequent\nweek, he will receive $1, plus an additional amount equal to the\nsum of all payments he’s received in previous weeks. How much\nmoney will Mitchell be paid in total during the summer, if he works\nfor the entire duration of the 8-week-long camp?"} +{"id": "book 2_p288_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 288, "page_end": 288, "topic_guess": "sequences_patterns", "text": "Solutions\n1. 5: The sequence starts with the number 4, then adds 5 to each\nsubsequent term. The math plays out in this way:\nThe sequence pattern is 4 + 5(n − 1), where n is the number of the term.\nThe 75th term in the sequence is therefore 4 + 5(75 − 1) = 4 + 5(74) = 374.\nIn order to find the remainder, first find the multiple of 9 closest to 374\nbut smaller than that number. 360 is a multiple of 9, and so is 369.\nTherefore, \n has a remainder of 0, so \n has a remainder of 5.\n2. x = 5; 1 ≤ y ≤ 9: Only the integer 5 can be raised to a power to result in a\nunits digit of 5. Any power of 5 will have a units digit of 5, other than\nzero, because 50 = 1. The integer y, on the other hand, can have any\nvalue except for 0.\n3. 0: The remainder when dividing an integer by 10 always equals the units\ndigit. Ignore all but the units digits and rephrase the question: “What is\nthe units digit of 317 + 713 ?”"} +{"id": "book 2_p289_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 289, "page_end": 289, "topic_guess": "sequences_patterns", "text": "The pattern for the units digits of 3 is [3, 9, 7, 1]. Every fourth term is the\nsame. The 17th power is 1 past the end of the repeat. Since 316 ends in 1,\n317 must end in 3.\nThe pattern for the units digits of 7 is [7, 9, 3, 1]. Every fourth term is the\nsame. The 13th power is 1 past the end of the repeat. Since 712 ends in 1,\n713 must end in 7.\nThe sum of these units digits is 3 + 7 = 10. Thus, the units digit is 0.\nFor problems 4 and 5, reference the following chart:\nFor integers ending in a certain digit, the table shows the units digit\npattern. For example, integers ending in 2 have a units digit pattern of 2,\n4, 8, 6. Integers ending in 9 have a units digit pattern of 9, 1.\n1 2 3 4 5 6 7 8 9 0\ny1 1 2 3 4 5 6 7 8 9 0\ny2 1 4 9 6 5 6 9 4 1 0\ny3 1 8 7 4 5 6 3 2 9 0\ny4 1 6 1 6 5 6 1 6 1 0\n4. (B):\n(1) INSUFFICIENT: According to the Units Digit Patterns chart above,\nboth 4 and 6 yield a units digit of 6 when raised to the second power.\n(Notice that you would only need to check even numbers for this"} +{"id": "book 2_p290_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 290, "page_end": 290, "topic_guess": "sequences_patterns", "text": "statement, as odd numbers to any power cannot end in a 6, an even\nnumber.)\n(2) SUFFICIENT: Only 5 yields a units digit of 5 when raised to any power.\nSince the units digit of y + 1 is 5, the units digit of y must be 4.\nThe correct answer is (B).\n5. (C): Begin with statement (2).\n(2) INSUFFICIENT: The statement indicates only that the units digit does\nnot equal 1, but it could still be any other digit.\n(1) INSUFFICIENT: Because both 1 and 9 yield a units digit of 1 when\nraised to the second power, there are two possible values for the units\ndigit.\n(1) AND (2) SUFFICIENT: The two statements together indicate that the\nunits digit of y must be 9.\nThe correct answer is (C).\n6. (E) A400 < A300 < A200 < A100: The answer choices refer to terms that are\nvery late and very far apart in the sequence. Since there isn’t time to\ncalculate that many terms, there must be a pattern. Write down what\nyou know about the first few terms in order to identify that pattern. \nEach term in the sequence is strictly between the two terms that come\nimmediately before it. To keep track of this, draw a number line on your"} +{"id": "book 2_p291_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 291, "page_end": 291, "topic_guess": "sequences_patterns", "text": "paper. Put the third term of the sequence somewhere between the first\ntwo terms:\nContinue to add terms to the number line, making sure to put each term\nbetween the two that came before it. The first eight terms have been\nplotted on the number line shown below: \nThe odd-numbered terms of the sequence appear in increasing order, so\nthat A1 < A3 < A5 < A7, and so on. The even-numbered terms are in\ndecreasing order so that A2 > A4 > A6 > A8, etc. \nThe answer choices only refer to even-numbered terms, so the terms\nshould be in decreasing order, with A100 being the greatest and A400\nbeing the least. \nThe correct answer is (E). \n7. (B): Because the question is asking about the remainder when x is\ndivided by 5, you only need to check the units digits 0 through 4. The\npattern will recycle for the next set of 5 (5 through 9) and for every set of\n5 a er that.\n(1) INSUFFICIENT:"} +{"id": "book 2_p292_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 292, "page_end": 292, "topic_guess": "sequences_patterns", "text": "x x2\n = remainder of 4\n = remainder of?\n2 4 Yes\n = 0 remainder 2\n3 9 Yes\n = 0 remainder 3\nThere are at least two possible values for the remainder.\n(2) SUFFICIENT:\nx x3\n = remainder of 2\n = remainder of?\n3 27 Yes\n = 0 remainder 3\nThere is only one possible remainder: 3.\nIf you aren’t sure that you only need to test these five cases, try the next\nfive values for x. Note the repeat of the remainder pattern. Only every\nthird term (3, 8, 13, and so on) will be valid based on the information\nfrom statement (2) and will always have a remainder of 3.\n0 0 No (invalid)\n1 1 No (invalid)\n0 0 No (invalid)\n1 1 No (invalid)\n2 8 No (invalid)\n4 64 No (invalid)"} +{"id": "book 2_p293_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 293, "page_end": 293, "topic_guess": "sequences_patterns", "text": "The correct answer is (B).\n8. (C): 5x will always end in 5.\n(1) INSUFFICIENT: If x is even, then 5x = 25, 625, and so on. This\nstatement provides no information about y, though. For example if y = 5,\nthen \n remainder 0. If y = 4, then \n remainder 1.\n(2) INSUFFICIENT: Test some different values for x.\nx 5x Remainder of \n1 5 2\n2 25 1\n3 125 2\n4 625 1\nA er testing the first two numbers, it’s clear that statement (2) is not\nsufficient. Because you will have to test the two statements together,\ncontinue testing another couple of numbers to see whether there is a\npattern. When 5 is raised to an odd power, the remainder is 2, but when\n5 is raised to an even power, the remainder is 1."} +{"id": "book 2_p294_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 294, "page_end": 294, "topic_guess": "number_theory", "text": "(1) AND (2) SUFFICIENT: When 5even integer is divided by 3, the remainder\nis always 1.\nThe correct answer is (C).\n9. 110: The remainder must always be smaller than the divisor. Thus, b\nmust be at least 10, and d must be at least 11. Therefore, bd must be at\nleast 110. The purpose of this problem is to remind you of these\nconstraints on remainders.\n10. 63: Each column contains a series of evenly spaced integers. To avoid\ncalculation, cancel out as many values as possible before summing the\nremaining values. \nIn each row, the sum of the integers in the first two columns is −1, and\nthe sum of the integers in the third and fourth columns is 1. Therefore,\nthe sum of the integers in the first four columns in each row is −1 + 1 = 0.\nOnly the integers in the fi h column need to be considered.\nThe integers in the fi h column can be rewritten as follows:\nThe sum of the integers is therefore 3(21) = 63. \n11. 165: Write out the first few terms of the sequence to find the pattern."} +{"id": "book 2_p295_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 295, "page_end": 295, "topic_guess": "number_theory", "text": "To simplify the calculation, these terms can be rewritten as follows:\nTherefore, the nth term in the sequence can be found by adding 10 a\ntotal of n times, then adding the sum of the integers from 1 to n – 1,\ninclusive. For n = 11, this value is equal to 10(11) + (1 + 2 + 3 + ... + 10),\nwhich equals 10(11) + 55 = 165.  \n12. (B) Between \n and 0: Compute the first few elements of Sn.\nn Sn\n1     \n2"} +{"id": "book 2_p296_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 296, "page_end": 296, "topic_guess": "word_problems", "text": "n Sn\n3     \n4\n5    \nUse a number line to track the sum:\nPlace the first term, \n , on the number line. The second term is + \n ,\nso the sum will move to the right (closer to 0) on the number line. The\nthird term is \n  so the sum will move to the le , but it can’t go as\nfar as \n again, because you’re only subtracting \n this time—the\ndistance is smaller than the first \n hop that you made.\nEach subsequent hop flips back and forth between positive and\nnegative but also keeps getting smaller and smaller, so you’ll never\n“break out” of the range \n to 0.\nThe correct answer is (B)."} +{"id": "book 2_p297_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 297, "page_end": 297, "topic_guess": "sequences_patterns", "text": "13. 9: Calculate the first several terms of the sequence to find the pattern:\nThe pattern should continue, so 8 will be the units digit 9 out of the first\n10 times.\n14. (B) 24: This is a counting pattern problem. In order for ♦ K♦ to be divisible\nby 411, it must be divisible by (22)11 = 222. Thus,  ♦ K♦ must contain 22\ntwos in its prime factorization.\nIf K = 10, for example, then ♦ K♦ equals 2 × 4 × 6 × 8 × 10. The number 2\nhas one 2 in its prime factorization; 4 has two 2’s; 6 has one 2 (and a 3); 8\nhas three 2’s; 10 has one 2 (and a 5). This amounts to a total of only eight\n2’s:\nNumber 2 4 6 8 10\nPrime Factor(s) 2 2, 2 2, 3 2, 2, 2 2, 5\nTotal 2’s in PF 1 2 1 3 1\nCumulative 2’s 1 3 4 7 8"} +{"id": "book 2_p298_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 298, "page_end": 298, "topic_guess": "number_theory", "text": "Keep adding even numbers to the result until you get to 22 twos in total:\nNumber 2 4 6 8 10 12 14 16 18 20 22 24\nPrime\nFactor(s)\n2 2,\n2\n2,\n3\n2,\n2,\n2\n2,\n5\n2,\n2,\n3\n2,\n7\n2, 2,\n2, 2\n2,\n3,\n3\n2,\n2,\n5\n2,\n11\n2, 2,\n2, 3\nTotal 2’s in\nPF\n1 2 1 3 1 2 1 4 1 2 1 3\nCumulative\n2’s\n1 3 4 7 8 10 11 15 16 18 19 22\nThus, the smallest possible value for K is 24. Notice the pattern in the\nnumber of 2’s in each even number: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1 …\nThe correct answer is (B).\n15. $255 (or 28 − 1): At the end of the first week, Mitchell receives $1. At the\nend of the second week, he gets $1, plus $1 for the total he had been\npaid up to that point, for a total of $2. At the end of the third week, he\ngets $1, plus ($1 + $2), or $3, for the total he had been paid up to that\npoint, so this third week’s total is $4. Put this in a table:\nWeek # Paid This Week($) Cumulative Pay Including This Week ($)\n1 1 1"} +{"id": "book 2_p299_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 299, "page_end": 299, "topic_guess": "sequences_patterns", "text": "Week # Paid This Week($) Cumulative Pay Including This Week ($)\n2 1 + 1 = 2     1 + 2 = 3\n3 1 + 3 = 4     3 + 4 = 7\n4 1 + 7 = 8     7 + 8 = 15\n5 1 + 15 = 16 15 + 16 = 31\n6 1 + 31 = 32 31 + 32 = 63\n7 1 + 63 = 64 63 + 64 = 127\n8 1 + 127 = 128 127 + 128 = 255\nThis calculation is not so bad, but you may notice that this payment\nschedule is a geometric sequence, 2n − 1, where n is the number of the\nweek in which Mitchell is being paid. Summing that sequence is\nequivalent to 2t − 1, where t is the total number of weeks. In other\nwords, the cumulative pay is one less than the next power of 2.\nThe correct answer is 28 − 1 = $255."} +{"id": "book 2_p300_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 300, "page_end": 300, "topic_guess": "algebra", "text": "CHAPTER 6\nCommon Terms & Quadratic\nTemplates"} +{"id": "book 2_p301_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 301, "page_end": 301, "topic_guess": "algebra", "text": "In This Chapter...\nCommon Terms\nQuadratic Templates\nQuadratic Templates in Disguise"} +{"id": "book 2_p302_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 302, "page_end": 302, "topic_guess": "algebra", "text": "Chapter 6\nCommon Terms & Quadratic\nTemplates\nYou are probably already familiar with the mechanics of algebraic\nmanipulations—what is allowed and what is not:\nBut of all the steps you could take, how do you decide which steps you\nshould take?\nTwo indicators can o en help you on the GMAT:\nYou can substitute one expression for another if they are equal.\nYou can add some number to one side of an equation as long as you do\nthe same on the other side of the equation.\nYou can cross-multiply to simplify equations with fractions on each side\nof the equals sign. And so on.\nCommon Terms1.\nQuadratic Templates2."} +{"id": "book 2_p303_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 303, "page_end": 303, "topic_guess": "algebra", "text": "Common Terms\nIf you spot common terms, you can o en spot the path all the way to the\nsolution. Common terms appear on the GMAT in three typical ways.\n1. ALGEBRA\nLook for terms that appear in the same form more than once. Those\nrecurring expressions might also appear in slightly modified form such as\nreciprocal, negative, or raised to a power:\n is the reciprocal of \n .\n is negative\n.\nIs \n ? where\n and 3y appear both\nsquared and multiplied\ntogether.\nThis expression is of the form:\n(More on Quadratic Templates\nlater in this chapter)"} +{"id": "book 2_p304_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 304, "page_end": 304, "topic_guess": "number_theory", "text": "Try-It #6-1\nIf y ≠ 3, simplify as much as possible: \nSpot the common term (3 − y). Note that (−3 + y) is −(3 − y), or −1 × (3 − y).\nFactor out the common term (3 − y) and cancel:\n.\nBy the way, the condition that y could not equal 3 just prevented you from\ndividing by 0 and ending up with an undefined number.\n2. EXPONENTS\nExponents can be manipulated when either bases or exponents are\ncommon. Also look for bases that have common factors, such as 3 and 12\n(common factor of 3). You can o en create a common base. For example:\n(16x)(42) =\n256\n4 is a factor of 4, 16, and 256.\nSimilarly, 4, 16, and 256 are all powers\nof 4."} +{"id": "book 2_p305_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 305, "page_end": 305, "topic_guess": "algebra", "text": "Try-It #6-2\nIf 3x + 243 = 2(3x), what is the value of x ?\nNote the common term 3x, and note the fact that 243 = 35:\n3. FACTORS AND MULTIPLES\nWhen many terms share a factor, pull that shared factor out to the side.\nThese can appear in algebraic or numerical expressions:\n→ get common denominators, then cross\nthem all off →\nx18 + 2x16 + x14 → x14 is a factor of each term → x14(x4 + 2x2 + 1) = x14(x2 +\n1)2"} +{"id": "book 2_p306_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 306, "page_end": 306, "topic_guess": "number_theory", "text": "Factorials are particularly noteworthy, as they o en have an abundance of\nshared factors. For any integer n, the factorial n! is calculated as follows: n!\n= n(n − 1)(n − 2)(n − 3) . . . 1. Thus, all the terms in 4! = (4)(3)(2)(1) are also\ncommon factors of 6! = (6)(5)(4)(3)(2)(1) = (6)(5)(4!).\nMore generally, factorials are “super multiples.” Without ever computing\ntheir precise value, you can tell that they’re divisible by all sorts of\nnumbers. For example:\nIf x is an integer\nbetween \n7! + 2 and 7! + 4, \ninclusive, is x prime?\nx is one of the following\nintegers:\n7! + 2 = (7)(6)(5)(4)(3)(2)(1) + 2\n7! + 3 = (7)(6)(5)(4)(3)(2)(1) + 3\n7! + 4 = (7)(6)(5)(4)(3)(2)(1) + 4\nx has one of the following\nfactors, if x is:\n7! + 2 = 2 × Integer\n7! + 3 = 3 × Integer\n7! + 4 = 4 × Integer…so x is\nnot prime!\nSometimes a common factor is just a random number buried inside a\ncouple of larger numbers. Find it and pull it out:\n10’s in the numerator\nline up with 5’s in the\ncorresponding digit\nplace of the\ndenominator:"} +{"id": "book 2_p307_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 307, "page_end": 307, "topic_guess": "number_theory", "text": "Try-It #6-3\nIf n is a positive integer and\n is a positive\ninteger, what is the value of n ?\nWhat would need to be true in order for the square root to be a positive\ninteger? The number under the square root symbol would have to be a\nperfect square. Rearrange the expression to determine whether there are\nany restrictions that could help narrow down the possibilities before going\nto the statements. Try to break the numbers down into primes to locate\nand pull out any existing perfect squares:\nn is prime.(1)\nn < 3(2)"} +{"id": "book 2_p308_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 308, "page_end": 308, "topic_guess": "general", "text": "Pull out common terms:\nIt turns out that you can also pull out the term 7n − 1. The term 7n + 1 = (7n − 1)\n(72):\nThe first two terms are both perfect squares and can be pulled out of the\nsquare root sign:\nWhat would need to be true in order for 7n − 1 to be pulled out of the square\nroot sign? It would also have to be a perfect square, so n – 1 must be even\nand n itself must be odd."} +{"id": "book 2_p309_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 309, "page_end": 309, "topic_guess": "general", "text": "The question is: “If n is odd, what is the value of n?”\n(1) INSUFFICIENT: This statement allows multiple possible odd values of n.\n(2) SUFFICIENT: The question stem indicates that n is a positive integer,\nand this statement specifies that n < 3. The only odd, positive integer less\nthan 3 is the number 1.\nThe correct answer is (B)."} +{"id": "book 2_p310_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 310, "page_end": 310, "topic_guess": "algebra", "text": "Quadratic Templates\nOn the GMAT, quadratic expressions take three common forms called the\nQuadratic Templates. Memorize these templates, and get comfortable\ntransforming back and forth between factored and distributed form:\n  Factored\n Distributed\nSquare of a Sum (a + b)2 = a2 + 2ab + b2\nSquare of a Difference (a − b)2 = a2 − 2ab + b2\nDifference of Two Squares (a + b) (a − b) = a2 − b2\nQUICK MANIPULATION\nExpressions with both squared and non-squared common terms should\nmake you suspect that you are looking at a Quadratic Template.\nTry-It #6-4\nFactor \n ."} +{"id": "book 2_p311_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 311, "page_end": 311, "topic_guess": "algebra", "text": "This problem requires you to manipulate a rather complicated expression.\nHowever, by using the common terms, you can put the problem in the\nmore basic template form to solve:\nOnce you are comfortable with Quadratic Templates, you can manipulate\neven complicated expressions quickly, as in the middle box above. Until\nthen, write down the templates and the substitution of the common terms,\nas in the box on the right.\nThe very same problem could have been presented in disguise:\nFactor \n .\nThe common terms are slightly harder to spot in this form. In such a case,\nstart with the squared terms, \n and 25y2. Then, try to untangle their\nsquare roots, \n and 5y, from the remaining term. The factored form is:"} +{"id": "book 2_p312_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 312, "page_end": 312, "topic_guess": "algebra", "text": "Consider all three of the common Quadratic Templates before deciding\nwhich one or ones are most convenient to use.\nTHE MIDDLE TERM: 2AB\nThe square of a sum and square of a difference templates have something\nin common: the middle term is ±2ab. The only difference is the sign of that\nmiddle term.\nWhen you add these two templates, the middle terms cancel, leaving the\nend terms:\n  Factored\n Distributed\nSquare of a Sum (a + b)2 = a2 + 2ab + b2\n+ Square of a Difference (a − b)2 = a2 − 2ab + b2\n\nAddition: (a + b)2 + (a − b)2 = 2a2 + 0 + 2b2\n2(a2 + b2)"} +{"id": "book 2_p313_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 313, "page_end": 313, "topic_guess": "algebra", "text": "In contrast, when you subtract these two templates, the end terms cancel,\nleaving the middle term:\n  Factored\n Distributed\nSquare of a Sum (a + b)2 = a2 + 2ab + b2\n+ Square of a Difference (a − b)2 = a2 − 2ab + b2\n\nSubtraction: (a + b)2 − (a − b)2 = 0 + 4ab + 0\nThis is handy for simplification. Also, whenever you see the sum of two\nsquares (a2 + b2), which is not itself a Quadratic Template, remember that it\ncan be derived from this sum of two templates.\nTry It #6-5\nWhat is the sum of 9,9992 and 10,0012?\n99,980,001(A)\n199,999,998(B)\n200,000,002(C)\n399,999,996(D)\n400,000,004(E)"} +{"id": "book 2_p314_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 314, "page_end": 314, "topic_guess": "algebra", "text": "These numbers all look seriously cumbersome. If only they had given the\neasier number of 10,000 instead . . .\nWhen wishful thinking pops up, try to use it to make the problem easier. If\nyou changed both of these numbers to the form 10,000 plus or minus a\nnumber, what would that be?\nThese are the Quadratic Templates! In this case, a = 10,000 and b = 1. If\nyou’ve memorized the sum term, plug these in:\nNow, notice something: the part in the parentheses is going to have a units\ndigit of 1. Multiply the number by 2, and the end result will have a units\ndigit of 2. Only answer choice (C) fits.\nNot sure about that? Go ahead and do the math:\nThe correct answer is (C).\nEven if you don’t memorize the sum term of the Quadratic Templates, the\nmath is still far easier to do in this rewritten form:"} +{"id": "book 2_p316_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 316, "page_end": 316, "topic_guess": "algebra", "text": "Quadratic Templates in Disguise\nQuadratic Templates can be disguised in arithmetic computations.\nTry-It #6-6\nWhat is 198 × 202 ?\nYou can round each number and quickly estimate the result to be about\n2002. Or, you could laboriously multiply two 3-digit numbers by hand to get\nan exact result. But if you need an exact result quickly, you can use a\nQuadratic Template, as shown in the previous section. You just need to\nturn 198 into (200 − 2) and 202 into (200 + 2) as shown here:\nAnother place to hide a Quadratic Template is in an advanced right-triangle\nproblem:"} +{"id": "book 2_p317_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 317, "page_end": 317, "topic_guess": "geometry", "text": "You know that \n and d2 = b2 + h2 (by the Pythagorean\ntheorem). Do the common terms b2, h2, and bh look familiar? Use the\nSquare of a Sum template:\nLikewise, there is a similar relationship based on the Square of a Difference\ntemplate:\nAn advanced GMAT problem can draw on these complicated relationships.\nFor instance, you can compute the area of a right triangle directly from the\nsum of the shorter sides and the hypotenuse:"} +{"id": "book 2_p318_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 318, "page_end": 318, "topic_guess": "algebra", "text": "You should absolutely not memorize these particular formulas. Rather, be\nable to recognize when the GMAT is indirectly testing these generic\nQuadratic Templates."} +{"id": "book 2_p319_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 319, "page_end": 319, "topic_guess": "fractions_decimals_percents", "text": "Problem Set\nFor the following problems, use the Pattern Recognition techniques\ndiscussed in this chapter to solve.\nFor problems 2–4, if x < –1, which of the following inequalities must be\ntrue?\n1. If xy > 0, is \n(1)\nx > 2y(2)\n2. Is x4 > x2 ?\n3. Is x3 + x4 > x3 + x2 ?\n4. Is x6 − x7 > x5 − x6 ?\n5. If, \n and |x| ≠ |y|, what is the ratio of x to y ?"} +{"id": "book 2_p320_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 320, "page_end": 320, "topic_guess": "number_theory", "text": "4 : 1(A)\n3 : 1(B)\n2 : 1(C)\n3 : 2(D)\n1 : 3(E)\n6. If \n , what is the value of \n7. If n is an integer and (−3)4n = 37n – 3, then n = ?\n8. If (ax + by)2 = (bx + ay)2, what is the value of a – b ?\nx2 > y2(1)\na and b are positive integers.(2)\n9. If x and k are both integers, x > k, and x–k = 625, what is x ?\n|k| is a prime number.(1)\nx + k > 20(2)"} +{"id": "book 2_p321_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 321, "page_end": 321, "topic_guess": "algebra", "text": "Distribute the expression in problems 10–14 without FOILing (doing the\nmath the long way). Use the Quadratic Templates.\nFactor problems 15–19 according to the Quadratic Templates.\n10. \n11. \n12. \n13. \n14. \n15. \n16. 4 + 4a + a2"} +{"id": "book 2_p322_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 322, "page_end": 322, "topic_guess": "algebra", "text": "For problems 20–23, simplify the expressions completely.\n17. 81 − x4\n18. \n19. 4x2 − 12xy + 9y2\n20. \n21. \n22. (111)(89)\n23. 3502 − 3202"} +{"id": "book 2_p323_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 323, "page_end": 323, "topic_guess": "geometry", "text": "24. \nIn the right triangle below, side a is 7 inches longer than side b. If\nthe area of the triangle is 30 inches2, what is the length of\nhypotenuse c ?"} +{"id": "book 2_p324_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 324, "page_end": 324, "topic_guess": "algebra", "text": "Solutions\n1. (C): From xy > 0, you know that neither x nor y equals 0, and they must\nhave the same sign (++ or − −). Combine the exponent in the question to\nget \nWhat is the significance of the inequality in the question stem? If the\nexponent equals 2, then 52 = 25. In order to be greater than 25, the\nexponent has to be greater than 2. The question can be rephrased as “Is\n?”\nYou may want to start with statement (2).\n(2) INSUFFICIENT: If y is positive, then \n . If, on the other hand, y is\nnegative, then \n .\n(1) INSUFFICIENT: Simplify the equation:"} +{"id": "book 2_p325_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 325, "page_end": 325, "topic_guess": "algebra", "text": "Therefore, y3 must equal 8, so y = 2. This provides no information about\nx, however.\n(1) AND (2) SUFFICIENT: If y = 2, then you can plug into statement (2) to\nfind the range of values for x. Since x > 2y and y is positive, you can\ndivide the expression by y, giving \n .\nThe correct answer is (C).\nFor questions 2–4: since x ≠ 0, divide by the common terms, making sure\nto flip the inequality sign if the common term is negative.\n2. \nTRUE:\n ?\nTrue\nNote: x2 is positive.\nDivide both sides by x2, leaving the sign as it is.\nThe square of a number smaller than −1 will be greater than\npositive 1.\n3. \nTRUE:\n\nTrue\nNote: x2 is positive.\nDivide both sides by x2, leaving the sign as it is.\nSubtract x from both sides.\n(Or, you might have subtracted x3 immediately.)\nTRUE:\n\nTrue\nNote: x5 is negative.\nDivide both sides by x5, flipping the inequality sign.\nGroup like terms."} +{"id": "book 2_p326_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 326, "page_end": 326, "topic_guess": "algebra", "text": "4. \n5. (C) 2 : 1: Try to find like terms in order to simplify the le -hand side of\nthe equation:\nThe ratio of x to y is 2 : 1.\nThe correct answer is (C).\n6. \n7. n = 1: Since n is an integer, 4n is even. An even exponent “hides the sign”\nof the base, so you can treat the (−3) base as a (3):"} +{"id": "book 2_p327_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 327, "page_end": 327, "topic_guess": "algebra", "text": "8. (C): Begin by simplifying the equation in the question stem, making it\nlook as similar as possible to the question itself:\nDo not divide by x2 – y2, since it may equal 0. Instead, subtract b2(x2 – y2)\nfrom both sides of the equation, then factor:\nSo either a2 – b2 = 0 (in which case a2 = b2), or x2 – y2 = 0 (in which case x2\n= y2). One or both of these must be true."} +{"id": "book 2_p328_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 328, "page_end": 328, "topic_guess": "number_theory", "text": "(1) INSUFFICIENT: If x2 > y2, then x2 cannot equal y2. Therefore, a2 = b2.\nCase 1: It is possible that a and b are equal, in which case a – b = 0, so\nthe answer is 0.\nCase 2: It is also possible that a and b are equal in absolute value but\nhave opposite signs. For example, a = –2 and b = 2. In this case, the\nanswer is a – b = –4.\nMore than one answer is possible, so this statement is not sufficient.\n(2) INSUFFICIENT: This statement only says that a and b are positive\nintegers, but does not give further indication of what they could be.\nThere are many possible answers for the value of a – b, so this statement\nis insufficient.\n(1) AND (2) SUFFICIENT: According to statement (1), a2 = b2. Thus, either\na = b or a = –b. According to statement (2), a and b are both positive, so it\nis not possible that a = –b. Therefore, a = b, and the answer to the\nquestion is a – b = 0. The statements together are sufficient.\nThe correct answer is (C). \n9. (A): The fact that x and k are both integers (and that x > k) significantly\nlimits the possible values for x and k. The possible pairings are as\nfollows:\nx k"} +{"id": "book 2_p329_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 329, "page_end": 329, "topic_guess": "number_theory", "text": "x k\n5 −4\n25 −2\n625 −1\n(1) SUFFICIENT: If the absolute value of k is prime, then only the second\npossibility works: x = 25 and k = −2.\n(2) INSUFFICIENT: The second and third possibilities both make this\nstatement true, so it isn’t possible to determine a single value for x.\n10. \n11. \n12. \n13. 4:\n14."} +{"id": "book 2_p330_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 330, "page_end": 330, "topic_guess": "general", "text": "15. \n16. \n17. \n18. \n19. \n20. \n21. 4:\n22. 9,879:\n23. 20,100:\n24. (a − b) = 7 and \n : From the Pythagorean theorem, a2 + b2 =\nc2.\nUse the Square of a Difference template:"} +{"id": "book 2_p331_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 331, "page_end": 331, "topic_guess": "general", "text": "Plug in the values:"} +{"id": "book 2_p332_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 332, "page_end": 332, "topic_guess": "general", "text": "CHAPTER 7\nVisual Solutions"} +{"id": "book 2_p333_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 333, "page_end": 333, "topic_guess": "statistics", "text": "In This Chapter...\nRepresenting Objects with Pictures\nRubber Band Geometry\nBaseline Calculations for Averages\nNumber Line Techniques for Statistics Problems"} +{"id": "book 2_p334_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 334, "page_end": 334, "topic_guess": "word_problems", "text": "Chapter 7\nVisual Solutions\nVisual interpretations—good pictures, essentially—can help you solve\ncertain types of GMAT problems. This chapter highlights some of these\ntypes of problems and demonstrates how you can use visual techniques to\nsolve these problems more confidently, accurately, and quickly.\nMany problems discussed in this chapter can be solved with other\ntechniques. Still, visual thinking is a powerful tool. It can expand your\ncomprehension of a topic. It may enable you to solve particular problem\ntypes more easily or “break through” on a difficult problem. In fact,\nvisualization is the only realistic way to approach certain problems. So it’s\nworth trying your hand with visual approaches.\nIn this chapter, we will discuss the following Visual Solution techniques:\nRepresenting Objects with Pictures: Many Word Problems and Geometry\nproblems do not provide a diagram alongside the problem. Drawing a\ngood picture will make the problem-solving process easier and less\nerror-prone.\nRubber Band Geometry: For Geometry questions involving both\nconstraints and flexibility (especially in Data Sufficiency), drawing"} +{"id": "book 2_p335_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 335, "page_end": 335, "topic_guess": "statistics", "text": "different “rubber band” scenarios according to those constraints and\nfreedoms can o en help you solve the problem without doing any\ncomputation.\nBaseline Calculations for Averages: Visual techniques can help you\ncompute averages (both basic and weighted) and can also foster a\nbetter understanding of those calculations.\nNumber Line Techniques for Statistics Problems: You can solve a variety\nof common problems involving statistics by using a number line to\nvisualize and manipulate the problem."} +{"id": "book 2_p336_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 336, "page_end": 336, "topic_guess": "word_problems", "text": "Representing Objects with Pictures\nFor Word Problems that describe a physical object, or for Geometry\nproblems that do not give a diagram alongside the problem, drawing a\npicture is o en the best approach. Sometimes it’s the only viable approach!\nEven if you are good at visualizing objects in your head, draw the picture\nanyway. It’s just too easy to make a mistake on many of these questions.\nTry-It #7-1\nA rectangular wooden dowel measures 4 inches by 1 inch by 1 inch.\nIf the dowel is painted on all surfaces and then cut into \n-inch\ncubes, what fraction of the resulting cube faces are painted?\nIf you draw a picture, this problem becomes a matter of counting:\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p337_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 337, "page_end": 337, "topic_guess": "fractions_decimals_percents", "text": "Total cubes = (4 inches × 2 cubes per inch) × (1 × 2) × (1 × 2) = 32 cubes\nTotal cube faces = 32 cubes × 6 faces per cube = 192 faces total\nNow, consider the faces that were painted on the front and back of the\ndowel, the top and bottom of the dowel, and the ends of the dowel. In the\ndiagram above, you can see 16 faces on the front, 16 faces on the top, and 4\nfaces on the end shown. Of course, there are other sides: the back, the\nbottom, and the other end. Now you can find the number of painted cubes:\nTherefore, the fraction of faces that are painted is \n .\nThe correct answer is (B).\nNotice that there is no shortcut to solving this kind of problem, so don’t\nwaste time looking for one—just draw the diagram and count.\nEven if you can easily picture 3-D shapes and objects in your head, it is still better to draw a\npicture on your scrap paper."} +{"id": "book 2_p338_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 338, "page_end": 338, "topic_guess": "algebra", "text": "The test will usually include among the incorrect answers numbers that you might get by\nlosing track of your progress as you process the object in your mind.\nThis kind of process can also help you with questions that deal with the\nrelative size of different objects.\nTry-It #7-2\nBucket A has twice the capacity of Bucket B, and Bucket A has \n the\ncapacity of Bucket C. Bucket B is full of water and Bucket C is half\nfull of water. When the water from Bucket B is poured into Bucket C,\nBucket C will be filled to what fraction of its capacity?\nYou could attempt to solve this problem algebraically, but the equations get\nmessy very quickly. Instead, try drawing buckets A, B, and C in correct\nproportion to one another. Then think through the problem:"} +{"id": "book 2_p339_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 339, "page_end": 339, "topic_guess": "algebra", "text": "Algebra Picture\nThe algebra and the picture say the same thing, but the picture has several\nadvantages:\nBased on this picture, you might pick a capacity of 1 for Bucket B, yielding a\ncapacity of 2 for Bucket A and 6 for Bucket C. Bucket B would contain 1 unit\nof water and Bucket C, 3 units. When the contents of B are poured into C,\nBucket C would then be \n full.\nIt’s much easier to comprehend at a glance.\nIt’s harder to mistake relative sizes (e.g., accidentally thinking A is\nsmallest).\nYou can easily represent both total capacity and amount of water visually.\nIt prompts you to pursue the smartest, easiest solution: picking numbers\nfor the capacities of the buckets."} +{"id": "book 2_p340_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 340, "page_end": 340, "topic_guess": "general", "text": "Notice that the buckets are not labeled in alphabetical order, even though\nthat would be easy to incorrectly assume. The GMAT frequently adds little\nlayers of disguise and complexity such as this to induce you to make a\nmistake. By drawing the buckets carefully, you minimize the chance that you\nwill fall into a trap on a problem such as this one."} +{"id": "book 2_p341_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 341, "page_end": 341, "topic_guess": "geometry", "text": "Rubber Band Geometry\nMany Geometry problems—particularly of the Data Sufficiency (DS) variety\n—describe objects for which only partial information is known. We call\nthese questions Rubber Band Geometry problems, because they\nsimultaneously involve constraints and flexibility. Some parts of the\ndiagram can stretch like a rubber band as you open or close angles.\nYour job is to figure out what specifics in the problem are constrained and\nwhat specifics are flexible.\nFor example, if a problem specifies that a line has a slope of 2, it will be\nsteep and upward-sloping. In fact, it will always “rise” 2 units for every unit\nof “run.” However, you don’t know where the line will appear. The line is\nconstrained in its slope, but it is flexible in that it can be moved up or down,\nright or le . You can draw many different lines with a slope of 2 (these lines\nwill all be parallel, of course).\nIf, however, the problem only specifies that a line must go through the point\n(4, 0) in the coordinate plane, then the line is “fixed” at that point. However,\nthe slope of the line would now be flexible. You could draw many different\nlines with different slopes that run through that point.\nIf you knew both of these specifications—that the line must have a slope of\n2 and must run through the point (4, 0)—then you would be able to"} +{"id": "book 2_p342_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 342, "page_end": 342, "topic_guess": "algebra", "text": "calculate the exact line that is being described. The slope of the line and a\npoint that the line goes through specify the line precisely—there is no\nremaining flexibility for either the slope or the location of the line. (Note\nthat in this example, the line is described by the equation y = 2x − 8.)\nGiven these specifications, every other feature of this line is also known: its\nx-intercept, its y-intercept, whether it goes through some fixed point, which\nquadrants it crosses, etc. In a DS problem, you could answer any such\nquestions about this line without actually calculating the answer.\nTherefore for these types of problems, your goal is to figure out what\ncombination of information “cements” the problem in place—in other\nwords, what combination of information removes all of the remaining\nflexibility. No flexibility means sufficiency. And by using rubber band\ngeometry thinking, you can o en do this without using any calculation or\nalgebra at all.\nFor each piece of information that you’re given in this type of problem,\nthink about what is fixed and what is flexible. Try to draw multiple versions\nof each object (if possible), testing the boundaries of this flexibility. The\nfollowing everyday objects may be useful as analogies in your thinking:\nRubber band: Determines a straight line segment. Can stretch\nbetween any two points.\nDrinking straw: Determines a straight line segment, but with fixed\nlength.\nThumbtack: Fixes a point, but can allow rotations through that point\nin many cases.\nWedge: Fixes an angle."} +{"id": "book 2_p343_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 343, "page_end": 343, "topic_guess": "general", "text": "Here are some common examples of how these objects can be used to help\nyou think through these problems:\nConstrained Flexible Analogous Object(s) Mental\nPicture/Simplified\nSketch\nA line passes\nthrough a\nspecified\npoint.\nSlope of\nthe line\nDrinking straw = line \nThumbtack = point\nLine free to spin about a\npoint.\nTwo lines\nintersect at a\nspecified\npoint.\nSlope of\nthe lines\nDrinking straw = line \nThumbtack = point\nBoth lines free to spin\nabout a point. The angle\nis free to change.\nA line passes\nthrough two\npoints.\nNothing\nflexible\nDrinking straw = line \nThumbtacks = points\nTwo points pin down a\nline—no flexibility."} +{"id": "book 2_p344_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 344, "page_end": 344, "topic_guess": "word_problems", "text": "Constrained Flexible Analogous Object(s) Mental\nPicture/Simplified\nSketch\nSpecified\ndistance\nbetween two\npoints\nAbsolute\nor\nrelative\nlocation\nof the\npoints\nThumbtacks = points \nDrinking straw = \ndistance between points\nFix one point\ntemporarily. Line (straw)\nfree to spin about one\npoint, tracing the circle\nof possible locations of\nthe other point.\nSlope of a line Location\nof the\nline, or\npoints\nthe line\nmay pass\nthrough\nDrinking straw = line\nLine is free to “float\naround” but not rotate.\nPoints are on\na line (either\nin the\ncoordinate\nplane or on a\nbasic number\nline).\nDistance\nbetween\nthe\npoints\nRubber band = stretched between\npoints\n “Stretchy” distance\nbetween points"} +{"id": "book 2_p345_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 345, "page_end": 345, "topic_guess": "geometry", "text": "Constrained Flexible Analogous Object(s) Mental\nPicture/Simplified\nSketch\nPoints on a\nline are a\nspecified\ndistance\napart.\nOrder of\nthe\npoints\n(le -to-\nmiddle-\nto-right)\nDrinking straws = fixed lengths\nbetween points\nManipulation will be determined by\nother stated constraints, but\nthinking of the points as the\nendpoints of rigid straws will ensure\nthat you do not forget the distance\nconstraint.\nLines could be laid end-\nto-end, overlapping,\nseparated, and flipped\nright-to-le .\nTriangle with\na fixed area\nand a fixed\nbase (and\ntherefore a\nfixed height)\nPosition\nof the\nthird\nvertex\nalong a\nline\nparallel\nto the\nbase\nStraw = fixed base \nRubber bands = other sides of the\ntriangle \nThumbtacks = endpoints of the base\nThis is not intended to be an exhaustive list. The idea is to show you a new\nway of thinking through some difficult Geometry problems."} +{"id": "book 2_p346_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 346, "page_end": 346, "topic_guess": "geometry", "text": "Is B inside the circle?\nTry-It #7-3\nA circle in a coordinate plane has a center at point A and a diameter\nof 6. If points B and C also lie in the same coordinate plane, is point\nB inside the circle?\nThe exact locations of points A, B,\nand C do not matter—only the\nrelative locations of the points\nmatter. Therefore, you can\narbitrarily assign point A to a\nspecific location (when possible,\nchoose the origin of the coordinate\nplane) and draw a circle with a\nradius of 3 units around it.\n(1) INSUFFICIENT: Statement (1)\ndoes not indicate anything about\npoint B, so it is not sufficient.\nHowever, it does indicate that A and C are 2 units apart, so statement (1)\nenables you to place point C anywhere along the gray circle.\n(2) INSUFFICIENT: Statement (2) does not indicate anything about point B\nrelative to point A, so it is not sufficient. However, it does constrain point B\nto be exactly 2 units away from wherever point C is. You could imagine\npoint C at the center of a circle of size 2, with point B somewhere on the\nThe distance between point A and point C equals 2.(1)\nThe distance between point B and point C equals 2.(2)"} +{"id": "book 2_p347_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 347, "page_end": 347, "topic_guess": "geometry", "text": "(1) The distance between A and C is\n2.\n(2) The distance between B and C is\n2.\ncircle around it (and this circle could\npick up and move anywhere on the\ncoordinate plane).\n(1) AND (2) INSUFFICIENT: Finally,\ncombine these two statements to\nsee that depending on where point\nC is drawn, point B may be inside\nthe dotted circle, and it may not be.\nThe correct answer is (E).\nAs long as you can achieve a visual\nproof of the answer, you don’t need\nto prove it algebraically.\nThat’s what rubber band geometry\nis all about: testing scenarios for\nGeometry problems without the\nneed to plug in numbers or use\nalgebra. All you need is a visual\nenvironment that can be\nmanipulated—one that preserves all\nkey constraints and freedoms in the\nproblem and allows you to see and\ntest them."} +{"id": "book 2_p348_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 348, "page_end": 348, "topic_guess": "statistics", "text": "Baseline Calculations for Averages\nBASIC AVERAGES\nTry-It #7-4\nWhat is the average (arithmetic mean) of 387, 388, and 389 ?\nWithout even calculating, you may be able to see that the average is 388. How did you\narrive at that answer?\nIt’s very unlikely that you calculated the average the classical way:\nYou probably noticed that the numbers are very close together and evenly spaced: 387 is\n1 less than 388, and 389 is 1 greater than 388. Thus, the average must be 388—right in the\nmiddle.\nWhether you realize it or not, you’re using a relatively advanced technique to solve this\nproblem: a baseline calculation. The baseline in this case is 388—the middle number.\nThis concept can be applied to more difficult calculations of averages, making the\ncalculation process much easier.\nTry-It #7-5\nA consumer finds that five bags of popcorn contain 257, 261, 273, 280, and 259\ncorn kernels per bag, respectively. What is the average (arithmetic mean)\nnumber of corn kernels per bag of popcorn?"} +{"id": "book 2_p349_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 349, "page_end": 349, "topic_guess": "statistics", "text": "First, note that all of the bags have at least 257 kernels, so the average must be greater\nthan 257. How much greater than 257? First, consider how much each term differs from\n257. Represent every number with a column rising above the baseline value (in this\nexample, 257). The biggest numbers rise the highest; a number equal to the baseline has\nno height. The height of the column thus represents the difference between the number\nand the baseline value:\nCalculate the sum of the differences: 0 + 4 + 16 + 23 + 2 = 45.\nDivide by the number of terms: the average difference is 45 ÷ 5 = 9.\nTherefore, the average number of kernels per bag equals baseline + average difference:\n257 + 9 = 266.\nSimply put, a baseline picture is a column chart. The columns don’t show the actual\nvalue of any number—rather, they show the difference between the baseline and the\nnumber.\n!\nThe baseline can be any convenient number. Consider the following when choosing a baseline:\nThe smallest term in the set\nThe largest term in the set\nThe median term in the set\nA round number near the range of values"} +{"id": "book 2_p350_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 350, "page_end": 350, "topic_guess": "statistics", "text": "For sets with apparent symmetry, choosing a baseline in the middle is a good way not\nonly to confirm the symmetry, but also to compute the average. In this scenario,\nrepresent numbers lower than the baseline with columns that drop below the baseline.\nAs before, the size of the column represents the difference between the number and the\nbaseline.\nUse trial and error to pick a possible average baseline, then adjust the drawing and\ncalculations if necessary.\nTry-It #7-6\nIf a small business paid quarterly taxes last year of $10,079, $10,121, $10,112,\nand $10,088, what was the average (arithmetic mean) quarterly tax payment\nlast year?\nIn this case, some of the numbers are below $10,100 and others are above $10,100, so\n$10,100 is a natural first guess:\nIf the baseline is the average, then the sum of the differences from the baseline will be\nzero. Since the differences from the baseline do in fact sum to zero, $10,100 is indeed the\naverage of this set.\nWEIGHTED AVERAGES"} +{"id": "book 2_p351_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 351, "page_end": 351, "topic_guess": "statistics", "text": "Some sets may have many terms, but each of those terms has one of only two possible\nvalues. Rather than add each individual term together, simplify the calculations by using\na weighted average calculation.\nFor weighted averages, use visualization to advance your understanding of the math.\nTwo real-life analogies can make it easier to remember how the relative weights of high\nand low values determine where the weighted average falls.\nTeeter-Totter Hanging Scale\nThese pictures represent the values in a set as horizontal positions—le -to-right, as on a\nnumber line, not as vertical columns. (It may help to imagine the balance beam or lever\nmarked off in equal units like a number line.) Each pin or weight corresponds to the\npresence of a value in a set. The X in each picture marks the equilibrium point—in other\nwords, the weighted average.\nBoth visual interpretations regard weighted averages as a kind of balancing act: the\nweighted average will be closer to the end of the range that has more weight. In the\npictures, there are more instances of the le -hand side number in the set, so the X is\nrelatively closer to the le -hand side. You can think of this point as the point where the\nweight would be “balanced” between the two sets.\nTry-It #7-7\nA convenience store stocks soda in 12-ounce and 24-ounce bottles. If the\naverage capacity of all the bottles in the store is 22 ounces, then what fraction of\nthe bottles in the store are 12 ounces?"} +{"id": "book 2_p352_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 352, "page_end": 352, "topic_guess": "statistics", "text": "Note that 22 is much closer to 24 than to 12. This implies that there will be many more\n24-ounce bottles than 12-ounce bottles. Because the question asks about 12-ounce\nbottles, you could strategically eliminate any answer greater than or equal to \n .\nUse your understanding of weighted averages as a balancing act to work backwards from\nthe weighted average to the ratio of high-to-low terms:\nThe total range between the high and low values (24 and 12) is 12 units. Mark off the\ndistance from each end to the average of 22. Because the weighted average is closer to\n24, that side of the teeter-totter is assigned the greater weight of 10 out of 12. The other\nside, 12, is assigned the smaller weight of 2 out of 12.\nTherefore, the 24-ounce bottles constitute \n of the total number of bottles. The\n12-ounce bottles constitute \n of the total number of bottles.\nWhen using this technique, it is important to remember that the weighted average is\ncloser to (i.e., fewer units away from) the side that has greater weight, so that side should\nalways be assigned the higher fraction. It is easy to reverse this logic accidentally when\nsolving a weighted average problem with this technique, so be very careful! Just\nremember: the weighted average point will be closer to the side with more weight.\nTry-It #7-8\nDarla decides to mix lemonade with limeade to make a new drink called\ncitrusade. The lemonade is 50% water, 30% lemon juice, and 20% sugar. The\nlimeade is 40% water, 28% lime juice, and 32% sugar. If the citrusade is more\nthan 45% water and more than 24% sugar, which of the following could be the\nratio of lemonade to limeade?\n3 : 1(A)"} +{"id": "book 2_p353_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 353, "page_end": 353, "topic_guess": "word_problems", "text": "The citrusade is more than 45% water. The lemonade is 50% water and the limeade is\n40% water, so the lemonade must be more heavily weighted:\nTherefore, the lemonade must make up more than 50% of the mixture. Eliminate\nanswers (D) and (E).\nNext, the citrusade is more than 24% sugar. The lemonade is 20% sugar and the limeade\nis 32% sugar, so what else can you figure out about the relative weighting?\nSince lemonade is more heavily weighted (the mixture is to the right of 50/50), the\nweighting of the mixture is somewhere between 24% and the 50/50 weighting of the\nsugar, 26%. What weighting does the 24% figure represent?\nThe weighting is at most \n lemonade, or \n . The weighting, then, must be between\n and \n lemonade and the rest limeade.\n7 : 3(B)\n3 : 2(C)\n4 : 5(D)\n3 : 4(E)"} +{"id": "book 2_p354_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 354, "page_end": 354, "topic_guess": "statistics", "text": "Answer (A) represents a weighting of \n lemonade. Answer (B) represents a weighting of\n lemonade. Only answer (C) offers a weighting in the correct range: \n or 3 : 5\nlemonade.\nThe correct answer is (C)."} +{"id": "book 2_p355_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 355, "page_end": 355, "topic_guess": "statistics", "text": "Number Line Techniques for Statistics\nProblems\nSeveral other common types of questions involving statistics can be solved with\nvisualization. Specifically, using a number line can help simplify the work for many of\nthese problems.\nMEDIAN RELATIVE TO MEAN\nMost questions involving the term median are really asking about the order of terms in\na set: you line up the terms in a set in order of size, then select the middle term. By\ncontrast, the average, or arithmetic mean, is the sum of all of the terms divided by the\nnumber of terms. It can be visualized as the balancing point of all the terms laid out on\nthe number line, as in the discussion of the balancing point for weighted averages in\nthe previous section. For Data Sufficiency questions involving median, you generally\nneed to picture the placement of the unknown terms relative to the given terms in the\nproblem.\nThis technique is similar to rubber band geometry discussed earlier in this chapter,\nexcept this technique applies to problems involving sets rather than problems involving\nthe coordinate plane. In this technique, you must place fixed terms in order from least\nto greatest (as you would on a number line), then move variable terms around\naccording to the constraints. By doing so, you can visualize what impact these changes\nhave on the answer.\nTry-It #7-9\nIf set S consists of the numbers n, −2, and 4, is the mean of set S greater than\nthe median of set S ?"} +{"id": "book 2_p356_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 356, "page_end": 356, "topic_guess": "statistics", "text": "The mean of set S is \n . The median depends on where n\nfalls relative to −2 and 4: below −2, between −2 and 4, or above 4. One approach to this\nquestion is to think through the potential answers for all possible n values in the likely\nrange (you can glance at the statements and other values in the list to get a sense of the\nrelevant range) and draw out the scenarios on a number line. Try numbers around the\nrelevant numbers, including a smaller and larger one at the far ends of the range; note\nthat, by definition, all numbers in a set are different, so n cannot be −2 or 4:\nThis requires a fair amount of up-front work, but evaluating the statements is fast as a\nresult. Statement (1) indicates that n > 2, which is not sufficient. If n = 5, for example,\nthe mean would equal \n and the median would equal 4. By contrast, if n = 10, the\nmean would equal 4 and the median would still equal 4.\nSimilarly, statement (2) indicates that n < 3, which is not sufficient. If n = −3, for\nexample, the mean would equal \n and the median would equal −2. By contrast, if n\n= −8, the mean would equal −2 and the median would still equal −2.\nTaken together, however, any number in the range of 2 < n < 3 would feature a median\ngreater than the mean.\nn > 2(1)\nn < 3(2)"} +{"id": "book 2_p357_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 357, "page_end": 357, "topic_guess": "statistics", "text": "The correct answer is (C).\nNotice from this problem that as you move the variable terms, the mean always\nchanges when the value of the variable terms change, but the median typically changes\nin jumps. The median can get stuck while the number you’re changing doesn’t affect\nwhich number is in the middle.\nCHANGES IN STANDARD DEVIATION\nThe GMAT will rarely (if ever) ask you to calculate the standard deviation of a list of\nnumbers. However, the exam will expect you to have some intuition about standard\ndeviations.\nOne way in which the GMAT might test your intuitive knowledge of standard deviations\nis by changing numbers within a list and asking you what the impact on standard\ndeviation would be. The relationship is relatively straightforward:\nYou might also see the term variance, which is also a measure of the spread of numbers\nin a set or list. Variance and standard deviation indicate the same information. A\nvariance of 0 indicates that all of the numbers are identical, as does a standard\ndeviation of 0. The larger the variance, or standard deviation, the more the numbers are\nspread out. (Variance and standard deviation are never negative.)\nTry-It #7-10\nLast Year 9 9.5 10 10 11 11 11 11 11 12.5 13 13\nThis Year 9 x 10 10 11 11 11 11 11 y 13 13\nMoving terms away from the mean increases the standard deviation of the list.\nMoving terms toward the mean decreases the standard deviation of the list."} +{"id": "book 2_p358_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 358, "page_end": 358, "topic_guess": "statistics", "text": "The monthly sales (in thousands of $) at a certain restaurant for the past two\nyears are given in the chart above. If the standard deviation of the monthly\nsales is greater this year than last year, which of the following are possible\nvalues for x and y ?\nExcept for x and y, the two lists of monthly sales numbers are identical, so focus\nexclusively on those terms that changed: 9.5 and 12.5 from last year were replaced by x\nand y this year. If this year’s standard deviation is greater, then this year’s numbers\nmust be more spread out from the mean than last year’s. The numbers are close\nenough together to indicate that the average should be somewhere around 11.\nVisually, here are the interesting terms from last year:\nThis problem does not require actual computation of the standard deviation using the\nnew x and y values. The math would be too complex to complete in two minutes.\nInstead, determine visually which x and y values increase the standard deviation: the\npair of x and y values that are farther from the mean than are 9.5 and 12.5 will increase\nthe standard deviation.\n9 and 12.5(A)\n10 and 11(B)\n10 and 12.5(C)\n11 and 11(D)\n11 and 12.5(E)"} +{"id": "book 2_p359_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 359, "page_end": 359, "topic_guess": "statistics", "text": "All of the sets have either one point or both points shi ed in toward the mean EXCEPT\n(A), which has one of the points shi ed away from the mean while the other is\nunchanged. Only in this case is the deviation greater.\nThe correct answer is (A).\nThe GMAT may also test you on standard deviations by adding numbers to a list. When\nnew terms are added, the GMAT will o en ask you to compare the old list to the new\nlist, or to compare various options for the new list, or to do both. You must have a\ntechnique to evaluate the standard deviation of different lists relative to one another.\nAgain, pictures make for great comparison tools!\nTry-It #7-11\nA list of 12 test scores has an average (arithmetic mean) of 500 and a standard\ndeviation of 50. Which of the following lists of additional test scores, when\ncombined with the original list of 12 test scores, must result in a combined list\nwith a standard deviation less than 50 ?\n6 test scores with average of 450 and standard deviation of 50(A)\n6 test scores with average of 500 and standard deviation of 25(B)\n6 test scores with average of 550 and standard deviation of 25(C)\n12 test scores with average of 450 and standard deviation of 25(D)\n2 test scores with average of 550 and standard deviation of 50(E)"} +{"id": "book 2_p360_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 360, "page_end": 360, "topic_guess": "statistics", "text": "It is not generally true that all of the terms in a list are within 1 standard deviation of the\nmean. However, standard deviation is a measure of the spread of the terms of a list, so\nyou could represent the original list of scores this way:\nThe oval spans ±1 standard deviation from the mean, where many of the scores will\nlikely be. This simplification is acceptable as long as you represent all of the other lists\nthe same way so that you can compare the relative effects of the new test scores\nsystematically.\nFor each of the answer choices, overlay the representative ovals for the new data on top\nof the oval for the original data:\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p361_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 361, "page_end": 361, "topic_guess": "statistics", "text": "Only answer (B) concentrates the list of scores closer to the original average of 500.\nThus, adding the data in answer (B) will result in a smaller standard deviation than that\nfound in the original data. The correct answer is (B).\nIn general, these are the rules for adding a single term to a list:\nNote that mathematically this is a slight oversimplification, but for the purpose of\nadding terms to a list of numbers on the GMAT, you can accept this simplification as\ntrue.\nFLOATING TERMS IN A SET\nOn GMAT Statistics problems involving elements (i.e., terms) in a list, you can usually\nfocus your attention on a single term or two. These terms could be considered the\nfloating terms—the terms that are unknown or not completely defined among a list of\nmore clearly defined terms.\nAs you approach a question of this type, try to rephrase the question quickly so that\nyou focus on the unknown, or floating, terms rather than on the known terms.\nTry-It #7-12\nList A contains 5 positive integers, and the average (arithmetic mean) of the\nintegers in the list is 7. If the integers 6, 7, and 8 are in List A, what is the range\nof List A ?\nAdding a new term more than 1 standard deviation from the mean generally\nincreases the standard deviation.\nAdding a new term less than 1 standard deviation from the mean generally\ndecreases the standard deviation.\nThe integer 3 is in List A.(1)"} +{"id": "book 2_p362_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 362, "page_end": 362, "topic_guess": "number_theory", "text": "The average of all five integers in the list is 7. Three of the integers in the list are given\n(6, 7, and 8), and they all have an average of 7. Therefore, the floating terms in this\nproblem must also have an average of 7. Assign x and y to represent these terms:\nThe rephrased question is thus, “Given that x + y = 14, what is either x or y?” Once you\nknow one of the values, you can solve for the other and thereby determine the range of\nthe list.\n(1) SUFFICIENT: If 3 is one of the unknown integers, the other must be 11. The range is\nthus 11 − 3 = 8.\n(2) SUFFICIENT: This statement might seem a little too vague to be sufficient, but by\nvisually listing the possible pairs that add up to 14, you can rule out pairs that don’t fit\nthe constraint from this statement:\nNotice that the pairings represent the constraint x + y = 14. Visually, this means that x\nand y are balanced around 7.\nThe largest term in List A is greater than 3 times and less than 4 times the\nsize of the smallest term.\n(2)"} +{"id": "book 2_p363_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 363, "page_end": 363, "topic_guess": "word_problems", "text": "Among these pairs:\nOnly one pair of integers results in a ratio strictly between 3 and 4. The unknown terms\nmust therefore be 3 and 11, and the range is 11 − 3 = 8.\nThe correct answer is (D).\nIn this problem, the constraint x + y = 14 is a fixed sum. Another common constraint is a\nfixed difference, such as a − b = 2. A fixed difference can be represented visually as a\nfixed distance between a and b on the number line, with a to the right because it is\nlarger. That distance could move le or right:\nMAXIMIZING (OR MINIMIZING) ONE TERM\nAnother visual technique for statistics involves maximizing (or minimizing) the value of\na term in a set or list of numbers, subject to some constraints. Such problems will\nusually employ the word maximum or minimum. For these problems, you o en should\nmaximize (or minimize) the term by minimizing (or maximizing) the other terms,\nbecause the constraints usually involve mathematical trade-offs.\nTry-It #7-13\n8 is 1.33 times the size of 6 (the ratio is too low).\n9 is 1.8 times the size of 5 (the ratio is too low).\n10 is 2.5 times the size of 4 (the ratio is too low).\n11 is 3.66 times the size of 3 (an acceptable ratio).\n12 is 6 times the size of 2 (the ratio is too high)."} +{"id": "book 2_p364_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 364, "page_end": 364, "topic_guess": "statistics", "text": "In a certain lottery drawing, five balls are selected from a tumbler in which\neach ball is printed with a different two-digit positive integer. If the average\n(arithmetic mean) of the five numbers drawn is 56 and the median is 60, what\nis the greatest value that the lowest number selected could be?\nThe goal is to maximize the value of the lowest-numbered ball. All balls contain a two-\ndigit positive integer, and none of the balls have the same number. The problem\nprovides enough information to calculate the sum and to lay out a visual listing of the\nnumbers:\nIn order to maximize the value of the first (lowest) number in the set, what do you need\nto do to the other numbers?\nYou’d want to minimize them. Select the smallest numbers that you can for the\nremaining slots:\nIn some problems, you might actually make different slots equal to each other (e.g., the\nthree largest numbers could be 60, 60, and 60). This problem, though, specifies that the\n43(A)\n48(B)\n51(C)\n53(D)\n56(E)"} +{"id": "book 2_p365_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 365, "page_end": 365, "topic_guess": "algebra", "text": "integers are all different.\nThe five numbers must sum to 280, so you can set up an algebraic equation and solve.\nCan you think of a way to minimize the arithmetic needed to solve in that way?\nThe three numerical values, 60, 61, and 62, are all larger than the average of 56.\nSpecifically, they are +4, +5, and +6 away from that average.\nThe other two numbers, then, need to make up for that overage of 4 + 5 + 6 = 15:\nThe numbers x and x + 1 are also consecutive, so they need to be −8 and −7 away from\nthe average of 56. The two remaining numbers are 48 and 49. (We call this the\nover/under approach, by the way.)\nThe correct answer is (B).\nTry-It #7-14\nThe average (arithmetic mean) of six numbers is 18 and the median of the six\nnumbers is 16. What is the minimum possible value for the greatest number in\nthe list?\n19(A)\n20(B)\n21(C)\n22(D)\n23(E)"} +{"id": "book 2_p366_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 366, "page_end": 366, "topic_guess": "word_problems", "text": "This time, the goal is to minimize the largest number:\nIn order to minimize the final term, you’d want to maximize all of the other terms. This\ntime, though, the list contains an even number of terms, so you can’t just set the\nmedian to the middle number. The two middle numbers average to 16.\nThe pair (16, 16) averages to 16, as does the pair (15, 17). The first pair, though, is better\nwhen the goal is to minimize the final term, since the terms to the right have to be\nequal to or greater than the terms to the le .\nIn other words, when trying to minimize the final term, it’s true that you want to\nmaximize the earlier terms, but you also have to think about how to do so in a way that\ndoesn’t make the final term too large (since it has to be larger than the earlier terms). In\nthis case, the first four terms are all 16:\nIn order to minimize the final term, set the last two terms equal to each other and solve\nalgebraically:\nOr you can use the over/under approach:"} +{"id": "book 2_p367_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 367, "page_end": 367, "topic_guess": "statistics", "text": "Those last two terms have to make up for the −8 on the other side, so each x must be 4\nover the average, or 22.\nTAKING AND GIVING\nAnother common scenario involves taking value from one term in a list and giving it to\nanother term. The relative value of the terms in the list will change, leading to some\ninteresting results.\nTry-It #7-15\nJake 51\nKeri 63\nLuke 15\nMia 38\nNora 22\nThe table above shows the number of points held by five players of a certain\ngame. If an integer number of Keri’s points were taken from her and given to\nLuke, and the median score of the five players increased, how many points\nwere transferred from Keri to Luke?\nThe key to this problem is that by taking enough points from Keri and giving them to\nLuke, the median of the list can change.\n23(A)\n24(B)\n25(C)\n26(D)\n27(E)"} +{"id": "book 2_p368_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 368, "page_end": 368, "topic_guess": "statistics", "text": "Set it up visually. Order the scores from low to high on a number line, and represent the\nchange in Luke’s score with x:\nThe current median is Mia’s 38, circled in the diagram. In order for the median to\nchange, Luke’s score must leap-frog those of Nora and Mia, pushing Mia into the bottom\ntwo scores and making Luke’s score the median. But be careful! You don’t want to\ndecrease Keri’s score so much that Luke and Mia surpass her, leaving Mia once again in\nthe median score position.\nIf 15 + x = 38, Luke would match the current median score. That is x = 23, and Keri’s new\nscore would be 63 − 23 = 40. However, the median score would remain 38, with both\nLuke and Mia having that score. Therefore, x must be greater than 23.\nTry x = 24. Luke’s new point value is 15 + 24 = 39. Keri’s new point value is 63 − 24 = 39.\nBoth Nora and Mia are below Luke (and Keri), so the new median is 39:\nThe correct answer is (B)."} +{"id": "book 2_p369_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 369, "page_end": 369, "topic_guess": "geometry", "text": "Problem Set\nUse the Visual Solutions techniques discussed in this chapter to solve the\nfollowing problems.\n1. Does a rectangular mirror have an area greater than 10 square\ncentimeters?\nThe perimeter of the mirror is 24 cm.(1)\nThe diagonal of the mirror is less than 11 cm.(2)\n2. A number line is numbered with the integers from 0 to 50 inclusive.\nAn ant walks along the number line as follows: First, it walks in the\npositive direction until it reaches a multiple of 5 that it hasn’t\npreviously reached. Then, it walks in the negative direction for at\nleast 1 unit, stopping when it reaches any multiple of 2. The ant\nstarts at 0 on the number line and repeats this process until it first\nreaches the point marked 50. How many units does it travel in\ntotal?"} +{"id": "book 2_p370_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 370, "page_end": 370, "topic_guess": "number_theory", "text": "60(A)\n65(B)\n76(C)\n80(D)\n90(E)\n3. If a, b, and c are positive, is \nOn the number line, a is closer to b than it is to c.(1)\nb > c(2)\n4. The length of one edge of a cube equals 4. What is the distance\nbetween the center of the cube and one of its vertices?\n5. A test is taken by 100 people and possible scores are the integers\nbetween 0 and 50, inclusive. For each of the following scenarios,\ndetermine whether the average (arithmetic mean) score would be\ngreater than 30 (answer Yes, No, or Uncertain).\nMore than 70 people scored 40 or higher.a)\n75 people scored 40 or higher.b)\nFewer than 10 people scored 50.c)\nNo more than 2 test-takers scored any given score.d)"} +{"id": "book 2_p371_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 371, "page_end": 371, "topic_guess": "coordinate_geometry", "text": "6. \nAs part of an experiment, a student repeatedly tests the\ntemperature of a light bulb. The bar graph below displays the\nnumber of readings the student recorded at various temperatures,\nmeasured in degrees Fahrenheit. What was the (arithmetic mean)\ntemperature reading of the light bulb?\n7. In a certain dance troupe, there are 55 women and 33 men. If all of\nthe women are 62 inches tall and all of the men are 70 inches tall,\nwhat is the average height of the dancers in the troupe?\n8."} +{"id": "book 2_p372_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 372, "page_end": 372, "topic_guess": "word_problems", "text": "Anton and Lena start at the same point on a circular track,\nmeasuring 10 meters in diameter, and begin walking\ncounterclockwise at the same time, with Anton walking more\nquickly than Lena. When Anton has traveled exactly halfway around\nthe track, they both stop walking. They then observe that the\ndistance between them along a straight line measures exactly \n meters, as shown above. What fraction of the track has Lena\ncovered? \n(A)\n(B)\n(C)\n(D)\n(E)\n9. Eddy, Mario, and Peter have $32, $72, and $98, respectively. They\npool their money and redistribute the entire amount among\nthemselves. If Eddy now has the median amount of money in the\ngroup, what is the greatest amount of money that Eddy could now\nhave?"} +{"id": "book 2_p373_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 373, "page_end": 373, "topic_guess": "coordinate_geometry", "text": "$72(A)\n  $85(B)\n  $98(C)\n$101(D)\n$202(E)\n10. \nA square is drawn in the coordinate plane with its vertices at the\npoints (–2, –2), (–2, 6), (6, 6), and (6, –2), and a non-vertical line is\ndrawn that passes through the point (0, 4). The portion of the\ncoordinate plane that lies within the square, but above the line, is\nthen shaded as shown above. If A is the area of the shaded region in\nsquare units, which of the following specifies all the possible values\nof A ?\n8 ≤ A ≤ 16(A)\n8 ≤ A < 48(B)\n16 ≤ A< 48(C)\n8 < A ≤ 32(D)\n16 < A ≤ 32(E)"} +{"id": "book 2_p374_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 374, "page_end": 374, "topic_guess": "geometry", "text": "Solutions\n1. (C): Because the formula for the area of a rectangle is A = lw, rephrase the\nquestion. Is lw > 10 ?\n(1) INSUFFICIENT: The perimeter is 24 cm. The area of a quadrilateral is\nmaximized when the quadrilateral is a square, so first try l = w = 6. In this case,\nthe area is 36 cm2 and the answer to the question is Yes.\nIf, on the other hand, l = 11.5 and w = 0.5, the perimeter is still 24, but the area\nis (11.5)(0.5) = 5.75 cm2. In this case, the answer to the question is No.\n(2) INSUFFICIENT: The diagonal of the rectangle is less than 11. If the diagonal\nis less than 11, then the sides must also be less than 11. If l = w = 6, then the\ndiagonal is shorter than 11 and, as last time, the area is 36. In this case, the\nanswer to the question is Yes.\nIf, on the other hand, l = 3 and w = 1, the diagonal is less than 11, but the area\nis (3)(1) = 3 cm2. In this case, the answer to the question is No.\n(1) AND (2) SUFFICIENT: The sides must be less than 11 and the perimeter\nmust be 24. The case of the square still maximizes the area: l = w = 6 and the\narea is 36.\nThe largest possible length is just under 11, making the width just over 1. (<11)\n(>1) = something larger than 10. The area must be greater than 10."} +{"id": "book 2_p375_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 375, "page_end": 375, "topic_guess": "word_problems", "text": "The correct answer is (C).\n2. (C) 76: The problem describes an ant walking along a number line in a\ncomplex way. It starts at 0, then walks to the right until it reaches a multiple of\n5:\nThen, it walks to the le until it reaches a multiple of 2. The problem specifies\nthat the ant must walk to the le at least one unit. Therefore, even if it is\nalready standing on a multiple of 2, it must walk to the next one. In this case, it\nwalks le until it reaches 4:\nThen it walks to the right again until it reaches a new multiple of 5:\nThe question asks how far the ant walks, in total, to reach the number 50 for\nthe first time. Since this is too many units to realistically count by hand, there\nmust be an underlying pattern.\nWrite down the first few distances that the ant walks, using your number line:\n5, 1, 6, 2, 7, 1, 6, 2, 7, 1 ..."} +{"id": "book 2_p376_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 376, "page_end": 376, "topic_guess": "statistics", "text": "There is a repeating cycle of four numbers. Starting at 5 on the number line,\nthe ant moves le 1, then right 6, then le 2, then right 7, which takes it to 15\non the number line. That is, the ant walks a total of 1 + 6 + 2 + 7 = 16 units to\nmove from 5 to 15. Then it walks another 16 units to move from 15 to 25, from\n25 to 35, and from 35 to 45. In total, the ant walks 4(16) = 64 units to move\nfrom 5 to 45.\nTo get from 45 to 50, the ant first walks one unit le to 44, then six units right\nto 50. In total, the ant walks 64 + 1 + 6 = 71 units to move from 5 to 50.\nFinally, add the five units it walks from the beginning: 71 + 5 = 76.  \nIt is also possible to estimate the answer. As established above, traveling 10\npoints on the number line (from 15 to 25, from 25 to 35, and so on) requires\nthe ant to travel 16 units. To begin at 0 and end at 50, the ant would travel\napproximately 5(16) = 80 units. This is enough to let you eliminate answers (A),\n(B), and (E). By estimation alone, the answer must be either (C) or (D).\nThe correct answer is (C). \n3. (C): The problem indicates that a, b, and c are positive and asks whether a is\ngreater than \n , which is the average (or arithmetic mean) of b and c.\nDraw a picture and rephrase the question: “On the number line, is a\npositioned to the right of the midpoint between b and c?”"} +{"id": "book 2_p377_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 377, "page_end": 377, "topic_guess": "word_problems", "text": "(1) INSUFFICIENT: It isn’t clear whether b or c is the larger value—the point on\nthe right—so a could be closer to the smaller number (making a less than the\naverage) or the larger number (making a greater than the average).\n(2) INSUFFICIENT: This statement indicates nothing about a, so it can’t be\nsufficient.\n(1) AND (2) SUFFICIENT: Together, the two statements indicate that b is the\npoint on the right, so a must be on the right side of the midpoint.\nThe correct answer is (C).\n4. \n : Draw it out!\nThe length of any side of the cube is 4, and the problem asks for the distance\nbetween the center of the cube and any of its vertices (corners). Chop up the\ncube into eight smaller cubes to see that the distance from the center of the 4\n× 4 × 4 cube to any corner is the diagonal of a 2 × 2 × 2 cube."} +{"id": "book 2_p378_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 378, "page_end": 378, "topic_guess": "statistics", "text": "You can find the diagonal of a cube in a variety of ways. Probably the fastest\n(besides applying a memorized formula) is to use the “Super Pythagorean”\ntheorem, which extends to three dimensions:\nIn the special case when the three sides of the box are equal, as they are in a\ncube, use this equation:\nSince s = 2, \n .\n5. To visualize this set of weighted averages problems, imagine a teeter-totter\nthat is 50 meters long, marked off from 0 to 50 to represent scores. One\nhundred people of equal weight sit on the teeter-totter at their respective\nscores. The weighted average is the position where the teeter-totter would\nbalance. Thus, to answer whether the average score is greater than 30, take\nextremes according to the given conditions and see whether you can swing\nthe balance to either side of 30 (or whether you are forced to balance on one\nside of 30 only)."} +{"id": "book 2_p379_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 379, "page_end": 379, "topic_guess": "general", "text": "(A) Uncertain: More than 70 people scored 40 or higher. If all 100 scored 40 (or\nhigher), then the average is above 30.\nIf, on the other hand, 71 people scored 40 and the other 29 scored 0, does the\naverage drop to 30 or lower?\nMatch up the 29 people who scored 0 with 29 of the people who scored 40.\nThese 58 people together have an average score of 20. The other 42 people\nhave a score of 40. If you average these two groups, the average must be\nbelow 30, since 58 is larger than 42.\n(B) Uncertain: Seventy-five people scored 40 or higher. If all 100 scored 40,\nthen the average is higher than 30. If 75 scored 40 and the other 25 scored 0,\nthen what?\nMatch the 25 people who scored 0 with 25 people who scored 40. These 50\npeople have an average score of 20. The remaining 50 people have an average\nscore of 40, so the overall average is exactly 30. It is possible, therefore, to\nhave an average that is higher than 30 or an average that is not higher than 30:"} +{"id": "book 2_p380_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 380, "page_end": 380, "topic_guess": "number_theory", "text": "(C) Uncertain: Fewer than 10 people scored 50. Say that 1 person scored 50\nand the other 99 scored 0. The average is definitely below 30. If on the other\nhand, 1 person scored 50 and the other 99 scored 40, then the average is\ndefinitely above 30.\n(D) No: Each score was achieved by no more than 2 people. There are 51\nintegers between 0 and 50, inclusive. In other words, there are 102 possible\nscores to spread among the test-takers. Since there are 100 test-takers, almost\nevery score is taken.\nThe highest possible average will occur when nobody scores 0 points. If 2 test-\ntakers score 1, 2 test-takers score 2, and so on up to 50, then the average score\nwill be approximately 25. As this is the highest allowable average for this\nscenario, it’s impossible for the average to be greater than 30.\n6. 112°: Eyeball the graph. The average looks like it’s in the 112° range. Calculate\nthe over/under with 112° as the assumed baseline.\nThe four 110° readings are each 2° below the baseline, and the three 111°\nreadings are each 1° below the baseline, for a total of 11° below baseline.\nOn the other side, the 113°, 114°, and 115° readings are a total of 11° above the\nbaseline. They balance perfectly! The average is exactly 112°.\n7. 65 inches: Algebraic solution:"} +{"id": "book 2_p381_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 381, "page_end": 381, "topic_guess": "word_problems", "text": "Visual solution:\n8. (B) \n : Draw a diagram showing the track, and label all of the given\ninformation. Because Anton has walked exactly halfway around the track, the\ndistance between Anton and the starting point is the diameter.\nThe answer choices are fractions that are fairly far apart from each other,\nsuggesting that visual estimation could be effective. If you use this approach,\ncarefully draw your diagram to scale.  Since \n  is approximately 5(1.7) =\n8.5, the line between Anton and Lena is only slightly shorter than the diameter\nof the circle. Therefore, Lena is closer to the starting point than she is to\nAnton, so Lena has traveled less than one-quarter of the circle: eliminate\nanswers (D) and (E). \nAt the other extreme, choice (A) is quite small, suggesting that the distance\nbetween Anton and Lena would be closer to 10 than it actually is. The only\nreasonable answer choices are (B) and (C)."} +{"id": "book 2_p382_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 382, "page_end": 382, "topic_guess": "geometry", "text": "To calculate the exact answer, observe that Anton, Lena, and the starting line\nform the three vertices of a triangle. Because one side of the triangle is the\ndiameter of the circle, the triangle is a right triangle with hypotenuse 10. The\nknown leg is equal to \n .\nYou can either use the Pythagorean theorem to calculate the missing leg (the\nstraight line between Lena and the starting point) or you can note that the\nknown leg and the hypotenuse have a ratio of \n , so the triangle\nmust be a 30–60–90 triangle and the remaining side must equal 5.\nThe arc of the circle that Lena traveled has a corresponding inscribed angle of\n30 degrees (the vertex at point A). The corresponding central angle is twice the\ninscribed angle, or 60 degrees.\nAlternatively, the distance between Lena and the starting line is equal to the\nradius of the circle. Therefore, the triangle with vertices at Lena, the starting\nline, and the center of the circle is an equilateral triangle, and all of its angles\nequal 60 degrees. So Lena has covered 60 degrees out of the 360 degrees in a\nfull circle, or a fraction of \n . \nThe correct answer is (B)."} +{"id": "book 2_p383_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 383, "page_end": 383, "topic_guess": "coordinate_geometry", "text": "9. (D) $101: The pool of money is $32 + $72 + $98 = $202. A er the redistribution,\neach person will have an amount between $0 and $202, inclusive. Call the\namounts L, M, and H (low, median, high). To maximize M (Eddy’s share),\nminimize L and H.\nThe correct answer is (D).\nMinimum L = $0\nMinimum H = M\nMaximum M = Total pool of money − Minimum L − Minimum H\nM = $202 − $0 − M\n2M = $202\nM = $101\n10. (B) 8 ≤ A < 48: On your paper, draw the square in the coordinate plane. The line\npasses through the point (0, 4), so label that point as well. Since the slope of\nthe line could vary, sketch several possible lines that pass through the point,\nbeing certain to test extreme cases. \nIf the slope is very positive, then the line is nearly vertical, and the area above\nthe line will be very close to 2(8) = 16:"} +{"id": "book 2_p384_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 384, "page_end": 384, "topic_guess": "geometry", "text": "As the slope decreases, the area above the line will decrease. If the slope is 1,\nthe area above the line is the area of a triangle with base 4 and height 4, which\nis equal to \n (4)(4) = 8.\nThen, as the slope continues to decrease, the area above the line will once\nagain increase steadily. The greatest area is found when the slope is very\nnegative, since the area will be slightly less than (6)(8) = 48."} +{"id": "book 2_p385_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 385, "page_end": 385, "topic_guess": "geometry", "text": "The area of the shaded region can be a minimum of exactly 8 and can be a\nmaximum of almost, but not equal to, 48 (since the line cannot be vertical).\nThe correct answer is (B)."} +{"id": "book 2_p386_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 386, "page_end": 386, "topic_guess": "general", "text": "CHAPTER 8\nHybrid Problems"} +{"id": "book 2_p387_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 387, "page_end": 387, "topic_guess": "sequences_patterns", "text": "In This Chapter...\nPop Quiz!\nHybrid Problems\nIdentify and Sequence the Parts\nWhere to Start\nMinor Hybrids"} +{"id": "book 2_p388_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 388, "page_end": 388, "topic_guess": "statistics", "text": "Chapter 8\nHybrid Problems\nPOP QUIZ!\nTry these four hybrid problems.\nTry-It #8-1\nSet A consists of four consecutive positive integers. Set B is\nconstructed as follows: each integer in set A is randomly either\nincreased by 10% or decreased by 10%, and the four resulting\nvalues compose set B. What is the range of set B ?\nTry-It #8-2\nSet S contains 100 consecutive integers. If the range of the negative\nelements of set S equals 80, what is the average (arithmetic mean)\nof the positive numbers in the set?\nThe smallest integer in set A was increased by 10% when set B\nwas constructed.\n(1)\nThe greatest integer in set A was increased by 10% when set B\nwas constructed.\n(2)"} +{"id": "book 2_p389_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 389, "page_end": 389, "topic_guess": "sets_probability_counting", "text": "Try-It #8-3\nIf a and b are consecutive positive integers, is ab divisible by 30 ?\nTry-It #8-4\nA carnival card game gives the contestant a one in three probability\nof successfully choosing the correct card and thereby winning the\ngame. If a contestant plays the game repeatedly, what is the\nminimum number of times that he must play the game so that the\nprobability that he never loses is less than 1% ?\na2 is divisible by 25.(1)\n63 is a factor of b2.(2)"} +{"id": "book 2_p390_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 390, "page_end": 390, "topic_guess": "geometry", "text": "Hybrid Problems\nHybrid problems blend topics together. They contain two or more\nqualitatively different kinds of obstacles that you must surmount on the\nway to the answer.\nSome hybrid problems feature content areas that are fairly closely related.\nOther hybrid problems feature content areas that share little in common;\nthese problems must be solved in separate steps.\nThe difficulty of a hybrid problem is related to the following questions:\nProblems with minor additional content areas are generally easier to solve\nthan hybrids that blend topics together in an unusual, fundamental, and\nHow closely related are the subjects being tested? Are the content\nareas covered in the same section of the Manhattan Prep\nAll the Quant guide? The more closely related the subjects, the\neasier it will be to navigate the problem.\nHow important is each of the subject areas? Are each of the subject\nareas fundamental to the problem or is one of them just a low-level\ndisguise that can be quickly disposed of? The more important each\ntopic is in solving the problem, the more difficult the problem will\nusually be."} +{"id": "book 2_p391_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 391, "page_end": 391, "topic_guess": "general", "text": "clever way. The best hybrids are one of a kind. You will have to bring your A\ngame to solve them.\nBut solve them you can. That’s why you’re reading this chapter!"} +{"id": "book 2_p392_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 392, "page_end": 392, "topic_guess": "number_theory", "text": "Identify and Sequence the Parts\nWhen you encounter a hybrid problem, first pick out all of the topics\ntested. Take a look at the first problem from the pop quiz:\nTry-It #8-1\nSet A consists of four consecutive positive integers. Set B is\nconstructed as follows: each integer in set A is randomly either\nincreased by 10% or decreased by 10%, and the four resulting\nvalues compose set B. What is the range of set B ?\nThis problem refers to consecutive integers, but it involves percents and\nalso asks for the range of a set of numbers, which is a statistics concept.\nWhich aspect of the problem should you tackle first? \nIn this problem, a set begins as four consecutive integers and is then\nchanged. So you might start your approach by jotting down some sets of\nconsecutive integers—either as numbers or in terms of a variable. Then,\nThe greatest integer in set A was increased by 10% when set B\nwas constructed.\n(1)\nThe smallest integer in set A was increased by 10% when set B\nwas constructed.\n(2)"} +{"id": "book 2_p393_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 393, "page_end": 393, "topic_guess": "fractions_decimals_percents", "text": "apply the percent change described in the problem and, finally, check the\nrange of the resulting set. \nIn general, start with wherever you feel the logical beginning is or\nwhichever part you feel is an easier, cleaner starting point."} +{"id": "book 2_p394_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 394, "page_end": 394, "topic_guess": "number_theory", "text": "Where to Start\nSTARTING AT THE BEGINNING\nAs you contemplate the logical order of steps, you might feel less confident\nwith the second stage than with the first. If so, go ahead and start at the\nbeginning. Just articulate very clearly, “What intermediate result will I get\nonce I’m finished with the first part of the problem?”\nBegin by defining set A. Set A consists of four consecutive integers, such as\n1, 2, 3, 4, or 10, 11, 12, 13. \nNext, choose a statement to work with first. Statement (1) says that the\nsmallest integer in set A is increased by 10%. The other integers might have\nbeen either increased or decreased. Using one of your cases, calculate\nwhat might happen to the set: \nSet A Scenarios Set B\n1, 2, 3, 4 All values increased by 10% 1.1, 2.2, 3.3, 4.4\n1, 2, 3, 4 Smallest value increased by 10%, other values decreased by 10% 1.1, 1.8, 2.7, 3.6\nFinally, calculate the range of the resulting set. The range of a set is the\ndifference between the greatest number and the smallest number in the\nset."} +{"id": "book 2_p395_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 395, "page_end": 395, "topic_guess": "statistics", "text": "Set A Scenarios Set B Range\n1, 2,\n3, 4\nAll values increased by 10% 1.1, 2.2, 3.3,\n4.4\n4.4—1.1 =\n3.3\n1, 2,\n3, 4\nSmallest value increased by 10%, other values\ndecreased by 10%\n1.1, 1.8, 2.7,\n3.6\n3.6—1.1 =\n2.7\nSince different ranges are possible, the statement is insufficient. Eliminate\nanswers (A) and (D). \nSTARTING AT THE END\nYou might decide that it is easier to start at the end and work backwards.\nThat’s fine. Just ask yourself, “What information do I need to have as a last\nstep before arriving at a solution to the question?” \nIn this problem, you might first consider what would cause set B to have a\ndifferent range. The range of a set is the difference between the greatest\nand smallest numbers in the set. If the difference between these two\nnumbers can vary, the range can vary as well. \nAccording to statement (2), the greatest integer in set A was increased by\n10%. However, the smallest integer might have increased by 10%, bringing\nit closer to the greatest integer, or it might have decreased by 10%,\nbringing it farther from the greatest integer. Therefore, the range of the set\ncan vary, so statement (2) is insufficient. Eliminate answer (B)."} +{"id": "book 2_p396_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 396, "page_end": 396, "topic_guess": "statistics", "text": "Now, put the two statements together. One option is to try a few different\ncases, making sure to increase the smallest integer in set A and decrease\nthe greatest integer in set A. One of the two cases used for statement (1)\nworks here as well:\nSet A Scenario Set B Range\n1, 2, 3, 4 All values increased by 10% 1.1, 2.2, 3.3, 4.4 4.4—1.1 = 3.3\nTry another case, using different numbers: \nSet A Scenario Set B Range\n10, 11, 12, 13 All values increased by 10% 11, 12.1, 13.2, 14.3 14.3—11 = 3.3\nThe range is the same, but this doesn’t mean that the statements are\nsufficient together. To see why, use a different approach. Set A always\ncontains the numbers x, x + 1, x + 2, and x + 3. When the smallest value in\nset A is increased by 10%, it goes from x to 1.1x. When the greatest value is\nincreased by 10%, it goes from x + 3 to 1.1x + 3.3. So these two values differ\nby 3.3. \nHowever, you can’t assume that the smallest and greatest values in set\nA always become the smallest and greatest values in set B. If two of the\nvalues switch places, the range might change. Here’s an example: \nSet A Scenario Set B Range"} +{"id": "book 2_p397_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 397, "page_end": 397, "topic_guess": "statistics", "text": "Set A Scenario Set B Range\n10, 11,\n12, 13\nSmallest and largest values increased by 10%, other\nvalues decreased by 10%\n11, 9.9,\n10.8, 14.3\n14.3—9.9\n= 4.4\nThe range of the set might vary, even when both statements are used. The\ncorrect answer is (E). \nLet’s look at another example:\nTry-It #8-2\nSet S contains 100 consecutive integers. If the range of the negative\nelements of set S equals 80, what is the average (arithmetic mean)\nof the positive numbers in the set?\nIn this problem, information is given about the range of numbers in a set.\nThus, knowing how to work with statistics techniques will be important in\nsolving the problem.\nSolving this problem also requires using consecutive integers techniques—\nnamely, counting consecutive integers and computing their average. These\ntwo topics are closely related but still cover different ideas.\nIf you start with the statistics piece of the problem, then you’ll be able to\nfind the highest and lowest negative numbers in the set of consecutive\nintegers. This will act as an input to the formulas for computing the largest\nand smallest positive integers in the set, and subsequently the average of\nthe positive integers in the set."} +{"id": "book 2_p398_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 398, "page_end": 398, "topic_guess": "number_theory", "text": "First, determine what set of consecutive negative integers will result in a\nrange of 80. Range is defined as the difference between the highest and\nlowest numbers in a set:\nThe high number among the negative terms is the largest negative integer,\n−1:\nTherefore, the lowest number in set S is −81. Use this result to jump to the\nconsecutive integers portion of the solution. In this case, −81 is the\nsmallest or “first” element in the set:\nTherefore, the highest or “last” number in set S is 18.\nFinally, calculate the average of the positive terms in the set using 1 as the\nsmallest positive integer (“first pos”) and 18 as the largest  positive integer\n(“last pos”):"} +{"id": "book 2_p399_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 399, "page_end": 399, "topic_guess": "number_theory", "text": "The average of the positive numbers in the set is 9.5.\nIf instead you decided to start with the second step, you would write the\nformula for the average of consecutive integers first, focusing on the\npositive integers:\nThe two unknowns you need to solve for are the greatest positive\ninteger and the least positive integer because the question asks only about\nthe positive integers in the set.\nSome of the integers in set S are negative and some are positive, so clearly\nthe least positive integer will be 1. Therefore, you only need to figure out\nwhat the greatest integer in the set will be. This is the information needed\nin the last step of solving the problem.\nIn order to calculate this number, you would now need to apply the\ndefinition of range:\nThe greatest negative integer (the “high” number) is −1:"} +{"id": "book 2_p400_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 400, "page_end": 400, "topic_guess": "number_theory", "text": "Therefore, the least number in the set is −81; plug this into the formula for\ncounting consecutive integers:\nTherefore, the highest number in the set is 18.\nFinally, plug 18 into the average formula:\nWhether you start solving this problem from the beginning or from the end,\nstudy how the steps hook together: the output of one step becomes the\ninput to another. These are the “turns” you have to make in solving any\nhybrid problem. O en, a number that plays one role in a particular formula\nor calculation plays a completely different role in the next step."} +{"id": "book 2_p401_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 401, "page_end": 401, "topic_guess": "number_theory", "text": "Minor Hybrids\nIn minor hybrid problems, one of the following conditions applies:\nThese problems can be easier to solve than major hybrids. However, you\nwill still benefit greatly from paying close attention to the turns as you\nmove through the stages of the solution.\nTry-It #8-3\nIf a and b are consecutive positive integers, is ab divisible by 30 ?\nThis problem tests your skill with both Divisibility & Primes and\nConsecutive Integers. Consecutive integer concepts o en lend themselves\nwell to questions about divisibility.\nThe content areas in the problem are closely related. For instance,\nthey are covered in the same section of the Manhattan Prep\nAll the Quant guide.\n1.\nOne of the content areas is a low-level disguise or some other minor\nfeature.\n2.\na2 is divisible by 25.(1)\n63 is a factor of b2.(2)"} +{"id": "book 2_p402_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 402, "page_end": 402, "topic_guess": "number_theory", "text": "To be divisible by 30, a number must have 2, 3, and 5 as prime factors.\nThus, the question becomes: “Does ab have 2, 3, and 5 as prime factors?”\nNext, use a concept from consecutive integers: a and b are consecutive\npositive integers. Thus, either a or b is an even number, which means that\nthe product ab is automatically divisible by 2. The question can be further\nsimplified: “Does the product ab have 3 and 5 as prime factors?”\n(1) INSUFFICIENT: Statement (1) indicates that 25 is a factor of a2, meaning\nthat 5 and 5 are prime factors of a2. Knowing that 5 is a prime factor of a\nindicates that 5 is a factor of ab, but does not indicate whether 3 is a factor.\n(2) INSUFFICIENT: Statement (2) indicates that 63 is a factor of b2, which\nmeans that 3, 3, and 7 are prime factors of b2 (3 × 3 × 7 = 63). Knowing that\n3 is a prime factor of b indicates that 3 is a factor of ab, but does not\nindicate whether 5 is a factor.\n(1) AND (2) SUFFICIENT: According to the combined statements, a is\ndivisible by 5 and b is divisible by 3. This is sufficient information to answer\nthe rephrased question.\nThe correct answer is (C).\nNotice that the solution never indicated exactly what consecutive integers\na and b are—this isn’t necessary. Also notice a potential trap in this\nproblem—assuming that a < b because of the way the question is phrased.\nThis assumption would not be fatal in this case, but if you realize that a and\nb can come in either order, then it’s easier to create an accurate list of\ndifferent scenarios for a and b."} +{"id": "book 2_p403_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 403, "page_end": 403, "topic_guess": "sets_probability_counting", "text": "Try-It #8-4\nA carnival card game gives the contestant a one in three probability\nof successfully choosing the correct card and thereby winning the\ngame. If a contestant plays the game repeatedly, what is the\nminimum number of times that he must play the game so that the\nprobability that he never loses is less than 1% ?\nThis problem primarily tests Probability theory. In addition, Exponents are\nneeded to represent the impact of playing multiple games on the\nprobability of the outcomes. The probabilities are given in fractions, yet the\nquestion is asked in terms of percents, so Fraction, Decimal, and Percent\n(FDP) connections are relevant. Finally, the question is phrased in terms of\nan inequality, so Inequalities come into play.\n“The probability that he never loses” can be rephrased as “the probability\nthat he always wins.” This probability can be expressed as \n , where\n is the chance of winning on a single play and n is the number of times\nthe contestant plays.\nTrack scenarios in a chart:\nNumber of Plays P(All Wins) Approx. Equivalent\n1\n 0.33 = 33%"} +{"id": "book 2_p404_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 404, "page_end": 404, "topic_guess": "sets_probability_counting", "text": "Number of Plays P(All Wins) Approx. Equivalent\n2\n 0.11 = 11%\n3\n = 0.04 = 4%\n4\n = 1.25%\n5\n  = 0.4% < 1%\nYou might also have noticed that only the denominator in the probabilities\nmattered, as the numerator was always 1. To have a probability of less than\n1%, the fractional probability must be “1 over something greater than 100.”\nIn order for 3n >100, n must be at least 5.\nThe correct answer is 5."} +{"id": "book 2_p405_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 405, "page_end": 405, "topic_guess": "number_theory", "text": "Problem Set\nSolve the following problems and identify the topics being tested.\n1. The average (arithmetic mean) of a list of six numbers is equal to 0.\nWhat is the positive difference between the number of positive\nnumbers in the list and the number of negative numbers in the list?\nEach of the positive numbers in the list equals 10.(1)\nEach of the negative numbers in the list equals −5.(2)\n2. Simplify: \n3. If a, b, and c are positive, is a > b ?\n(1)\nb + c < a(2)\n4. If c is randomly chosen from the integers 20 to 99, inclusive, what is\nthe probability that c3 − c is divisible by 12 ?"} +{"id": "book 2_p406_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 406, "page_end": 406, "topic_guess": "number_theory", "text": "5. If x and y are positive integers greater than 1 such that x − y and \nare both even integers, which of the following numbers must be\nnon-prime integers?\nxI.\nx + yII.\nIII.\nI only(A)\nII only(B)\nIII only(C)\nI and II only(D)\nI, II, and III(E)\n6. A number cube with faces numbered 1 through 6 has an equal\nchance of landing on any face when rolled. If the number cube is\nrolled twice, what is the probability that the sum of the two rolls is\na prime number?\n7. The function {x} is defined as the area of a square with diagonal of\nlength x. If x > 0 and {x2} = x2, what is the value of x  ?"} +{"id": "book 2_p407_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 407, "page_end": 407, "topic_guess": "number_theory", "text": "1(A)\n(B)\n(C)\n2(D)\n4(E)\n8. A circular microchip with a radius of 2.5 centimeters is\nmanufactured by following a circular diagram. The scale of the\ndiagram is such that a measurement of 1 centimeter on the\ndiagram corresponds to a measurement of 0.5 millimeters on the\nmicrochip. What is the area of the diagram, in square centimeters?\n(1 centimeter = 10 millimeters)\n9. Three consecutive integers are selected from the integers 1 to 50,\ninclusive. What is the sum of the remainders that result when each\nof the three integers is divided by x  ?\nWhen the greatest of the consecutive integers is divided by\nx, the remainder is 0.\n(1)\nWhen the least of the consecutive integers is divided by\nx, the remainder is 1.\n(2)"} +{"id": "book 2_p408_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 408, "page_end": 408, "topic_guess": "number_theory", "text": "10. If x, y, and z are all distinct positive integers, and the percent\nincrease from x to y is equal to the percent increase from y to z,\nwhat is x  ?\ny is prime.(1)\nz = 9(2)"} +{"id": "book 2_p409_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 409, "page_end": 409, "topic_guess": "general", "text": "Solutions\n1. (E): Since the average of the six numbers in the list is 0, the sum of the\nsix numbers is 0. There could be positive numbers and negative\nnumbers in the set. Zero is not mentioned, but this does not rule it out.\nIn order for the sum of the numbers in the set to be 0, either all the\nterms are 0 or there are some positives and some negatives.\n(1) INSUFFICIENT: Statement (1) indicates that the list contains at least\none positive number and that each positive term is 10. The list could be\n{−2, −2, −2, −2, −2, 10}, and the positive difference between the number\nof positive terms and the number of negative terms would be 4.\nAlternatively, the list could be {−20, −20, 10, 10, 10, 10}, and the positive\ndifference would be 2.\n(2) INSUFFICIENT: Statement (2) indicates that the list contains at least\none negative number and that each negative term is −5. The set could be\n{−5, 1, 1, 1, 1, 1}, and the positive difference between the number of\npositive terms and the number of negative terms would be 4. The set\ncould be {−5, −5, −5, 5, 5, 5}, and the positive difference would be 0.\n(1) AND (2) INSUFFICIENT: The statements together suggest that the set\nhas twice as many −5 terms as 10 terms, in order to maintain a sum of 0.\nIf every term is negative or positive, then the set would have to be {−5,\n−5, −5, −5, 10, 10}, and the positive difference would be 2. However, zero"} +{"id": "book 2_p410_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 410, "page_end": 410, "topic_guess": "general", "text": "terms are possible, so the set could be {−5, −5, 0, 0, 0, 10}, and the\npositive difference would be 1.\n2. 30: You could simplify the numerator arithmetically (multiply out the\nterms and then add). Alternatively, factor a 22 out of the numerator.\nThen, distribute the denominator (which becomes the difference of\nsquares).\n3. (D): Statement (2) appears to be a bit easier to work with so begin there.\n(2) SUFFICIENT: If a, b, and c are all positive, then a > b + positive.\nTherefore, a must be greater than b. You could also prove this fact by\nTesting Cases.\nCase b + c < a Is this case possible according to (2)?\na > b b + positive < number greater than b Possible\na = b b + c < b Impossible, since c is positive\na < b b + c < number less than b Impossible, since c is positive"} +{"id": "book 2_p411_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 411, "page_end": 411, "topic_guess": "algebra", "text": "Only the a > b case is possible, so the answer is a definite Yes.\n(1) SUFFICIENT: Statement (1) can be cross-multiplied without flipping\nthe inequality sign, since the denominators are positive.\na(a + c) > b(b + c)  \n a2 + ac > b2 + bc  \n a2 + ac − b2 − bc > 0  \n a2 − b2 + ac − bc > 0 Group similar terms to simplify.\n [a2 − b2] + [ac − bc] > 0 Note the Quadratic Template.\n (a − b)(a + b) + c(a − b) > 0\n (a − b)[(a + b) + c] > 0 Factor out a − b.\nTherefore, a + b + c is positive, because all three additive terms are\npositive. So (a − b)(positive) > 0. By Number Properties sign rules, (a − b)\nmust also be positive in order for the product to be greater than 0. Thus,\na > b.\nThis algebra is very tough; it is hard to see where to begin or what series\nof manipulations will be productive. If you did not see this, you could try\nTesting Cases to see which are allowed by statement (1). Note: LT = less\nthan and GT = greater than. GTb means “a number greater than b.”\nCase\n Is this case possible according to (1)?\na > b\n Possible. The le is greater than the right."} +{"id": "book 2_p412_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 412, "page_end": 412, "topic_guess": "number_theory", "text": "Case\n Is this case possible according to (1)?\na = b\n Impossible; the two expressions are equal.\na < b\n Impossible; the le side is actually less than the right\nside, not greater than as (1) requires.\nThe correct answer is (D).\n4. \n : The words divisible and probability are used, so this question is\nabout Divisibility & Primes and Probability.\nProbability is \n . There are 99 − 20 + 1 = 80\npossible values for c, so the unknown is how many of these c values\nyield a c3 − c that is divisible by 12.\nThe prime factorization of 12 is 2 × 2 × 3. There are at least two ways to\nthink about this: numbers are divisible by 12 if they are divisible by 3\nand by 2 twice, or if they are multiples of both 4 and 3.\nThe expression involving c can be factored.\nThese are consecutive integers. It may help to put them in increasing\norder: (c − 1)c(c + 1). Thus, this question has a lot to do with Consecutive\nIntegers, and not only because the integers 20 to 99 themselves are\nconsecutive."} +{"id": "book 2_p413_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 413, "page_end": 413, "topic_guess": "number_theory", "text": "In any set of three consecutive integers, a multiple of 3 will be included.\nThus, (c − 1)c(c + 1) is always divisible by 3 for any integer c. This takes\ncare of part of the 12. So the question becomes: “How many of the\npossible (c − 1)c(c + 1) values are divisible by 4?” Since the prime factors\nof 4 are 2’s, it makes sense to think in terms of Odds and Evens.\nFor example,  (c − 1)c(c + 1) could be (E)(O)(E), which is definitely\ndivisible by 4, because the two evens would each provide at least one\nseparate factor of 2. Thus, c3 − c is divisible by 12 whenever c is odd,\nwhich are the cases c = 21, 23, 25 … 95, 97, 99. That’s\n possibilities.\nAlternatively, (c − 1)c(c + 1) could be (O)(E)(O), which will only be\ndivisible by 4 when the even term itself is a multiple of 4. Thus, c3 − c is\nalso divisible by 12 whenever c is a multiple of 4, which are the cases c =\n20, 24, 28, . . . , 92, 96. That’s \npossibilities.\nThe probability is thus \n5. (D) I and II only: x cannot equal y, as that would make \n ≠ even.\nSo either x > y or y > x.\nx and y are both positive and \n is an integer, so x > y.\nSince x − y is even, either x and y are both even or they are both odd.\nSince \n = an even integer, x = y × (even integer)."} +{"id": "book 2_p414_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 414, "page_end": 414, "topic_guess": "number_theory", "text": "Therefore, x is an even integer, as is y.\nThe correct answer is (D).\nTRUE. x and y are both positive even integers and x > y. No even\nnumber greater than 2 is prime, so x can’t be prime.\nI.\nTRUE. x and y are each positive even integers and x > y. Thus, x +\ny is even and the least possible value of x + y = 4 + 2 = 6. All even\nnumbers greater than or equal to 6 are non-prime.\nII.\nFALSE. It could be that x = 4 and y = 2, so \n , which is not\nprime, but is also not an integer. In fact, if \n an even integer,\nthen \nIII.\n6. \n : First, think about the prime numbers less than 12, the maximum\nsum of the numbers. These primes are 2, 3, 5, 7, 11.\nThe probability of rolling 2, 3, 5, 7, or 11 is equal to the number of ways\nto roll any of these sums divided by the total number of possible\noutcomes. The total number of possible outcomes is 6 × 6 = 36.\nSum of 2 can happen 1 way: 1 + 1\nSum of 3 can happen 2 ways: 1 + 2, 2 + 1\nSum of 5 can happen 4 ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1\nSum of 7 can happen 6 ways: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1"} +{"id": "book 2_p415_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 415, "page_end": 415, "topic_guess": "coordinate_geometry", "text": "Sum of 11 can happen 2 ways: 5 + 6, 6 + 5\nThat’s a total of 1 + 2 + 4 + 6 + 2 = 15 ways to roll a prime sum.\nThus, the probability is \n .\n7. (B)\n : The problem defines the function in words; you’ll need to\ntranslate into math.\n“The function {x} is defined as the area of a square with diagonal of\nlength x.”\n{x} = s2, where s is the side of the square.\nIf the diagonal equals x, then the side of the square equals \n .\nThe question stem also indicates that {x2} = x2. In other words, applying\nthe defined function to x2 will result in an answer of x2. First, find {x2}.\nTherefore:"} +{"id": "book 2_p416_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 416, "page_end": 416, "topic_guess": "coordinate_geometry", "text": "Now, set that equal to x2 and solve.\n(Note: It is acceptable to divide by x2 because the question stem\nindicates that x ≠ 0.) Therefore, x is equal to \n  or \n . Since the\nquestion stem indicates that x > 0, only \n is a valid solution.\nAlternatively, you could work backwards from the answers. Start with\n(B) or (D).\n(B) If \n , then x2 = 2. Next, the question specifies that that\n becomes {2}, so what does this function\nreturn?\n“The function {x} is defined as the area of a square with diagonal\nof length x.”\nThe function {2} is defined as the area of a square with diagonal length\n2. The side length of this square is \n , so the area of this square is\n. Therefore, {2} = 2, which matches the question\nstem specification that {x2} = x2."} +{"id": "book 2_p417_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 417, "page_end": 417, "topic_guess": "number_theory", "text": "The correct answer is (B).\n8. 2,500π: Microchip radius = (2.5 cm)(10 mm/cm) = 25 mm\n9. (C): When consecutive integers are divided by the same number, the\nremainders follow a repeating pattern. For instance, when consecutive\nintegers are divided by 4, the remainders form a repeating pattern of [1,\n2, 3, 0], with every fourth integer being divisible by 4. \n(1) INSUFFICIENT: This statement indicates that the greatest of the three\nintegers is divisible by x. However, the sum of the remainders depends\non the value of x. For instance, if x = 10, the three remainders would be 8,\n9, and 0, respectively, and their sum would be 17. If x = 5, the three\nremainders would be 3, 4, and 0, with a sum of 7."} +{"id": "book 2_p418_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 418, "page_end": 418, "topic_guess": "number_theory", "text": "(2) INSUFFICIENT: Statement (2) indicates that the least of the three\nintegers is one greater than a multiple of x. However, the sum of the\nremainders again depends on the value of x. If x = 10, the three\nremainders would be 1, 2, and 3, with a sum of 6. If x = 2, the three\nremainders would be 1, 0, and 1, with a sum of 2. \n(1) AND (2) SUFFICIENT: Call the three consecutive integers y, y + 1, and\ny + 2. The remainder when y is divided by x is 1 and the remainder when\ny + 2 is divided by x is 0. Write out the known remainders and write out\nenough other terms to see where the pattern repeats. In this case, the\nterm y – 1 also has a remainder of 0, so the pattern repeats every three\nterms.\ny − 1 y y + 1 y + 2\n0 1 ? 0\nThe only way for the remainders to have a consistent repeating pattern\nis for the missing remainder to equal 2 (in which case x must equal 3).\nTherefore, the sum of the three remainders is 1 + 2 + 0 = 3. \n10. (A): The question asks for the value of x. Translate the equation given in\nthe question stem."} +{"id": "book 2_p419_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 419, "page_end": 419, "topic_guess": "number_theory", "text": "The problem stem specifies that x is an integer, so the right-hand side of\nthe equation must also be an integer. Therefore, z must be a factor of y2.\nThe question asks for the value of x.\n(1) SUFFICIENT: If y is prime, then there are two possible scenarios.\nCase 1: y = z. This isn’t allowed, though, because the question stem\nindicates that the variables represent three different positive integers.\nCase 2: y2 = z, in which case x must equal 1. This is the only possible\ncase, so statement (1) is sufficient.\n(2) INSUFFICIENT: If z = 9, then y2 is a multiple of 9, so z must be a\nmultiple of 3. It’s possible that z = 9, y = 3, and x = 1. It’s also possible\nthat z = 9, y = 36, and x = 4.\nThe correct answer is (A)."} +{"id": "book 2_p420_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 420, "page_end": 420, "topic_guess": "general", "text": "PART THREE\nPractice"} +{"id": "book 2_p421_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 421, "page_end": 421, "topic_guess": "general", "text": "CHAPTER 9\nWorkout Sets"} +{"id": "book 2_p422_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 422, "page_end": 422, "topic_guess": "general", "text": "In This Chapter...\nWorkout Set 1\nWorkout Set 2\nWorkout Set 3\nWorkout Set 4\nWorkout Set 5\nWorkout Set 6\nWorkout Set 7\nWorkout Set 8\nWorkout Set 9\nWorkout Set 10\nWorkout Set 11\nWorkout Set 12\nWorkout Set 13\nWorkout Set 14\nWorkout Set 15\nWorkout Set 16"} +{"id": "book 2_p423_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 423, "page_end": 423, "topic_guess": "geometry", "text": "CHAPTER 9 Workout Sets\nWorkout Set 1\n1. \nFigure not drawn to scale.\nThe circle with center O has a circumference of \n . If AC is a\ndiameter of the circle, what is the length of BC ?"} +{"id": "book 2_p424_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 424, "page_end": 424, "topic_guess": "number_theory", "text": "(A)\n3(B)\n(C)\n9(D)\n(E)\n2. A batch of widgets costs p + 15 dollars for a company to produce\nand each batch sells for p(9 − p) dollars. For which of the following\nvalues of p, does the company make a profit?\n3(A)\n4(B)\n5(C)\n6(D)\n7(E)\n3. If K is the sum of the reciprocals of the consecutive integers from 41\nto 60 inclusive, which of the following is less than K ?\nI."} +{"id": "book 2_p425_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 425, "page_end": 425, "topic_guess": "sets_probability_counting", "text": "II.\nIII.\nNone(A)\nI only(B)\nII only(C)\nI and II only(D)\nI, II, and III(E)\n4. Triplets Adam, Bruce, and Charlie enter a triathlon. There are nine\ncompetitors in the triathlon. If every competitor has an equal\nchance of winning, and three medals will be awarded, what is the\nprobability that at least two of the triplets will win a medal?"} +{"id": "book 2_p426_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 426, "page_end": 426, "topic_guess": "word_problems", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n5. Floyd started at one end of a trail at 8:07 a.m. and hiked the entire 5\nmiles to the other end of the trail at an average rate of 3 miles per\nhour. Connie started at the same end of the trail at 8:30 a.m. and\nhiked the entire length of the trail at an average rate of 2.5 miles per\nhour. Finally, Karen started her hike in the same place at 10:00 a.m.\nand hiked the entire length of the trail at an average rate of 4 miles\nper hour. During approximately what percent of the 4-hour period\nfrom 8:00 a.m. to noon were at least two of the three hikers on the\ntrail?"} +{"id": "book 2_p427_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 427, "page_end": 427, "topic_guess": "word_problems", "text": "34%(A)\n45%(B)\n55%(C)\n59%(D)\n68%(E)\n6. Half an hour a er Car A started traveling from Newtown to\nOldtown, a distance of 62 miles, Car B started traveling along the\nsame road from Oldtown to Newtown. The cars met each other on\nthe road 15 minutes a er Car B started its trip. If Car A traveled at a\nconstant rate that was 8 miles per hour greater than Car B’s\nconstant rate, how many miles had Car B driven when they met?\n14(A)\n12(B)\n10(C)\n  9(D)\n  8(E)\n7. If x = 2b − (88 + 86), for which of the following b values is x closest to\nzero?\n20(A)\n24(B)\n25(C)\n30(D)\n42(E)"} +{"id": "book 2_p428_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 428, "page_end": 428, "topic_guess": "fractions_decimals_percents", "text": "8. If k > 1, which of the following must be equal to\n ?\n2(A)\n(B)\n(C)\n(D)\n(E)\n9. If y is 20% less than 90% of x and z is 25% more than 130% of y,\nthen z is what percent of x ?\n  72%(A)\n  92.5%(B)\n108.5%(C)\n117%(D)\n135%(E)\n10. Set A consists of three consecutive positive multiples of 3, and set B\nconsists of five consecutive positive multiples of 5. If the sum of the"} +{"id": "book 2_p429_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 429, "page_end": 429, "topic_guess": "number_theory", "text": "integers in set A is equal to the sum of the integers in set B, what is\nthe least number that could be a member of set A ?\n69(A)\n72(B)\n75(C)\n78(D)\n81(E)"} +{"id": "book 2_p430_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 430, "page_end": 430, "topic_guess": "general", "text": "Workout Set 1 Answer Key\n1. D\n2. B\n3. D\n4. B\n5. B\n6. A\n7. B\n8. E\n9. D\n10. B"} +{"id": "book 2_p431_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 431, "page_end": 431, "topic_guess": "geometry", "text": "Workout Set 1 Solutions\n1. (D) 9: If AC is a diameter of the circle, then triangle ABC is a right triangle,\nwith angle ABC = 90 degrees. The shortest side of a triangle is across\nfrom its least angle and the longest side of a triangle is across from its\ngreatest angle. Therefore, AC > BC > AB.\nThe circumference of the circle = \n , so\n. Thus, AC ≈ 10.2 and BC < 10.2.\nAnswer (E) is too great and answer (D), while less than 10.2, is\nquestionable, because side BC is opposite a 60-degree angle. The value\nshould be less than 9.\nSince angle ACB is 30 degrees, angle CAB is 60 degrees.\nThe sides in a 30–60–90 triangle have the ratio \n , so use\nthe ratio to compute the desired side, BC.\nFind the length of BC."} +{"id": "book 2_p432_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 432, "page_end": 432, "topic_guess": "strategy", "text": "Therefore, BC has length 9.\nThe correct answer is (D).\n2. (B) 4: You can work backwards from the answers. Start with (B) or (D).\nEliminate (D) and try (B) next.\nThe company makes a profit when p = 4.\nAlternatively, profit equals revenue minus cost. The company’s profit is:\n(D) p = 6\nCost: 6 + 15 = 21\nRevenue: 6(9 − 6) = 18\nProfit: 18 − 21 = −3              Loss, not profit!\n(B) p = 4\nCost: 4 + 15 = 19\nRevenue: 4(9 − 4) = 20\nProfit: 20 − 19 = 1"} +{"id": "book 2_p433_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 433, "page_end": 433, "topic_guess": "fractions_decimals_percents", "text": "Profit will be zero if p = 5 or p = 3, which eliminates answers (A) and (C).\nFor p > 5, both (p − 5) and (p − 3) are positive. In that case, the profit is\nnegative (i.e., the company loses money). The profit is only positive if (p\n− 5) and (p − 3) have opposite signs, which occurs when 3 < p < 5.\nThe correct answer is (B).\n3. (D) I and II only: The sum \n has 20\nfractional terms. It is impossible to compute this by hand in two\nminutes. Instead, look at the maximum and minimum possible values\nfor the sum.\nMaximum: The greatest fraction in the sum is \n  . K is definitely less\nthan 20 × \n , or \n  , which is less than \n  .\nMinimum: The least fraction in the sum is \n  . K is definitely greater\nthan 20 × \n = \n  .\nTherefore, \n < K < \n  .\n < \nI.\n\nII."} +{"id": "book 2_p434_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 434, "page_end": 434, "topic_guess": "sets_probability_counting", "text": "The correct answer is (D).\n\nIII.\n4. (B) \n : With nine competitors and only three medals awarded, only \nof the competitors will win overall. Although a simplification, it is\nreasonable for each competitor to see his or her chance of winning a\nmedal as \n , or to expect to win \n of a medal (pretending for a\nmoment that medals can be shared).\nThe question asks for the probability that at least two of the triplets will\nwin a medal. In other words, you want \n to \n of the triplets to win\nmedals or for each triplet to win \n to \n of a medal. Since \n and \nare both greater than \n , you are looking for the probability that the\ntriplets will win medals at a rate greater than that expected for\ncompetitors overall. In other words, this would be an unusual outcome.\nThus, the probability should be less than \n . Eliminate (D) and (E). You\ncould then guess from among the remaining answers with a 1 in 3\nchance of guessing correctly.\nTo solve, use the probability formula and combinatorics."} +{"id": "book 2_p435_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 435, "page_end": 435, "topic_guess": "general", "text": "First, find the total number of outcomes for the triathlon. There are nine\ncompetitors; three will win medals and six will not. Set up an anagram\ngrid where Y represents a medal, N no medal.\nCompetitor C1 C2 C3 C4 C5 C6 C7 C8 C9\nMedal Y Y Y N N N N N N\nTherefore, the number of ways three medals can be awarded is\n.\nNow, determine the number of instances when at least two brothers win\na medal. Practically speaking, this could happen when 1) exactly three\nbrothers win or 2) exactly two brothers win.\nStart with all three triplets winning medals, where Y represents a medal.\nTriplet A B C Non-triplet C1 C2 C3 C4 C5 C6\nMedal Y Y Y Medal N N N N N N"} +{"id": "book 2_p436_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 436, "page_end": 436, "topic_guess": "sets_probability_counting", "text": "The number of ways this could happen is \n . This makes\nsense, as there is only one instance in which all three triplets would win\nmedals and all of the other competitors would not. (If you recognize this\nimmediately, no need to write out the math.)\nNext, calculate the instances when exactly two of the triplets win\nmedals.\nTriplet A B C Non-triplet C1 C2 C3 C4 C5 C6\nMedal Y Y N Medal Y N N N N N\nSince both triplets and non-triplets win medals in this scenario, consider\nthe possibilities for both sides of the grid. For the triplets, the number of\nways that two could win medals is \n .\nFor the non-triplet competitors, the number of ways that one could win\nthe remaining medal is \n .\nMultiply these two numbers to get the total number of instances: 3 × 6 =\n18.\nThe brothers win at least two medals in 18 + 1 = 19 cases. The total\nnumber of cases is 84, so the probability is \n ."} +{"id": "book 2_p437_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 437, "page_end": 437, "topic_guess": "word_problems", "text": "The correct answer is (B).\n5. (B) 45%: To answer the question, you’ll need to identify the periods of\ntime during which different numbers of hikers are on the trail. To\norganize your work, start by drawing out a diagram showing the four\nhours between 8:00 a.m. and noon.\nNext, identify the time period during which each hiker was on the trail.\nFloyd traveled the 5 miles at a rate of 3 miles per hour (mph), so he\nspent \n = 1 hour and 40 minutes on the trail. Therefore, Floyd\nwas on the trail from 8:07 to 9:47.\nConnie traveled the 5 miles at a rate of 2.5 mph, so she spent\n = 2 hours on the trail. Connie was on the trail from 8:30 to\n10:30.\nFinally, Karen traveled the 5 miles at a rate of 4 mph, so she spent\n = 1 hour 15 minutes on the trail. Karen was on the trail from\n10:00 to 11:15.\nPlot each of these time periods on your drawing."} +{"id": "book 2_p438_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 438, "page_end": 438, "topic_guess": "word_problems", "text": "There were no periods of time during which all three hikers were on the\ntrail. However, there were two periods during which exactly two of the\nthree hikers were on the trail. Floyd and Connie were both on the trail\nfrom 8:30 to 9:47, and Connie and Karen were both on the trail from\n10:00 to 10:30. This represents a total of 1 hour and 47 minutes, or\napproximately 1.75 hours.\n1.75 hours is more than a third of 4 hours, but less than half of 4 hours.\nTherefore, the only reasonable answer is (B). It is also possible to\nsimplify the fraction more precisely.\nThe correct answer is (B).\n6. (A) 14: Draw a diagram to illustrate the moment at which Car A and Car B\npass each other moving in opposite directions."} +{"id": "book 2_p439_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 439, "page_end": 439, "topic_guess": "word_problems", "text": "Try Working Backwards from the answers, starting with (B) or (D).\n  B’s Distance\n(miles)\nB’s Rate\n(mph)\n=\n\n= \nA’s Rate\n(mph)\n= B’s Rate +\n8\nA’s Distance\n(miles)\n= R × T\n= R × 0.75\nTotal Distance =\n62?\n(B) 12 48 56 42 54\nThe total distance is not 62, so answer (B) is incorrect. Furthermore, a\ndistance of 54 is too short, so the answer must be (A). If you aren’t sure,\nconfirm by checking answer (A).\n  B’s Distance\n(miles)\nB’s Rate\n(mph)\n=\n\n= \nA’s Rate\n(mph)\n= B’s Rate +\n8\nA’s Distance\n(miles)\n= R × T\n= R × 0.75\nTotal Distance =\n62?\n(A) 14 56 64 48 62\nAlternatively, you could solve algebraically, using a rate–time–distance\n(RTD) chart. Note that you must convert 15 minutes to 0.25 hours.\n  Rate Time Distance"} +{"id": "book 2_p440_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 440, "page_end": 440, "topic_guess": "word_problems", "text": "Rate Time Distance\nCar A (r + 8) mph 0.75 hours (0.75) (r + 8) miles\nCar B r mph 0.25 hours 0.25r miles\nTotal     62 miles\nSet up and solve an equation for the total distance.\nTherefore, in 15 minutes, Car B traveled a distance of 0.25r = (0.25)(56) =\n14 miles.\nThe correct answer is (A).\n7. (B) 24: Testing the choices would be a natural way to solve this problem,\nsince the question doesn’t ask you to solve for b in general, but rather\n“for which of the following is x closest to zero?” However, numbers\nbetween 220 and 242 are too great to plug and compute. Instead,\nmanipulate the terms with base 8 to see how they might balance with\n2b."} +{"id": "book 2_p441_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 441, "page_end": 441, "topic_guess": "algebra", "text": "The correct answer is (B).\n8. \n Since there are variables in the\nanswer choices, choose a smart number to solve. If k = 2, then\n,\nwhich is less than 1. Now, test the answer choices and try to match the\ntarget; stop if you can tell that an answer won’t equal the target.\n(A) 2 Too high. Eliminate.\n(B)\n Too high. Eliminate.\n(C)\n Too high. Eliminate.\n(D)\n Too high. Eliminate.\n(E)\n Correct!\nAlternatively, solve the problem algebraically. The expression given is of\nthe form \n , where \n and \n ."} +{"id": "book 2_p442_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 442, "page_end": 442, "topic_guess": "algebra", "text": "You need to either simplify or cancel the denominator, as none of the\nanswer choices have the starting denominator and most of the choices\nhave no denominator at all. First, try to eliminate the radical signs\nentirely, leaving only a2 and b2 in the denominator. To do so, multiply by\na fraction that is a convenient form of 1.\nNotice the “difference of two squares” special product created in the\ndenominator.\nSubstituting for a and b:\nThe correct answer is (E).\n9. (D) 117%: An algebraic setup would be fairly ugly on this problem, but a\ncombination of two other techniques work quite nicely: choose smart\nnumbers and estimate.\nThere are no real values for the variables, so Choosing Smart Numbers is\na valid approach. Furthermore, the answer choices are far enough apart"} +{"id": "book 2_p443_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 443, "page_end": 443, "topic_guess": "algebra", "text": "that you can estimate if the Smart Numbers get a little messy.\nOf the three variables, it is easiest to pick for x. If you were to pick for y\nor z, you would have to do “reverse” calculations to find the other\nvariables. Since this is a percent problem, try 100 first.\ne.g., Since x = 100, 90% of x is equal to 90. Therefore, you is equal\nto 20% less than 90 = 90 − 18 = 72\nThis number is a little ugly, so round down to 70. Therefore, y = 70, so\n130% of y is equal to 70 + 21 = 91.\nThe next step requires you to take 25% of that number, so 91 is going to\nget messy. Estimate again, but this time round up a little bit (to offset\nthe error you introduced when you rounded down earlier). Round to the\nnearest number that is divisible by 4, which is 92. Next, find 25% more\nthan 92.\nz = 92 + 23 = 115\nIf x = 100 and z ≈ 115, then the final calculation is \nwhich is closest to 117%.\nThe correct answer is (D).\n10. (B) 72: The values in set A and set B are unknown. Jot them down in\nterms of variables. If x is a positive integer, then x, x + 1, and x + 2 are\nconsecutive integers, and 3x, 3(x + 1), and 3(x + 2) are three consecutive"} +{"id": "book 2_p444_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 444, "page_end": 444, "topic_guess": "number_theory", "text": "multiples of 3. Similarly, set B contains 5y, 5(y + 1), 5(y + 2), 5(y + 3), and\n5(y + 4), where y is a positive integer. Do not use the same variable for\nboth sets, since it is unclear whether or how they are related.\nSum the numbers in each set.\nSet A: 3x + 3(x + 1) + 3(x + 2) = 3(3x + 3)\nSet B: 5y + 5(y + 1) + 5(y + 2) + 5(y + 3) + 5(y + 4) = 5(5y + 10)\nThe sums of the sets are equal. Set them equal to each other and\nsimplify as much as possible.\n3(3x + 3) = 5(5y + 10)\n9(x + 1) = 25(y + 2)\nBecause x and y are positive integers, the right side of the equation is a\nmultiple of 25 and the le side is a multiple of 9. In other words, each\nside of the equation must be a multiple of both 9 and 25. Because 9 and\n25 have no common factors, x + 1 must be a multiple of 25 and y + 2\nmust be a multiple of 9. The least possible value for x is x = 24, and the\nleast number that could be a member of set A is 3x = 3(24) = 72.\nAlternatively, the sum of the integers in set A can be calculated by\nWorking Backwards from the choices. Manually add them up or note\nthat for an evenly spaced set the sum is (# of terms) × (the median term).\nThe integers in set B are multiples of 5, so the sum of set B is a multiple\nof 5. By extension, the (equivalent) sum of set A must be a multiple of 5."} +{"id": "book 2_p445_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 445, "page_end": 445, "topic_guess": "general", "text": "Try the answer choices; since the problem asks for the least value, start\nwith answer (A).\n(A) {69, 72, 75} sums to 3(72) = not a multiple of 5.\nEliminate.\n(B) {72, 75, 78} sums to 3(75) = a multiple of 5.\nCorrect.\nThe problem asks for the least value that could be a member of set A, so\nthe correct answer is (B)."} +{"id": "book 2_p446_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 446, "page_end": 446, "topic_guess": "general", "text": "Workout Set 2\n11. If a ≠ b, is \n   ?\n|a| > |b|(1)\na < b(2)\n12. If m = 4n + 9, where n is a positive integer, what is the greatest\ncommon factor of m and n ?\nm = 9s, where s is a positive integer(1)\nn = 4t, where t is a positive integer(2)\n13. A museum sold 30 tickets on Saturday. Some of the tickets sold\nwere $10 general exhibit tickets and the rest were $70 special\nexhibit tickets. How many general exhibit tickets did the museum\nsell on Saturday?\nThe museum’s total revenue from ticket sales on Saturday was\ngreater than $1,570 and less than $1,670.\n(1)\nThe museum sold more than 20, but fewer than 25, special\nexhibit tickets on Saturday.\n(2)"} +{"id": "book 2_p447_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 447, "page_end": 447, "topic_guess": "number_theory", "text": "14. If x and y are integers, is xy = yx ?\nx − y = 2(1)\nxy = 8(2)\n15. If ab3c4 > 0, is a3bc5 > 0 ?\nb > 0(1)\nc > 0(2)\n16. What is the perimeter of isosceles triangle ABC ?\nThe length of side AB is 9.(1)\nThe length of side BC is 4.(2)\n17. If x, y, and z are integers and 2x 5y z = 6.4 × 106, what is the value of\nxy ?\nz = 20(1)\nx = 9(2)\n18. Is x > y ?\ny2 < z2 < x2(1)\nz < x(2)"} +{"id": "book 2_p448_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 448, "page_end": 448, "topic_guess": "number_theory", "text": "19. A school has a students and b teachers. If a < 150, b < 25, and\nclasses have a maximum of 15 students, can the a students be\ndistributed among the b teachers so that each class has the same\nnumber of students? (Assume that any student can be taught by\nany teacher.)\nIt is possible to divide the students evenly into groups of 2, 3, 5,\n6, 9, 10, or 15.\n(1)\nThe greatest common factor of a and b is 10.(2)\n20. A set of five distinct positive integers has a median of 3 and a range\nof 12. What is the mean of the set of integers?\nThe product of the integers in the set is a multiple of 14.(1)\nThe sum of the integers in the set is a multiple of 13.(2)"} +{"id": "book 2_p449_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 449, "page_end": 449, "topic_guess": "general", "text": "Workout Set 2 Answer Key\n11. E\n12. A\n13. A\n14. C\n15. B\n16. C\n17. A\n18. C\n19. E\n20. D"} +{"id": "book 2_p450_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 450, "page_end": 450, "topic_guess": "general", "text": "Workout Set 2 Solutions\n11. (E): You can cross-multiply the inequality, as long as you consider both\ncases.\nThis is a conditional rephrased question—importantly, one with\ncompletely opposite questions as possibilities. Any statement that\ndoesn’t at least answer the question of whether a − b is positive or\nnegative is unlikely to be sufficient. You would also have to carry each\nstatement through both questions, so this path is not very efficient. It’s\nbetter to use the original inequality to test cases in order to eliminate\nthe incorrect answers.\n(1) INSUFFICIENT: Test cases to determine whether this statement is\nsufficient.\n  a b\n Is \nCase 1 2 1 ✓\n No\nCase 2 2\n ✓\n Yes\nIf a − b is negative, the question becomes “Is 1 < ab(a − b)?”\n(FLIPPED inequality sign)\nIf a − b is positive, the question becomes “Is 1 > ab(a − b)?”\n(ORIGINAL inequality sign)"} +{"id": "book 2_p451_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 451, "page_end": 451, "topic_guess": "general", "text": "(2) INSUFFICIENT: Test cases again.\na b a < b Is \n ?\nCase 1\n 1 ✓\n  Yes\nCase 2\n ✓\n ? No\n(1) AND (2) INSUFFICIENT: Whenever possible, reuse cases you’ve\nalready tested (you can only do this when the case makes both\nstatements true). The pairs tested for statement (2) are also valid for\nstatement (1).\na b Valid? Is \n ?\nCase 1\n 1 ✓\n ? Yes\nCase 2\n ✓\n ? No\nOnce again, two valid sets of numbers return two different answers, so\neven together, the statements are insufficient. The correct answer is (E)."} +{"id": "book 2_p452_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 452, "page_end": 452, "topic_guess": "number_theory", "text": "12. (A): The greatest common factor (GCF) of any two numbers is given by\nthe product of the shared or overlapping primes in the two numbers.\nAny such question can be rephrased accordingly: “What exactly are the\noverlapping factors of m and n?”\n(1) SUFFICIENT: This statement indicates that m must be a multiple of 9.\nIf m is a multiple of 9 and m = 4n + 9, then this equation is really saying.\nmultiple of 9 = 4n + 9\nIn order for this to be true, 4n must also be a multiple of 9. The number 4\ndoes not contain any factors of 9, so n itself must be a multiple of 9.\nIf both m and n are multiples of 9, and m is exactly 9 units away from 4n,\nthen the greatest possible common factor is 9, so this statement is\nsufficient.\n(If you’re not sure about the logic of that last part, test out a couple of\nreal numbers. Remember that n must be a multiple of 9. If n = 9, then m\n= 4(9) + 9 = 45. The GCF of the numbers 9 and 45 is 9. If n = 18, then m =\n4(18) + 9 = 81. The GCF of the numbers 18 and 81 is still 9.)\nAlternatively, methodically list possible values of s, m, and n. Ignore any\ncase for which the statement is not true; for the remaining cases,\ndetermine the answer to the question. \ns = pos int m = 9s\n n = pos int GCF of \nm and n?"} +{"id": "book 2_p453_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 453, "page_end": 453, "topic_guess": "fractions_decimals_percents", "text": "s = pos int m = 9s\n n = pos int GCF of \nm and n?\n1 9\n Zero ✗\n2 18\n Fraction ✗\n3 27\n Fraction ✗\n4 36\n Fraction ✗\n5 45\n ✓ 9\n6 54\n Fraction ✗\n7 63\n Fraction ✗\n8 72\n Fraction ✗\n9 81\n ✓ 9\n10 90\n Fraction ✗"} +{"id": "book 2_p454_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 454, "page_end": 454, "topic_guess": "sequences_patterns", "text": "In this case, n is only a positive integer when (s − 1) is a multiple of 4.\nAlso, m and n are both multiples of 9. Here’s the proof:\nTry the next number in the pattern, 9 + 4 = 13, to be confident that the\napparent GCF pattern will continue.\ns = pos int m = 9s\n n = pos int GCF of\nm and n?\n5 (9)(5) = 45\n ✓ 9\n9 (9)(9) = 81\n ✓ 9\n13 (9)(13) = 117\n ✓ 9\nThe variables m and n always share an overlapping factor of 9, but there\nis never any overlap between their remaining factors. The GCF is always\n9.\n(2) INSUFFICIENT: n = 4t, where t is a positive integer. In this case, n must\nbe a multiple of 4. This does not allow you to deduce anything\nconsistent about m, as statement (1) did."} +{"id": "book 2_p455_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 455, "page_end": 455, "topic_guess": "statistics", "text": "t = pos int n = 4t m = 4n+ 9 GCF of \nm and n ?\n1 4 16 + 9 = 25 1\n2 8 32 + 9 = 41 1\n3 12 48 + 9 = 57 3\nThere is more than one possible GCF, so the statement is not sufficient.\nThe correct answer is (A).\n13. (A):\n(1) SUFFICIENT: Use the integer constraint to test possible cases. Let g\nequal the number of general exhibit tickets sold. Because the special\ntickets are so much more expensive, begin by choosing numbers for\nwhich (70)(# of special tickets) is approximately $1,600.\nIf g = 7, total revenue = 10(7) + 70(23) = 70 + 1,610 = 1,680. Too high.\nIf g = 8, total revenue = 10(8) + 70(22) = 80 + 1,540 = 1,620. Ok.\nIf g = 9, total revenue = 10(9) + 70(21) = 90 + 1,470 = 1,560. Too low.\nOnly g = 8 gives a total revenue in the given range.\nDon’t assume that having a range of values automatically means a\nstatement is insufficient to answer a Value question. At times, a"} +{"id": "book 2_p456_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 456, "page_end": 456, "topic_guess": "algebra", "text": "constraint (such as an integer constraint) may limit the number of cases\nto just one possibility.\nAlternatively, you can use algebra, though that path is a bit long on this\nproblem. Let g equal the number of general exhibit tickets sold. Then\n(30 − g) represents the number of special exhibit tickets sold. Set up and\nsolve the following inequality:\nThe only integer between 7.2 and 8.8 is 8, so g must be 8. The museum\nsold 8 general exhibit tickets.\n(2) INSUFFICIENT: This statement indicates that the museum sold 21, 22,\n23, or 24 special exhibit tickets. Since the museum sold a total of 30\ntickets, this means that it sold 9, 8, 7, or 6 general exhibit tickets.\nThe correct answer is (A).\n14. (C):\n(1) INSUFFICIENT: Test cases on this theory problem. If you can generate\nboth Yes and No cases, you will prove insufficiency.\nx y xy yx Does xy = yx ?"} +{"id": "book 2_p457_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 457, "page_end": 457, "topic_guess": "number_theory", "text": "x y xy yx Does xy = yx ?\nCase 1 2 0 20 = 1 02 = 0 No\nCase 2 3 1 31 = 3 13 = 1 No\nCase 3 4 2 42 = 16 24 = 16 Yes\n(2) INSUFFICIENT: Continue to test cases.\nx y xy yx Does xy = yx ?\nCase 1 8 1 81 = 8 18 = 1 No\nCase 2 4 2 42 = 16 24 = 16 Yes\n(1) AND (2) SUFFICIENT: Test whether any cases work for both\nstatements. The case x = 4 and y = 2 was already used for both\nstatements and returned a Yes answer. Can any of the No cases work for\nboth statements?\nThey don’t. Now you have a choice. If you are good with numbers, you\ncan try to find another pair of integers that will work with both\nequations. Since x and y have to be integers, there are only a limited\nnumber of possibilities. The two numbers need to multiply to 4 and"} +{"id": "book 2_p458_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 458, "page_end": 458, "topic_guess": "algebra", "text": "have a difference of 2. In addition to 4 and 2, the pair −4 and −2 fit the\nbill, as long as x = −2 and y = −4.\nAlternatively, solve algebraically to determine the possible values for x\nand y.\nTherefore, x = −2 or x = 4.\nYou have already tested the case where x = 4 (and y = 2), so test the other\ncase.\nx y xy yx Does xy = yx ?\n−2 −4\n Yes\nIn either case, xy = yx. The correct answer is (C).\n15. (B): Odd exponents do not “hide the sign” of the base. If ab3c4 > 0, then a\nand b must have the same sign so that their product is positive. In that"} +{"id": "book 2_p459_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 459, "page_end": 459, "topic_guess": "geometry", "text": "case, a3b must also be positive. As a result, in order for the inequality\na3bc5 to be positive, c5 must be positive. The rephrased question is, “Is c\n> 0?”\n(1) INSUFFICIENT: No information about the sign of c.\n(2) SUFFICIENT: Answers the rephrased question directly.\nThe correct answer is (B).\n16. (C): The perimeter of a triangle is equal to the sum of the three sides.\n(1) INSUFFICIENT: Knowing the length of one side of the triangle is not\nenough to find the sum of all three sides.\n(2) INSUFFICIENT: Knowing the length of one side of the triangle is not\nenough to find the sum of all three sides.\n(1) AND (2) SUFFICIENT: Triangle ABC is an isosceles triangle, which\nmeans that two of the sides are equal in length. The statements provide\ntwo of the side lengths, so the third side, AC, must equal one of the\ngiven sides.\nThere is a hidden constraint in this problem: the triangle must be valid.\nRecall that the sum of the lengths of any two sides of a triangle must be\ngreater than the length of the third side.\nAB BC AC Perimeter Valid triangle?\n9 4 4 9 + 4 + 4 = 17 No: 4 + 4 < 9"} +{"id": "book 2_p460_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 460, "page_end": 460, "topic_guess": "geometry", "text": "AB BC AC Perimeter Valid triangle?\n9 4 9 9 + 4 + 9 = 22 Yes: 4 + 9 > 9\nA “triangle” with three sides of 4, 4, and 9 is not really a triangle, as it\ncannot be drawn with those dimensions.\nTherefore, the actual sides of the triangle must be AB = 9, BC = 4, and\nAC = 9. The perimeter is 22.\nThe correct answer is (C).\n17. (A): Express both sides of the equation in terms of prime numbers.\nThe right side of the equation is composed of only 2’s and 5’s. The le \nside of the equation has x number of 2’s and y number of 5’s along with\nsome factor z. This unknown factor z must be composed of only 2’s\nand/or 5’s, or it must be 1 (i.e., with no prime factors).\nIf z = 1, then x = 11 and y = 5."} +{"id": "book 2_p461_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 461, "page_end": 461, "topic_guess": "number_theory", "text": "If z = 2?5?, where the exponents are not 0, then x and y will depend on\nthe value of those exponents.\nThe rephrased question is thus, “How many factors of 2 and 5 are in z?”\n(1) SUFFICIENT: If z = 20 = 2251, then this answers the rephrased\nquestion. Incidentally, this implies that 2x5y(2251 ) = 21155, so x = 9 and y\n= 4, making xy = 36 (though you do not have to solve for x and y).\n(2) INSUFFICIENT: x = 9, but the statement doesn’t indicate anything\nabout the value of y.\nThe correct answer is (A).\n18. (C): This problem deals with both squares and inequalities. Remember\nthat negative numbers become positive when squared and that\nsquaring a number less than 1 results in a value that is even closer to 0.\n(1) INSUFFICIENT:  Without any more information about z, this\nstatement only tells you that x2 is greater than y2. However, x could be\neither greater than y or less than y, depending on whether x is positive\nor negative.\nCase 1: x = –100, y = 1. Since all squares are positive, x2 is much greater\nthan y2. However, x is less than y. \nCase 2: x = 100, y = 1. In this case, x2 is greater than y2 and x is greater\nthan y."} +{"id": "book 2_p462_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 462, "page_end": 462, "topic_guess": "word_problems", "text": "(2) INSUFFICIENT: This statement does not provide any information\nabout y, so y could be greater than, less than, or equal to x. \n(1) AND (2) SUFFICIENT: According to statement (2), x is greater than z.\nAccording to statement (1), x2 is greater than z2. This implies that |x| is\ngreater than |z|.\nThere are only two situations in which x is greater than z and the\nabsolute value of x is greater than the absolute value of z. One\npossibility is that x is positive and z is a lesser positive number. Another\npossibility is that x is positive and z is a negative number with an\nabsolute value that is less than x. These scenarios are illustrated below.\nThere are no valid cases in which x is negative because in order to fit\nstatement (2), z would have to be even more negative than x, in which\ncase |z| would be greater than, not less than, |x|. Therefore, x is definitely\npositive.\nStatement (1) says that x2 > y2, which implies that |x| > |y|. Since x is\npositive, |x| = x, so x > |y|. If x is greater than the absolute value of y, x\nmust also be greater than y itself, so the answer to the question is\ndefinitely Yes.\nThe correct answer is (C)."} +{"id": "book 2_p463_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 463, "page_end": 463, "topic_guess": "number_theory", "text": "19. (E): In order for the a students to be distributed evenly among the b\nteachers, a must be divisible by b. This is a factor question in disguise:\n“Is a divisible by b?”\n(1) INSUFFICIENT: The given numbers represent factors of a. If the a\nstudents can be divided evenly into a group of 2, then 2 is a factor of a.\nLikewise, 3 and 5 must be factors of a.\nIf 2 and 3 are already factors of a, then 6 must also be a factor (since 2 × 3\n= 6); you can ignore it. Likewise, you’ve already counted one factor of 3,\nso you need only one more factor of 3 to make 9. You already have the\nnecessary factors to make both 10 and 15, so ignore those numbers as\nwell. The final list of factors is: 2, 3, 5, and 3.\nAs a result, a must equal (2)(3)(5)(3) = 90 or a multiple of 90. Since the\nquestion stem indicated that a < 150, a must actually be 90. This\nstatement provides no information about the teachers, however, so it is\ninsufficient.\n(2) INSUFFICIENT: If the greatest common factor (GCF) of a and b is 10,\nthen a and b must both be multiples of 10. Since the question stem\nindicated that b < 25, b must be 10 or 20. The value of a is any multiple of\n10 up to 140, inclusive. This is not enough to determine whether a is\ndivisible by b. For example, if a = 50 and b = 10, then a is divisible by b. If\na = 50 and b = 20, then a is not divisible by b.\n(1) AND (2) INSUFFICIENT: From statement (1), a = 90. From statement\n(2), b = 10 or 20. If b = 10, then the 90 students can be divided up evenly\ninto classes of 9 students each. If b = 20, then the 90 students cannot be\ndivided up evenly."} +{"id": "book 2_p464_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 464, "page_end": 464, "topic_guess": "number_theory", "text": "The correct answer is (E).\n20. (D): All five of the integers in the set are positive, and all of the integers\nare distinct, or different from each other. Jot down blanks to represent\nthe values in the set.\n_ _ _ _ _\nThe median of the set is 3.\n_ _ 3 _ _ \nThe least two integers in the set must be less than the median, different\nfrom each other, and positive: the first two numbers must be 1 and 2.\n1 2 3 _ _\nFinally, the range of the set, or the difference between the greatest and\nleast integers, is 12. Therefore, the greatest integer in the set is 13.\n1 2 3 _ 13\nThe only unknown value in the set is the second-greatest integer, which\nmust be between 3 and 13, exclusive. Finding this integer will give you\nenough information to find the mean of the set.\n(1) SUFFICIENT: For a number to be a multiple of 14, it must be a\nmultiple of both 2 and 7. The product of the four known values is already\na multiple of 2, since the set already contains the number 2. However,\nthere are no multiples of 7 among the known values. Therefore, the"} +{"id": "book 2_p465_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 465, "page_end": 465, "topic_guess": "statistics", "text": "missing fourth value must be a multiple of 7. The only multiple of 7\nbetween 3 and 13 is 7 itself, so this is the missing integer and the mean\ncan now be calculated.\n(2) SUFFICIENT: The four known integers sum to 19. Since the unknown\ninteger must be positive, the sum of the entire set must be greater than\n19. The next multiples of 13 are 13(2) = 26 and 13(3) = 39, which would\nimply that the missing integer is 26 – 19 = 7 or 39 – 19 = 20, respectively.\n(For greater multiples of 13, the missing integer would have to be even\ngreater than 20.) However, the missing integer must be between 3 and\n13, so 20 is too great. The missing integer must be 7, and the mean can\nbe calculated.\nThe correct answer is (D)."} +{"id": "book 2_p466_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 466, "page_end": 466, "topic_guess": "number_theory", "text": "Workout Set 3\n21. Let f(x) equal the sum of all of the integers from 1 to x, inclusive,\nwhere x ≥ 1. If\n, which of\nthe following is true of g ?\n0 ≤ g < 5(A)\n5 ≤ g < 10(B)\n10 ≤ g < 15(C)\n15 ≤ g < 20(D)\n20 ≤ g < 25(E)\n22. If x and y are positive integers, what is the remainder when xyis\ndivided by 10 ?\nx = 26(1)\nyx = 1(2)\n23. For all positive integers n, the sequence An is defined by the\nfollowing relationship:"} +{"id": "book 2_p467_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 467, "page_end": 467, "topic_guess": "number_theory", "text": "What is the sum of all the terms in the sequence from A1 through\nA10, inclusive?\n(A)\n(B)\n(C)\n(D)\n(E)\n24. If x and y are positive integers and n = 5x + 7y + 3, what is the units\ndigit of n ?\ny = 2x − 16(1)\ny is divisible by 4.(2)"} +{"id": "book 2_p468_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 468, "page_end": 468, "topic_guess": "sequences_patterns", "text": "25. At a certain store, all notebooks have the same price and all pencils\nhave the same price. The price of four notebooks and three pencils\nis more than $12 and less than $13. The price of two notebooks and\nfive pencils is more than $8 and less than $9. If a notebook costs x\nmore than a pencil, which of the following could be the value of x ?\n$0.60(A)\n$1.05(B)\n$1.30(C)\n$2.20(D)\n$2.70(E)\n26. A certain sequence is defined by the following rule: Sn = k(Sn − 1),\nwhere k is a constant. If S1 = 2 and S13 = 72 , what is the value of S7  ?\n  6(A)\n12(B)\n24(C)\n36(D)\n37(E)\n27. If y =90 + 91 + 92 + … + 9n and integer n is greater than 2, what is the\nremainder when y is divided by 5 ?\nn is divisible by 3.(1)\nn is odd.(2)"} +{"id": "book 2_p469_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 469, "page_end": 469, "topic_guess": "sequences_patterns", "text": "28. In a certain sequence, each term a er the first is twice the previous\nterm. If the first term of the sequence is 3, what is the sum of the\n14th, 15th, and 16th terms in the sequence?\n3(216)(A)\n9(215)(B)\n21(214)(C)\n9(214)(D)\n21(213)(E)\n29. n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined\nas tn = tn − 1 + n. If t0 = 3, is tn even?\nn + 1 is divisible by 3.(1)\nn − 1 is divisible by 4.(2)\n30. To complete a 120-mile trip, a train first travels at a constant rate of\nx miles per hour (mph) for 80 miles. Then it travels the remaining 40\nmiles at a constant rate of y mph. If the train had instead completed\nthe entire trip at a constant rate of z mph, the trip would have taken\n1 hour longer. Which of the following is the value of z in terms of x\nand y ?"} +{"id": "book 2_p470_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 470, "page_end": 470, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p471_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 471, "page_end": 471, "topic_guess": "general", "text": "Workout Set 3 Answer Key\n21. B\n22. A\n23. C\n24. B\n25. D\n26. B\n27. B\n28. E\n29. B\n30. C"} +{"id": "book 2_p472_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 472, "page_end": 472, "topic_guess": "number_theory", "text": "Workout Set Solutions\n21. (B) 5 ≤ g < 10: f(x) is the sum of all integers from 1 to x, inclusive.\nTherefore, f(x) equals the average of the integers from 1 to x, which is\n, multiplied by the number of integers from 1 to x, which is x.\nTherefore, f(x) =   \n .\nThe question asks for the sum of the values of \n , \n , and so on\nup to \n . Instead of summing these values by hand, look for a\npattern, since the problem only asks for a range and not a specific\nanswer. \nOne possibility is to use algebra. \n  can be simplified as follows:"} +{"id": "book 2_p473_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 473, "page_end": 473, "topic_guess": "sequences_patterns", "text": "Using this formula, the first few terms of the sum are \n ,\n, \n , \n , and so on, until the\ntenth term, which is \n . All of these terms are greater than \n and no\ngreater than 1. If all of the terms were 0.5, then the sum would be 10(0.5)\n= 5. However, if all of the terms were 1, then the sum would be 10.\nTherefore, the sum must be between 5 and 10.\nThis problem can also be solved without using a formula for f(x).\nManually calculate the first few values of f(x), and look for a pattern. \nx f(x)\n1 1\n = 1\n2 3\n3 6\n4 10\n5 15\n6 21\n7 28\netc. etc. etc."} +{"id": "book 2_p474_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 474, "page_end": 474, "topic_guess": "sequences_patterns", "text": "The values appear to be decreasing and approaching \n . If each term in\nthe sum is greater than \n but no greater than 1, their sum will be\ngreater than 5 but no greater than 10. \nThe correct answer is (B).\n22. (A): The question asks for the remainder when xy is divided by 10. The\nremainder of any number when divided by 10 is equal to the units digit\nof that number, so the question is really asking for the units digit of xy.\n(1) SUFFICIENT: The tendency is to deem statement (1) insufficient\nbecause no information is given about the value of y. But 26 has a units\ndigit of 6, and remember that 6any positive integer has a units digit of 6 (the\npattern is a single-term repeat).\n61 = 6\n    62 = 36\n    63 = 216\n    etc.\nThus, 26 raised to ANY positive integer power will also have a units digit\nof 6 and therefore a remainder of 6 when divided by 10.\n(2) INSUFFICIENT: Given that yx = 1, there are a few possible scenarios:"} +{"id": "book 2_p475_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 475, "page_end": 475, "topic_guess": "number_theory", "text": "x y yx = 1\n0 anything nonzero (anything nonzero)0 = 1 ✓\nanything 1 1anything = 1 ✓\neven −1 −1even = 1 ✓\nHowever, the question stem indicates that x and y are POSITIVE integers,\nso eliminate the first and third scenarios.\nThe remaining scenario indicates that y = 1 and x = any positive integer.\nWithout more information about x, you cannot determine the remainder\nwhen xy is divided by 10.\nSince statement (1) indicates the value of x and statement (2) indicates\nthe value of y (y = 1), the temptation might be to combine the\ninformation to arrive at an answer of (C). This is a common trap on\ndifficult Data Sufficiency problems; in this case, you don’t need both\nstatements because statement (1) is sufficient by itself.\nThe correct answer is (A).\n23. (C) \n : Set up a table to list the first few terms of the sequence\nand also the cumulative sum."} +{"id": "book 2_p476_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 476, "page_end": 476, "topic_guess": "sequences_patterns", "text": "n\n Cumulative Sumn\n Cumulative Sum\n1\n 0\n2\n3\n4\nAs you build the table, compare the input column values (n) with the\noutput column values (cumulative sum), looking for a pattern. The\ndenominator of the cumulative sum is n! and the numerator is one less\nthan n!. Therefore:\nSum of terms through \nSubstitute to find the sum through A10.\nSum of terms through \nThe correct answer is (C)."} +{"id": "book 2_p477_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 477, "page_end": 477, "topic_guess": "sequences_patterns", "text": "24. (B): The units digit of n is determined solely by the units digit of the\nexpressions 5x and 7y + 3 because when two numbers are added\ntogether, the units digit of the sum is determined solely by the units\ndigits of the added numbers.\nSince x is a positive integer, and 5any positive integer always has a units digit\nof 5, 5x always ends in a 5. However, the units digit of 7y + 3 is not certain,\nas the units digit pattern for the powers of 7 is a four-term repeat: [7, 9,\n3, 1].\nThe question can be rephrased as, “What is the units digit of 7y + 3?”\nNote: Determining y would be one way of answering the question above,\nbut don’t rephrase to, “What is y?” Because the units digits of the\npowers of 7 have a repeating pattern, you might get a single answer for\nthe units digit of 7y + 3 despite having multiple values for y.\n(1) INSUFFICIENT: This statement indicates neither the value of y nor the\nunits digit of 7y + 3, as y depends on the value of x, which could be any\npositive integer. For example, if x = 9, then y = 2, and 7y + 3 has a units\ndigit of 7. By contrast, if x = 10, then y = 4, and 7y + 3 has a units digit of 3.\n(Note: The statement does indicate that y is an even number, so the\nexponent y + 3 must be odd. As a result, there are only two possible\nvalues for the units digit of the desired term: 7 or 3. That could be useful\nto know if you later need to combine the two statements.)\n(2) SUFFICIENT: Regardless of what multiple of 4 you pick, 7y + 3 will have\nthe same units digit."} +{"id": "book 2_p478_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 478, "page_end": 478, "topic_guess": "sequences_patterns", "text": "y y + 3 Units Digit of 7y + 3y y + 3 Units Digit of 7y + 3\n4 7 3\n8 11 3\n12 15 3\nUltimately, this means that n has a units digit of 5 + 3 = 8.\nThe correct answer is (B).\n25. (D) $2.20: The answer choices in this problem are relatively simple\nnumbers, but they represent the difference between two prices, rather\nthan a single price. Therefore, Working Backwards may not be possible.\nInstead, start by translating the information in the problem into math.\nLet n be the price of one notebook, and p be the price of one pen. The\nfirst piece of information in the problem translates to the following\ninequality:\nThe second piece of information translates to this inequality:"} +{"id": "book 2_p479_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 479, "page_end": 479, "topic_guess": "statistics", "text": "The question asks for the difference between the cost of a notebook and\nthe cost of a pencil. In other words, what is the value of n – p ?\nOne option is to use the previous two inequalities to create a third\ninequality, which includes the combo n – p. One way to accomplish this\nis to multiply the second inequality by –1, taking care to flip the signs.\nThen, reorder the result so that the inequality signs point in the\nexpected direction:\nNow, sum the two inequalities. You can add inequalities as long as the\ninequality signs are facing the same direction:\nThis simplifies to 1.5 < n – p < 2.5. The difference between the cost of a\nnotebook and the cost of a pencil must be between $1.50 and $2.50, so\nthe only answer choice that is in the correct range is answer (D). \nIt is also possible to “solve” the two inequalities for n and for p,\nseparately, then to use the resulting ranges to find the possible\ndifferences between n and p."} +{"id": "book 2_p480_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 480, "page_end": 480, "topic_guess": "sequences_patterns", "text": "The correct answer is (D).\n26. (B) 12: To form each new term of the sequence, multiply the previous\nterm by k. If S1 = 2, then S2 = 2k, and S3 = 2k2, and Sn = 2kn − 1.\nThus, S13 = 2k12. Since the problem indicates that S13 = 72, set up an\nequation to solve for k.\nS7 is equal to 2k6. Plug the value of k into that expression to calculate\nS7. \nThe correct answer is (B).\n27. (B): Remember the units digit pattern for 9x, where x is an integer. The\nunits digit of 9x is 9 if x is odd, but the units digit is 1 if x is even: a\nrepeating pattern of [9, 1].\nNow, consider the sums of the powers of 9 up to 9n:"} +{"id": "book 2_p481_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 481, "page_end": 481, "topic_guess": "sequences_patterns", "text": "n y = 90 + 91 + 92 + . . . + 9n Units Digit of yn y = 90 + 91 + 92 + . . . + 9n Units Digit of y\n1 1 + 9 = 10 0\n2 1 + 9 + 81 = 91 1\n3 [1 + 9 + 81] + 729 = [units digit of 1] + units digit of 9 0\n4 Units digit of 0 + units digit of 1 1\nOdd Units digit: pairs of (1 + 9) 0\nEven Units digit: pairs of (1 + 9) + 1 1\nThe alternating 1’s and 9’s in the units digits pair to a sum of 10, or a\nunits digit of 0. Thus, the units digit of the sum displays another two-\nterm repeating pattern. The units digit of y is 0 if n is odd, but 1 if n is\neven.\nThe remainder when y is divided by 5 depends only on the units digit\nand will be either 0 or 1 as well. The rephrased question is, “Is n odd or\neven?”\n(1) INSUFFICIENT: If n is a multiple of 3, it may be either odd or even.\n(2) SUFFICIENT: If n is odd, the units digit of y is 0, and the remainder is 0\nwhen y is divided by 5."} +{"id": "book 2_p482_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 482, "page_end": 482, "topic_guess": "sequences_patterns", "text": "The correct answer is (B).\n28. (E) 21(213): The sequence is 3, 6, 12, 24, 48, and so on. You could write\nout the first 16 terms and add the 14th, 15th, and 16th together, but\nsuch an approach would be prone to error and time-consuming.\nAdditionally, you don’t need to calculate the sum explicitly: the answer\nchoices all have some power of 2 as a factor, providing a hint at the best\nsolution method.\nWrite the sequence in terms of the powers of 2:\n3, 6, 12, 24, 48, and so on\n    3(20), 3(21), 3(22), 3(23), 3(24), and so on\nSo Sn, the nth term of the sequence, equals 3(2n − 1).\nThus, the sum of the 14th, 15th, and 16th terms equals 3(213) + 3(214) +\n3(215).\nAll of the terms share the common factors 3 and 213, so factor those\nterms out:"} +{"id": "book 2_p483_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 483, "page_end": 483, "topic_guess": "sequences_patterns", "text": "The correct answer is (E).\n29. (B): Sequence problems are o en best approached by charting out the\nfirst several terms of the given sequence. In this case, keep track of n, tn,\nand whether tn is even or odd.\nn tn Is tn even or odd?\n0 3 Odd\n1 3 + 1 = 4 Even\n2 4 + 2 = 6 Even\n3 6 + 3 = 9 Odd\n4 9 + 4 = 13 Odd\n5 13 + 5 = 18 Even\n6 18 + 6 = 24 Even\n7 24 + 7 = 31 Odd\n8 31 + 8 = 39 Odd\nNotice that beginning with n = 1, a four-term repeating cycle of [even,\neven, odd, odd] emerges for tn. Thus, a statement will be sufficient only"} +{"id": "book 2_p484_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 484, "page_end": 484, "topic_guess": "algebra", "text": "if it indicates how n relates to a multiple of 4 (i.e., n = a multiple of 4 ±\nknown constant).\n(1) INSUFFICIENT: This statement does not indicate how n relates to a\nmultiple of 4. If n + 1 is a multiple of 3, then n + 1 could be 3, 6, 9, 12, 15,\netc. This means that n could be 2, 5, 8, 11, 14, etc. From the chart, if n = 2\nor n = 5, then tn is even. However, if n = 8 or n = 11, then tn is odd.\n(2) SUFFICIENT: This statement indicates exactly how n relates to a\nmultiple of 4. If n − 1 is a multiple of 4, then n − 1 could be 4, 8, 12, 16, 20,\netc., and n could be 5, 9, 13, 17, 21, etc. From the chart (and the\ncontinuation of the four-term pattern), tn must be even.\nThe correct answer is (B).\n30. (C) \n : The answer choices are complex and include\ntwo different variables. If you choose to use Smart Numbers, select your\nnumbers cautiously to simplify the arithmetic.\nFor instance, if x is equal to y, the problem may be much simpler to\nsolve, although some of the answer choices may then yield the same\nresult. Choose x = y = 40, so that the train completes the entire trip at 40\nmph, taking \n  = 3 hours.\nIf the train had instead traveled at z mph, it would have taken one more\nhour, or 4 hours in total. Therefore, z = \n = 30 mph."} +{"id": "book 2_p485_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 485, "page_end": 485, "topic_guess": "word_problems", "text": "The correct answer will yield a value of z = 30 mph when x = 40 and y =\n40 are plugged in.\n(A) \n is a fraction that is significantly less\nthan 1. Eliminate.\n(B) \n , which\nis too low. Eliminate.\n(C) \nCorrect!\n(D) \n is much greater than 30.\nEliminate. \n(E) \n  is not an integer, so it cannot equal 30. Eliminate. \nAlternatively, solve algebraically by creating an RTD chart.\nRate Time Distance\nx mph\n  hours 80 miles"} +{"id": "book 2_p486_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 486, "page_end": 486, "topic_guess": "word_problems", "text": "Rate Time Distance\ny mph\n  hours 40 miles\n(Original speed)\n  hours 120 miles\nz mph\n hours 120 miles\nUse the last row of the chart to create an equation, which can then be\nsolved for z. \nThe correct answer is (C)."} +{"id": "book 2_p487_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 487, "page_end": 487, "topic_guess": "number_theory", "text": "Workout Set 4\n31. If \n , and x and y are integers such that x < y, what is\nthe value of y − x ?\n−1(A)\n  1(B)\n  3(C)\n  5(D)\n  7(E)\n32. If Sn = 5n + 94 and K = (S80 + S82 + S84) − (S81 + S83 + S85), what is the\nvalue of K ?\n−282(A)\n  −84(B)\n  −30(C)\n  −15(D)\n    −3(E)\n33. Is xy + xy < xy ?"} +{"id": "book 2_p488_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 488, "page_end": 488, "topic_guess": "sequences_patterns", "text": "(1)\n(2)\n34. If y ≠ x, then \n is the equivalent of\nwhich of the following?\n(x − 1)2y(A)\n(x + 1)2(B)\nx2 + x + 1(C)\n(x2 + x + 1)y(D)\n(x2 + x + 1)(x − y)(E)\n35. In the sequence gn defined for all positive integer values of n, g1 = g2\n= 1 and, for n ≥ 3, gn = gn − 1 + 2n − 3. If the function \n equals the\nsum of the terms g1, g2 . . . gi, what is"} +{"id": "book 2_p489_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 489, "page_end": 489, "topic_guess": "number_theory", "text": "g3(A)\ng8(B)\n(C)\n(D)\n(E)\n36. If 3k + 3k = (39)39\n − 3k, then what is the value of k ?\n(A)\n(B)\n242(C)\n310(D)\n311 − 1(E)\n37. If x and y are positive integers, what is the value of xy ?"} +{"id": "book 2_p490_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 490, "page_end": 490, "topic_guess": "number_theory", "text": "x! = 6y!(1)\n(2)\n38. If k is an integer and \n is divisible by 6k, what is the maximum\npossible value of k ?\n3(A)\n4(B)\n5(C)\n6(D)\n7(E)\n39. In a certain sequence, the term Sn is given by the formula Sn = (n +\n1)! for all integers n ≥ 1. Which of the following is equivalent to the\ndifference between S100 and S99 ?\n101!(A)\n100!(B)\n992(98!)(C)\n1002(99!)(D)\n(100!)2(E)"} +{"id": "book 2_p491_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 491, "page_end": 491, "topic_guess": "general", "text": "40. If \n and \n , what is the value of \n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p492_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 492, "page_end": 492, "topic_guess": "general", "text": "Workout Set 4 Answer Key\n31. B\n32. D\n33. E\n34. C\n35. A\n36. E\n37. C\n38. D\n39. D\n40. C"} +{"id": "book 2_p493_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 493, "page_end": 493, "topic_guess": "algebra", "text": "Workout Set 4 Solutions\n31. (B) 1: Simplify the right side of the equation.\nThis still needs to be manipulated in order to match the le side of the\nequation, since the bases need to match their exponents. The number 33\nis okay, but 28 is not. Try turning it into a base of 4 instead.\nThat’s it: x = 3 and y = 4, since the problem states that y > x. So y − x = 4 −\n3 = 1.\nThe correct answer is (B).\n32. (D) −15: Note that the expression can be rewritten as follows by pairing\nup consecutive terms in the sequence.\nIn each of these pairs, the 94 term cancels out."} +{"id": "book 2_p494_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 494, "page_end": 494, "topic_guess": "algebra", "text": "Therefore, the value of K is as follows.\nThe correct answer is (D).\n33. (E): First, rephrase the question stem by subtracting xy from both sides:\n“Is xy < 0?” The question is whether xy is negative, or “Do x and y have\nopposite signs?”\nBe careful! Do not rephrase as follows:\nIs xy + xy < xy ?\nIs 2xy < xy ?\nIs \n ? Incorrect! Dividing by variables is the mistake:\nWhat if xy = 0? What if xy < 0 ?\nIs 2 < 1? Incorrect as a result of the mistake in the previous step."} +{"id": "book 2_p495_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 495, "page_end": 495, "topic_guess": "fractions_decimals_percents", "text": "Not only does this rephrase make the statements moot (2 is definitely\nnot less than 1, no matter what the statements say), but it also ignores\nsome special cases. If xy = 0, then dividing by xy yields an undefined\nvalue. If xy < 0, you should have flipped the sign of the inequality.\nInstead, use the correct rephrasing: “Do x and y have opposite signs?”\n(1) INSUFFICIENT: If \n , then x2 and y must have opposite signs.\nSince x2 must be positive, y must be negative. However, x could be either\npositive or negative.\n(2) INSUFFICIENT: If x3y3 < (xy)2, then you can divide both sides by (xy)2\n(since that quantity is positive). The simplified inequality is xy < 1, which\nis not sufficient to answer the question.\n(1) AND (2) INSUFFICIENT: Each statement indicates that xy could be\neither positive or negative. The statements are equally insufficient and\nneither provides any additional information to the other.\nThe correct answer is (E).\n34. (C) x2 + x + 1: First, distribute the numerator.\nNone of the answer choices are fractions, so the x − y in the denominator\nmust be canceled by a x − y in the numerator. Group the numerator\nterms with x and −y in mind."} +{"id": "book 2_p496_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 496, "page_end": 496, "topic_guess": "strategy", "text": "Alternatively, you could choose smart numbers. If x = 2 and y = 3, then.\nPlug the selected values into the choices. The choice that equals 7 is the\ncorrect answer.\nThe correct answer is (C).\n(x − 1)2y = (2 − 1)2(3) = 3 Eliminate.(A)\n(x + 1)2 = (2 + 1)2 = 9 Eliminate.(B)\nx2 + x + 1 = 4 + 2 + 1 = 7 Correct!(C)\n(x2 + x + 1)y = (4 + 2 + 1)(3) = Too great. Eliminate.(D)\n(x2 + x + 1)(x − y) = (4 + 2 + 1)(2 − 3) = Negative. Eliminate.(E)\n35. (A) g3: Begin by listing some values of gn to get a sense for how gn\nprogresses."} +{"id": "book 2_p497_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 497, "page_end": 497, "topic_guess": "algebra", "text": "For n ≥ 3, gn = 2n − 2 .\nNow, look for a pattern in the sums defined by \n .\nEach value is double the previous value: \n .\nTherefore:\nNow, all you need to do is scan the answer choices to find an expression\nthat equals 2. You have already discovered that g3 = 2, so you can select\ng3 as the answer.\nThe correct answer is (A).\n36. (E) 311 − 1: The common term in this problem is the recurring base of 3.\nGroup like terms (i.e., all the terms with k on the le side, all the other"} +{"id": "book 2_p498_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 498, "page_end": 498, "topic_guess": "number_theory", "text": "powers of 3 on the right side), then simplify each power of 3 using\nexponent rules.\nThe correct answer is (E).\n37. (C): Note the constraints that x and y are positive integers.\n(1) INSUFFICIENT: If y = 5 and x = 6, then this statement is true: 6! = 6 × 5!.\nAre these the only possible positive integer values for x and y ?\nThe constant 6 is equal to 3!, so there’s one more: when y = 1 and x = 3,\nthe statement is also true: 3! = 6 × 1!.\n(2) INSUFFICIENT: You can multiply both sides by y to eliminate that\nvariable entirely; in other words, this statement doesn’t indicate\nanything about y."} +{"id": "book 2_p499_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 499, "page_end": 499, "topic_guess": "number_theory", "text": "(1) AND (2) SUFFICIENT: Continue to simplify the equation from\nstatement (2).\nThe quadratic produces two solutions, but only x = 6 is valid, since x\nmust be a positive integer. From statement (1), if x = 6, then y = 5. This is\nsufficient to calculate a value for xy.\nThe correct answer is (C).\n38. (D) 6: The question asks for the greatest value of k such that \n is\ndivisible by 6k. Since 6 = 3 × 2, the greatest value of k will equal the\nnumber of 3 × 2 pairs among the prime factors of \n ."} +{"id": "book 2_p500_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 500, "page_end": 500, "topic_guess": "number_theory", "text": "To count the number of times 3 appears as a factor of \n , rewrite the\nexpression, pulling out any factor(s) of 3 from each term.\n   (33)(32)(31)(30)(29)(28)(27)(26)(25)(24)(23)\n   = (3 × 11)(32)(31)(3 × 10)(29)(28)(33)(26)(25)(3 × 8)(23)\nThere are six factors of 3 in \n .\nTo count the number of times 2 appears as a factor of \n , rewrite the\nexpression, pulling out any factor(s) of 2 from each term.\n(33)(32)(31)(30)(29)(28)(27)(26)(25)(24)(23)\n= (33)(25)(31)(2 × 15)(29)(22 × 7)(27)(2 × 13)(25)(23 ×3)(23)\nThere are twelve factors of 2 in \n .\nSince there are twelve 2’s but only six 3’s, there are only six 3 × 2 pairs\namong the prime factors of \n . In general, focus on the greatest prime\nin the divisor (in this case, the six maximum possible factors of 3 in 6k),\nas it will be the limiting factor.\nThe correct answer is (D)."} +{"id": "book 2_p501_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 501, "page_end": 501, "topic_guess": "general", "text": "39. (D) (1002)(99!): S100 = 101! and S99 = 100!.\nFactor the difference.\nThe correct answer is (D).\n40. (C) \n Not only does \n , but you can also factor \n out of p.\nThus, \n , and \n . Watch out! This is one of the\nincorrect answers."} +{"id": "book 2_p502_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 502, "page_end": 502, "topic_guess": "general", "text": "The question asks for \n , which is \n .\nThe correct answer is (C)."} +{"id": "book 2_p503_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 503, "page_end": 503, "topic_guess": "number_theory", "text": "Workout Set 5\n41. If x and y are positive integers, what is the value of \n ?\nx2 = 2xy − y2(1)\n2xy = 8(2)\n42. What is the remainder when (47)(49) is divided by 8 ?\n1(A)\n3(B)\n4(C)\n5(D)\n7(E)\n43. If a = 4x2 + 4xy and b = 4y2 + 4xy, which of the following is equivalent\nto x + y ?"} +{"id": "book 2_p504_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 504, "page_end": 504, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n44. What is the value of \n ?\na + b = 7(1)\nab = 12(2)\n45. What is the value of \n ?"} +{"id": "book 2_p505_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 505, "page_end": 505, "topic_guess": "number_theory", "text": "48(A)\n(B)\n(C)\n1(D)\n(E)\n46. If x is the square of an integer, is y the square of an integer?\nxy is the square of an integer.(1)\n is the square of an integer.(2)\n47. If x and y are positive integers such that x > y and\n, which of the following is equivalent\nto 2b ?"} +{"id": "book 2_p506_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 506, "page_end": 506, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n48. If (x + y)2 = 16 and x2 − y2 = 16, what is the value of 2x2 ?\n  8(A)\n16(B)\n18(C)\n32(D)\n50(E)\n49. Which of the following is equal to 5,9952 – 4,7962 ?\n1,1992(A)\n2,6872(B)\n3,4252(C)\n3,5972(D)\n4,8202(E)"} +{"id": "book 2_p507_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 507, "page_end": 507, "topic_guess": "general", "text": "50. If \n , what is the value of xy ?\ny > x(1)\nx < 0(2)"} +{"id": "book 2_p508_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 508, "page_end": 508, "topic_guess": "general", "text": "Workout Set 5 Answer Key\n41. A\n42. E\n43. E\n44. B\n45. D\n46. B\n47. B\n48. D\n49. D\n50. A"} +{"id": "book 2_p509_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 509, "page_end": 509, "topic_guess": "general", "text": "Workout Set 5 Solutions\n41. (A):\n(1) SUFFICIENT: Move all terms to one side.\nFactor this “square of a difference” special product and simplify.\nSince x = y, the value of \n is 1.\n(2) INSUFFICIENT: If 2xy = 8, then xy = 4. Test different values for x and y.\nx (positive) y (positive) xy = 4\n2 2 2 × 2 = 4 ✓ 1"} +{"id": "book 2_p510_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 510, "page_end": 510, "topic_guess": "algebra", "text": "x (positive) y (positive) xy = 4\n1 4 1 × 4 = 4 ✓\nAlternatively, manipulate the equation xy = 4 to get \n on one side.\nThe expression \n does not equal a constant, so \n could take on many\ndifferent values.\nThe correct answer is (A).\n42. (E) 7: One way to solve would be to multiply (47)(49), then either divide\nthe result by 8 or repeatedly subtract known multiples of 8 from the\nresult until you are le with a remainder less than 8.\nAn alternative is to rewrite the given product as an equivalent yet easier-\nto-manipulate product. Note that 47 and 49 are equidistant from 48, a\nmultiple of 8. Write each of the original factors as terms in the form (a + b)\nor (a − b)."} +{"id": "book 2_p511_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 511, "page_end": 511, "topic_guess": "general", "text": "(47)(49) = (48 + 1)(48 − 1)\nThis form is the difference of two squares special product, (a + b)(a − b) =\na2 − b2; continue to simplify.\n(48 + 1)(48 − 1) = (482 − 12)\nForty-eight is a multiple of 8, and therefore so is 482. Thus, (482 − 12) is 1\nless than a multiple of 8. All such numbers (e.g., 7, 15, 23, 31) have a\nremainder of 7 when divided by 8.\nThe correct answer is (E).\n43. (E) \n : Add a and b to get a + b = 4x2 + 8xy + 4y2 = 4(x2 + 2xy + y2).\nThe right side is the square of a sum, so factor and solve.\nNote that you could safely take the square root of both sides because any\nsquare is non-negative."} +{"id": "book 2_p512_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 512, "page_end": 512, "topic_guess": "algebra", "text": "Alternatively, choose smart numbers. For example, if x = 2 and y = 3, then\nthe final answer x + y = 5.\nFor values to plug into the choices, first compute a and b.\nNext, test each answer choice; the one that equals 5 is the correct answer.\nThe correct answer is (E).\n Eliminate.(A)\n(B)\n(C)\n(D)\n Correct!(E)\n44. (B): Manipulate the question expression, noting the special products in\nthe exponents."} +{"id": "book 2_p513_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 513, "page_end": 513, "topic_guess": "general", "text": "The rephrased question is, “What is the value of ab?” Knowing the values\nof a and b individually would be sufficient, of course, but the individual\nvalues are not required as long as you can determine ab.\n(1) INSUFFICIENT: It is impossible to manipulate a + b = 7 to get ab, nor\ncan you solve for a and b individually.\n(2) SUFFICIENT: This statement answers the rephrased question directly.\nThe correct answer is (B).\n45. (D) 1:\nNotice the special product of the form (a + b)(a − b) = a2 − b2 under the\nsquare root symbol.\nThe correct answer is (D).\n46. (B): Jot down the information provided in the question stem. If x is the\nsquare of an integer, then x = i2, where i is an unknown integer. Is y a"} +{"id": "book 2_p514_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 514, "page_end": 514, "topic_guess": "general", "text": "perfect square?\n(1) INSUFFICIENT: x is a perfect square, and when multiplied by the\nunknown value y, it turns into a (possibly different) perfect square.\nCase 1: This situation can occur if y is also a perfect square. For example,\nif x = 22 = 4, and y = 32 = 9, then xy = 36 = 62. Since y is the square of an\ninteger, the answer is Yes.\nCase 2: If you only consider integer values for y, it seems as though y\nmust be a square. For instance, if x = 22, y cannot equal 2, 3, 5, 6, or any\nnon-square, since 2(22), 3(22), 5(22), etc., are not squares.\nHowever, the question does not restrict y to integer values. To find a non-\ninteger value of y that fits the known information, start by choosing\nsquare values for both x and xy. Suppose that x = 22 = 4 and xy = 32 = 9. In\nthis case, y = \n , which is not a perfect square. Therefore, the answer is\nNo.\n(2) SUFFICIENT: \n  is the square of an integer, so \n = j2, for some\nunknown integer j. From the question stem, x = i2, for some unknown\ninteger i. Simplify."} +{"id": "book 2_p515_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 515, "page_end": 515, "topic_guess": "algebra", "text": "Therefore, y must be the square of an integer. The answer is definitely\nYes, and this statement is sufficient.\nThe correct answer is (B).\n47. (B) \n : To solve this problem, isolate b in the equation.\nNote that it is okay to divide by \n , since x > y, which implies\nthat \n .\nThe question asks for 2b, but the result does not match any of the answer\nchoices. Most of the choices are not fractions, so try to cancel the\ndenominator. Since x − y is a well-disguised difference of two squares,\nfactor the numerator and denominator."} +{"id": "book 2_p516_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 516, "page_end": 516, "topic_guess": "algebra", "text": "Cancel \n in the numerator and denominator to get\n.\nThe correct answer is (B).\n48. (D) 32: Each of the expressions given is equal to 16, so set them equal to\neach other. Note that you have the square of a sum and the difference of\ntwo squares special products. Put them both in distributed form, then\nsimplify.\nSince 2y(x + y) = 0, it must be true that either 2y or (x + y) is equal to 0.\nHowever, (x + y) cannot equal 0, since the problem indicates that (x + y)2 =\n16 . So it must be that 2y = 0, and therefore y = 0.\nPlug in 0 for y in x2 – y2 = 16 to get x2 = 16. Thus, 2x2 = 2(16) = 32.\nAlternatively, you can solve this problem by Working Backwards from the\nanswer choices. Try (B) or (D) first.\n(D) If 32 = 2x2, then x = ±4. Use this to solve for y in the second equation."} +{"id": "book 2_p517_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 517, "page_end": 517, "topic_guess": "sequences_patterns", "text": "Do x = ±4 and y = 0 also work in the first equation? Yes!\nThe correct answer is (D).\n49. (D) 3,5972: There are several ways to approach this problem without\nextensive calculation.\nUnits Digit\nOne option is to note that four of the five answer choices have different\nunits digits. The units digit of 5,9952 must equal the units digit of 52,\nwhich is 5. Similarly, the units digit of 4,7962 must equal the units digit of\n62, which is 6.\nSubtracting a number with a units digit of 6 from a number with a units\ndigit of 5 results in a units digit of 9.\nHowever, this does not mean that answer (A) is correct! Since the answer\nchoices are squared, the correct answer must have a units digit of 9 a er\nbeing squared. Therefore, the correct answer is either (B) or (D), since 72\nhas a units digit of 9. See the estimation approach below for one way to\ndifferentiate between choices (B) and (D).\nEstimation\nIt is also possible to estimate the answer: 5,995 is approximately equal to\n6,000, and 4,796 is approximately equal to 5,000."} +{"id": "book 2_p518_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 518, "page_end": 518, "topic_guess": "sequences_patterns", "text": "6,0002 – 5,0002 = 36,000,000 – 25,000,000 = 11,000,000\nSince 3,0002 = 9,000,000, and 4,0002 = 16,000,000, the correct answer\nmust be between 3,0002 and 4,0002. Eliminate any answer other than\nanswers (C) and (D). Combined with the units digit approach above, this\nyields the correct answer, choice (D).\nFactoring\nThe number 5,995 is a multiple of 5 just under 6,000, which equals\n5(1,200).  The number 4,796 is a multiple of 4 just under 4,800, which\nequals 4(1,200). This “coincidence” suggests factoring.\n5,995 = 6,000 – 5 = 5(1,200 – 1) = 5(1,199)\n4,796 = 4,800 – 4 = 4(1,200 – 1) = 4(1,199)\nThe shared factor of 1,199 means that simplification is possible.\nThe correct answer is (D)."} +{"id": "book 2_p519_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 519, "page_end": 519, "topic_guess": "algebra", "text": "50. (A): Begin by simplifying the equation given in the question.\nAt this stage, you might be tempted to divide both sides of the equation\nby xy, in order to arrive at the simpler equation y2 + x2 = 2xy. However,\nthat would be a mistake—never divide both sides of an equation by an\nunknown (in this case, xy), unless you are certain that the unknown\ncannot equal zero. (Division by zero is undefined, and can lead to\nnonsensical results.) So rather than divide by xy, subtract 2x2y2 from both\nsides to group all terms on one side of the equals sign.\nThis last line implies that either xy = 0 or y – x = 0. In other words, either\nxy = 0 or y = x.\n(1) SUFFICIENT: If y > x, then it is impossible that y = x. Therefore, xy = 0.\n(2) INSUFFICIENT: If x < 0, then it is possible that xy = 0 (i.e., if y = 0) or that\ny = x (i.e., y is negative, too). If y = x = any negative number, then there are\ninfinitely many solutions for xy."} +{"id": "book 2_p520_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 520, "page_end": 520, "topic_guess": "general", "text": "The correct answer is (A)."} +{"id": "book 2_p521_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 521, "page_end": 521, "topic_guess": "coordinate_geometry", "text": "Workout Set 6\n51. Is it possible for a wooden block in the shape of a rectangular solid\nmeasuring l by w by h centimeter to pass through a square hole\nwith sides of length 4 centimeter?\nThe volume of the wooden block is 16 cm3.(1)\nL > w > h > 1(2)\n52. Three of the four vertices of a rectangle in the xy-coordinate plane\nare (–5, 1), (–4, 4), and (8, 0). What is the fourth vertex?\n(–4.5, 2.5)(A)\n(–4, 5)(B)\n(6, –2)(C)\n(7, –3)(D)\n(10, 1)(E)\n53. In the coordinate plane, circle C has radius r and its center is at\npoint (x, y). Is at least 50% of the area of circle C contained in a\nsingle quadrant of the coordinate plane?\nx > y > 0(1)"} +{"id": "book 2_p522_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 522, "page_end": 522, "topic_guess": "statistics", "text": "x > r(2)\n54. A group of men and women gathered to compete in a marathon.\nEach competitor was weighed before the competition; the average\nweight of the women was 120 pounds, and the average weight of\nthe men was greater than 120 pounds. If the average weight of the\nentire group was w pounds, what percent of the competitors were\nwomen?\nThe average weight of the men was 150 pounds.(1)\nThe difference between w and the average weight of the\nwomen was twice the difference between w and the average\nweight of the men. \n(2)\n55. What is the average (arithmetic mean) of these numbers: 12; 13; 14;\n510; 520; 530; 1,115; 1,120; and 1,125 ?\n419(A)\n551(B)\n601(C)\n620(D)\n721(E)\n56. A set of 5 numbers has an average (arithmetic mean) of 50. The\ngreatest element in the set is 5 greater than 3 times the least"} +{"id": "book 2_p523_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 523, "page_end": 523, "topic_guess": "statistics", "text": "element in the set. If the median of the set equals the mean, what is\nthe greatest possible value in the set?\n85(A)\n87(B)\n88(C)\n92(D)\n93(E)\n57. \nIn the triangle above, DE is parallel to AC. What is the length of DE ?\nAC = 14(1)\nBE = EC(2)\n58. 5, 2, 4, m, 9, 5\nFor the list of numbers above, what is the median?\nThe median is an integer.(1)"} +{"id": "book 2_p524_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 524, "page_end": 524, "topic_guess": "statistics", "text": "m = 8(2)\n59. Set M contains seven consecutive integers, and set N contains three\nvalues chosen from set M. Is the standard deviation of set N greater\nthan the standard deviation of set M ?\nSet N contains the median of set M.(1)\nThe range of set M and set N are equal.(2)\n60. Four different children have jelly beans: Aaron has 5, Bianca has 7,\nCallie has 8, and Dante has 11. How many jelly beans must Dante\ngive to Aaron to ensure that each child has within 1 jelly bean of all\nthe other children?\n2(A)\n3(B)\n4(C)\n5(D)\n6(E)"} +{"id": "book 2_p525_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 525, "page_end": 525, "topic_guess": "general", "text": "Workout Set 6 Answer Key\n51. C\n52. D\n53. B\n54. B\n55. B\n56. D\n57. C\n58. D\n59. B\n60. B"} +{"id": "book 2_p526_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 526, "page_end": 526, "topic_guess": "word_problems", "text": "Workout Set 6 Solutions\n51. (C): Start by considering the different types of blocks that could or could\nnot fit through the hole.\nIf all dimensions of the block are less than 4 centimeters (cm), the block\ncan definitely fit through the hole.\nIf exactly one of the three dimensions is greater than 4 cm, the block can\nstill definitely fit through the hole. To do so, rotate the block so that the\ntwo shorter dimensions are aligned with the sides of the hole."} +{"id": "book 2_p527_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 527, "page_end": 527, "topic_guess": "general", "text": "However, if at least two of the dimensions of the block are greater than 4\ncm, further investigation is required. It may be possible to fit the block\nthrough the hole by rotating it to take advantage of the longest\ndimension of the hole, which is the diagonal of \n ,\nbut if the dimensions of the block are too great, this may be impossible. \n(1) INSUFFICIENT: The block can have any dimensions such that Lwh =\n16. A block with dimensions 1, 1, and 16 can fit through the hole as\nshown in the second diagram above. However, a block with dimensions\n8, 8, and 0.25 cannot fit through the hole. Since the answer can be Yes or\nNo, this statement is insufficient.\n(2) INSUFFICIENT: A block with dimensions 2, 3, and 4 can fit through the\nhole, but a block with dimensions 100, 101, and 102 cannot.\n(1) and (2) SUFFICIENT: Lwh = 16 and h is greater than 1, so Lw is less\nthan 16. Therefore, L and w cannot both be greater than 4. Since h is the\nleast dimension, it must also be less than 4. Therefore, the block has at\nleast two dimensions that are less than 4, and it can fit through the hole\nas shown in the second diagram.\nThe correct answer is (C).\n52. (D) (7, −3): Your GMAT scratch pad has a grid; use it to plot the diagram to\nscale."} +{"id": "book 2_p528_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 528, "page_end": 528, "topic_guess": "geometry", "text": "“Eyeball” solution: Complete the rectangle with the dashed lines shown.\nThe fourth point must be located approximately where the bigger dot is\ndrawn. Answers (A), (B), and (E) must be incorrect. The closest answer\nchoice is the point (7, −3). Alternatively, you could plot the remaining\ntwo answer choice points to see which one “works” with the three given\npoints.\nAlternatively, compute the location of the fourth point, using the fact\nthat the short sides have the same slope. The known short side connects\nthe points (−5, 1) and (−4, 4). In other words, the bottom le corner is 1\nto the le and 3 down from the top le corner. The unknown bottom\nright corner should therefore be 1 to the le and 3 down from the top\nright corner, or x = 8 − 1 = 7 and y = 0 − 3 = −3, corresponding to the point\n(7, −3).\nThe correct answer is (D).\n53. (B): The question asks whether at least half of the circle is contained in\none quadrant. However, it does not specify a specific quadrant. If at\nleast half of the circle is contained in any one of the four quadrants, the\nanswer to the question will be Yes. \nNote that if circle C lies in exactly one or two quadrants, then the answer\nto the question must be Yes, because at least half of the circle must be in"} +{"id": "book 2_p529_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 529, "page_end": 529, "topic_guess": "geometry", "text": "one of those quadrants.\n(1) INSUFFICIENT: If x = 2, y = 1, and r = 0.5, then the answer to the\nquestion is Yes, since the entire circle is contained in the first quadrant.\nHowever, if x = 2, y = 1, and r is much greater than 2, then approximately\na quarter of the circle is contained in each quadrant and the answer to\nthe question is No.\n(2) SUFFICIENT: Since the x coordinate is greater than the radius of the\ncircle, no part of the circle can be in the second or third quadrants."} +{"id": "book 2_p530_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 530, "page_end": 530, "topic_guess": "geometry", "text": "Therefore, at least half of the circle must be in the first quadrant or at\nleast half of the circle must be in the fourth quadrant. The answer to the\nquestion is definitely Yes, and this statement is sufficient.\nThe correct answer is (B). \n54. (B): The question stem indicates that the average weight of the women\nwas 120 and the average weight of the men was greater than 120. In\norder to determine what percent of the competitors were women, you\nwould need to know more about the weight of the men and you’d also\nneed to know something about the relative number of women vs. men.\nIf there were not an equal number of each, then this is a weighted\naverage question.\n(1) INSUFFICIENT: Statement (1) provides the average weight of the men\nbut does not indicate whether there was an equal number of men and\nwomen.\n(2) SUFFICIENT: This does not provide the weight of the men or even the\nvalue of w, but it does indicate where the weighted average falls\nbetween the two values."} +{"id": "book 2_p531_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 531, "page_end": 531, "topic_guess": "statistics", "text": "If the weighted average is twice as far from the women’s end of the line,\nthen the men are responsible for \n of the total weight and the women\nare responsible for \n of the total weight. Therefore, \n , or\napproximately \n of the competitors were women. (Note: You do\nnot need to calculate this figure. You can stop whenever you understand\nthat this figure can be calculated.)\nThe correct answer is (B).\n55. (B) 551: The simple average formula\n applies to this problem."} +{"id": "book 2_p532_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 532, "page_end": 532, "topic_guess": "word_problems", "text": "However, the chance of computational error is high on a problem with\nthis many terms of such great value.\nTry grouping the similar terms.\nGroup A: 12, 13, 14 (equidistant terms with an average of 13, the\nmiddle term)\nGroup B: 510, 520, 530 (equidistant terms with an average of 520, the\nmiddle term)\nGroup C: 1,115; 1,120; 1,125 (equidistant terms with an average of\n1,120, the middle term)\nSince each group of terms consists of three values (and are therefore\nequally weighted in the set of nine terms), the average of all nine\noriginal terms is the average of the respective averages of Groups A, B,\nand C.\nAverage = \nYou can make that division easier by breaking 1,653 into lesser numbers\nthat are divisible by 3: 1,653 = 1,500 + 150 + 3. Divide each separately by\n3 to get 500 + 50 + 1 = 551."} +{"id": "book 2_p533_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 533, "page_end": 533, "topic_guess": "statistics", "text": "The correct answer is (B).\n56. (D) 92: Two techniques will help you efficiently interpret the information\ngiven in the question. First, draw a number line with five dots\nrepresenting the five numbers in the set. Second, label these numbers A,\nB, C, D, and E, with the understanding that A ≤ B ≤ C ≤ D ≤ E.\nThe problem indicates that:\nA + B + C + D + E = 250 (The set of five numbers has an average of 50.)\nE = 5 + 3A (The greatest element is five greater than three times the\nleast element in the set.)\nC = 50 (The median of the set equals the mean.)\nYou’re asked to maximize E. Arrange the dots on the number line such\nthat you obey the constraints, yet also note where you have some\nflexibility.\nPoint D can be anywhere on the line from point C to point E. Given the\ninformation from the question stem, you can maximize E by minimizing\nD. Therefore, make D = C = 50."} +{"id": "book 2_p534_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 534, "page_end": 534, "topic_guess": "geometry", "text": "Similarly, point B can be anywhere on the line from point A to point C.\nMaximize E by minimizing B, so make B = A.\nTherefore, E = 5 + 3A = 5 + 3(29) = 5 + 87 = 92.\nThe correct answer is (D).\n57. (C):\n(1) INSUFFICIENT: Many elements in this triangle could vary; you don’t\neven know the placement of B relative to AC, so the triangle itself might\nstretch. Even for a fixed triangle, DE could slide up or down, so various\nlengths are possible for DE, as shown here:"} +{"id": "book 2_p535_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 535, "page_end": 535, "topic_guess": "geometry", "text": "(2) INSUFFICIENT: You don’t know the lengths of any sides of the\ntriangle. The side that most affects the length of DE is AC, so stretch that\nside. Stretching the triangle out to the right stretches DE.\n(1) AND (2) SUFFICIENT: AC must be 14 and DE must be parallel to AC\nand halfway between AC and B, in order to maintain BE = EC. Even\nthough vertex B is free to move, DE will always be the average of the\nwidth of the triangle at AC (14) and the width at B (0). Thus, DE must be\n7, no matter how the picture shi s."} +{"id": "book 2_p536_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 536, "page_end": 536, "topic_guess": "statistics", "text": "The correct answer is (C).\n58. (D): To find the median of a set of numbers, line them up in order of\nvalue. The question of interest is, “Where is m relative to the other\nvalues?” This set has six values, an even number of terms, so the median\nis the average of the two middle terms.\nThese are the three scenarios shown above.\nIf m ≤ 4, then the two middle terms are 4 and 5, and the median is\n4.5.\nIf 4 < m < 5, then the two middle terms are m and 5, and the median\nis \nIf m ≥ 5, then the two middle terms are 5 and 5, and the median is 5.\n(1) SUFFICIENT: For the case where 4 < m < 5, the median is\n\nThus, if the median is an integer, it must be 5.\n(2) SUFFICIENT: If m = 8, then m ≥ 5, and the median is 5.\nThe correct answer is (D)."} +{"id": "book 2_p537_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 537, "page_end": 537, "topic_guess": "statistics", "text": "59. (B): Standard deviation is a measure of the “spread” of a group of\nnumbers.\n(1) INSUFFICIENT: If set M contains the numbers {1, 2, 3, 4, 5, 6, 7}, then 4\nis the median. Set N could be {3, 4, 5}, which has a lesser standard\ndeviation than set M because the two sets have the same average but N\nis not as spread out as M.\nAlternatively, N could be {1, 4, 7}, which has a greater standard deviation\nthan set M because the two sets still have the same average, but N is\nnow more spread out than M. (It doesn’t have additional values that are\ncloser to the average.)\n(2) SUFFICIENT: If set M contains the numbers {1, 2, 3, 4, 5, 6, 7}, and set\nN has the same range, then N must contain 1 and 7. If set N contains {1,\n4, 7}, then N has a greater standard deviation than M. If set N contains {1,\n2, 7}, the standard deviation actually increases even more (because the\nnumbers are no longer evenly distributed}. No matter what combination\nyou try, the standard deviation of N has to be greater than the standard\ndeviation of M.\nThe correct answer is (B).\n60. (B) 3: Conceptually, the transfer of jelly beans from Dante to Aaron\nreduces the range of the number of jelly beans held by individual\nchildren. The constraint is that the final distribution represents a range\nof just 1 jelly bean, a condition Bianca and Callie already satisfy.\nDraw a picture (a number line) to visualize the scenario."} +{"id": "book 2_p538_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 538, "page_end": 538, "topic_guess": "algebra", "text": "Aaron and Dante must end up with a number of jelly beans that is either\n7 or 8. If either Aaron or Dante has a number of jelly beans other than 7\nor 8, he will differ too much from either Bianca’s or Callie’s number. You\ncan count out the necessary change on the number line above or you\ncan write out the algebra.\nThe solution to both equations is x = 3: Add 3 to Aaron to get 8, and\nsubtract 3 from Dante to get 8. The resulting number of jelly beans is A =\n8, B = 7, C = 8, and D = 8.\nThe correct answer is (B)."} +{"id": "book 2_p539_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 539, "page_end": 539, "topic_guess": "fractions_decimals_percents", "text": "Workout Set 7\n61. Is rectangle R a square?\nAt least one side of rectangle R has an integer length.(1)\nThe diagonals of rectangle R have integer lengths.(2)\n62. A group of friends charters a boat for $540 and each person\ncontributes equally to the cost. They determine that if they can get\nthree more of their friends to join them, every person in the group\nwill pay $9 less. If they find three more friends to join them, what is\nthe total number of people renting the boat?\n  6(A)\n  9(B)\n15(C)\n18(D)\n21(E)\n63. In Smithtown, the ratio of right-handed people to le -handed\npeople is 3 to 1 and the ratio of men to women is 3 to 2. If the\nnumber of right-handed men is maximized, then what percent of all\nthe people in Smithtown are le -handed women?"} +{"id": "book 2_p540_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 540, "page_end": 540, "topic_guess": "fractions_decimals_percents", "text": "50%(A)\n40%(B)\n25%(C)\n20%(D)\n10%(E)\n64. The sum of the interior angle measures for any n-sided polygon\nequals 180(n − 2). If Polygon A has interior angle measures that\ncorrespond to a set of consecutive integers, and if the median angle\nmeasure for Polygon A is 140°, what is the least angle measure in\nthe polygon?\n130°(A)\n135°(B)\n136°(C)\n138°(D)\n140°(E)\n65. When the positive integer x is divided by the positive integer y, the\nquotient is 2 and the remainder is z. When x is divided by the\npositive integer a, the quotient is 3 and the remainder is b. Is z > b ?\nThe ratio of y to a is less than 3 to 2.(1)\nThe ratio of y to a is greater than 2 to 3.(2)"} +{"id": "book 2_p541_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 541, "page_end": 541, "topic_guess": "number_theory", "text": "66. If a and b are odd integers, a ∆ b represents the product of all odd\nintegers between a and b,inclusive. If y is the least prime factor of (3\n∆ 47) + 2, which of the following must be true?\ny > 50(A)\n30 ≤ y ≤ 50(B)\n10 ≤ y < 30(C)\n3 ≤ y < 10(D)\ny = 2(E)\n67. Set S is the set of all prime integers between 0 and 20. If three\nnumbers are chosen randomly from set S, and no number is chosen\nmore than once, what is the probability that the sum of all three\nnumbers is odd?\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p542_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 542, "page_end": 542, "topic_guess": "geometry", "text": "68. Sets A and B each consist of three terms selected from the first five\nprime integers. No term appears more than once within a set, but\nany integer may be a term in both sets. If the average of the terms\nin set A is 4 and the product of the terms in set B is divisible by 22,\nhow many terms are shared by both sets?\nThe product of the terms in set B is not divisible by 5.(1)\nThe product of the terms in set B is divisible by 14.(2)\n69. \nIn the figure above, SAND and SURF are squares, and O is the center\nof the circle. If Q is the sum of the areas of squares SAND and SURF\nand C is the area of the circle, then the fraction \n is"} +{"id": "book 2_p543_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 543, "page_end": 543, "topic_guess": "number_theory", "text": "less than \n(A)\nbetween \n and \n(B)\nbetween \n and \n(C)\nbetween \n and 1(D)\ngreater than 1(E)\n70. Set S consists of n consecutive positive integers. If n > 3, what is the\nvalue of n ?\nThe number of multiples of 2 contained in set S is equal to the\nnumber of multiples of 3 contained in set S.\n(1)\nn is odd.(2)"} +{"id": "book 2_p544_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 544, "page_end": 544, "topic_guess": "general", "text": "Workout Set 7 Answer Key\n61. C\n62. C\n63. C\n64. C\n65. A\n66. A\n67. D\n68. D\n69. C\n70. E"} +{"id": "book 2_p545_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 545, "page_end": 545, "topic_guess": "geometry", "text": "Workout Set 7 Solutions\n61. (C): A rectangle is a square if and only if its length equals its width.\n(1) INSUFFICIENT: If the length is 1 and the width is 2, the rectangle is\nnot a square. If the length and width are both 2, the rectangle is a\nsquare.\n(2) INSUFFICIENT: Since R is a rectangle, both of its diagonals have the\nsame length. It is only necessary to examine one of the diagonals. Since\nthe diagonal and two of the sides form a right triangle, the length of the\ndiagonal is always equal to\n .\nThe diagonal of the rectangle below has an integer length, but the\nrectangle is not a square:"} +{"id": "book 2_p546_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 546, "page_end": 546, "topic_guess": "fractions_decimals_percents", "text": "However, in the rectangle below, the diagonal has an integer length and\nthe rectangle is a square. Note that it is not necessary to calculate the\nexact length of the sides of the square, since this is a Data Sufficiency\nproblem. However, \n  in this case. \nThe answer could be Yes or No, so this statement is insufficient.\n(1) AND (2) SUFFICIENT: According to the statements, the rectangle has\nat least one integer side, and it also has integer diagonals.\nIn a square, the legs and hypotenuse must be in the ratio \n. To be a square, a rectangle must have a diagonal that is \n  times as\nlong as its sides.\nHowever, \n  is an irrational number, so if x is an integer, then \n  is\nnot an integer (and vice versa). It is impossible for the rectangle\ndescribed by these statements to be a square, and the answer to the\nquestion is definitely No. The statements together are sufficient. \nThe correct answer is (C)."} +{"id": "book 2_p547_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 547, "page_end": 547, "topic_guess": "algebra", "text": "62. (C) 15: Call the initial number of friends in the group f and the initial cost\nc. The problem allows you to write two equations.\nThe question asks for f + 3. These equations can be solved algebraically,\nbut the math is going to result in having to solve a quadratic. The\nnumbers in the answer choices are fairly small; try plugging them into\nthe problem to find the correct answer. Start with (B) or (D).\n  f + 3 f Orig Cost \n New Cost \n Are the costs $9 apart?\n(B) 9 6\n No\nThe two costs are much too far apart. More people need to join the\ngroup in order to bring the costs closer together. Eliminate (A) and (B).\nTry (D) next.\nf + 3 f Orig Cost \n New Cost \n Are the costs $9 apart?\n(D) 18 15\n No\nThis time, the two costs are only $6 apart. They’re too close. The answer\nmust be between 9 and 18; therefore, the answer must be (C)."} +{"id": "book 2_p548_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 548, "page_end": 548, "topic_guess": "number_theory", "text": "Check the math if you’re not sure, but do practice this technique enough\nthat you know when you can actually tell what the answer must be\n(without doing the math).\nThe correct answer is (C).\n63. (C) 25%: Use a double-set matrix to solve this problem.\nRight-Handed Le -Handed Total\nMen 3x\nWomen 2x\nTotal 3y y 4y = 5x\nThere is a hidden constraint: the number of people must be an integer.\nThus, both x and y are integers. Moreover, the total number of people\nmust be a multiple of 4 and of 5 in order for the given ratios to be\npossible. From this constraint, there are two ways to solve.\n1. Algebraic Solution\nSince the question specifies that the number of right-handed men be as\ngreat as possible, assume that all the men are right-handed; of course,\nthat means that none of the men are le -handed. Because each column\nin a double-set matrix must total, you can also fill in the number of le -\nhanded women (the group you want)."} +{"id": "book 2_p549_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 549, "page_end": 549, "topic_guess": "fractions_decimals_percents", "text": "Right-Handed Le -Handed Total\nMen 3x 0 3x\nWomen y 2x\nTotal 3y y 4y = 5x\nThus, le -handed women represent \n of the total\npopulation.\n2. Smart Number Solution\nSince the total number of people in Smithtown must be a multiple of 20,\nset the total to 20 and determine the subtotals of men, women, le -\nhanded, and right-handed based on the ratios given in the problem.\nRight-Handed Le -Handed Total\nMen 12\nWomen 8\nTotal 15 5 20"} +{"id": "book 2_p550_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 550, "page_end": 550, "topic_guess": "number_theory", "text": "To maximize the number of right-handed men, assign all the men to the\n“right-handed men” cell, and fill in the remaining cells.\n  Right-Handed Le -Handed Total\nMen 12 0 12\nWomen 3 5 8\nTotal 15 5 20\nTherefore, le -handed women represent \n of the\npopulation.\nThe correct answer is (C).\n64. (C) 136°: If the median angle measure is 140° and the interior angle\nmeasures correspond to a set of consecutive integers, then the average\nangle measure must equal 140°. Since the sum of the angles must equal\n180(n − 2), the average angle must equal \n ."} +{"id": "book 2_p551_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 551, "page_end": 551, "topic_guess": "number_theory", "text": "Therefore, the polygon has nine sides and nine interior angles, and the\nmeasures of these angles are equal to a set of consecutive integers\ncentered at 140. The set of nine consecutive integers must therefore be:\n{136, 137, 138, 139, 140, 141, 142, 143, 144}\nThe least angle measure is 136°.\nThe correct answer is (C).\n65. (A): If \n  has a quotient of 2 and a remainder of z, then x is z more than\n2y. Mathematically, x = 2y + z. Therefore, z = x − 2y.\nIf \n has a quotient of 3 and a remainder of b, then x is b more than 3a.\nMathematically, x = 3a + b. Therefore, b = x − 3a.\nThe question asks whether z > b, and the statements give information\nabout \n . Simplify the question by replacing z and b with their\nequivalents and solving for the combination \n ."} +{"id": "book 2_p552_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 552, "page_end": 552, "topic_guess": "number_theory", "text": "Note that the inequality sign flipped in the last step because of the\ndivision by −2. The variable a is a positive integer, so no additional flip is\nrequired for that manipulation. Rephrase the question as, “Is \n?”\n(1) SUFFICIENT: This directly answers the rephrased question: “Yes,\n.” Therefore, z > b.\n(2) INSUFFICIENT: This indicates only that \n . The answer might\nbe Yes if \n . However, the answer might be No if \n is\ngreater than \n .\nThe correct answer is (A).\n66. (A) y > 50: The function (3 ∆ 47) equals the product (3)(5)(7) … (43)(45)\n(47). This product is a very large odd number, as it is the product of only\nodd numbers and thus does not have 2 as a factor. Therefore, (3 ∆ 47) + 2\n= Odd + Even = Odd, and (3 ∆ 47) + 2 does not have 2 as a factor either.\nEvery odd prime number between 3 and 47 inclusive is a factor of (3 ∆\n47), since each of these primes is a component of the product. For\nexample, (3 ∆ 47) is divisible by 3, since dividing by 3 yields an integer\nminus the product (5)(7)(9) … (43)(45)(47).\nNow consider the sum (3 ∆ 47) + k, where k is an integer. The sum will\nonly be divisible by 3 if k is also divisible by 3. In other words, when you"} +{"id": "book 2_p553_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 553, "page_end": 553, "topic_guess": "number_theory", "text": "divide (3 ∆ 47) + k by 3, you are evaluating \n . Because\n is an integer, \n must also be an integer to yield an integer\nsum.\nIn this problem, k = 2, which is not divisible by any of the odd primes\nbetween 3 and 47. Since (3 ∆ 47) IS divisible, but 2 is NOT divisible, the\nsum (3 ∆ 47) + 2 is NOT divisible by any of the odd primes between 3 and\n47.\nSo (3 ∆ 47) + 2 is not divisible by any prime number less than or equal to\n47. The least prime factor of (3 ∆ 47) + 2 must be greater than 47. Thus,\nthe minimum possible prime factor is 53, since that is the least prime\ngreater than 47.\nThe correct answer is (A).\n67. (D): If set S is the set of all prime integers between 0 and 20, then S = {2,\n3, 5, 7, 11, 13, 17, 19}.\nThere are seven odd terms and one even term in set S. If the even term is\namong those selected, the sum will be even (E + O + O = E). The sum will\nbe odd if all three terms selected are odd (O + O + O = O).\nThe probability of selecting three odd terms is \n .\nThe correct answer is (D)."} +{"id": "book 2_p554_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 554, "page_end": 554, "topic_guess": "number_theory", "text": "68. (D): The first five prime integers are 2, 3, 5, 7, and 11. These are the only\nterms that can appear in sets A and B. There are some other restrictions\non the sets:\nSet A: The average of the terms in set A is 4, so the sum of the terms is (4)\n(3) = 12. There is only one way for three of the first five primes to sum to\n12: 2 + 3 + 7. Set A is {2, 3, 7}.\nSet B: The product of the terms in set B is divisible by 22, so 2 and 11 are\nterms in set B. Set B is {2, 11, x}, where x can be 3, 5, or 7, but not 2 or 11\n(no duplicates).\nSets A and B share at least one term: the 2. If x is either 3 or 7, the sets\nwill share two terms. If x is 5, the sets will only share one term.\nThe rephrased question is, “Is x = 5?” A definite Yes or No answer leads to\na definite value answer for the number of shared terms (i.e., Yes = one\nshared term, No = two shared terms).\n(1) SUFFICIENT: If the product of the terms in set B is not divisible by 5, x\n≠ 5 and the answer to the rephrased question is a definite No.\n(2) SUFFICIENT: If the product of the terms in set B is divisible by 14,\nthen 2 and 7 are terms in B. Therefore, x = 7 and the answer to the\nrephrased question is a definite No.\nThe correct answer is (D)."} +{"id": "book 2_p555_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 555, "page_end": 555, "topic_guess": "geometry", "text": "69. (C) between \n and \n : The area of a square is equal to the length of\none side squared. The area of a circle is equal to πr2. The question asks\nfor the fraction \n .\nInscribed angle USA cuts off a diameter (UA) of the circle, so angle USA\nmust be a right angle. Therefore, the triangle is a right triangle with\nhypotenuse UA.\nThe Pythagorean theorem indicates that US2 + SA2 = UA2.\nUS2 also represents the area of the smaller square. SA2 also represents\nthe area of the larger square. The sum of the two equals the quantity Q\nmentioned in the question stem, so Q = UA2.\nUA is a diameter of the circle, so the quantity Q = (2r)2 = 4r2. The fraction\nis:\nThe value of π is approximately 3.14, which falls between 3 and 3.5.\nTherefore:"} +{"id": "book 2_p556_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 556, "page_end": 556, "topic_guess": "general", "text": "The correct answer is (C).\n70. (E): The value of n must be 4 or greater. Start with statement (2), as it is\nmuch easier than statement (1).\n(2) INSUFFICIENT: If n is odd, it could equal 3, 5, 7, or so on.\n(1) INSUFFICIENT: Test cases to figure out what is possible.\nn Set S # Multiples 2 = # Multiples 3?\n4 3, 4, 5, 6 Yes: 2 multiples of each\n5 3, 4, 5, 6, 7 Yes: 2 multiples of each\n(1) AND (2) INSUFFICIENT: The case of n = 5 was already proven in the\nlast step. Is there a way to have n = another odd number? Keep testing.\nn Set S # Multiples 2 = # Multiples 3?\n5 3, 4, 5, 6, 7 Yes: 2 multiples of each\n7 3, 4, 5, 6, 7, 8, 9 Yes: 3 multiples of each"} +{"id": "book 2_p557_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 557, "page_end": 557, "topic_guess": "general", "text": "Since there are still at least two possible values for n, none of the\ninformation is sufficient to answer the question.\nThe correct answer is (E)."} +{"id": "book 2_p558_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 558, "page_end": 558, "topic_guess": "word_problems", "text": "Workout Set 8\n71. \nEnergy usage (units) 11 10 8 7\nNumber of days   4   5 n 3\nThe table above shows daily energy usage for an office building and\nthe number of days that amount of energy was used. If the average\n(arithmetic mean) daily energy usage was greater than the median\ndaily energy usage, what is the least possible value for n ?\n2(A)\n3(B)\n4(C)\n5(D)\n6(E)\n72. A painting crew painted 80 houses. They painted the first y houses\nat a rate of x houses per week. Then, more painters arrived and\neveryone worked together to paint the remaining houses at a rate\nof 1.25x houses per week. How many weeks did it take to paint all\n80 houses, in terms of x and y  ?"} +{"id": "book 2_p559_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 559, "page_end": 559, "topic_guess": "fractions_decimals_percents", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n73. a = x + y and b = x − y. If a2 = b2, what is the value of y ?\n(1)\n(2)\n74. A herd of 33 sheep is sheltered in a barn with 7 stalls, each of which\nis labeled with a unique letter from A to G, inclusive. Is there at least\none sheep in every stall?\nThe ratio of the number of sheep in stall C to the number of\nsheep in stall E is 2 to 3.\n(1)"} +{"id": "book 2_p560_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 560, "page_end": 560, "topic_guess": "number_theory", "text": "The ratio of the number of sheep in stall E to the number of\nsheep in stall F is 5 to 2.\n(2)\n75. If (s × 10q) − (t × 10r) = 10r, where q, r, s, and t are positive integers\nand q > r, then what is the units digit of t ?\n0(A)\n1(B)\n5(C)\n7(D)\n9(E)\n76. Is \n ?\n|y − 3| ≤ 1(1)\ny × |y| > 0(2)\n77. What is the greatest prime factor of 21054 − 21352 + 214 ?\n  2(A)\n  3(B)\n  7(C)\n11(D)\n13(E)"} +{"id": "book 2_p561_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 561, "page_end": 561, "topic_guess": "fractions_decimals_percents", "text": "78. A decimal is called a “shrinking number” if its value is between 0\nand 1 and each digit to the right of the decimal is not less than the\ndigit to its immediate right. For instance, 0.86553221 is a shrinking\nnumber. If x is a shrinking number, which of the following must be\ntrue?\n is a shrinking number.I.\n is a shrinking number.II.\n is a shrinking number. III.\nI only(A)\nII only(B)\nI and II only(C)\nIII only(D)\nI, II, and III(E)\n79. Each of the seven paintings in an art gallery has a different price. Is\nit possible to purchase at least three paintings for no more than\n$1,800 in total?\nThe median price of the seven paintings is $550.(1)\nIt is possible to purchase four paintings at the gallery for a total\nof $2,300.\n(2)"} +{"id": "book 2_p562_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 562, "page_end": 562, "topic_guess": "coordinate_geometry", "text": "80. An (x, y) coordinate pair is to be chosen at random from the xy-\nplane. What is the probability that y ≥ |x| ?\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p563_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 563, "page_end": 563, "topic_guess": "general", "text": "Workout Set 8 Answer Key\n71. E\n72. B\n73. B\n74. C\n75. E\n76. A\n77. C\n78. B\n79. D\n80. E"} +{"id": "book 2_p564_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 564, "page_end": 564, "topic_guess": "statistics", "text": "Workout Set 8 Solutions\n71. (E) 6: Here’s what you know about the arithmetic mean:\nIn contrast, the median depends on n, but not in a linear way. To find the\nmedian, you order the terms and pick the middle one, so try various n\nvalues (i.e., vary the number of 8’s in the list). This implies the eventual\nneed to plug n values into the mean formula above, so draw a picture to\nhelp eliminate some answers first.\nThe “low” and “high” grouping is a fast way to find the relationship\nbetween median and mean for this set."} +{"id": "book 2_p565_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 565, "page_end": 565, "topic_guess": "statistics", "text": "If the number of “lows” and “highs” are equal, the median is the average\nof the middle terms. That is, if n = 6, then the median = 9.\nWhen n = 6, the mean must be greater than 9. Why? Pairs of 8 and 10\nterms average to 9, but there is one “extra” 8. Pairs of 7 and 11 terms\naverage to 9, but there is one “extra” 11. These “extra” terms differ from\n9 by −1 and +2 respectively, for a total difference of +1. That positive\ndifference implies that (mean > 9), or (mean > median).\nTo prove that n = 5 is too low, you could take a more conventional\napproach. If n = 5, the number of terms is 3 + 5 + 5 + 4 = 17. The median is\nthe ninth term in this ordered set: 7, 7, 7, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10,\n11, 11, 11, 11. Thus, the median is 10, while the mean is closer to 9.\nThat is, if n = 5, then median > mean.\nIf n = 6, then mean > median.\nThe correct answer is (E)."} +{"id": "book 2_p566_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 566, "page_end": 566, "topic_guess": "word_problems", "text": "72. (B) \n : This is a combined work problem, so use the work\nformula: rate × time = work. The work and rates are given, but you need\nto calculate time, so manipulate the formula: \n . This\nproblem also has variables in the answer choices, so it is efficient to\nchoose smart numbers.\nThere are 80 houses, y houses are painted at a rate of x houses per week,\nand the rate increases to 1.25x houses per week for the remaining 80 − y\nhouses. To make the math easier, choose values such that x and 1.25x\nare integers (i.e., x is a multiple of 4) and y and 80 − y are divisible by x\nand 1.25x, respectively.\nVariable Value Units\nTotal Houses 80 80 Houses\nHouses Painted at Slower Rate y 20 Houses\nHouses Painted at Faster Rate 80 − y 60 Houses\nInitial Rate x 4 Houses/Week\nIncreased Rate 1.25x 5 Houses/Week\nThe total painting time is:"} +{"id": "book 2_p567_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 567, "page_end": 567, "topic_guess": "word_problems", "text": "20 houses painted at a rate of 4 houses/week = 5 weeks\n60 houses painted at a rate of 5 houses/week = 12 weeks\nTotal time for 80 houses = 5 + 12 = 17 weeks\nThe correct answer is (B).\n Eliminate.(A)\n Correct!(B)\n Not an integer.\nEliminate.\n(C)\n Not an integer.\nEliminate.\n(D)\n Eliminate.(E)\n73. (B): If a2 = b2, then (x + y)2 = (x − y)2. Distribute both sides using the\nsquare of a sum and square of a difference special products, then\nsimplify."} +{"id": "book 2_p568_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 568, "page_end": 568, "topic_guess": "general", "text": "There are three basic scenarios.\nx y xy = 0\n0 Any nonzero ✓\nAny nonzero 0 ✓\n0 0 ✓\n(1) INSUFFICIENT: This statement indicates that x and y must be non-\nnegative for their square roots to be real values. The statement also\neliminates the last scenario, in which x = y = 0. But y could still be 0 or\nany positive value.\nx y xy = 0\n0 Any positive ✓ ✓\nAny positive 0 ✓ ✓\n0 0 ✓ ✗ \n(2) SUFFICIENT: This statement indicates that x and y must be non-\nnegative for their square roots to be real values. This statement"} +{"id": "book 2_p569_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 569, "page_end": 569, "topic_guess": "fractions_decimals_percents", "text": "eliminates the last scenario. If x and y were both 0, \n would\nequal 0. It also eliminates the first scenario. If \n , then\n. Therefore, \n . Thus, you can conclude that y = 0.\nx y xy = 0\n0 Any positive ✓ ✗ \nAny positive 0 ✓ ✓\n0 0 ✓ ✗ \nThe correct answer is (B).\n74. (C):\n(1) INSUFFICIENT: Set up a table and assign sheep to stalls.\nStall A B C D E F G\n# of Sheep 2x 3x\nSince a fractional sheep is not possible in this problem, x must be a\npositive integer. Suppose x = 2, so there are 4 sheep in C and 6 sheep in\nE. With 23 sheep remaining, it is possible for each of the other stalls to"} +{"id": "book 2_p570_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 570, "page_end": 570, "topic_guess": "number_theory", "text": "hold at least 1 sheep (a Yes answer). However, the 23 other sheep might\nall be in stall B, leaving stalls A, D, F, and G empty (a No answer).\n(2) INSUFFICIENT: Set up a table and assign sheep to stalls.\nStall A B C D E F G\n# of Sheep 5y 2y\nIf y = 1, there are 5 sheep in E and 2 sheep in F. With 26 sheep remaining,\nit is possible for each of the other stalls to hold at least 1 sheep (a Yes\nanswer). However, 13 sheep might be in both stalls A and B, leaving\nstalls C, D, and G empty (a No answer).\n(1) AND (2) SUFFICIENT: Set up a table and assign sheep to stalls.\nStall A B C D E F G\n# of Sheep 2x 3x = 5y 2y\nSince a fractional sheep is not possible in this problem, x and y must\nboth be positive integers that satisfy the equation 3x = 5y. The only\npossibility is x = 5 and y = 3, since higher multiples would require more\nthan 33 sheep total. Thus, 31 sheep are allocated among these stalls as\nfollows:"} +{"id": "book 2_p571_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 571, "page_end": 571, "topic_guess": "sequences_patterns", "text": "Stall A B C D E F G\n# of Sheep 10 15 6\nThirty-one of the 33 sheep are in three of the 7 pens. With only 2 sheep\nunaccounted for, there is no way to place at least 1 sheep in each of the\nremaining four pens (a definite No answer).\nThe correct answer is (C).\n75. (E) 9: Group the 10r terms together.\nNow, solve for y.\nSince q > r, the exponent on 10q − r is positive. Since s is a positive\ninteger, s × 10q − r is a multiple of 10 and therefore ends in 0. Any multiple\nof 10 minus 1 yields an integer with a units digit of 9."} +{"id": "book 2_p572_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 572, "page_end": 572, "topic_guess": "algebra", "text": "The correct answer is (E).\n76. (A): The complicated expression in the question stem leads to a\ndisguised Positive/Negative problem. In general, \n . Think\nabout this relationship with a real example.\nIn both cases (positive or negative 3), the end result is 3. Thus, in\ngeneral, \n will always result in a positive value, or |x|. Rephrase the\noriginal question using the absolute value symbol in place of the\n“square root of the square” symbols; then try to make the right-hand\nside look more like the le .\nIs |y − 4| = 4 − y ? becomes Is |y − 4| = −(y − 4) ?\nSince the absolute value of y − 4 must be positive or zero, you can\nrephrase the question further.\nIs −(y − 4) ≥ 0 ? becomes Is (y − 4) ≤ 0 ? and then Is y ≤ 4 ?\n(1) SUFFICIENT: The absolute value |y − 3| can be interpreted as the\ndistance between y and 3 on a number line. Thus, y is no more than 1\nunit away from 3 on the number line, so 2 ≤ y ≤ 4. Thus, y ≤ 4."} +{"id": "book 2_p573_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 573, "page_end": 573, "topic_guess": "number_theory", "text": "(2) INSUFFICIENT: If y × |y| > 0, then y × |y| is positive. This means y and\n|y| must have the same sign. The term |y| is non-negative, so y must be\npositive. However, knowing that y is positive is not enough to indicate\nwhether y ≤ 4.\nThe correct answer is (A).\n77. (C) 7: A base of 2 is common to each term, and 10 is the smallest\nexponent appearing with that base. Factor 210 out from all of the terms\nin the expression.\nClearly, 2 is a prime factor, but is it the greatest? Examine (54 − 2352 + 24)\nto determine whether it has a larger prime factor. The expression is of\nthe form x2 − 2xy + y2, where:\nWrite the expression in factored form.\nThe prime factors of 21 are 3 and 7, so the largest prime factor of the\noriginal expression is 7."} +{"id": "book 2_p574_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 574, "page_end": 574, "topic_guess": "fractions_decimals_percents", "text": "Alternatively, if you did not see the quadratic template in (54 − 2352 + 24),\nyou could also perform the computation and factor the result.\nThe correct answer is (C).\n78. (B) II only: According to the text, a “shrinking number” is a decimal\nbetween 0 and 1 in which each digit to the right of the decimal is no\nsmaller than the digit to its immediate right.\nThat is, in a shrinking number, each digit is either larger than the one to\nits right, or the same as the one to its right. As you go through the digits\nfrom le to right, the digits will either stay the same or get smaller. For\ninstance, 0.5331 is a shrinking number, but 0.37654 is not a shrinking\nnumber.\nThis Roman Numeral problem asks which of three options must be true.\nTest cases and eliminate any options that are not necessarily true.\nOne shrinking number is x = 0.1000 . . . This is a shrinking number\nbecause the digits do not get larger a er the decimal point. Since this"} +{"id": "book 2_p575_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 575, "page_end": 575, "topic_guess": "fractions_decimals_percents", "text": "number is relatively easy to do math with, start by using x = 0.1 to test\ncases.\nI. As a fraction, \n , so\n. However, this is not a\nshrinking number, since 9 is greater than 0. So statement I is not\nnecessarily true. Eliminate choices (A), (C), and (E).\nII. \n , which is a shrinking number.\nStatement II could be true. \nIn fact, statement II must always be true. Adding 9 and then dividing by\n10 is equivalent to inserting the digit 9 to the immediate right of the\ndecimal. Since 9 is the largest possible digit, it will never be lower than\nthe digit to its immediate right, so the decimal will still be a shrinking\nnumber.\nIII. \n , which is not a shrinking number, since the digit\ntwo places to the right of the decimal is greater than the digit\nimmediately to the right of the decimal. Statement III is not necessarily\ntrue. Eliminate choice (D).\nThe correct answer is (B)."} +{"id": "book 2_p576_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 576, "page_end": 576, "topic_guess": "statistics", "text": "79. (D): A gallery has seven paintings with different prices. Is it possible to\npurchase at least three of the paintings for no more than $1,800? This is\npossible only if the three least expensive paintings at the gallery have a\ntotal price of $1,800 or less. The question can be rephrased as, “Is the\ntotal price of the three least expensive paintings $1,800 or less?” \n(1) SUFFICIENT: The median price of the seven paintings is $550; since\nall of the paintings have different prices, the three least expensive\npaintings have prices below the median. Therefore, the total price of the\nthree least expensive paintings is less than $550 + $550 + $550 = $1,650.\nThe answer is definitely Yes, so this statement is sufficient.\n(2) SUFFICIENT: This statement implies that the four least expensive\npaintings have a total price of $2,300 or less. Jot down some possible\ncases.\nCase 1: $1, $2, $3, and $2,000. The three least expensive paintings can be\npurchased for $1,800, and the answer to the question is Yes.\nCase 2: To find a No answer, try to maximize the lowest three prices out\nof the four. To do so, make the highest price as low as possible, or only\nslightly more than one-fourth of the total. Since one-fourth of $2,300 is\n$575, a good case to test is $573, $574, $576, and $577. However, the\nsum of the lowest three prices is $1,723, which is still well below $1,800,\nand the answer is still Yes.\nEven when maximizing the lowest of the three prices, the three least\nexpensive paintings can still be purchased for no more than $1,800, and\nthe answer to the question is definitely Yes, so this statement is\nsufficient."} +{"id": "book 2_p577_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 577, "page_end": 577, "topic_guess": "coordinate_geometry", "text": "The correct answer is (D).\n80. (E) \n : Graph the equation y = |x|.\nThe inequality y > |x| represents everything above the line (on either side\nof the y-axis)—that is, the shaded region. Since the equation y = |x| forms\na 45-degree angle from the x-axis, there are 90 degrees above the line\n(on both sides of the y-axis). This represents one-fourth of the xy-plane.\nTherefore, if a random pair of (x, y) coordinates is chosen from the\nplane, the probability is \n that the point will fit the criterion y ≥ |x|.\nThe correct answer is (E)."} +{"id": "book 2_p578_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 578, "page_end": 578, "topic_guess": "geometry", "text": "Workout Set 9\n81. \nAs shown in the figure, square ABCD is inscribed in a circle with\ncircumference \n . What is the area of the shaded region in the\ndiagram above?\n2x(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p579_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 579, "page_end": 579, "topic_guess": "number_theory", "text": "82. If 3a + 3a − 2 = (90)(3b), what is b in terms of a ?\na − 4(A)\na − 2(B)\na + 4(C)\n3a + 2(D)\n3a + 4(E)\n83. If xy ≠ 0, what is the value of \n ?\ny = 4 − x(1)\nx(x − 6y) = −9y2(2)\n84. When one new integer is added to an existing list of six integers,\ndoes the median of the list change?\nThe mean of the original six numbers is 50.(1)\nAt least two of the numbers in the original list were 50.(2)\n85. Let abc and dcb represent three-digit positive integers. If abc + dcb\n= 598, then which of the following must be equivalent to a ?"} +{"id": "book 2_p580_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 580, "page_end": 580, "topic_guess": "number_theory", "text": "d − 1(A)\nd(B)\n3 − d(C)\n4 − d(D)\n5 − d(E)\n86. For nonzero integers a, b, c and d, is \n negative?\nad + bc = 0(1)\nabcd = −4(2)\n87. Set A consists of 8 distinct prime numbers. If x is equal to the range\nof set A and y is equal to the median of set A, is the product xy\neven?\nThe smallest integer in the set is 5.(1)\nThe largest integer in the set is 101.(2)\n88. If x3 < 16x, which of the following CANNOT be true?\n|x| > 4(A)\nx > −4(B)\nx < −4(C)\nx > 4(D)\nx < 4(E)"} +{"id": "book 2_p581_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 581, "page_end": 581, "topic_guess": "general", "text": "89. Is the two-digit positive integer n divisible by 3 ?\nIf the digits of n are reversed to produce the two-digit integer\nm, then m is divisible by 3.\n(1)\nIf the digits of n are reversed to produce the two-digit integer\nm, then m + n is divisible by 3.\n(2)\n90. Of all the sheets of paper in the tray of Printer A, 20% are removed\nand transferred to the tray of Printer B. A er the transfer, are there\nmore sheets of paper in the tray of Printer B than in the tray of\nPrinter A ?\nThe transfer increases the number of sheets of paper in the tray\nof Printer B by more than 25%.\n(1)\nThe transfer increases the number of sheets of paper in the tray\nof Printer B by less than 30%.\n(2)"} +{"id": "book 2_p582_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 582, "page_end": 582, "topic_guess": "general", "text": "Workout Set 9 Answer Key\n81. B\n82. A\n83. B\n84. E\n85. D\n86. D\n87. A\n88. D\n89. D\n90. B"} +{"id": "book 2_p583_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 583, "page_end": 583, "topic_guess": "geometry", "text": "Workout Set 9 Solutions\n81. (B) πx − 2x: The circumference of the circle is equal to \nso \n .\nThe area of the circle is equal to \n .\nThe area of a square is side2. The diagonal of this square is the diameter\nof the circle, which is equal to \n . The diagonal of a square is\nalways \n , so the side of the square is equal to \n .\nTherefore, side \n and the area of square ABCD is:\nThe shaded area is the area of the circle minus the area of the square =\nπx − 2x."} +{"id": "book 2_p584_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 584, "page_end": 584, "topic_guess": "algebra", "text": "The correct answer is (B).\n82. (A) a − 4: Because the problem never provides real values for a or b, you\ncan choose smart numbers to solve. Choose something for a that will\nmake both exponents on the le side positive. If a = 3, then:\nTherefore, b = −1. Plug a = 3 into the answers and look for −1.\nAlternatively, you can solve algebraically. Some manipulation is\nrequired to determine b in terms of a, but spend a minute thinking\nstrategically about what manipulations to do. It would help to isolate\nterms with an a by factoring 3a out on the le side. This will let you\ncompare 3a to 3b, and thus compare a to b, once you clean up the\nconstant terms that are le over. Also, 90 is a multiple of 9, which is a\n3 – 4 = −1 Correct.(A)\n3 – 1 = 1 Eliminate.(B)\n3 + 4 = Too big. Eliminate.(C)\n3(3) + 2 = Too big. Eliminate.(D)\n3(3) + 4 = Too big. Eliminate.(E)"} +{"id": "book 2_p585_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 585, "page_end": 585, "topic_guess": "algebra", "text": "power of 3. Ninety is also the sum of 81 and 9, both powers of 3\nthemselves. It is likely that some constant terms will cancel, as shown\nbelow.\nTherefore, b = a − 4.\nThe correct answer is (A).\n83. (B): The constraint in the question stem indicates that neither x nor y\nequals zero.\n(1) INSUFFICIENT: You can prove insufficiency by Testing Cases. If y = 1,\nthen x = 3, in which case \n is 3. If y = 2, then x = 2, in which case \n is 1.\n(2) SUFFICIENT: Testing Cases might not be such a great idea for this\nstatement because there are multiple instances of each variable. Begin\nby distributing the le -hand side of the equation."} +{"id": "book 2_p586_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 586, "page_end": 586, "topic_guess": "statistics", "text": "Now, you’ve got a quadratic, so solve that way.\nThe correct answer is (B).\n84. (E): You can test cases to try to prove or disprove the statements.\n(1) INSUFFICIENT: If the original set is 50, 50, 50, 50, 50, 50, the mean and\nmedian were 50 originally. The new term may be any value and the\nmedian will not change.\nIf, on the other hand, the original set is 0, 0, 0, 100, 100, 100, the mean\nand median were 50 originally. If the new term is 75, the median\nincreases to 75. More generally for this set, if the new term is not equal\nto the original median of 50, the median will change.\n(2) INSUFFICIENT: If the original set is 50, 50, 50, 50, 50, 50, the median\nwas 50 originally. The new term may be any value and the median will\nnot change.\nIf, on the other hand, the original set is 0, 0, 0, 50, 50, 50, the median was\n25 originally. If the new term is 40, the median increases to 40. More"} +{"id": "book 2_p587_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 587, "page_end": 587, "topic_guess": "statistics", "text": "generally for this set, if the new term is not equal to the original median\nof 25, the median will change.\n(1) AND (2) INSUFFICIENT: The mean must be 50 and at least two terms\nin the original set must be 50.\nIf the original set is 50, 50, 50, 50, 50, 50, the mean and median were 50\noriginally. The new term may be any value and the median will not\nchange.\nIf the original set is 0, 0, 0, 50, 50, 200, the mean was 50 and the median\nwas 25 originally. If the new term is greater than 25, the median\nincreases. If the new term is less than 25, the median decreases.\nThe correct answer is (E).\n85. (D) 4 − d: Set up the addition and extract several equations by summing\neach digit place individually.\nNote that both the ones digit and tens digit come from the sum c + b.\nSince the result is different in the ones digit (8) and the tens digit (9), a 1\nmust be carried from the ones to the tens digit. Thus, c + b ≠ 8; instead, c\n+ b = 18.\nIf c + b = 18, then both c and b must be 9 (the largest digit). Place an 8 in\nthe ones digit of the sum and carry a 1 to the tens place. The sum in the"} +{"id": "book 2_p588_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 588, "page_end": 588, "topic_guess": "number_theory", "text": "tens digit is thus 1 + b + c = 1 + 18 = 19. Next, place a 9 in the tens digit of\nthe sum and carry a 1 to the hundreds place.\nIn the hundreds place, the sum is 1 + a + d = 5.\nNow, solve for a.\nThe correct answer is (D).\n86. (D): If an even number (0, 2, or 4) of the integers a, b, c, and d is negative,\neach pair of negatives will cancel, because (−1)(−1) = +1 and\n. This would yield a positive result for \n .\nThus, a way to rephrase the question is, “Among the integers a, b, c, and\nd, are an odd number (one or three) of them negative?”\n(1) SUFFICIENT: This statement can be rephrased as ad = −bc.\na d b c ad = −bc Odd number of negatives?\n+ + − + ✓ Yes\n+ − + + ✓ Yes"} +{"id": "book 2_p589_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 589, "page_end": 589, "topic_guess": "number_theory", "text": "a d b c ad = −bc Odd number of negatives?\n− − − + ✓ Yes\n+ − − − ✓ Yes\nThough the table doesn’t list all possibilities, it lists enough to realize\nthat, in order for the signs of ad and bc to be opposite one another,\neither one or three of the four integers must be negative.\n(2) SUFFICIENT: You might recognize that the (−1)(−1) = +1 property\nimplies that abcd is only negative when there are non-paired negatives\namong the integers. That is, an odd number (one or three) of the\nintegers a, b, c, and d must be negative. If not, you could list a few cases\nto see the pattern.\na b c d abcd = negative Odd number of negatives?\n+ + − + ✓ Yes\n+ − + + ✓ Yes\n− + − − ✓ Yes\n− − + − ✓ Yes"} +{"id": "book 2_p590_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 590, "page_end": 590, "topic_guess": "statistics", "text": "The correct answer is (D).\n87. (A): The product xy will be even if x is even, y is even, or both are even.\nThe prime numbers include 2, 3, 5, 7, 11, 13, 17, 19, etc. The smallest\npossible term in set A is 2, which is the only even prime.\nx = the range of set A = largest term − smallest term in set A. If the\nsmallest term in set A is 2, then x = odd − even = odd. If the smallest term\nin set A is odd (i.e., not 2), then x = odd − odd = even.\nThe median of set A is the average of the two middle terms, since the\nnumber of terms in the set is even. Thus,\n. However, y could be\neither even (e.g., when the middle terms are 11 and 13) or odd (e.g.,\nwhen the middle terms are 7 and 11).\nA useful rephrase of this question is, “Is either x or y even?”\n(1) SUFFICIENT: If the smallest prime in the set is 5, x = even, and\ntherefore xy is even.\n(2) INSUFFICIENT: If the largest integer in the set is 101, the range of the\nset can be odd or even (e.g., 101 − 3 = 98 or 101 − 2 = 99). The median of\nthe set can also be odd or even, as discussed. Therefore, xy can be either\nodd or even.\nThe correct answer is (A)."} +{"id": "book 2_p591_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 591, "page_end": 591, "topic_guess": "algebra", "text": "88. (D) x > 4: It may be tempting to simplify this way:\nThe first step of this solution is wrong because you can’t divide by x\nwithout knowing its sign. If x is negative, you would have to flip the sign.\nCheck both cases.\nx x3 < 16x becomes: Take square root: Solve for x:\nPositive x2 < 16 (don’t flip) |x| < 4 0 < x < 4\nNegative x2 > 16 (flip) |x| > 4 x < −4\nThere are two ranges of solutions for x. The question asks for something\nthat is definitely NOT true. Check the answer choices for an answer that\ndoes NOT include any values in these two ranges. All of the answer\nchoices include some of the values above, except for (D): if x must be\neither between 0 and 4, or less than −4, it cannot be the case that x > 4.\nHowever, it is possible, for certain values of x, for each of the other\nanswer choices to be true.\nAlternatively, there is a way to do algebra with inequalities containing\nvariables. Instead of dividing by the variable, subtract it from both sides,\nthen factor."} +{"id": "book 2_p592_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 592, "page_end": 592, "topic_guess": "number_theory", "text": "In order for the product of three numbers to be negative, either all three\nneed to be negative, or exactly one of the three needs to be negative.\nThis cannot occur if x is greater than 4, since all three values would be\npositive. Therefore, (D) cannot be true. \nThe correct answer is (D).\n89. (D): Before diving into the statements, remind yourself of the divisibility\nrules. One way to check for divisibility by 3 is to add the digits. If their\nsum is divisible by 3, then the number itself is divisible by 3.\n(1) SUFFICIENT: If m is divisible by 3, then the sum of the digits of m\nmust be divisible by 3 as well. Since m has the same digits as n, the sum\nof the digits of n must also be a multiple of 3. Therefore, n itself is\ndivisible by 3.\n(2) SUFFICIENT: This statement requires a bit more work because it’s not\nimmediately clear what would happen when you add m and n. Because\nthe question asks you to move digits around to different places (the tens\nplace, the units place), rewrite the information using place-value\nnotation, where a = the tens digit of n and b = the units digit of n."} +{"id": "book 2_p593_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 593, "page_end": 593, "topic_guess": "algebra", "text": "The statement indicates that the sum m + n is divisible by 3, so 11(a + b)\nmust also be divisible by 3. Eleven itself is not divisible by 3, so the sum\na + b must be divisible by 3. This sum represents the sum of the two\ndigits of the number m as well as the sum of the two digits of the\nnumber n; if the sum of n’s two digits is divisible by 3, then n is also\ndivisible by 3.\nThe correct answer is (D).\n90. (B): You can solve this problem algebraically or by Testing Cases. Both\nmethods are shown.\n(1) INSUFFICIENT: If Printer A starts out with 100 sheets, then 20 sheets\nare moved to Printer B, and Printer A ends up with 80 sheets. If B started\nout with 75 sheets, then an increase of 20 sheets would be more than\n25% of the starting number, and B would end up with 95 sheets. In this\ncase, B has more sheets than A a er the transfer.\nIf, instead, B started out with 40 sheets, then an increase of 20 sheets\nwould be more than 25% of the starting number, and B would end up\nwith 60 sheets. In this case, B has fewer sheets than A a er the transfer.\n(2) SUFFICIENT: If Printer A starts out with 100 sheets and 20 are moved\nto Printer B, then A ends up with 80 sheets. If B started out with 2,000\nsheets, then an increase of 20 sheets would be less than 30% of the\nstarting number, and B would end up with many more sheets than A."} +{"id": "book 2_p594_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 594, "page_end": 594, "topic_guess": "algebra", "text": "If B started out with 80 sheets, then a 20-sheet increase would still be\nless than 30% of the starting number, and again B would end up with\nmore sheets than A.\nHow low can B go? If the 20 sheets represent a less-than-30% increase,\nthen use 30% as the limiting figure. If 20 represented exactly 30% of B’s\nsheets, then B would have to start with \n sheets (ignore the fact\nthat this isn’t an integer). In this case, B would end up with \nsheets, which is still greater than A’s 80 sheets. A er the transfer, B can’t\ndrop below A.\nAlgebraically, use a for the original number of sheets in Printer A and use\nb for the original number of sheets in Printer B.\nA er the transfer, A has a – 0.2a = 0.8a sheets. B has b + 0.2a sheets.\nRephrase the question: “Is b + 0.2a > 0.8a?” Or, “Is b > 0.6a?”\n(1) INSUFFICIENT: Translate the statement into algebra.\nFlip the statement around to compare it more easily to the rephrased\nquestion: b < 0.8a. Is it true that b > 0.6a? Maybe.\n(2) SUFFICIENT: Translate the statement into algebra."} +{"id": "book 2_p595_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 595, "page_end": 595, "topic_guess": "fractions_decimals_percents", "text": "Therefore, \n or \n .\nThe decimal equivalent of \n is \n , so b is indeed always greater than\n0.6a.\nThe correct answer is (B)."} +{"id": "book 2_p596_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 596, "page_end": 596, "topic_guess": "word_problems", "text": "Workout Set 10\n91. If |x| ≠ |y|, xy ≠ 0, \n , and \n , then \n is\nequal to which of the following?\n(A)\n(B)\n(C)\n(D)\n(E)\n92. If a certain culture of bacteria increases by a constant factor of x\nevery y minutes, how long will it take for the culture to increase to\n10,000 times its original size?"} +{"id": "book 2_p597_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 597, "page_end": 597, "topic_guess": "number_theory", "text": "(1)\nIn two minutes, the culture will increase to 100 times its\noriginal size.\n(2)\n93. A cylinder of height h is \n full of water. When all of the water is\npoured into an empty cylinder whose radius is 25% larger than that\nof the original cylinder, the new cylinder is \n full. The height of the\nnew cylinder is what percent of h ?\n  25%(A)\n  50%(B)\n  60%(C)\n  80%(D)\n100%(E)\n94. If a and b are distinct positive integers, what is the units digit of\n2a8b4a + b ?\nb = 24 and a < 24(1)\nThe greatest common factor of a and b is 12.(2)\n95. Employees of a certain company are each to receive a unique\nseven-digit identification code consisting of the digits 0, 1, 2, 3, 4, 5,\nand 6 such that no digit is used more than once in any given code."} +{"id": "book 2_p598_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 598, "page_end": 598, "topic_guess": "sets_probability_counting", "text": "In valid codes, the second digit in the code is exactly twice the first\ndigit. How many valid codes are there?\n  42(A)\n120(B)\n210(C)\n360(D)\n840(E)\n96. An n-sided die has sides labeled with the numbers 1 through n,\ninclusive, and has an equal probability of landing on any side when\nrolled. If the die is rolled twice, what is the probability of rolling 1\nboth times?\nIf the die is rolled twice, the probability that the two rolls are\ndifferent is 80%.\n(1)\nIf the die is rolled twice, the probability that the two rolls are\nequal is \n .\n(2)\n97. If w, x, y, and z are positive integers and \n , what is\nthe proper order, increasing from le to right, of the following\nquantities: \n ?"} +{"id": "book 2_p599_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 599, "page_end": 599, "topic_guess": "number_theory", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n98. What is the value of |a + b| ?\n(a + b + c + d) (a + b − c − d) = 16(1)\nc + d = 3(2)\n99. Set A consists of all the integers between 10 and 21, inclusive. Set B\nconsists of all the integers between 10 and 50, inclusive. If x is a\nnumber chosen randomly from set A, y is a number chosen\nrandomly from set B, and y has no factor z such that 1 < z < y, what\nis the probability that the product xy is divisible by 3 ?"} +{"id": "book 2_p600_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 600, "page_end": 600, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n100. At a birthday party, x children will be seated at two different tables.\nAt the table with the birthday cake on it, exactly y children will be\nseated, including the birthday girl, Sally. How many different\ngroups of children may be seated at the birthday cake table?"} +{"id": "book 2_p601_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 601, "page_end": 601, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p602_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 602, "page_end": 602, "topic_guess": "general", "text": "Workout Set 10 Answer Key\n91. D\n92. D\n93. D\n94. B\n95. D\n96. A\n97. B\n98. C\n99. B\n 00. E"} +{"id": "book 2_p603_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 603, "page_end": 603, "topic_guess": "general", "text": "Workout Set 10 Solutions\n91. (D) \n : If x = 2 and y = 3, then \n and \n .\nThe target number for testing the answer choices is \n .\n(A)\n(B)\n Eliminate.(C)\nCorrect.\n(D)\n Greater\nthan 1. Eliminate.\n(E)"} +{"id": "book 2_p604_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 604, "page_end": 604, "topic_guess": "algebra", "text": "This problem can be solved algebraically, but this path is not\nrecommended.\nNow that the fractions have a common denominator of x, subtract one\nfrom the other.\nYou are not alone if the algebraic solution was not obvious to you!\nAlgebraic “false starts” are common in this type of problem. In addition,\nthere are other, equally valid algebraic paths whose final forms would not\nmatch any of the answers."} +{"id": "book 2_p605_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 605, "page_end": 605, "topic_guess": "algebra", "text": "When you encounter a tough “pure algebra” problem that has variables in\nthe answer choices, picking numbers and testing the answer choices is\no en the best approach; it’s fast, easy, and correct.\nThe correct answer is (D).\n92. (D): To understand the question stem, pick some numbers: the bacteria\nculture begins with an initial quantity of I = 100 and increases by a factor\nof x = 2 every y = 3 minutes. Construct a table to track the growth of the\nbacteria.\nTime (min.) Bacteria\n3                100(2) = 100(2)1\n6            100(2)(2) = 100(2)2\n9       100(2)(2)(2) = 100(2)3\n12 100(2)(2)(2)(2) = 100(2)4\n… …\n3n = t 100(2)n\nn represents the number of growth periods, and \n where t is time\nin minutes. For example, the fourth growth period in the table above"} +{"id": "book 2_p606_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 606, "page_end": 606, "topic_guess": "algebra", "text": "ended at 12 minutes, and \n .\nFrom this example, construct a general formula for the quantity of\nbacteria, F.\nThis question asks how long it will take for the bacteria to grow to 10,000\ntimes their original amount. In other words, “What is t when F = 10,000I?”\nThe rephrased question is, “What is t when \n ?”\n(1) SUFFICIENT: Note that the yth root of x is equivalent to x to the \npower. This statement indicates that \n . If you plug this value\ninto the equation, you can solve for t (though stop the calculation at the\npoint that you can tell that you can solve for t)."} +{"id": "book 2_p607_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 607, "page_end": 607, "topic_guess": "fractions_decimals_percents", "text": "(2) SUFFICIENT: The culture grows one-hundredfold in 2 minutes. In other\nwords, the sample grows by a factor of 102. Since exponential growth is\ncharacterized by a constant factor of growth (i.e., by a factor of x every y\nminutes), in another 2 minutes, the culture will grow by another factor of\n102. Therefore, a er a total of 4 minutes, the culture will have grown by a\nfactor of 102 × 102 = 104, or 10,000.\nThe correct answer is (D).\n93. (D) 80%: The answer choices contain percentages, and the problem never\noffers a real value for h. Choose a smart number.\nAlso note that the problem mentions the volume of the cylinder (V =\nπr2h), so you’re going to need a radius. That radius needs to be increased\nby 25% later in the problem, so pick a small integer that is easy to\nincrease by 25%, such as 4.\nThe problem also contains two fractions, \n and \n . Since one number is\nalready a multiple of 4, make the other number a multiple of 5."} +{"id": "book 2_p608_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 608, "page_end": 608, "topic_guess": "general", "text": "Now, solve the problem using these numbers. Remember that the original\ncylinder is only \n full.\nThe amount of water in the new cylinder has to be the same 60π.\nRemember that it is only \n full.\nThe new height is 4. The original height was 5, so the new height is \n , or\n80%, of the original height.\nThe correct answer is (D).\n94. (B): This problem contains a common trap seen in many difficult DS\nquestions. Answer choice (C) is a tempting short-cut answer as the"} +{"id": "book 2_p609_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 609, "page_end": 609, "topic_guess": "sequences_patterns", "text": "combined statements would provide the values of both a and b, which\ncould be plugged into the expression to answer the question.\nRephrase the question:\nRemembering the units digit patterns for powers of 2 will help on this\nproblem:\nThe units digits for powers of 2 is a repeating pattern of [2, 4, 8, 6].\nIf you can determine the relationship of 3a + 5b to a multiple of 4 (i.e.,\nwhere 23a + 5b is in the predictable four-term repeating pattern of units\ndigits), you will be able to answer the question. This question can be\nrephrased as, “What is the remainder when 3a + 5b is divided by 4?”"} +{"id": "book 2_p610_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 610, "page_end": 610, "topic_guess": "sets_probability_counting", "text": "(1) INSUFFICIENT: If b = 24, then 5b is a multiple of 4. However, a could be\nany integer less than 24. Possible remainders when 3a is divided by 4 are\n0, 1, 2, or 3.\n(2) SUFFICIENT: If the greatest common factor of a and b is 12, then 12\nmust be a factor of both variables. That is, both a and b are multiples of 12\nand thus also multiples of 4. As a result, 3a and 5b will be multiples of 4 as\nwell, so the remainder will be 0 when 3a + 5b is divided by 4.\nThe correct answer is (B).\n95. (D) 360: Valid codes must have a second digit that is exactly twice the first\ndigit. There are three ways to do this with the available digits.\nScenario A: 12XXXXX\nScenario B: 24XXXXX\nScenario C: 36XXXXX\nFor each of these basic scenarios, there are 5! ways to shuffle the\nremaining five numbers (represented by X’s above).\nThus, the total number of valid codes is 3 × 5! = 3 × 120 = 360.\nThe correct answer is (D).\n96. (A): When an n-sided die is rolled once, the probability of rolling any\nparticular number (such as 1) is equal to \n . Therefore, the probability of"} +{"id": "book 2_p611_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 611, "page_end": 611, "topic_guess": "sets_probability_counting", "text": "rolling 1 both times is equal to \n . Rephrase the\nquestion as, “What is the value of n ?”\n(1) SUFFICIENT: The probability that the two rolls are different is 80%, or\n.\nWhen rolling an n-sided die twice, the first roll can have any value from 1\nto n, inclusive. Regardless of the value of the first roll, the probability is\n that the second roll will have that same value. Therefore, the overall\nprobability of rolling the same value twice is \n , and the probability of\nrolling two different values is \n .\nTherefore, according to this statement, \n . The value of n is\n5, and this statement is sufficient.\n(2) INSUFFICIENT: The probability that two rolls of an n-sided die will\nhave the same value is \n , regardless of the value of n. This statement\nwould be true for any value of n, so the value of n cannot be determined\nand this statement is insufficient.\nThe correct answer is (A)."} +{"id": "book 2_p612_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 612, "page_end": 612, "topic_guess": "algebra", "text": "97. (B) \n : It would require a lot of tricky\nwork to solve this algebraically, so test cases instead. Make sure to pick\nvalues for the unknowns such that \n holds true. For\nexample, if w = 1, x = 2, y = 3, and z = 4, then \n is true.\nBefore plugging those values in for the quantities, check the answers. All\nof them begin with 1 and \n , so don’t bother to test \n . In addition, check\nthe answer choices a er each term that you evaluate.\n\nNow, place in ascending order: \n .\nThe third term is \n , so cross off answers (A), (C), and (D). The\nfourth term is \n , so cross off answer (E).\nThe correct answer is (B)."} +{"id": "book 2_p613_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 613, "page_end": 613, "topic_guess": "algebra", "text": "98. (C): The question asks for the absolute value of a + b, so try to manipulate\nthe statements to isolate that combination of variables, a + b. Statement\n(2) looks much easier, so start there.\n(2) INSUFFICIENT: This provides information about c and d, and the\nrelationship between them, but no information about a or b.\n(1) INSUFFICIENT: Manipulate the equation to group (a + b) and (c + d)\nterms.\nNote that this is of the form (x + y)(x − y), where x = (a + b) and y = (c + d).\nThis is the difference of squares special product, (x + y)(x − y) = x2 − y2. Use\nthis to transform this expression.\nThis is not enough to determine the value of (a + b)2.\n(1) AND (2) SUFFICIENT: From statement (2), (a + b)2 = 16 + (c + d)2. From\nstatement (1), c + d = 3. You can substitute for c + d and solve. The\nsolution is shown below, but note that you don’t have to do that math:\nThe question asked for the absolute value of a + b, so the fact that the\nequation has (a + b)2 doesn’t matter."} +{"id": "book 2_p614_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 614, "page_end": 614, "topic_guess": "number_theory", "text": "The correct answer is (C).\n99. (B) \n : If y has no factor z such that 1 < z < y, then y must be prime.\nExamine a few examples to see why this is true.\nBecause y is selected from set B, it is a prime number between 10 and 50,\ninclusive. The only prime number that is divisible by 3 is 3, so y is\ndefinitely not divisible by 3.\nThus, xy is only divisible by 3 if x itself is divisible by 3. Rephrase the\nquestion: “What is the probability that a multiple of 3 will be chosen\nrandomly from set A?”\nThere are 21 − 10 + 1 = 12 terms in set A. Of these, 4 terms (12, 15, 18, and\n21) are divisible by 3.\n6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime.\n15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime.\n3 has NO factor between 1 and 3: 3 IS prime.\n7 has NO factor between 1 and 7: 7 IS prime."} +{"id": "book 2_p615_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 615, "page_end": 615, "topic_guess": "statistics", "text": "Thus, the probability that x is divisible by 3 is \n .\nThe correct answer is (B).\n 00. (E) \n : This problem contains variables in the answer\nchoices, so Choose Smart Numbers.\nThere are 8 children at the party, and 5 will sit at the table with the cake.\nSally must sit at the birthday cake table, so pick 5 − 1 = 4 of the other 8 − 1\n= 7 children to sit at that table with her. How many different ways can you\nchoose 4 from a group of 7? Set up an anagram grid, where Y means “at\nthe cake table” and N means “at the other table.”\nA B C D E F G\nY Y Y Y N N N\nNow, you can calculate the number of possible groups:\nNote that the answer choices are in factorial form; it may be the case that\nyou will find the unsimplified factorial form, rather than 35. Test each\nanswer choice by plugging in x = 8 and y = 5."} +{"id": "book 2_p616_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 616, "page_end": 616, "topic_guess": "algebra", "text": "As an alternative to testing all five choices, you could use a hybrid\napproach to determine the formula using variables.\nThe number of possible groups was \n , but remember that this\nformula took Sally into account.\nThe 3 came from the difference between these numbers: 7 − 4 = (x − 1) − (y\n− 1) = (x − y).\n(A)\n(B)\n(C)\n(D)\n(E)"} +{"id": "book 2_p617_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 617, "page_end": 617, "topic_guess": "algebra", "text": "Substitute these variable expressions in place of the numbers.\nThe correct answer is (E)."} +{"id": "book 2_p618_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 618, "page_end": 618, "topic_guess": "word_problems", "text": "Workout Set 11\n101. A casino pays players with chips that are either turquoise or violet\ncolored. If each turquoise-colored chip is worth t dollars and each\nviolet-colored chip is worth v dollars, where t and v are integers,\nwhat is the combined value of four turquoise-colored chips and two\nviolet-colored chips?\nThe combined value of six turquoise-colored chips and three\nviolet-colored chips is $42.\n(1)\nThe combined value of five turquoise-colored chips and seven\nviolet-colored chips is $53.\n(2)\n102. Jonas and Amanda stand at opposite ends of a straight road and\nstart running toward each other at the same moment. Their rates\nare randomly selected in advance so that Jonas runs at a constant\nrate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant\nrate of 4, 5, 6, or 7 miles per hour. What is the probability that Jonas\nhas traveled farther than Amanda by the time they meet?"} +{"id": "book 2_p619_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 619, "page_end": 619, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n103. If p is a positive integer, is p2 divisible by 96 ?\np is a multiple of 8.(1)\np2 is a multiple of 12.(2)"} +{"id": "book 2_p620_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 620, "page_end": 620, "topic_guess": "geometry", "text": "104. \nIn the figure above, the trapezoid ABCD is inscribed in a circle.\nParallel sides AB and CD are 7 inches apart and 6 and 8 inches long,\nrespectively. What is the radius of the circle in inches?\n4(A)\n5(B)\n7(C)\n(D)\n(E)\n105. If 9y + 3b = 10(3b), then 2y is equal to which of the following?"} +{"id": "book 2_p621_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 621, "page_end": 621, "topic_guess": "geometry", "text": "b − 2(A)\nb − 1(B)\nb(C)\nb + 1(D)\nb + 2(E)\n106. The radius of a circle is r yards. Is the area of the circle at least r\nsquare yards? (1 yard = 3 feet)\nThe diameter of the circle is more than 2 feet.(1)\nIf the radius of the same circle is f feet, the area of the circle is\nmore than 2f square feet.\n(2)\n107. For how many values of x from 1 to 300, inclusive, is the sum of the\nintegers from 1 to x, inclusive, divisible by 3 ?\n  30(A)\n  99(B)\n100(C)\n199(D)\n200(E)\n108. If x and y are integers and |xy| = 8, what is the value of |x + y| ?\nx and y are both divisible by 2.(1)"} +{"id": "book 2_p622_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 622, "page_end": 622, "topic_guess": "number_theory", "text": "xy > 0(2)\n109. Set X consists of exactly four distinct integers that are greater than\n1. For each integer in the set, all of that integer’s unique prime\nfactors are also in the set. For example, if 20 is in set X, then 2 and 5\nmust also be in set X. How many of the distinct integers in set X are\nprime?\nThe product of the integers in set X is divisible by 36.(1)\nThe product of the integers in set X is divisible by 60.(2)\n110. If p = (22)(3x) and r = (22)(3y), where x and y are prime numbers and\nx ≠ y, which of the following represents the least common multiple\nof p and r ?\n12xy(A)\n6xy(B)\nxy(C)\n12(D)\n6(E)"} +{"id": "book 2_p623_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 623, "page_end": 623, "topic_guess": "general", "text": "Workout Set 11 Answer Key\n 01. D\n 02. A\n 03. C\n 04. B\n 05. E\n 06. A\n 07. E\n 08. C\n 09. B\n 10. A"} +{"id": "book 2_p624_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 624, "page_end": 624, "topic_guess": "number_theory", "text": "Workout Set 11 Solutions\n 01. (D): The question asks for the value of 4t + 2v, where t and v represent the values\nof the turquoise and violet chips, respectively. Note that the question asks for a\ncombination of variables, or a combo; it may not be necessary to be able to solve\nfor the individual values of t and v.\n(1) SUFFICIENT: Translate and simplify the statement.\nYou can’t solve for t and v, so this statement might look sufficient—but remember\nwhat the question is asking! You need to find the value of 4t + 2v. Multiply the\nequation by 2: 4t + 2v = 28.\n(2) SUFFICIENT: Translate the statement.\nThere isn’t a way to simplify this one, but notice all those prime numbers. Primes\ntend to minimize the number of allowable scenarios, especially when the\nquestion also specifies that the variables have to be integers. See what\ncombinations of integers would actually work here.\nThe 5t term could be 5, 10, 15, 20, and so on. This term can contribute only\nnumbers that end in 5 or 0. If that’s the case, the 7v term must have a units digit\nof either 3 or 8. List out the possibilities for 7v, and try only the ones that end in 3\nor 8."} +{"id": "book 2_p625_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 625, "page_end": 625, "topic_guess": "word_problems", "text": "7v: 7, 14, 21, 28, 35, 42, 49\nOnly one ends in 3 or 8! If 7v = 28, then v = 4 and the 5t term must equal 25, so t =\n5. This one scenario works and it is the only possible scenario for this equation,\ngiven that v and t must be integers.\nThe correct answer is (D).\n 02. (A) \n : If Jonas and Amanda run at the same rate, they will meet each other\nexactly in the middle. Jonas will only run farther than Amanda if Jonas’ rate is\ngreater than Amanda’s. In math terms, Distance = rate × time, and since Jonas\nand Amanda run for the same time, their relative distances depend solely on\ntheir relative rates. Rephrase the question as, “What is the probability that Jonas\nran faster than Amanda?”\nThere are four possible rates for Jonas (3, 4, 5, and 6) and four for Amanda (4, 5, 6,\nand 7). In total, there are (4)(4) = 16 possible rate scenarios.\nOf Jonas’ four possible rates, only two (5 and 6) are greater than some of\nAmanda’s possible rates (4 and 5). List the three rate scenarios that result in a\nfaster speed (greater distance) for Jonas.\nSince there are 16 possible combinations of rates, the probability that Jonas ran\nfarther than Amanda is \n .\nThe correct answer is (A).\nJonas ran 5 miles per hour (mph), and Amanda ran 4 mph.\nJonas ran 6 mph, and Amanda ran 4 mph.\nJonas ran 6 mph, and Amanda ran 5 mph."} +{"id": "book 2_p626_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 626, "page_end": 626, "topic_guess": "number_theory", "text": "03. (C): The prime factorization of 96 is (2)(2)(2)(2)(2)(3) = (25)(31). In order for p2 to be\ndivisible by 96, p2 would have to have the prime factors 2531 in its prime box. The\nrephrased question is therefore, “Does p2 have at least five 2’s and one 3 in its\nprime box?”\n(1) INSUFFICIENT: If p is a multiple of 8 = (2)(2)(2), p has 23 in its prime box.\nTherefore, p2 has (23)2 = 26 in its prime box, and thus has the required five 2’s.\nHowever, it is uncertain whether p2 has at least one 3 in its prime box.\n(2) INSUFFICIENT: If p2 is a multiple of 12 = (2)(2)(3), p2 has two 2’s and one 3 in its\nprime box. It is uncertain whether p2 has at least five 2’s total as there may or may\nnot be three more 2’s in the prime box.\n(1) AND (2) SUFFICIENT: Statement (1) indicates that p2 has 26 in its prime box\nand statement (2) indicates that p2 has a 3 in its prime box. Therefore, it is certain\nthat p2 has at least five 2’s and one 3 in its prime box.\nThe correct answer is (C).\n 04. (B) 5: Redraw the figure as closely to scale as possible (remember that the official\nscrap paper is graph paper!), labeling the known dimensions and the radius in\nquestion."} +{"id": "book 2_p627_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 627, "page_end": 627, "topic_guess": "geometry", "text": "In order for the trapezoid vertices to lie on the circle, the trapezoid must be\nsymmetrical about the dotted line, which passes through the center of the circle.\nDraw this vertical and the radii to points B and C to create two right triangles,\nallowing you to use the Pythagorean theorem.\nIn fact, you might play an educated hunch that the triangles are 3–4–5 common\nright triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3\nand 4 side. The unknown vertical sides are thus 4 and 3, which sum to 7 as they\nmust.\nAlgebraically, set up the following equations from the picture.\nSetting the two equations for r2 equal."} +{"id": "book 2_p628_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 628, "page_end": 628, "topic_guess": "algebra", "text": "Since (x + y)(x − y) = 7, (x − y) = 1.\nSolve for x and y.\nThe radius of the circle is 5, because r2 = 32 + 42 = 25.\nThe correct answer is (B).\n 05. (E) \n : First, simplify the given equation.\nThe correct answer is (E).\n 06. (A): First, translate the question stem: Is πr2 ≥ r ? Simplify."} +{"id": "book 2_p629_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 629, "page_end": 629, "topic_guess": "sequences_patterns", "text": "Note that the question is stated in yards, but the statements use feet, so convert\nthe question stem to feet.\n(1) SUFFICIENT: D > 2 feet. Therefore, r > 1 foot. The value of \n is approximately\n, or a little bit less than 1. Yes, \n .\n(2) INSUFFICIENT: According to this statement, πf2 > 2f. Simplify.\nIn other words, \n or approximately \n . Therefore, f could be smaller\nthan \n feet (just a bit smaller than 1) but it could also be larger.\nThe correct answer is (A).\n 07. (E) 200: Start at x = 1, then x = 2, and so on, looking for a pattern."} +{"id": "book 2_p630_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 630, "page_end": 630, "topic_guess": "number_theory", "text": "x Sum of the integers from 1 to x, inclusive Is the sum divisible by 3 ?x Sum of the integers from 1 to x, inclusive Is the sum divisible by 3 ?\n1 1 No\n2 3 Yes\n3 6 Yes\n4 10 No\n5 15 Yes\n6 21 Yes\n7 28 No\nIt appears that the pattern is No, Yes, Yes, No, Yes, Yes, etc., repeating every three\nterms. Therefore, 2 out of every 3 values of x yield a sum that is divisible by 3. So\n200 of the 300 values of x will yield a sum that is divisible by 3. The answer is 200.\nThe correct answer is (E).\n 08. (C): If the two variables must be integers and the absolute value of their product\nis 8, then x and y have to represent either the factor pair (1, 8) or the factor pair (2,\n4), in either order and with either sign (positive or negative). This narrows the\npossible number of cases considerably.\n(1) INSUFFICIENT: List out the possible cases allowed by this statement. If x and y\nare both divisible by 2, then you’re dealing with the factor pair (2, 4). A er you try\nyour first case, ask yourself what different case would be most likely to return a\ndifferent answer to the question.\nx y |xy| = 8 |x + y| = ?\n2 4 ✓ 6"} +{"id": "book 2_p631_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 631, "page_end": 631, "topic_guess": "general", "text": "x y |xy| = 8 |x + y| = ?\n−2 4 ✓ 2\nIn this case, just reversing the numbers (4, 2) won’t make a difference to the final\nquestion, so don’t list that case second. Try one of the cases that allows a\nnegative value. Now that you have two different values, you know the statement\nis not sufficient.\n(2) INSUFFICIENT: Follow the same process. This statement indicates that the two\nvalues have the same sign: both positive or both negative.\nx y |xy| = 8 |x + y| = ?\n1 8 ✓ 9\n2 4 ✓ 6\nAgain, list cases until you have two contradictory results (or until you’ve tried all\ncases and realized that there is only one possible answer).\n(1) AND (2) SUFFICIENT:\nx y Both divisible by 2?\nSame signs? |xy| = 8 |x + y| = ?\n2 4 ✓\n✓\n✓ 6"} +{"id": "book 2_p632_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 632, "page_end": 632, "topic_guess": "number_theory", "text": "x y Both divisible by 2?\nSame signs? |xy| = 8 |x + y| = ?\n−2 −4 ✓\n✓\n✓ 6\nOnly four possible cases are allowed: the variables must match the (2, 4) factor\npair and the signs have to be the same. The chart above shows two of the four\npossible cases, but you could also reverse the order: (4, 2) and (−4, −2). Since\nyou’re just adding at the end, though, the order of the two variables doesn’t\nmatter. In all four cases, |x + y| = 6.\nThe correct answer is (C).\n 09. (B): Set X contains exactly four distinct integers; distinct means the four integers\nmust be different. Also, if a number is contained in the set, all of its unique prime\nfactors must also be in the set. If a prime number, such as 3, is in the set, no\nadditional numbers are required in the set: the only prime factor of 3 is 3 itself,\nwhich is already included. However, if the composite number 8 is in the set, its\none unique prime factor, 2, must also be included.\nTo better understand the question, jot down some four-integer sets that fit the\ndescription. Specifically, look for sets that have different numbers of primes.\n{2, 3, 5, 7} is a valid set that contains four primes.\n{2, 4, 8, 16} is a valid set that contains only one prime, since all of the terms are\npowers of 2.\n(1) INSUFFICIENT: The question deals with primes and divisibility; start by\nfactoring 36."} +{"id": "book 2_p633_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 633, "page_end": 633, "topic_guess": "number_theory", "text": "The prime factorization of 36 is 36 = 2 × 2 × 3 × 3. For the product of the four\nintegers in the set to be divisible by 36, the four integers must include 2, 2, 3, and\n3 as prime factors, in some combination. Therefore, based on the definition of\nthe set, 2 and 3 must both be included in the set. The remaining two numbers\nmust include at least one 2 and at least one 3 as prime factors between them.\nHowever, there are several different valid sets that fit this criterion. \nCase 1: {2, 3, 6, 9} has a product divisible by 36, since it includes at least two 2’s\nand at least two 3’s in its prime factors. Also, it is a valid set since it includes all of\nthe unique prime factors of 6 (2 and 3) and all of the unique prime factors of 9\n(only 3). This set has two primes.\nCase 2: The primes 2 and 3 must be included in the set; include a third prime as\nwell, such as 5. The final number in the set must include 2 and 3 as prime factors\nin order for the product of the whole set to be divisible by 36. {2, 3, 5, 6} is one\nexample of a valid set that contains three primes and has a product divisible by\n36. This set has three primes.\nBecause the set might have either two primes or three primes, the statement is\ninsufficient. \n(2) SUFFICIENT: Based on similar reasoning to that from statement (1), note that\n2, 3, and 5 must be included in the set. The product of 2, 3, and 5 is 30, so the\nproduct of the set will definitely be divisible by 30. \nIn order for the product to be divisible by 60, there must be at least one\nadditional multiple of 2 in the set. Because this number is a multiple of 2, and\nbecause 2 itself is already included in the set, it is not a prime. Therefore, the only\nprimes in the set are 2, 3, and 5, and the answer to the question is 3. \nThe correct answer is (B)."} +{"id": "book 2_p634_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 634, "page_end": 634, "topic_guess": "number_theory", "text": "10. (A) 12xy: p and r share three prime factors: 2, 2, and 3. Each also has an additional\nprime factor that is not shared by the other. Draw overlapping circles in which to\nplace the shared and non-shared prime factors of p and r. To find the least\ncommon multiple (LCM), multiply from le to right and include all the common\nfactors in the product.\nThe correct answer is (A)."} +{"id": "book 2_p635_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 635, "page_end": 635, "topic_guess": "algebra", "text": "Workout Set 12\n111. \nWhich of the following equations represents a line parallel to line ℓ in the\nfigure above?\n2y − 3x = 0(A)\n2y + 3x = 0(B)\n2y − 3x = 6(C)\n3y + 2x = 6(D)\n3y − 2x = 9(E)\n112. If y = |x − 1| and y = 3x + 3, then x must be between which of the following\nvalues?"} +{"id": "book 2_p636_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 636, "page_end": 636, "topic_guess": "word_problems", "text": "2 and 3(A)\n1 and 2(B)\n0 and 1(C)\n−1 and 0(D)\n−2 and −1(E)\n113. If a is a positive integer, is a2 a multiple of 8 ?\na3 is a multiple of 16.(1)\n(a + 4)2 is a multiple of 8. (2)\n114. \nA road crew painted two black lines across a road, as shown in the figure\nabove, to mark the start and end of a one-mile stretch. Between the two\nblack lines, they will paint across the road a red line at each third of a\nmile, a white line at each fi h of a mile, and a blue line at each eighth of a\nmile. What is the smallest distance (in miles) between any of the painted\nlines on this stretch of highway?"} +{"id": "book 2_p637_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 637, "page_end": 637, "topic_guess": "number_theory", "text": "0(A)\n(B)\n(C)\n(D)\n(E)\n115. Set S consists of n consecutive integers, where n > 1. What is the value of\nn ?\nThe sum of the integers in set S is divisible by 7.(1)\nThe sum of the integers in set S is 14.(2)\n116. A trapezoid is symmetrical about a vertical center line. If a circle is drawn\nsuch that it is tangent to exactly three points on the trapezoid and is\nenclosed entirely within the trapezoid, what is the diameter of the circle?\nThe parallel sides of the trapezoid are 10 inches apart.(1)\nOf the parallel sides of the trapezoid, the shorter side is 15 inches\nlong.\n(2)"} +{"id": "book 2_p638_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 638, "page_end": 638, "topic_guess": "sets_probability_counting", "text": "117. Three boys are ages 4, 6, and 7, respectively. Three girls are ages 5, 8, and\n9, respectively. If two of the boys and two of the girls are randomly\nselected and the sum of the selected children’s ages is z, what is the\ndifference between the probability that z is even and the probability that\nz is odd?\n(A)\n(B)\n(C)\n(D)\n(E)\n118. If two distinct integers x and y are randomly selected from the integers\nbetween 1 and 10, inclusive, what is the probability that \n  is a\npositive odd integer?"} +{"id": "book 2_p639_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 639, "page_end": 639, "topic_guess": "sequences_patterns", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n119. Positive integers a and b are less than or equal to 9. If a and b are\nassembled into the six-digit number ababab, which of the following must\nbe a factor of ababab ?\n3(A)\n4(B)\n5(C)\n6(D)\nNone of the above(E)\n120. The two-digit positive integer s is the sum of the two-digit positive\nintegers m and n. Is the units digit of s less than the units digit of m ?\nThe units digit of s is less than the units digit of n.(1)"} +{"id": "book 2_p640_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 640, "page_end": 640, "topic_guess": "general", "text": "The tens digit of s is not equal to the sum of the tens digits of m and\nn.\n(2)"} +{"id": "book 2_p641_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 641, "page_end": 641, "topic_guess": "general", "text": "Workout Set 12 Answer Key\n 11. B\n 12. D\n 13. D\n 14. D\n 15. E\n 16. C\n 17. A\n 18. A\n 19. A\n 20. D"} +{"id": "book 2_p642_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 642, "page_end": 642, "topic_guess": "algebra", "text": "Workout Set 12 Solutions\n 11. (B) 2y + 3x = 0: Parallel lines have the same slope. The slope of line ℓ is\n. Work backwards from the\nanswers by rearranging into slope-intercept form: \n form,\nwhere the slope is m. Since only one answer can be correct, only one\nanswer will have a slope of \n ; stop when you find the one with the\nmatching slope.\n(A) 2y = 3x\n(B) 2y = − 3x\nAlternatively, use the slope to write the equation, manipulate, and look for\na match.\nOnly answer (B) is a match for the x and y portions.\nThe correct answer is (B)."} +{"id": "book 2_p643_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 643, "page_end": 643, "topic_guess": "number_theory", "text": "12. (D) −1 and 0: Set the two equations for y equal and algebraically solve |x − 1|\n= 3x + 3 for x. This requires two solutions: one for the case that x − 1 is\npositive, the other for the case that x − 1 is negative.\nx – 1 is positive: x – 1 is negative:\nIn the x − 1 is positive case, the answer is x = −2. In this case, x − 1 is not\nactually positive, so this is a false case. Discard it. As a result, there is only\none solution: \n .\nThe correct answer is (D).\n 13. (D): In order for a2 to be a multiple of 8, the prime factorization of a2 must\ninclude all of the prime factors of 8. That is, to be a multiple of 8, a2 would\nhave to include 2, 2, and 2 among its prime factors.\nWhen an integer is squared, all of its prime factors are duplicated.\nTherefore, if a only includes a single 2 among its prime factors, a2 might\nonly include two 2’s. In order to guarantee that a2 includes at least three 2’s,\na must include at least two 2’s in its prime factors.\nThe question translates to: “Does the prime factorization of a include at\nleast two 2’s?” Or, “Is a divisible by 4?”"} +{"id": "book 2_p644_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 644, "page_end": 644, "topic_guess": "algebra", "text": "(1) SUFFICIENT: The positive integer a, when cubed, is a multiple of 16. Odd\nintegers, when cubed, will always yield an odd result, so a cannot be odd.\nWhat positive even values could a have? \nIf a = 2, then a3 = 8, which is not a multiple of 16. Neither is 63 or 103.\nHowever, if a = 4, then a3 = 64, which is a multiple of 16, as are 83 = (43)(23)\nand 123 = (43)(33). Therefore, this statement indicates that a is a multiple of\n4. When a multiple of 4 is squared, the result is always a multiple of 8, so\nthe answer is definitely Yes.\n(2) SUFFICIENT: Start by multiplying out the expression in the statement.\n(a + 4)2 is a multiple of 8.\na2 + 8a + 16 is a multiple of 8.\na2 + 8(a + 2) is a multiple of 8.\nThe second part of the expression, 8(a + 2), is a multiple of 8 for any value of\ninteger a. In order for the entire expression to be a multiple of 8, the first\npart of the expression must be a multiple of 8 as well. So according to this\nstatement, a2 is a multiple of 8. The answer is a definite Yes, so this\nstatement is sufficient.\nThe correct answer is (D).\n 14. (D) \n : When comparing fractional pieces of a whole, find a common\ndenominator. In this case, the one-mile stretch is divided into thirds, fi hs,\nand eighths. The smallest common denominator of 3, 5, and 8 is 120. If the"} +{"id": "book 2_p645_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 645, "page_end": 645, "topic_guess": "statistics", "text": "one-mile highway is divided into 120 equal increments, where will the red,\nwhite, and blue marks fall?\nRed (thirds): 40, 80 (out of 120 increments)\nWhite (fi hs): 24, 48, 72, 96 (out of 120 increments)\nBlue (eighths): 15, 30, 45, 60, 75, 90, 105 (out of 120 increments)\nThe smallest distance between two marks is 75 − 72 = 3 or 48 − 45 = 3. This\nequates to \n , or \n miles.\nThe correct answer is (D).\n 15. (E): Both statements provide information about the sum of the set, so\nrephrase with this in mind.\nFor n = odd, the median is the middle term, an integer. For n = even, the\nmedian is the average of the two middle terms, a non-integer of the form\n“integer + 0.5.” You can determine n if you can determine both the median\nof set S and the sum of the integers in set S.\nNext, glance at the statements. Statement (2) is actually a subset of\nstatement (1): if the sum is exactly 14, then the sum is divisible by 7. As a\nresult, the answer can’t be (A), because if statement (1) is sufficient, then\nstatement (2) would also have to be sufficient. The answer also can’t be (C),\nbecause statement (2) doesn’t add new information to statement (1). Cross\noff answers (A) and (C) and start with statement (2).\nSum of consecutive set = (Median)(Number of terms)\nSum of consecutive set = (Median)(n)"} +{"id": "book 2_p646_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 646, "page_end": 646, "topic_guess": "statistics", "text": "(2) INSUFFICIENT: Since n must be an integer, use divisibility rules to\nnarrow down possible median values.\nCheck some possible median and n values (remember that n > 1).\nThere are at least two possible values for n.\n(1) INSUFFICIENT: The two possible cases for statement (2) also apply to\nstatement (1), since the numbers chosen for statement (2) must also work\nin statement (1).\n(1) AND (2) INSUFFICIENT: As noted, the final two cases tested for statement\n(2) are also allowed by statement (1), so there are still at least two possible\nvalues for n.\nThe correct answer is (E).\nn = 2 and median = 7: Set S can’t have an integer median if there are only\ntwo terms. Ignore.\nn = 7 and median = 2: Set S is {−1, 0, 1, 2, 3, 4, 5}, which has a sum of 14.\nOkay.\nn = 4 and median = 3.5: Set S is {2, 3, 4, 5}, which has a sum of 14. Okay.\n 16. (C): There are four basic ways this picture could look as there are four sides\nof the trapezoid that could serve as the non-tangent side."} +{"id": "book 2_p647_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 647, "page_end": 647, "topic_guess": "geometry", "text": "(1) INSUFFICIENT: If the circle is tangent to both of the parallel sides (Figure\nA or B), then the diameter must be 10. If the circle is tangent to only one of\nthe parallel sides (Figure C or D), then the diameter is less than 10. Since\nthere are multiple possibilities for the diameter of the circle, statement (1)\ndoes not contain enough information to answer the original question.\n(2) INSUFFICIENT: Just knowing the length of the shorter parallel side is not\nenough to determine which of the basic figures above describes the correct\nsituation. If Figure A or B represents the correct situation, the diameter of\nthe circle is determined solely by the distance between the parallel sides;\nthe diameter is independent of the length of the shorter parallel side. If\nFigure C or D describes the correct situation, then the diameter would\ndepend on not only the 15-inch side but also the longer parallel side, which\nhas an unknown length.\n(1) AND (2) SUFFICIENT: In Figures C and D, the diameter must be less than\n10 but also greater than the length of the smaller parallel side. If the length\nof the smaller parallel side is 15, then the diameter would have to be"} +{"id": "book 2_p648_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 648, "page_end": 648, "topic_guess": "geometry", "text": "greater than 15 for Figures C and D, but this is impossible (since the\ndiameter has to be less than 10). Therefore, Figure A or B represents the\ncorrect situation, and the diameter of the circle must equal 10.\nAlternatively, notice that Figure D could not represent the situation since\nthe circle would have to have a diameter larger than 15 inches in order to\nbe tangent to the short parallel side and the non-parallel sides of the\ntrapezoid. The parallel sides of the trapezoid are only 10 inches apart, so\nthe circle would be too large to be drawn entirely within the trapezoid as\nrequired. Similar logic explains why Figure C is also impossible when\nconsidering (1) and (2) together.\nThe correct answer is (C).\n 17. (A) \n : List the possible cases. Wherever possible, avoid computing; use\nOdd & Even principles to reduce your computation.\nBoys Girls Sum z\n4, 6 = Even 5, 8 = Odd E + O = O\n4, 6 = Even   5, 9 = Even  E + E = E\n4, 6 = Even 8, 9 = Odd  E + O = O\n4, 7 = Odd 5, 8 = Odd  O + O = E\n4, 7 = Odd 5, 9 = Even  O + E = O"} +{"id": "book 2_p649_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 649, "page_end": 649, "topic_guess": "sets_probability_counting", "text": "Boys Girls Sum z\n4, 7 = Odd 8, 9 = Odd  O + O = E\n6, 7 = Odd 5, 8 = Odd  O + O = E\n6, 7 = Odd 5, 9 = Even  O + E = O\n6, 7 = Odd 8, 9 = Odd  O + O = E\nOf the nine scenarios listed, five yield an even z and four yield an odd z.\nThe difference between the probability that z is even and the probability\nthat z is odd is therefore \n .\nThe correct answer is (A).\n 18. (A) \n : This problem involves both probability and special quadratics. In\norder to find the probability that \n is a positive odd integer, you\nmust find the number of values of x and y for which this is the case. The\nproblem is really asking you to count the values of x and y for which\n is a positive odd integer.\nBecause x2 — y2 is a special quadratic, the expression can be simplified to  \n."} +{"id": "book 2_p650_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 650, "page_end": 650, "topic_guess": "algebra", "text": "This expression is an integer when (x — y)(x + y) is divisible by 4. Note that\nx – y and x + y are always either both even or both odd. Whenever x – y and\nx + y are both even, their product will be divisible by 4. In other words,\nwhen x and y are either both even or both odd, \n is an\ninteger. But when x and y are different in parity (one even and the other\nodd), \n is not an integer.\nThe expression also needs to be positive. This will occur whenever x is\ngreater than y. Otherwise, x – y would not be positive and the entire\nexpression would be negative or zero.\nFinally, when is \n an ODD integer? Test a few cases to\ndetermine when this occurs. Based on the reasoning above, you only need\nto test cases in which x is greater than y and in which x and y are either both\neven or both odd.\nx y\n Odd or even?\n5 3\n Even\n7 3\n Even"} +{"id": "book 2_p651_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 651, "page_end": 651, "topic_guess": "general", "text": "x y\n Odd or even?\n9 3\n Even\n3 1\n Even\n7 1\n Even\nIt appears that whenever x and y are both odd, \n will be\neven. Therefore, x and y cannot both be odd. Test some cases in which x\nand y are both even:\nx y\n Odd or even?\n4 2\n Odd\n6 2\n Even"} +{"id": "book 2_p652_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 652, "page_end": 652, "topic_guess": "general", "text": "x y\n Odd or even?\n8 2\n Odd\n10 2\n Even\n6 4\n Odd\n8 4\n Even\nIf x and y are both even, then whether the case is valid depends on whether\nx and y differ by a multiple of 4. Whenever x – y is a multiple of 4 in the table\nabove, so is x + y, and vice versa. Therefore, whenever x – y is a multiple of\n4, \n is an even integer.\nIn summary:\nx and y must both be even.\nx must be greater than y.\nx – y must NOT be a multiple of 4."} +{"id": "book 2_p653_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 653, "page_end": 653, "topic_guess": "number_theory", "text": "That leaves only a few cases to count:\nx y\n10 8\n10 4\n8 6\n8 2\n6 4\n4 2\nThere are six valid cases. In total, there are (10)(9) = 90 ways to choose two\ndistinct values for x and y from the integers between 1 and 10, inclusive. So\nthe probability is \n .\nThe correct answer is (A).\n 19. (A) 3: This problem is most efficiently solved by Working Backwards from\nthe answers.\n3: The sum of the digits of ababab is 3(a + b). This must be a multiple of\n3. On the test, stop here and select answer (A). Below, you’ll find\nexplanations for the other answers.\n(A)\n4: An integer is divisible by 4 if its last two digits represent a two-digit\nnumber that is itself divisible by 4. It is uncertain whether the two-digit\ninteger ab is divisible by 4.\n(B)\n5: An integer is divisible by 5 if the last digit is 0 or 5. It is uncertain\nwhether the positive integer b is 5.\n(C)"} +{"id": "book 2_p654_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 654, "page_end": 654, "topic_guess": "sequences_patterns", "text": "The correct answer is (A).\n6: An integer is divisible by 6 if it is even and divisible by 3. Answer (A)\nestablished that ababab is divisible by 3, but it is uncertain whether the\nlast digit b is even, a requirement for ababab to be even. Note that you\ncan also use logic to eliminate this answer. If ababab were divisible by 6,\nit would also have to be divisible by 3, but that would lead to two correct\nanswers!\n(D)\n 20. (D): Try out a few randomly chosen numbers to help you understand the\nquestion.\nIf m = 10 and n = 10, then s = 20 and the units digits of m and s are\nequal.\nIf m = 11 and n = 12, then s = 23 and the units digit of s is greater\nthan the units digit of m.\nIf m = 19 and n = 11, then s = 30 and the units digit of s is less than\nthe units digit of m.\nWhat’s going on with all of these cases? Basically, if the units digits of m and\nn are small enough, then the units digit of s will be equal to or greater than\nthe units digit of m (and of n).\nOn the other hand, if the units digits of m and n are large enough to cause\nyou to carry over a 1 to the tens digit, then the units digit of s will end up\nbeing smaller than the units digit of m (and of n).\n(1) SUFFICIENT: If the units digit of s is definitely less than the units digits of\none of the smaller numbers, then the carryover situation must apply, in"} +{"id": "book 2_p655_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 655, "page_end": 655, "topic_guess": "sequences_patterns", "text": "which case the units digits of both m and n must be larger than the units\ndigit of s.\n(2) SUFFICIENT: There are only two ways in which the tens digit will not\nequal the tens digits of the two smaller numbers:\nCase 1: The tens digits of the two smaller numbers result in a number that\nneeds to carry over into the hundreds digit. This is impossible for this\nproblem because s is also a two-digit number.\nCase 2: The units digits of the two smaller numbers result in a number that\ncarries over into the tens digit. This must be what is happening in this case.\nIf so, then the units digit of the larger number, s, must be smaller than the\nunits digits of the two smaller numbers, m and n.\nThe correct answer is (D)."} +{"id": "book 2_p656_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 656, "page_end": 656, "topic_guess": "number_theory", "text": "Workout Set 13\n121. If x and y are positive integers such that x2 – y2 = 48, how many\ndifferent values of y are possible?\nTwo(A)\nThree(B)\nFour(C)\nFive(D)\nSix(E)\n122. If the number 200! is written in the form p × 10q, where p and q are\nintegers, what is the maximum possible value of q ?\n40(A)\n48(B)\n49(C)\n55(D)\n64(E)\n123. If x and y are integers, and x ≠ 0, what is the value of xy ?\n|x| = 2(1)"} +{"id": "book 2_p657_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 657, "page_end": 657, "topic_guess": "geometry", "text": "64x62x + y = 482x(2)\n124. The three sides of a triangle have lengths p, q, and r, each an\ninteger. Is this triangle a right triangle?\nThe perimeter of the triangle is an odd integer.(1)\nIf the triangle’s area is doubled, the result is not an integer.(2)\n125. If x is positive, what is the least possible value of \n ?\n(A)\n1(B)\n2(C)\n3(D)\n4(E)\n126. The average (arithmetic mean) cost of three computer models is\n$900. If no two computers cost the same amount, does the most\nexpensive model cost more than $1,000 ?"} +{"id": "book 2_p658_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 658, "page_end": 658, "topic_guess": "statistics", "text": "The most expensive model costs 25% more than the model\nwith the median cost.\n(1)\nThe most expensive model costs $210 more than the model\nwith the median cost.\n(2)\n127. \nWhich of the following equations represents a line perpendicular to\nline k in the figure above?\n3y + 2x = −12(A)\n2y + x = 0(B)\n2y − x = 0(C)\ny + 2x = 12(D)\ny − 2x = 12(E)\n128. Is x > 0 ?\n|2x − 12| < 10(1)"} +{"id": "book 2_p659_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 659, "page_end": 659, "topic_guess": "number_theory", "text": "x2 − 10x ≥ −21(2)\n129. K-numbers are positive integers with only 2’s as their digits. For\nexample, 2, 22, and 222 are K-numbers. The K-weight of a number n\nis the minimum number of K-numbers that must be added together\nto equal n.For example, the K-weight of 50 is 5, because 50 = 22 + 22\n+ 2 + 2 + 2. What is the K-weight of 600 ?\n10(A)\n11(B)\n12(C)\n13(D)\n14(E)\n130. If the reciprocals of two consecutive positive integers are added\ntogether, what is the sum in terms of the greater integer x ?"} +{"id": "book 2_p660_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 660, "page_end": 660, "topic_guess": "general", "text": "(A)\nx2 − x(B)\n2x − 1(C)\n(D)\n(E)"} +{"id": "book 2_p661_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 661, "page_end": 661, "topic_guess": "general", "text": "Workout Set 13 Answer Key\n 21. B\n 22. C\n 23. B\n 24. D\n 25. C\n 26. D\n 27. D\n 28. A\n 29. A\n 30. E"} +{"id": "book 2_p662_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 662, "page_end": 662, "topic_guess": "number_theory", "text": "Workout Set 13 Solutions\n 21. (B) Three: The choices themselves, as well as how many in the question text, suggest\nthat y could have more than one value. To begin, simplify the given information using\nalgebra and logical reasoning.\nSince x and y are integers, both x – y and x + y are integers. The product of two integers\nis 48. One of those integers is equal to x + y and the other is equal to x – y. Since x2 – y2\nis positive and x and y are both positive, x must be greater than y. Therefore, x + y and\nx – y are a pair of positive factors of 48, with x + y being the greater of the two factors.\nList the pairs of factors of 48 and solve for the values of x and y. \nNote one thing first. Since the factors are y above and y below x, the value of x equals\nthe average. Therefore, if for a particular case the average of (x – y) and (x + y) is not an\ninteger, that case is not valid. In order for the average of the two terms to be an integer,\nthe sum has to be even, so if the sum is odd, eliminate that possibility immediately.\nx – y x + y x y\n1 48 n/a (sum not even)\n2 24 13 11\n3 16 n/a (sum not even)\n4 12   8   4\n6   8   7   1"} +{"id": "book 2_p663_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 663, "page_end": 663, "topic_guess": "number_theory", "text": "Of the five factor pairs, three yield integer values for x and y. Double-check that they fit\nthe equation. \nThe correct answer is (B).\n 22. (C) 49: To maximize the value of q, you would need to constrain p to NOT be a multiple\nof 10. In other words, q should count every factor of 10 in 200!, and p would be the\nproduct of the remaining factors of 200!.\nTo count factors of 10 in 200!, you could start by counting the multiples of 10 between\n1 and 200, inclusive. But this method would undercount the number of 10’s that are\nfactors of 200! because it would miss other pairs that can create 10, such as:\nTherefore, this problem is really about counting the number of 2 × 5 factor pairs that\ncan be made from the factors of 200.\nOf 2 and 5, there will be fewer 5’s overall, so count the number of 5’s found among the\nfactors of 200!.\nNumber of Multiples between 1 and 200, Inclusive\nMultiples of 5 (= 51)\nMultiples of 25 (= 52)\n200! has 2 and 5 as factors, which multiply to 10, creating another factor of 10.\n200! has 6 and 15 as factors, which multiply to 90, creating another factor of 10."} +{"id": "book 2_p664_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 664, "page_end": 664, "topic_guess": "number_theory", "text": "Number of Multiples between 1 and 200, Inclusive\nMultiples of 125 (= 53) 125 is the only one: 1\nThe higher multiples contribute more than one factor of 5 to the total.\nThere are eight multiples of 25 in the range (namely, 25, 50, 75, 100, 125, 150, 175, and\n200). Each of these is also a multiple of 5, so each has already been counted once, and\nthus each of the eight multiples of 25 contributes one additional factor of 5. Finally,\n125 contributes a total of three 5’s to the count—but two have already been counted,\nleaving only one additional factor of 5 to include.\nTherefore, the prime factorization for 200! includes 40 + 8 + 1 = 49 factors of 5.\nThe correct answer is (C).\n 23. (B): Since x ≠ 0, x y does not equal 0 y or 0.\n(1) INSUFFICIENT: This statement indicates that x is equal to 2 or −2. The statement\nindicates nothing about y.\n(2) SUFFICIENT: Simplify using exponent rules, noting the common factors of 6 and 8\non each side of the equation.\nSince y = 0 and x ≠ 0 (as stated in the question stem), this information is sufficient to\nconclude that x y = x0 = 1."} +{"id": "book 2_p665_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 665, "page_end": 665, "topic_guess": "geometry", "text": "The correct answer is (B).\n 24. (D): The question stem specifies that the triangle’s sides all have integer lengths.\n(1) SUFFICIENT: Perimeter is calculated by summing the three side lengths. In order for\nthat sum to be odd, either all three numbers have to be odd or one of the three\nnumbers has to be odd.\nIf the triangle is a right triangle, then the Pythagorean theorem (a2 + b2 = c2) holds. If\nall three side lengths are odd, this would mean odd2 + odd2 = odd2. This is impossible,\nhowever; odd + odd = even.\nIf one side length is odd and the other two are even, then there are two possibilities:\nEvery right triangle possibility is impossible, so the triangle cannot be a right triangle.\n(2) SUFFICIENT: A = \n bh, or 2A = bh. If, as this statement indicates, 2A is not an\ninteger, then A itself is not an integer. As a result, at least one of b and h is also not an\ninteger.\nFor right triangles, b and h are the two shorter sides of the triangle. The question stem\nspecifies that all three sides are integers, so if the triangle were a right triangle, then b\nand h would both have to be integers. This is impossible, so the triangle cannot be a\nright triangle.\nThe correct answer is (D).\nO + O + O = odd\nO + E + E = odd\nodd2 + even2 = even2 (impossible: odd + even = odd)\neven2 + even2 = odd2 (also impossible: even + even = even)"} +{"id": "book 2_p666_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 666, "page_end": 666, "topic_guess": "algebra", "text": "25. (C) 2: The problem specifies that x has to be positive, so try some small positive\nintegers to see how the expression \n works.\nIf x = 1, then the expression becomes \n  . If x = 2, then the expression\nbecomes 1 + 1 = 2. Since x can equal 2, eliminate answers (D) and (E).\nWhat if x is a fraction, such as \n  ? In that case, the expression becomes\n … interesting. It’s even bigger than the integer options\nabove. How come?\nThe expressions \n and \n are reciprocals of each other. As long as x is positive, there\nare two broad scenarios possible:\nAnswers (A) and (B), then, are impossible.\nThe correct answer is (C).\nIf one expression is less than 1, then the other has to be greater than 1.1.\nIf one expression equals 1, then the other also equals 1.2.\n 26. (D): If the average of the cost of the three models is $900, then the sum is $2,700. Call\nthe three models a, b, and c, in order from least expensive to most.\n(1) SUFFICIENT: The statement indicates that c = 1.25b, so a + b + 1.25b = 2,700. This\nisn’t enough to solve for specific values, but it might be enough to tell whether c >\n$1,000. Keep working."} +{"id": "book 2_p667_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 667, "page_end": 667, "topic_guess": "word_problems", "text": "By definition, a < b, so plug LTb (less than b) into the equation for a.\nFrom here, estimate: 13 goes into 2,600 a total of 200 times and it goes into 100\napproximately 7 times. Therefore, LTb = (207)(4) = 828.\nIf b is approximately 830, then 125% of b is definitely larger than 1,000. Therefore, the\nmost expensive model does cost more than $1,000.\n(2) SUFFICIENT: Follow a similar path: c = b + 210. Therefore:\nIf b is a bit more than 830, then b + 210 is definitely over 1,000. Therefore, the most\nexpensive model does cost more than $1,000.\nThe correct answer is (D).\n 27. (D) y + 2x = 12: The product of the slopes of two perpendicular lines is −1. The slope of\nline k is:"} +{"id": "book 2_p668_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 668, "page_end": 668, "topic_guess": "algebra", "text": "Thus, the slope of a line perpendicular to k is −2.\nUse the desired slope to create a slope-intercept equation and look for a match among\nthe answers.\nOnly answer (D) offers a match for the le side of the equation.\nAlternatively, put each choice in slope-intercept form: y = mx + b, where the slope is m.\nStop when you find the correct answer.\n(A) 3y + 2x = −12\n Slope = −\n Not a match. Eliminate.\n(B) 2y + x = 0\n Slope = −\n Not a match. Eliminate.\n(C) 2y − x = 0\n Slope = \n Not a match. Eliminate.\n(D) y + 2x = 12 y = −2x + 12 Slope = −2 Correct.\nThe correct answer is (D).\n 28. (A):  The question stem asks a Yes/No question: Is x positive?\n(1) SUFFICIENT: Manipulate the absolute value expression to represent this inequality\non a number line."} +{"id": "book 2_p669_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 669, "page_end": 669, "topic_guess": "algebra", "text": "This can be interpreted as, “The distance between x and 6 is less than 5.” On a number\nline, this is the region between 1 and 11.\nAll possible values for x are positive, so the answer to the question is a definite Yes.\n(2) INSUFFICIENT: Manipulate the inequality to get 0 on one side, then factor the\nresulting quadratic.\nThe factored quadratic on the le side will equal 0 when x = 3 and 7. These are the\nboundary points. On a number line, check the regions on either side of these\nboundary points to determine the valid region(s) for x.\nNote that when 3 < x < 7, (x − 7)(x − 3) = (neg)(pos) = neg.\nSince both positive and negative values are possible for x, the answer is Maybe.\nThe correct answer is (A).\n 29. (A):"} +{"id": "book 2_p670_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 670, "page_end": 670, "topic_guess": "number_theory", "text": "Since you are looking for the minimum number of K-numbers that sum to 600, a\npractical place to start is with the largest K-number less than 600, or 222. There are\nbetween two and three multiples of 222 in 600, so subtract out the two whole\nmultiples.\nNow, the next largest K-number is 22. Again, subtract as many whole multiples as\npossible.\nThe next largest K-number is 2.\nThus, 600 = 222(2) + 22(7) + 2(1), and the K-weight of 600 is 2 + 7 + 1 = 10.\nThe correct answer is (A).\n 30. (E): If the greater of the two integers is x, then the two integers can be expressed as x −\n1 and x. The sum of the reciprocals would therefore be:\nAlternatively, choose smart numbers. For example, let the larger number x = 3. The\nsmaller number would therefore be 3 − 1 = 2. The sum of the reciprocals would be:"} +{"id": "book 2_p671_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 671, "page_end": 671, "topic_guess": "general", "text": "Plug x = 3 into the answers, and find the one that equals \n .\nThe correct answer is (E).\n Eliminate.(A)\nx2 − x = An integer. Eliminate.(B)\n2x − 1 = An integer. Eliminate.(C)\n Eliminate.(D)\n Correct.(E)"} +{"id": "book 2_p672_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 672, "page_end": 672, "topic_guess": "fractions_decimals_percents", "text": "Workout Set 14\n131. The sequence A is defined as follows: A1 = 1, and An = An − 1 + (−1)n +\n1(n2) for all integer values n > 1. What is the value of A15 − A13 ?\n  14(A)\n  29(B)\n169(C)\n196(D)\n421(E)\n132. If d represents the hundredths digit and e represents the\nthousandths digit in the decimal 0.4de, what is the value of this\ndecimal rounded to the nearest tenth?\nd − e is a positive perfect square.(1)\n(2)\n133. If x is a positive integer greater than 1, and y is the smallest positive\ninteger that is evenly divisible by every integer between 1 and x,\ninclusive, what is the value of x ?\ny = 10x(1)"} +{"id": "book 2_p673_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 673, "page_end": 673, "topic_guess": "fractions_decimals_percents", "text": "y = 60(2)\n134. A chain is comprised of 10 identical links, each of which\nindependently has a 1% chance of breaking under a certain load. If\nthe failure of any individual link means the failure of the entire\nchain, what is the probability that the chain will fail under the load?\n(0.01)10(A)\n10(0.01)10(B)\n1 − (0.10)(0.99)10(C)\n1 − (0.99)10(D)\n1 − (0.99)(10 × 9)(E)\n135. City A to City B City B to City C City C to City A\nFirst Bus Departs 5:00 5:00 5:00\nLength of Trip 25 minutes 35 minutes 60 minutes\nSubsequent\nDepartures\nEvery 20\nminutes\nEvery 30\nminutes\nEvery 20\nminutes\nThe table above shows the schedule for the bus routes between\nthree pairs of cities. If Ash plans to travel by bus from City A, to City\nB, to City C, and back to City A, what is the shortest possible\nduration, in minutes, of his entire trip?"} +{"id": "book 2_p674_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 674, "page_end": 674, "topic_guess": "number_theory", "text": "120(A)\n130(B)\n140(C)\n150(D)\n160(E)\n136. Are the positive integers x and y consecutive?\nx2 − y2 = 2y + 1(1)\nx2 − xy − x = 0(2)\n137. What is the value of y − x2 − x  ?\ny = −3x(1)\ny = −4(x + 1)(2)\n138. If f(x) = (x + 6)2 and g(x) = 9x, which of the following specifies all the\npossible values of x for which f(g(x)) < g(f(x)) ?\n–2 < x < 2(A)\nx > 2(B)\n0 < x < 3(C)\n–4 < x < 2(D)\n6 < x < 9(E)"} +{"id": "book 2_p675_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 675, "page_end": 675, "topic_guess": "general", "text": "139. \nIn the figure above, ABCD is a rectangle, and each of AP and CQ is\nperpendicular to BD. If DP = PQ = QB = 1, what is the length of AB  ?\n(A)\n (B)\n (C)\n(D)\n(E)"} +{"id": "book 2_p676_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 676, "page_end": 676, "topic_guess": "coordinate_geometry", "text": "140. \nIn the rectangular coordinate system above, point A is not shown. If\nthe area of triangle OAB is at least 16, which of the following could\nNOT be the coordinates of point A  ?\n(1, 8)(A)\n(4, 14)(B)\n(5, −4)(C)\n(8, −1)(D)\n(14, 4)(E)"} +{"id": "book 2_p677_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 677, "page_end": 677, "topic_guess": "general", "text": "Workout Set 14 Answer Key\n 31. B\n 32. E\n 33. A\n 34. D\n 35. C\n 36. D\n 37. C\n 38. A\n 39. C\n 40. A"} +{"id": "book 2_p678_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 678, "page_end": 678, "topic_guess": "sequences_patterns", "text": "Workout Set 14 Solutions\n 31. (B) 29: Generate the first several values of the sequence, using the given\nrelationship. Notice that a component such as (−1)n + 1 switches the sign of the\nadditive term back and forth. Use a table to keep your work organized.\nn An = An − 1 + (−1)n + 1(n2)\n1 1\n2 1 + (−1)2 + 1(22) = 1 − 4 = −3\n3 −3 + (−1)3 + 1(32) = −3 + 9 = 6\n4 6 + (−1)4 + 1(42) = 6 − 16 = −10\n5 −10 + (−1)5 + 1(52) = −10 + 25 = 15\n6 15 + (−1)6 + 1(62) = −15 − 36 = −21\nWhat is the pattern? The sign alternates between positive and negative. Ignoring\nthe sign of the terms (taking absolute values) might help to determine the full\npattern.\nn An = An − 1 + (−1)n + 1(n2) |An| Change from Previous Term\n1       1    1"} +{"id": "book 2_p679_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 679, "page_end": 679, "topic_guess": "fractions_decimals_percents", "text": "n An = An − 1 + (−1)n + 1(n2) |An| Change from Previous Term\n2    −3   3 +2\n3     6  6 +3\n4 −10 10 +4\n5   15 15 +5\n6 −21 21 +6\nThat is, the absolute value of each term equals the absolute value of the previous\nterm plus n. Following this pattern, |A15| is 15 greater than |A14|, and |A14| is 14\ngreater than |A13|. Thus, |A15| − |A13| = +15 + 14 = 29.\nSince the odd-numbered terms in the original sequence are positive, the absolute\nvalue of any odd-numbered term equals the term itself, so |A15| − |A13| = A15 − A13 =\n29.\nThe correct answer is (B).\n 32. (E): When rounding to the nearest tenth, use the hundredths digit, or d in this\ncase. If d ≥ 5, the decimal 0.4de is rounded up to 0.5. If d ≤ 4, the decimal 0.4de is\nrounded down to 0.4. Thus, a rephrase of this question is, “Is d < 5 (or,\nequivalently, is d ≥ 5)?”\n(1) INSUFFICIENT: Since d − e is positive, d > e. Since d and e are digits (i.e., 0, 1, 2 .\n. . 7, 8, 9), there is a maximum value for the difference: d − e ≤ 9. There are only\nthree perfect squares less than or equal to 9: d − e = 1, 4, or 9."} +{"id": "book 2_p680_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 680, "page_end": 680, "topic_guess": "statistics", "text": "d e d − e = Perfect Square Is d < 5?d e d − e = Perfect Square Is d < 5?\n9 0 9 − 0 = 9 ✓ No\n1 0 1 − 0 = 1 ✓ Yes\nIt is possible for d to round up or down, so this statement isn’t sufficient.\n(2) INSUFFICIENT: Since d and e are positive, you can square each side of the\ninequality without worrying about flipping the sign: d > e4. Since d and e are digits\n(i.e., 0, 1, 2 . . . 7, 8, 9), 9 ≥ d > e4, which means that e can only be 0 or 1. (Note that\n24 = 16, which is too large.) Test cases again; whenever possible, reuse numbers\nthat you tried in the last statement, assuming those numbers are allowed by the\nnew statement.\nd e\n Is d < 5?\n9 0 3 > 0 ✓ No\n1 0\n ✓ Yes\nIt is again possible for d to round up or down, so this statement isn’t sufficient.\n(1) AND (2) INSUFFICIENT: Taking both statements together, e must be 0 or 1 and d\n− e must equal 9, 4, or 1.\nYou’ve already tested the d = 9 and d = 1 scenarios for both of the statements\nindividually, so you don’t need to retest them here. Even together, the two\nstatements allow d to round up or down."} +{"id": "book 2_p681_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 681, "page_end": 681, "topic_guess": "number_theory", "text": "The correct answer is (E).\n 33. (A): List a few values of x and y to gain understanding; doing so also prepares you\nto most easily interpret the statements.\n  x        y\n2       2\n3      6\n4    12\n5    60\n6   60\n7 420\n(1) SUFFICIENT: Look in the table above for cases in which y = 10x. There is only\none: x = 6 and y = 60.\nTo be completely confident that this is the only valid case, approach the problem\ntheoretically. In order for y to equal 10x, 10x would have to be divisible by 1, 2, 3 . .\n.   (x – 1), and x. However, x itself cannot have any common factors with x – 1, so 10\nmust be divisible by x – 1. So x – 1 equals 1, 2, 5, or 10, so x must be 2, 3, 6, or 11.\nTherefore, x = 2 and x = 3 have already been eliminated as possibilities and x = 6\nhas already been confirmed to work. The only remaining case that needs to be\nchecked is x = 11. Since 110 is not divisible by every number smaller than 11 (e.g.,\n110 is not divisible by 3 or 4), x = 11 is not a valid case. The only possible answer is\nx = 6, so this statement is sufficient.\n(2) INSUFFICIENT: Based on the table above, y = 60 when x = 5 or 6. There are two\npossible answers, so the statement is insufficient.\nThe correct answer is (A)."} +{"id": "book 2_p682_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 682, "page_end": 682, "topic_guess": "word_problems", "text": "34. (D) 1 − (0.99)10: Qualitatively, many failure scenarios could occur.\nGiven the complexity of the failure scenarios, it is easier to look at the opposite\nscenario.\nFor each of the links, the probability that it will not fail is 1 − 0.01 = 0.99. The\nprobability that all 10 will not fail is thus (0.99)10, since the probability that all 10\nwill not fail equals the product of the probabilities of the individual links not\nfailing.\nTherefore, the probability that at least one link will fail equals 1 − (0.99)10.\nThe correct answer is (D).\nNone of the links will fail.\nExactly one of the links will fail.\nExactly two of the links will fail.\nAnd so on.\nProbability that at least one link will fail = 1 − probability that No links will fail\n 35. (C) 140: Ash’s round trip consists of two parts: the time spent on the bus and the\ntime spent waiting for the next bus. No matter which buses he takes, the time\nspent on the bus will always be equal: 25 + 35 + 60 minutes = 120 minutes. This\nquestion is really about minimizing the time that Ash spends waiting. Note the\nmismatch between the departure frequencies and the trip durations; it is unlikely\nthat the waiting time is 0, so eliminate (A).\nBecause buses leave each city every 20 or 30 minutes, and both of these time\nperiods divide evenly into one hour, the pattern of departures will always repeat\nevery hour. Therefore, it is only necessary to check three different scenarios: Ash\nwill leave City A either on the hour, 20 minutes a er the hour, or 40 minutes a er\nthe hour."} +{"id": "book 2_p683_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 683, "page_end": 683, "topic_guess": "word_problems", "text": "Then, to minimize waiting, Ash should take the first available buses in City B and\nCity C. For each of the three possible times that Ash could leave City A, note the\nnext available buses for each leg of his trip.\nLeave A Arrive in B Leave B Arrive in C Leave C    Arrive in A     Total Time\n5:00 5:25 5:30 6:05 6:20 7:20 2 hr 20 min\n5:20 5:45 6:00 6:30 6:40 7:40 2 hr 20 min\n5:40 6:05 6:30 7:05 7:20 8:20 2 hr 40 min\nTherefore, the shortest possible duration of the entire trip is 2(60) + 20 = 120 + 20 =\n140 minutes.\nThe correct answer is (C).\n 36. (D): Consecutive integers differ by exactly 1. For x and y to be consecutive, either x\n= y + 1 or x = y − 1. Rephrase the question as, “Does x equal either (y + 1) or (y −\n1)?”\n(1) SUFFICIENT: There’s only one term with x but a couple with y, so it is easiest to\nsolve for x.\nSince x and y are both positive, you can drop the absolute value signs and\nconclude that x = y + 1."} +{"id": "book 2_p684_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 684, "page_end": 684, "topic_guess": "algebra", "text": "(2) SUFFICIENT: Factor out a common term.\nTherefore, x = 0 or x − y − 1 = 0. The question stem indicates that x is positive, so x\ncannot equal 0. As a result, it must be the case that x − y − 1 = 0, or x = y + 1. The\ntwo integers are consecutive.\nThe correct answer is (D).\n 37. (C):\n(1) INSUFFICIENT: Solve for the expression in question by adding (−x2 − x) to both\nsides.\nThe answer depends on the value of x, which is not given.\n(2) INSUFFICIENT: Solve for the expression in question by adding (−x2 − x) to both\nsides.\nThe answer depends on the value of x, which is not given."} +{"id": "book 2_p685_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 685, "page_end": 685, "topic_guess": "coordinate_geometry", "text": "(1) and (2) SUFFICIENT: Each statement provides a different expression equal to y\n− x2 − x. Set these two equal to each other and solve.\nOnce the squared terms drop out, the equation is solvable.\nStop when you know you can solve!\nThe correct answer is (C).\n 38. (A) –2 < x < 2: This problem asks you to compare two “nested” functions: f(g(x))\nand g(f(x)). To simplify a nested function, replace the variable term in the outer\nfunction with the entire inner function.\nTo find the values of x for which f(g(x)) < g(f(x)), simplify the following inequality.\nOne possibility is to multiply out both quadratics and then simplify as much as\npossible."} +{"id": "book 2_p686_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 686, "page_end": 686, "topic_guess": "coordinate_geometry", "text": "In turn, this simplifies to –2 < x < 2. \nAnother approach involves using the multiples of 9 in the initial inequality. Factor\nout a 9 on the le side of the inequality. \nThen, simplify:\nThe correct answer is (A).\n 39. (C) \n : Start by labeling the three line segments with length 1. Label the\nunknown length as x."} +{"id": "book 2_p687_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 687, "page_end": 687, "topic_guess": "geometry", "text": "From here, there are two ways to approach the problem. One approach is to use\nthe Pythagorean theorem to label the lengths of the missing sides. Triangle ABP is\na right triangle, so the following is true:\nSimilarly, APD is a right triangle, and the following is true: \nFinally, since BAD is a right triangle, the Pythagorean theorem can be used to\nsolve for the value of x."} +{"id": "book 2_p688_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 688, "page_end": 688, "topic_guess": "geometry", "text": "The other solution uses the fact that triangles APD and BPA are similar. To prove\nthis, note that APD is a right angle. Therefore, angles PAD and ADP sum to 90\ndegrees. According to the diagram, angles PAD and BAP also sum to 90 degrees.\nTherefore, ADP and BAP have the same degree measure. Since triangles APD and\nBPA have two pairs of matching angles, the third pair of angles must match as\nwell, and the two triangles are similar. The sides line up as follows:\nTriangle PAD Triangle PBA\nShortest Side side PD side AP\nMiddle Side side AP side BP\nLongest Side side AD side AB\nSince PD = 1, BP = 2, and AP = AP, you can set up a proportion to find the length of\nside AP."} +{"id": "book 2_p689_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 689, "page_end": 689, "topic_guess": "geometry", "text": "Finally, use the Pythagorean theorem to determine the length of AB.\nAlternatively, since the diagram is drawn to scale, you can try to estimate. Side\nAB appears to be longer than BP, but shorter than BD. Its length should be\nbetween 2 and 3. Estimate the values of the answer choices.\nOnly answers (C) and (D) are close to the correct number.\nThe correct answer is (C).\n = Less than 2. Eliminate.(A)\n = Less than 2. Eliminate.(B)\n  = between 2 and 3.(C)\n = between 2 and 3.(D)\n = More than 3. Eliminate.(E)\n 40. (A) (1, 8): First, identify any answer choices that will be easy to eliminate before\nyou begin calculating the areas of triangles. Answers (B) and (E) are symmetrical;"} +{"id": "book 2_p690_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 690, "page_end": 690, "topic_guess": "geometry", "text": "the triangles they yield will have the same area. Therefore, neither can be the\ncorrect answer. Eliminate answers (B) and (E). \nThe question is asking for triangles with area less than 16 square units, so first try\ndrawing a triangle with an area of exactly 16 square units. If the base of the\ntriangle is OB, it has a length \n . Use the area formula to find the height.\nThe height is also \n . Draw the height of the triangle perpendicular to OB\nthrough the origin in either quadrant II or quadrant IV. The vertex of the height is\nat either (–4, 4) or (4, –4).\nAny triangle with the same height will have the same area. Therefore, if A lies on\nthe line y = x + 8 or on the line y = x – 8, the triangle will have an area of 16. If A lies\nanywhere between those two lines, then the triangle will have an area less than\n16. Of the answers, only (1, 8) lies between the two lines.\nAlternatively, plot the triangles formed by the remaining answer choices on your\npaper."} +{"id": "book 2_p691_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 691, "page_end": 691, "topic_guess": "geometry", "text": "Rather than calculating the exact area of each triangle, notice that each one is\nfairly close to a triangle whose area is easier to calculate. \nIn choice (A), the triangle has a slightly smaller area than the triangle with vertices\nat (0, 8), (4, 4), and (0, 0). The latter triangle has an area of exactly 16 square units.\nTherefore, triangle (A) has an area smaller than 16 square units, so answer (A) is\ncorrect.\nIn choice (C), the triangle has a slightly greater area than the triangle with vertices\nat (4, 4), (4, –4), and (0, 0). The latter triangle has an area of exactly 16 square\nunits. Therefore, the area of triangle (C) is slightly greater than 16. \nIn choice (D), the triangle has a slightly greater area than the triangle with vertices\nat (4, 4), (8, 0), and (0, 0). The latter triangle has an area of exactly 16 square units.\nTherefore, the area of triangle (D) is slightly greater than 16. \nThe correct answer is (A)."} +{"id": "book 2_p692_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 692, "page_end": 692, "topic_guess": "number_theory", "text": "Workout Set 15\n141. If \n , where a and b are integers and a\ndoes not equal 1 or −2, which of the following could be the value of\nb  ?\n1I.\n2II.\n3III.\nI only(A)\nII only(B)\nI and II only(C)\nI and III only(D)\nI, II, and III(E)\n142. The currencies of three countries are called credits, units, and bells.\nIf 12 credits can be exchanged for a total of exactly 8 units and 5\nbells, what is the smallest integer number of units that can be\nexchanged for an integer number of bells, with no change le over?\n15 credits can be exchanged for 4 units and 25 bells, with no\nchange le over.\n(1)"} +{"id": "book 2_p693_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 693, "page_end": 693, "topic_guess": "number_theory", "text": "25 bells can be exchanged for 8 units, with no change le over.(2)\n143. What is the area of the quadrilateral bounded by the lines\n  48(A)\n  64(B)\n  96(C)\n100(D)\n140(E)\n144. Each digit 1 through 5 is used exactly once to create a five-digit\ninteger. If the 3 and the 4 cannot be adjacent digits in the integer,\nhow many five-digit integers are possible?\n48(A)\n66(B)\n72(C)\n78(D)\n90(E)\n145. For positive integers a, b, and c, b is 120% greater than a, and c is\n25% less than the sum of a and b. Which of the following could be"} +{"id": "book 2_p694_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 694, "page_end": 694, "topic_guess": "sequences_patterns", "text": "the value of |a – c| ?\n  6(A)\n  7(B)\n  8(C)\n  9(D)\n10(E)\n146. A soccer competition consisted of two rounds. In each round, the 20\nteams competing were divided randomly into 10 pairs, and each\npair played a single game. None of the games resulted in a tie, and\nof the teams that lost in the first round, 7 also lost in the second\nround. How many teams won both of their games?\n3(A)\n4(B)\n5(C)\n6(D)\n7(E)\n147. What is the units digit of the positive integer x  ?\n, where y is a positive integer.(1)\n, where z is a positive integer.(2)"} +{"id": "book 2_p695_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 695, "page_end": 695, "topic_guess": "geometry", "text": "148. If x and y are positive integers, what is the value of x + y ?\n(x + y − 1)! < 100(1)\ny = x2 − x + 1(2)\n149. \nIn the figure above, two lines are tangent to a circle at points A and\nB. What is x ?\nThe area of the circle is 81π.(1)\nThe length of arc ADB is 7π.(2)\n150. For all n such that n is a positive integer, the terms of a certain\nsequence B are given by the following rules:\nBn = Bn − 1 + 5 if n is odd and greater than 1\nBn = −Bn − 1 if n is even\nB1 = 3"} +{"id": "book 2_p696_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 696, "page_end": 696, "topic_guess": "sequences_patterns", "text": "What is the sum of the first 65 terms in the sequence?\n−5(A)\n  0(B)\n  3(C)\n  5(D)\n  8(E)"} +{"id": "book 2_p697_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 697, "page_end": 697, "topic_guess": "general", "text": "Workout Set 15 Answer Key\n 41. C\n 42. D\n 43. C\n 44. C\n 45. B\n 46. E\n 47. C\n 48. E\n 49. C\n 50. C"} +{"id": "book 2_p698_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 698, "page_end": 698, "topic_guess": "algebra", "text": "Workout Set 15 Solutions\n 41. (C) I and II only: Factor the numerator:\nSince the fraction equals 0, either ab + 6 = 0 or ab − 3 = 0. Thus, ab = −6 or 3.\nSince a and b are integers, and ab = −6 or ab = 3, it must be the case that b\nis a factor of either −6 or 3. Note that a ≠ 1 and a ≠ −2.\nThe correct answer is (C).\nPOSSIBLE: If b = 1, then a = −6 or a = 3, which are both allowed.I.\nPOSSIBLE: If b = 2, then a = −3 or a = \n . One of those (a = −3) is allowed.II.\nIMPOSSIBLE: If b = 3, then a = −2 or a = 1, neither of which is allowed.III.\n 42. (D): Use three variables to represent the unknown values in this problem.\nLet c represent the value of one credit, u represent the value of one unit,\nand b represent the value of one bell. The given information translates to\nthe following equation:\n12c = 8u + 5b\nThe question asks about exchanging an integer number of units for an\ninteger number of bells. Therefore, your goal is to find the relationship\nbetween the value of a unit and the value of a bell."} +{"id": "book 2_p699_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 699, "page_end": 699, "topic_guess": "algebra", "text": "(1) SUFFICIENT: This statement can be rewritten as the following equation,\nusing the same variables as above.\n  15c = 4u + 25b\nNext, subtract one equation from the other and simplify.\nTherefore, 8 units are worth the same amount as 25 bells. Because the\nratio 8 : 25 cannot be reduced further, 8 is the smallest number of units\nthat can be exchanged for an integer number of bells. For instance, 4 units\ncould only be exchanged for 12.5 bells, which is not an integer. The answer\nto the question is 8. \nCombine this with the equation from the question stem. Because the\ngoal is to find the relationship between b and u, focus on eliminating\nthe other variable, c. First, multiply the equations by 5 and 4,\nrespectively, so that the coefficient of c is 60 in both."} +{"id": "book 2_p700_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 700, "page_end": 700, "topic_guess": "number_theory", "text": "(2) SUFFICIENT: If 8 units can be exchanged for 25 bells, then because 8\nand 25 have no common factors, no smaller integer number of units can\nbe exchanged for bells, as described above. The answer to the question is\n8, and this statement is sufficient.\nThe correct answer is (D).\n 43. (C) 96: All of these line equations are of the form y = mx + b, where m is the\nslope and b is the y-intercept. Two of these lines have a slope of \n and are\nthus parallel to each other. The other two lines are parallel to one another\nwith a slope of \n . Two of the lines have a y-intercept of 6 while the\nother two lines have a y-intercept of −6.\nSketch the lines."} +{"id": "book 2_p701_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 701, "page_end": 701, "topic_guess": "sets_probability_counting", "text": "In each quadrant, there is a triangle with the dimensions 6–8–10, a\nmultiple of the common 3–4–5 right triangle.\nAlternatively, recognize that the quadrilateral is a rhombus (four equal\nsides of length 10) and use the formula for the area of a rhombus:\n, where D indicates the length of the diagonals.\nThe correct answer is (C).\n 44. (C) 72: In a constrained combinatorics question such as this one, it is o en\neasier to consider the “violating” cases instead of the “okay” cases.\n# of permutations that obey the constraint = # of permutations total\n− # of permutations that violate the constraint\nAt first glance, ignore the constraint that 3 and 4 cannot be adjacent to\ndetermine the total number of five-digit integers possible:\n# of permutations total = 5! = (5)(4)(3)(2)(1) = 120"} +{"id": "book 2_p702_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 702, "page_end": 702, "topic_guess": "sets_probability_counting", "text": "Now, consider the basic ways that 3 and 4 might be adjacent to each other\nin a five-digit number:\n3 4 x x x\nx 3 4 x x\nx x 3 4 x\nx x x 3 4\nFor each of these four base cases, there are two ways to order the 3 and 4\n(3, 4 and 4, 3), as well as 3! ways the other digits (1, 2, and 5) can be\narranged in the x positions. Thus:\n# of permutations that violate the constraint = 4 × 2 × 3! = (4)(2)(3)\n(2)(1) = 48\nTherefore, the number of permutations that do not violate the constraint\nequals 120 − 48 = 72.\nThe correct answer is (C).\n 45. (B) 7: Start by translating the text into math. Note that b is 120% greater\nthan a, not 120% of a."} +{"id": "book 2_p703_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 703, "page_end": 703, "topic_guess": "number_theory", "text": "There are three variables and only two equations, so you won’t be able to\nsolve in the traditional sense, but you may be able to determine\nsomething about these integers. Substitute for b in the second equation.\nBoth a and c are integers, and 5 and 12 share no prime factors. So a must\nbe divisible by 5 and c must be divisible by 12. Try a = 5 and c = 12 to see\nwhether b is a positive integer in this case.\nThis is a valid case because b is a positive integer. In this case, the value of\n|a – c| = |5 – 12| = |–7| = 7, which is one of the answer choices. \nThe correct answer is (B).\n 46. (E) 7: Each team either won or lost the first game and either won or lost the\nsecond game. Therefore, this problem can be approached using\noverlapping sets. Create a Double-Set Matrix."} +{"id": "book 2_p704_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 704, "page_end": 704, "topic_guess": "general", "text": "Won first game Lost first game Total\nWon second game\nLost second game 7\nTotal 20\nInitially, it appears that there is not enough information to determine how\nmany teams won both games. However, note that there were no ties and\neach round consisted of exactly 10 games.\nSince each of the 10 games yielded one winner and one loser, there were\nexactly 10 teams that won in the first round and 10 teams that lost in the\nfirst round. Similarly, there were 10 teams that won in the second round\nand 10 teams that lost in the second round. \nBased on this information, fill in the remainder of the matrix. \nWon first game Lost first game Total\nWon second game   7   3 10\nLost second game   3   7 10\nTotal 10 10 20\nSeven teams won both of their games.\nThe correct answer is (E).\n 47. (C):\n(1) INSUFFICIENT: Manipulate the statement."} +{"id": "book 2_p705_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 705, "page_end": 705, "topic_guess": "sequences_patterns", "text": "Thus, x is 1 greater than a multiple of 5. Since all multiples of 5 end in\neither 0 or 5, x must end in either 1 or 6.\nAlternatively, you could list numbers. Since y is a positive integer, 5y could\nbe 5, 10, 15, 20, etc. Thus, x could be 6, 11, 16, 21, etc. The units digit of x\ncould be 1 or 6.\n(2) INSUFFICIENT: Manipulate the statement.\nThis indicates that x is odd, because 2z is an even number. Any odd single-\ndigit integer is a possible units digit for x: 1, 3, 5, 7, or 9.\n(1) AND (2) SUFFICIENT: Statement (1) indicates that x must end in either 1\nor 6. Because statement (2) indicates that x is odd, x must end in 1 and\ncannot end in 6.\nThe correct answer is (C).\n 48. (E): The question stem asks for the value of the combo x + y."} +{"id": "book 2_p706_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 706, "page_end": 706, "topic_guess": "number_theory", "text": "(1) INSUFFICIENT: (x + y − 1)! < 100. To see the possible values of (x + y − 1),\nlist the factorials of the first few integers.\nTherefore, (x + y − 1) ≤ 4, and (x + y) ≤ 5.\n(2) INSUFFICIENT: Add x to both sides of y = x2 − x + 1 to create (x + y) on\none side of the equation.\nThe exact value of (x + y) is unknown, as it depends on the value of x2,\nwhich could be any positive integer.\n(1) AND (2) INSUFFICIENT: (x + y) ≤ 5 and x + y = x2 + 1 combine to indicate\nthat x2 + 1 ≤ 5. There are two integer solutions: x = 1 or x = 2. Therefore:\nThe correct answer is (E).\n1! = 1\n2! = 2\n3! = 6\n4! = 24\n5! = 120 (too large)\nIf x = 1, then y = 12 − 1 + 1 = 1 and x + y = 1 + 1 = 2.\nIf x = 2, then y = 22 − 2 + 1 = 3 and x + y = 2 + 3 = 5.\n 49. (C): The size of the angle depends on two things."} +{"id": "book 2_p707_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 707, "page_end": 707, "topic_guess": "geometry", "text": "The size of the circle\n The distance of C from the circle (inversely related to arc\nlength ADB)\nLarger circle → Larger x Farther C → smaller x\nLarger arc ADB → smaller x\n(1) INSUFFICIENT: The rubber band picture on the right indicates that for a\ncircle of fixed size, x can still vary with the length of arc ADB (i.e., x varies\nwith the placement of C).\n(2) INSUFFICIENT: Draw some cases to prove that x can vary for a given arc\nADB.\nFor a circle with circumference 28π, the arc ADB is \n of the\ncircle, so x is 90°. For a circle with circumference 15π, the arc ADB is"} +{"id": "book 2_p708_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 708, "page_end": 708, "topic_guess": "geometry", "text": ", or nearly half of the circle, and the lines tangent to the\ncircle at A and B will join at a smaller angle x.\n(1) AND (2) SUFFICIENT: If the area of the circle is πr2 = 81π, then r = 9. The\ncircumference of the circle is 2πr = 18. Thus, arc ADB is \n  of\nthe circumference of the circle. There is only one way to draw the lines\ntangent to the circle at A and B, so x must be one specific value.\nThe correct answer is (C).\n 50. (C) 3: List the first few terms of the sequence according to the given rules."} +{"id": "book 2_p709_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 709, "page_end": 709, "topic_guess": "sequences_patterns", "text": "Note that the pattern is a four-term repeat: 3, −3, 2, −2. Also note that the\nsum of this repeating group is (3) + (−3) + (2) + (−2) = 0. This repeating\ngroup will occur 16 times through term number 64. (Note: Because the\nsum of the four terms is 0, you don’t actually have to figure out how many\ntimes the pattern repeats.) Thus, the sum of the first 64 terms will be 0.\nThis leaves the 65th term, which will have the same value as B1: 3.\nTherefore, the sum of the first 65 terms is 3.\nThe correct answer is (C)."} +{"id": "book 2_p710_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 710, "page_end": 710, "topic_guess": "statistics", "text": "Workout Set 16\n151. Model Profit as a Percent of Model’s Total Sales\nModel A 12%\nModel B 16%\nModel C 20%\nA bicycle store earned a different percent profit on each of the three\nmodels of bicycles it sells, as shown in the table above. If the store\nmade a total profit of 16% on sales of these three models of\nbicycles last month, and these three models represented the\nentirety of the store’s profits, what percent of its total profits last\nmonth came from sales of Model A bicycles?\nThe store earned 20% of its profits last month from sales of\nModel B bicycles.\n(1)\nThe store earned 50% of its profits last month from sales of\nModel C bicycles.\n(2)\n152. A beaker contains 100 milligrams of a solution of salt and water that\nis x% salt by weight such that x < 90. If the water evaporates at a"} +{"id": "book 2_p711_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 711, "page_end": 711, "topic_guess": "number_theory", "text": "rate of y milligrams per hour, how many hours will it take for the\nconcentration of salt to reach (x + 10)%, in terms of x and y ?\n(A)\n(B)\n(C)\n(D)\n(E)\n153. Set A consists of five consecutive positive integers. Set B consists of\nevery integer that is in set A as well as the sum of every pair of two\ndistinct integers from set A, discarding any numbers that already\nappear in set B. What is the greatest integer in set B  ?\nThe range of set B is 13.(1)\nThere are exactly 12 integers in set B.(2)"} +{"id": "book 2_p712_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 712, "page_end": 712, "topic_guess": "number_theory", "text": "154. A survey included two questions, each of which could be answered\nwith either Yes or No. A total of 100 respondents took the survey,\nand every respondent answered both questions. How many\nrespondents answered both questions with a Yes?\nOf the respondents who answered Yes to the first question,\n40% answered Yes to the second question.\n(1)\nOf the respondents who answered Yes to at least one question,\n30% answered Yes to both questions.\n(2)\n155. If x is a positive two-digit integer, is x divisible by 17 ?\nx is divisible by 7.(1)\nx has exactly 3 unique prime factors.(2)\n156. Green paint is made by mixing blue paint with yellow paint in a\nratio of 2 to x. Turquoise paint is made by mixing green paint with\nblue paint in a ratio of y to 2. In terms of x and y, how many gallons\nof yellow paint are required to make 10 gallons of turquoise paint?"} +{"id": "book 2_p713_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 713, "page_end": 713, "topic_guess": "general", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n157. Erin writes a list of every sixth integer in increasing order, starting\nwith the positive integer x. Harry writes a list of every ninth integer\nin increasing order, starting with the positive integer y. Will any\ninteger appear on both Erin’s list and Harry’s list?\nx is a multiple of y.(1)\nx – y is a multiple of 3.(2)\n158. The students in a certain college class took a midterm exam and a\nfinal exam, both of which were scored out of a total of 100 points."} +{"id": "book 2_p714_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 714, "page_end": 714, "topic_guess": "word_problems", "text": "What percent of the students earned a higher score on the final\nthan on the midterm?\nOf the students in the class, 28% scored at least 6 points higher\non the final than on the midterm.\n(1)\nOf the students who scored higher on the final than on the\nmidterm, 40% scored at least 6 points higher on the final.\n(2)\n159. A triathlon consists of three legs: a swim, a bike ride, and a run. A\nrace organizer analyzes the results of a certain triathlon and\nobserves that the ratio of swim distance to bike distance to run\ndistance was 1 to 50 to 12, and that the ratio of average swim time\nto average bike time to average run time for all participants was 3\nto 10 to 6. If the average speed of racers in each leg of the race does\nnot change, but the race organizer changes the distance of each leg\nof the race so that the average swim time, average bike time, and\naverage run time are now equal to each other, what fraction of the\ntotal distance of the race will be represented by the swim leg of the\ntriathlon?"} +{"id": "book 2_p715_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 715, "page_end": 715, "topic_guess": "geometry", "text": "(A)\n(B)\n(C)\n(D)\n(E)\n160. An unpainted wall with a total area of 100 square feet needed to be\npainted with three coats of paint. Nelson began the job by painting\nan area of 60 square feet. Then, Jamaica painted a total area of 70\nsquare feet. Finally, Bryan painted a total area of 80 square feet.\nA er Bryan finished painting, an area of q square feet of the wall\nwas covered in three coats of paint. If the entire surface area of the\nwall had received at least one coat of paint, which of the following\nspecifies all the possible values of q  ?\n0 ≤ q ≤ 50(A)\n0 ≤ q ≤ 60(B)\n10 ≤ q ≤ 50(C)\n10 ≤ q ≤ 55(D)\n10 ≤ q ≤ 60(E)"} +{"id": "book 2_p716_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 716, "page_end": 716, "topic_guess": "general", "text": "Workout Set 16 Answer Key\n 51. D\n 52. D\n 53. A\n 54. E\n 55. D\n 56. A\n 57. B\n 58. C\n 59. B\n 60. D"} +{"id": "book 2_p717_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 717, "page_end": 717, "topic_guess": "statistics", "text": "Workout Set 16 Solutions\n 51. (D): A store sold three different models of bicycles last month. For each of these\nmodels, the profit earned by the store represents a different percentage of the sales\nof that model. The question stem also states that the total profit was 16% of the\ntotal sales.\nOne way to handle the question stem is to use variables A, B, and C to represent the\ntotal sales of Model A, B, and C bicycles, respectively. Then, the profit from Model A\nbicycles is 0.12A, the profit from Model B bicycles is 0.16B, and the profit from Model\nC bicycles is 0.20C. Also, the total dollar amount of sales is A + B + C, so the following\nis true and can be simplified:\nYou might also notice that 16% is exactly in the middle of the three percentages\nprovided in the question stem. The problem is effectively a weighted average\nproblem: if more dollars worth of Model A bicycles were sold, then the average\nprofit would be pulled downwards, closer to 12%. If more dollars worth of Model C\nbicycles were sold, then the average profit would be pulled upwards, closer to 20%.\nSince the actual profit was in the middle, at 16%, the dollar value of the Model A\nbicycles sold and the dollar value of the Model C bicycles sold must be equal."} +{"id": "book 2_p718_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 718, "page_end": 718, "topic_guess": "statistics", "text": "The question asks about the share of profits that came from Model A bicycles. This\nquestion can be rephrased as, “What is \n ?” which can then\nbe simplified as follows:\nWhat is \n ?\nWhat is \n ?\nWhat is \n ?\nThe variables would cancel entirely if you could determine a known ratio of A to B.\n(1) SUFFICIENT: 20% of the store’s profits came from Model B bicycles, which can be\ntranslated and simplified as follows, remembering that A = C and the total profit\nequals 0.16(A + B + C).\nSubstituting A = 2B into the question causes all unknowns to cancel out, yielding an\nanswer of \n , or 30%."} +{"id": "book 2_p719_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 719, "page_end": 719, "topic_guess": "word_problems", "text": "(2) SUFFICIENT: The store earned 50% of its profits from sales of Model C bicycles,\nwhich can be translated and simplified as follows, remembering that A = C and the\ntotal profit equals 0.16(A + B + C).\nThis is the same result that statement (1) produced. The answer is again 30%, as\nshown above, so this statement is sufficient.\nThe correct answer is (D).\n 52. (D) \n : The answer choices in this mixture problem include variables.\nTherefore, it is possible to use Smart Numbers.\nSuppose that x = 10 (initial concentration) and y = 2 (water evaporation rate in\nmilligrams [mg] per hour). Therefore, the beaker initially contains 10% of 100 mg =\n10 mg of salt. The question asks how many hours it will take for the concentration of\nsalt to reach (x + 10)%, or 20%.\nThe beaker will always contain 10 mg of salt, since the salt does not evaporate. The\n10 mg of salt will represent 20%, or \n , of the total weight when the total solution\nweight has been reduced to 50 mg. To get to this point, 50 mg of water must\nevaporate. This will take a total of \n = 25 hours.\nPlug in x = 10 and y = 2 to see which answer choice equals 25 hours."} +{"id": "book 2_p720_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 720, "page_end": 720, "topic_guess": "fractions_decimals_percents", "text": "(A)   \n Negative. Eliminate.\n(B)   \n = A non-integer close to 50. Eliminate.\n(C)   \n = 110. Eliminate.\n(D)   \n = 25. Correct.\n(E)    \n . Eliminate.\nAlternatively, approach the problem with algebra. Initially, the beaker contains x mg\nof salt in a total solution of 100 mg. A er h hours, hy mg of the solution will have\nevaporated, so the solution will have a total weight of (100 – hy) mg, and the\namount of salt will still be x mg. So the new concentration, as a percentage, will be\n. This percent should equal (x + 10)%. Set up an equation\nand solve for h, the number of hours."} +{"id": "book 2_p721_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 721, "page_end": 721, "topic_guess": "number_theory", "text": "The correct answer is (D).\n 53. (A): To better understand how the sets are constructed, write out an example.\nSuppose that set A consists of the integers 1, 2, 3, 4, and 5. Then, set B will initially\ninclude 1, 2, 3, 4, and 5, and more numbers will be added to it as follows, discarding\nany numbers that already appear in set B.\n1 + 2 = 3 discard\n1 + 3 = 4 discard\n1 + 4 = 5 discard\n1 + 5 = 6 will be added to set B\n2 + 3 = 5 discard\n2 + 4 = 6 discard\n2 + 5 = 7 will be added to set B\n3 + 4 = 7 discard\n3 + 5 = 8 will be added to set B\n4 + 5 = 9 will be added to set B"} +{"id": "book 2_p722_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 722, "page_end": 722, "topic_guess": "number_theory", "text": "Thus, if set A contains the integers from 1 to 5, set B consists of 1, 2, 3, 4, 5, 6, 7, 8,\nand 9. The question asks for the greatest integer in set B.\n(1) SUFFICIENT: The range of a set is the difference between the greatest and the\nsmallest numbers in that set. The smallest number in set B is the smallest number\nfrom set A, since all of the additional numbers in set B are constructed by summing\nnumbers together and will therefore be greater. Call this number x.\nThe greatest number in set B is created by summing the two greatest numbers from\nset A. Those numbers are x + 3 and x + 4, and their sum is 2x + 7.\nTherefore, the range of set B is (2x + 7) – x. Set this equal to the given range of 13,\nand solve for x:\nThe smallest number in set A is 6, and both sets can be listed with certainty. There is\nonly one answer to the question, so this statement is sufficient.\n(2) INSUFFICIENT: There are exactly 12 integers in set B, and you will likely have to\ntest cases to determine how this could happen. Above, it was determined that if set\nA contains the integers 1, 2, 3, 4, and 5, then set B contains 9 integers in total: the\nintegers from 1 to 9, inclusive. This is not enough integers: Find a case with fewer\nduplicates to discard. Try larger numbers.\nCase 1: If set A contains the integers 5, 6, 7, 8, and 9, then set B contains 5, 6, 7, 8, 9,\n11, 12, 13, 14, 15, 16, and 17. This set B contains 12 integers, and the greatest integer\nis 17."} +{"id": "book 2_p723_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 723, "page_end": 723, "topic_guess": "number_theory", "text": "Case 2: If set A contains the integers 100, 101, 102, 103, and 104, then set B contains\n100, 101, 102, 103, 104, 201, 202, 203, 204, 205, 206, and 207. This set B again\ncontains 12 integers, but its greatest integer is 207. \nThe correct answer is (A). \n 54. (E): This is an Overlapping Sets problem. Create a Double-Set Matrix and fill out the\nknown information.\nYes to Question 1 No to Question 1 Total\nYes to Question 2\nNo to Question 2\nTotal 100\nThe question asks for the number of respondents who gave two Yes answers. Try to\ndetermine the number that belongs in the top le of the matrix.\n(1) INSUFFICIENT: Let x represent the number of people who answered Yes to the\nfirst question. The matrix can be filled out as follows:\nYes to Question 1 No to Question 1 Total\nYes to Question 2 0.4x\nNo to Question 2 0.6x\nTotal x 100 – x 100\nIf x = 5, then the answer is 0.4(5) = 2. If x = 90, then the answer is 0.4(90) = 36. More\nthan one value is possible for the answer, so this statement is not sufficient.\n(2) INSUFFICIENT: Let y represent the number of people who answered No to both\nquestions. Then, the number of people who answered Yes to at least one question is"} +{"id": "book 2_p724_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 724, "page_end": 724, "topic_guess": "general", "text": "100 – y. Since 30% of these people answered Yes to both questions, the matrix can\nbe filled out as shown.\nYes to Question 1 No to Question 1 Total\nYes to Question 2 0.3(100 – y)\nNo to Question 2 y\nTotal 100\nIf y = 10, the answer is 0.3(90) = 27. If y = 80, the answer is 0.3(20) = 6. Since more\nthan one value is possible for the answer, this statement is insufficient.\n(1) AND (2) INSUFFICIENT: Using the information in both statements, you can\nconclude that 0.3(100 – y) = 0.4x. This simplifies to 0.4x + 0.3y = 30, but both x and y\ncan still have multiple values.\nAccordingly, the matrix can be filled out with numbers in several different ways.\nYes to Question 1 No to Question 1 Total\nYes to Question 2 24 20 44\nNo to Question 2 36 20 56\nTotal 60 40 100\nYes to Question 1 No to Question 1 Total\nYes to Question 2 12 10 22\nNo to Question 2 18 60 78\nTotal 30 70 100\nThe correct answer is (E)."} +{"id": "book 2_p725_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 725, "page_end": 725, "topic_guess": "number_theory", "text": "55. (D): The question stem includes a constraint that x is positive and has two digits.\nWithout this constraint, the answer would be (E), since knowing whether x is\ndivisible by 7 or has other prime factors does not help you determine whether x is\ndivisible by 17. \nUse the constraint as you solve the problem. Since x is a positive two-digit integer, x\nis between 10 and 99, inclusive. The question asks whether x is divisible by 17. Since\nthere are only a few integers in this range that are divisible by 17, jot them down for\nlater reference: 17, 34, 51, 68, and 85. The question can be rephrased as, “Is x equal\nto 17, 34, 51, 68, or 85 ?”\n(1) SUFFICIENT: Positive two-digit integer x is a multiple of 7. Since none of the\nnumbers listed earlier (17, 34, 51, 68, 85) are multiples of 7, x is definitely not one of\nthese numbers. Therefore, x is not a multiple of 17, and the answer is definitely No.\nAlternatively, use logical reasoning to prove that the statement is sufficient without\ndoing too much arithmetic. Seven and 17 are both prime numbers, so the smallest\nnumber that is divisible by both of them is their product, (7)(17) = 119. However, you\nalready know that x is smaller than 119, since x only has two digits. Therefore, x\ncannot be a multiple of both 7 and 17, so the answer is definitely No.\n(2) SUFFICIENT: None of the numbers listed earlier have exactly three unique prime\nfactors: 17 is prime, while 34, 51, 68, and 85 each have exactly two unique prime\nfactors. Therefore, x is not one of these numbers, so x is not a multiple of 17 and the\nanswer to the question is No.\nIt is also possible to show that x cannot be a multiple of 17 while doing minimal\narithmetic. If x has exactly three unique prime factors, then x will only be divisible by\n17 if 17 is among those factors. However, the smallest number that has three unique\nprime factors that include 17 is (2)(3)(17) = 102, which is too large to be the value of\ntwo-digit integer x. Therefore, x can not be divisible by 17 and the answer is\ndefinitely No.\nThe correct answer is (D)."} +{"id": "book 2_p726_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 726, "page_end": 726, "topic_guess": "fractions_decimals_percents", "text": "56. (A)  \n : Since the answer choices include variables, choose smart\nnumbers. It will be simpler to choose y first, since the value of y is more closely\nrelated to the 10 gallons of turquoise paint. Choose y = 8, so that the sum of y and 2\nin the ratio of the paints used to make turquoise can be 8 + 2 = 10. In this case, it\ntakes 8 gallons of green paint and 2 gallons of blue paint to make 10 gallons of\nturquoise paint. \nNext, choose a value for x. Because the total amount of green paint should be 8\ngallons, choose x = 6, so that the of x and 2 in the ratio to make green paint is\n2 + 6 = 8. In total, it will take 6 gallons of yellow paint to create the turquoise paint,\nso the answer to the question is 6. \nPlug x = 6 and y = 8 into the answer choices. The correct answer will simplify to 6.\n(A) \n . Correct.\n(B) \n Not an integer. Eliminate. \n(C) \n Eliminate. \n(D) \n Eliminate.\n(E) \n = Not an integer. Eliminate. \nThe correct answer is (A). \n 57. (B): Erin’s list starts with an unknown positive integer and then includes every sixth\ninteger. So Erin’s list could be any of the following, depending on the value of x:"} +{"id": "book 2_p727_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 727, "page_end": 727, "topic_guess": "number_theory", "text": "x Erin’s Listx Erin’s List\n1 1, 7, 13, 19, 25 . . .\n2 2, 8, 14, 20, 26 . . .\n3 3, 9, 15, 21, 27 . . .\netc. etc.\nSimilarly, Harry’s list could be any of the following: \ny Harry’s List\n1 1, 10, 19, 28, 37 . . .\n2 2, 11, 20, 29, 38 . . .\n3 3, 12, 21, 30, 39 . . .\netc. etc.\n(1) INSUFFICIENT: \nCase 1: If x = y = 1, then x is a multiple of y, and the number 1 appears on both lists.\nTherefore, the answer to the question is Yes.\nCase 2: If x = 3, all of the numbers on Erin’s list are multiples of 3. If y = 1, none of the\nnumbers on Harry’s list are multiples of 3. Therefore, x is a multiple of y, but no\nnumbers appear on both lists, so the answer to the question is No. \n(2) SUFFICIENT: If x – y is a multiple of 3, then x – y = 3n, for some integer n. So x = 3n\n+ y. The integers on the two lists are as follows:  \nErin’s list: 3n + y, 3n + y + 6, 3n + y + 12, 3n + y + 18 . . .\nHarry’s list: y, y + 9, y + 18, y + 27 . . ."} +{"id": "book 2_p728_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 728, "page_end": 728, "topic_guess": "fractions_decimals_percents", "text": "If n is a multiple of 3, then 3n is a multiple of 9, so 3n + y appears on both lists. If n is\n1 more than a multiple of 3, then 3n + 6 = 3(n + 2) = 3(multiple of 3) is a multiple of 9,\nso y + 3n + 6 appears on both lists. Finally, if n is 2 more than a multiple of 3, then\n3n + 12 = 3(n + 4) = 3(multiple of 3) is a multiple of 9 so y + 3n + 12 appears on both\nlists. Regardless of the value of n, some number appears on both Erin’s and Harry’s\nlists, so the answer to the question is Yes.\nThe correct answer is (B). \n 58. (C): According to the question stem, each student in the class earned a certain score\non the midterm and a certain score on the final. A student’s final score could have\nbeen higher than, equal to, or lower than his or her midterm score. To answer the\nquestion, you’ll need to determine what percent of the students fell into the first of\nthese three categories.\n(1) INSUFFICIENT: It is possible that all of the remaining students scored between 1\nand 5 points higher on the final than on the midterm, which would make the answer\nto the question 100%. It is also possible that all of the remaining students scored\nhigher on the midterm than on the final, which would make the answer to the\nquestion 28%.\n(2) INSUFFICIENT: This statement relates the percent of students who scored higher\non the final to the percent of students who scored at least 6 points higher on the\nfinal. However, without any information about either of these percents, the\nquestion cannot be answered definitively. \nFor instance, 50% of the students could have scored higher on the final, and 40% of\nthis 50% could have scored at least 6 points higher. Or, only 10% of the students\ncould have scored higher on the final, and 40% of this 10% could have scored at\nleast 6 points higher.\n(1) AND (2) SUFFICIENT: According to statement (1), the following is true:\n0.28(total students) = students who scored 6+ points higher on the final"} +{"id": "book 2_p729_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 729, "page_end": 729, "topic_guess": "word_problems", "text": "According to statement (2), the following is true:\n0.4(students who scored higher on the final) = students who scored 6+ points\nhigher on the final\nCombine these two equations.\n0.28(total students) = 0.4(students who scored higher on the final)\nSimplify this equation to calculate the percent of students who scored higher on the\nfinal.\nSince an exact answer to the question can be calculated, the statements together\nare sufficient. \nThe correct answer is (C). \n 59. (B) \n : This problem provides two pieces of information. The first is the ratio of the\nswim distance to the bike distance to the run distance: 1 to 50 to 12. The second\npiece of information is the ratio of the average swim time to average bike time to\naverage run time: 3 to 10 to 6.\nUse an unknown multiplier for each of these ratios.\nDistance Time\nSwim d 3t\nBike 50d 10t\nRun 12d 6t"} +{"id": "book 2_p730_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 730, "page_end": 730, "topic_guess": "word_problems", "text": "Now, use an RTD (rate/time/distance) chart to calculate the average speed for\ncompetitors in each of the three legs, in terms of d and t. \nRate Time Distance\nSwim\n 3t d\nBike\n 10t 50d\nRun\n 6t 12d\nThe question asks about a scenario in which the average swim time, bike time, and\nrun time are equal, but the rates stay the same. In this scenario, the following will be\nthe case:\nRate Time Distance\nSwim\n t ?\nBike\n t ?\nRun\n t ?\nUse the RTD formula to fill in the missing column of this chart.\nRate Time Distance\nSwim\n t\nBike\n t 5d\nRun\n t 2d"} +{"id": "book 2_p731_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 731, "page_end": 731, "topic_guess": "word_problems", "text": "The distances are in a ratio of \n to 5 to 2, or 1 to 15 to 6. The swim distance\nrepresents \n of the total distance. \nThe correct answer is (B). \n 60. (D) 10 ≤ q ≤ 55: The problem does not specify the extent to which each person paints\nover the area that was already painted by the previous person. For instance, when\nJamaica painted 70 square feet (sq ) of the wall, she might have painted over the\nentire 60 sq already painted by Nelson, or she might have painted over only some\nof it.\nHowever, by the time Bryan finished painting, the entire wall had been painted with\nat least one coat of paint, and some of it might have been painted with as many as\nthree coats. The question asks how much of the wall had three coats, and the\nanswer choices represent ranges from minimum to maximum area. \nFirst, make q as large as possible. In order to do so, the amount of paint used on the\nremainder of the wall must be minimized. The area q will be as large as possible\nwhen the remaining 100 – q square foot area is covered in only a single coat."} +{"id": "book 2_p732_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 732, "page_end": 732, "topic_guess": "geometry", "text": "The entire amount of paint applied by all three painters is 60 + 70 + 80 = 210 sq . If\nq sq of the wall has three coats of paint, and the remaining 100 – q has only one\ncoat, the following is true:\n3(area with 3 coats) + 1(area with 1 coat) = total paint applied\nTurn this into an equation and solve for q. \nThe greatest possible value of q is 55 sq . Therefore, answer (D) must be correct.\nIt is also possible to determine the greatest possible value of q by Working\nBackwards from the choices, testing their extreme values using pictures. If the\npainting is done as shown below, then 55 sq of the wall will have three coats of\npaint and the rest will have only one coat."} +{"id": "book 2_p733_c1", "source_name": "Manhattan Prep GMAT Advanced Quant", "source_file": "book 2.pdf", "source_type": "pdf", "page_start": 733, "page_end": 733, "topic_guess": "geometry", "text": "To find the smallest possible number of square feet with three coats, do the\nopposite: Assume that the rest of the wall has as much paint as possible. In that\ncase, the remaining 100 – q sq of wall will have two coats of paint, so the following\nwill be true:\nThe smallest possible value of q is 10. Here is another way to visualize the\nminimization of the three-coat area.\nThe correct answer is (D)."} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p1_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 1, "page_end": 1, "topic_guess": "flashcards", "text": "These flashcards feature screenshots from \nGMAT Prep Now’s video lessons \nwww.GMATPrepNow.com \nGMAT Prep Now Quantitative Reasoning Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p2_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 2, "page_end": 2, "topic_guess": "word_problems", "text": "If, at any time, you’d like to watch the video \nrelated to a certain flashcard, just click on \nthe link at the top of that page, and you’ll \nbe taken to the corresponding video* \nGMAT Prep Now Quantitative Reasoning Flashcards \n*Many of our videos are free, but you must register \nan account to have access to all videos \nwww.GMATPrepNow.com"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p3_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 3, "page_end": 3, "topic_guess": "algebra", "text": "topic \narithmetic \npowers & roots \nalgebra & equation-solving \nword problems \ngeometry \ninteger properties \nstatistics \ncounting \nprobability \n- slide \n- 4 \n- 27 \n- 42 \n- 80 \n- 100 \n- 124 \n- 139 \n- 144 \n- 153 \nTABLE OF CONTENTS \nGMAT Prep Now Quantitative Reasoning Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p4_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 4, "page_end": 4, "topic_guess": "flashcards", "text": "(watch video) \nArithmetic Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p5_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 5, "page_end": 5, "topic_guess": "flashcards", "text": "(watch video) \nArithmetic Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p6_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 6, "page_end": 6, "topic_guess": "flashcards", "text": "(watch video) \nArithmetic Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p7_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 7, "page_end": 7, "topic_guess": "flashcards", "text": "(watch video) \nArithmetic 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154, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p155_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 155, "page_end": 155, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p156_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 156, "page_end": 156, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p157_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 157, "page_end": 157, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p158_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 158, "page_end": 158, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p159_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 159, "page_end": 159, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p160_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 160, "page_end": 160, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p161_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 161, "page_end": 161, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT Math Flashcards from GMAT Prep Now_p162_c1", "source_name": "GMAT Prep Now Quantitative Reasoning Flashcards", "source_file": "GMAT Math Flashcards from GMAT Prep Now.pdf", "source_type": "pdf", "page_start": 162, "page_end": 162, "topic_guess": "flashcards", "text": "(watch video) \nProbability Flashcards"} +{"id": "GMAT QUANT_p1_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 1, "page_end": 1, "topic_guess": "general", "text": "GMAT QUANTITATIVE REVIEW"} +{"id": "GMAT QUANT_p2_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 2, "page_end": 2, "topic_guess": "fractions_decimals_percents", "text": "GMAT STRATEGY/TIPS \nDO NOT assume that a number is an integer unless explicitly stated in the problem or if the object of a problem has a physical restriction of being divided (such as votes, cards, pencils etc.)\nExample: if p < q and p < r, is pqr < p? \nStatement (1): pq < 0 \nStatement (2): pr < 0 \nStatement (1) INSUFFICIENT: We learn from this statement that either p or q is negative, but since we know from the question that p < q, p must be negative. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet only 2 conditions: p must be negative and q must be positive. However, r could be either positive or negative, and hence, the product could be either negative (may or may not be less than p) or positive (greater than p)\nStatement (2) INSUFFICIENT: We learn from this statement that either p or r is negative, but since we know from the question that p < r, p must be negative and r must be positive. However, q could be either positive or negative. \nIf we look at both statements together, we know that p is negative and that both q and r are positive. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet 3 conditions: p must be negative, q must be positive, and r must be positive. For instance, if p = -2 , q = 10, r = 5; pqr = -100, which is less than p. Taking another example, if p = -2 , q = 7, r = 4; pqr = -56, which is also less than p. Therefore, at first glance, it may appear that we will always get a \"YES\" answer. But don't forget to test out fractional (decimal) values as well. The problem never specifies that p, q, and r must be integers. If p = -2, q = 0.3 and r = 0.4, pqr = -0.24, which is NOT less than p. Hence, the correct answer is E – Both statements together are insufficient"} +{"id": "GMAT QUANT_p2_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 2, "page_end": 2, "topic_guess": "algebra", "text": "ers. If p = -2, q = 0.3 and r = 0.4, pqr = -0.24, which is NOT less than p. Hence, the correct answer is E – Both statements together are insufficient\n\nGMAT can also hide POSITIVE CONSTRAINTS. Sides of a square, number of votes, etc. will always be positive. When you have a positive constraint, you can:\nEliminate negative solutions from a quadratic function\nMultiply or divide an inequality by a variable\nCross-multiply inequalities: x/y < y/x  x² < y²\nChange an inequality sign for reciprocals: x 1/y\n\nDO NOT ASSUME THE SIGN OF A VARIABLE\nExample: Is T/S > F/G?\n(1) T < S\n(2) F > G\nIf one assumes the signs for all four variables to be positive, then T/S (an improper fraction) is NOT greater than F/G (an improper fraction). However, if negative signs are taken into account, we cannot conclusively ascertain whether T/S > F/G"} +{"id": "GMAT QUANT_p3_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 3, "page_end": 3, "topic_guess": "word_problems", "text": "GMAT STRATEGY/TIPS"} +{"id": "GMAT QUANT_p4_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 4, "page_end": 4, "topic_guess": "word_problems", "text": "GMAT STRATEGY/TIPS"} +{"id": "GMAT QUANT_p5_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 5, "page_end": 5, "topic_guess": "word_problems", "text": "GMAT STRATEGY/TIPS"} +{"id": "GMAT QUANT_p6_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 6, "page_end": 6, "topic_guess": "word_problems", "text": "GMAT STRATEGY/TIPS"} +{"id": "GMAT QUANT_p6_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 6, "page_end": 6, "topic_guess": "number_theory", "text": "GMAT STRATEGY/TIPS\n\n64 is the only number (besides 1) under 100 that is the square of an integer and the cube of an integer\n16 and 81 are the only numbers (again, besides 1) that are both the square of an integer and the square of a square\n30 and 42 are the smallest integers with at least three prime factors\nThe only integers with exactly three factors are the squares of prime numbers\nWhen odd number n is doubled, 2n has twice as many factors as n\nWhen even number is doubled, 2n has 1.5 more factors as n\nAll prime numbers above 3 are of the form 6n – 1 or 6n + 1 , because all other numbers are divisible by 2 or 3\nXn – yn is always divisible by x – y and is divisible by x + y if n is even \nGenerally the last digit of 1! + 2! … + N!   can take ONLY 3 values:\u000bA. N=1 --> last digit 1;\u000bB. N=3 --> last digit 9;\u000bC. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33,   the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).\nFinding the number of powers of a prime number P in n! \nn/p + n/p2 + n/p3 …. Till px < n \nFor example, what is the power of 2 in 25!?\n25/2 + 25/4 + 25/8 + 25/16  12+ 6 + 3 + 1  22 \nFinding the power of non-prime in n!\nFor example, how many powers of 900 are in 50!?\nStep 1: Determine the prime factorization the number: PF of 900 is 22*32*52\nStep 2: Find the powers of these prime numbers in n! (refer above)\nPower of 2 = 47, power of 3 = 22 and power of 5 = 12 \nStep 3: We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!"} +{"id": "GMAT QUANT_p7_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 7, "page_end": 7, "topic_guess": "sets_probability_counting", "text": "OVERLAPPING SETS"} +{"id": "GMAT QUANT_p8_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 8, "page_end": 8, "topic_guess": "sets_probability_counting", "text": "2 OVERLAPPING SETS – VENN DIAGRAMS\nX\nFormula: Total – Neither = A + B – X\nFormula for elements who are AT LEAST one of the two groups: A + B – X \nFormula for elements who are in ONLY one of the two groups: A + B – 2X \nOnly in group A = a\nOnly in group B = b\nBoth in group A & B = X\nNeither = N"} +{"id": "GMAT QUANT_p9_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 9, "page_end": 9, "topic_guess": "general", "text": "DOUBLE SET MATRIX – 2 MUTUALLY EXCLUSIVE OPTIONS FOR A DECISION \nFor GMAT problems involving only two categorizations or decisions, the most efficient tool is the Double Set Matrix: a table whose rows correspond to the options for one decision and whose columns correspond to the options for the other decision. \nThe rows should correspond to the mutually exclusive options for one decision. Likewise, the columns should correspond to the mutually exclusive options for the other. For instance, if a problem deals with students getting either right or wrong answers on problems 1 and 2, the columns should not be \"problem 1\" and \"problem 2,\" and the rows should not be \"right\" and \"wrong.\" Instead, the columns should list options for one decision-problem 1 correct, problem 1 incorrect, total-and the rows should list options for the other decision-problem 2 correct, problem 2 incorrect, total"} +{"id": "GMAT QUANT_p10_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 10, "page_end": 10, "topic_guess": "sets_probability_counting", "text": "3 OVERLAPPING SETS \nFormula: Total – Neither = A + B + C – (a+b+c+2x)\n\nFormula for elements who are AT LEAST one of the three groups: A + B + C – (a + b + c + 2x) \nIt is important to note that “a” represents elements that belong ONLY to A and C (not B)\n\nTotal = A + B + C – (a+ x + b + x + c + x) + x\n\nFormula for elements in any two of the groups: a+ b + c\n\nNote: Work from the inside out\n\nOnly A = A – a – b – x \nOnly B = B – b – c – x \nOnly C = C – a – c – x\n\nx\na\nc\nb"} +{"id": "GMAT QUANT_p11_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 11, "page_end": 11, "topic_guess": "sets_probability_counting", "text": "3 OVERLAPPING SETS"} +{"id": "GMAT QUANT_p12_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 12, "page_end": 12, "topic_guess": "statistics", "text": "STATISTICS"} +{"id": "GMAT QUANT_p13_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 13, "page_end": 13, "topic_guess": "statistics", "text": "MEAN – THE AVERAGE \nMEAN = MEDIAN in a set of consecutive integers\nIf each one of the given numbers is increased (or decreased) by K, their average is increased (or decreased) by k\nIf each one of some given numbers is multiplied by K, their average is multiplied by K\nThe sum of first “n” natural numbers is given by n (n + 1)/2\nIf the average of a few consecutive integers is 0, then either all the numbers are zero or there will be an odd number of integers\nDETERMINING THE MEAN\nMean = Sum of terms/ number of terms\nFor an evenly spaced set (i.e., series where the difference between any two consecutive numbers is the same – 3, 6, 9), or a set of consecutive integers\nMean = (first term + last term) / 2 \nMean = middle term (if the number of terms in the set is even, then the mean is the average of the middle two terms)\nIMPORTANT: The mean of a set of consecutive integers will be an INTEGER only if the number of elements in the set is ODD or if the difference between/sum of the “consecutive” terms is divisible by 2. For example, in the case of 2,4,6 and 8, the average is 5\n\nCHANGE IN MEAN\nChange in mean = New term – Old mean/ New number of terms"} +{"id": "GMAT QUANT_p13_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 13, "page_end": 13, "topic_guess": "statistics", "text": "is divisible by 2. For example, in the case of 2,4,6 and 8, the average is 5\n\nCHANGE IN MEAN\nChange in mean = New term – Old mean/ New number of terms\n\nIf a set currently has 9 terms and a mean value of 56, and you add a tenth term of 71, then the mean will increase by (71 - 56)/10 = 1.5\nYou can interpret this formula conceptually as taking the \"excess\" or \"deficit\" in the new term relative to the mean and redistributing that excess (or deficit) evenly to all of the terms in the set, including that new term. For instance, in the example above, the new term (71) exceeds the average by 71 - 56 = 15. When you add the new term to the set, those 15 \"extra points\" are distributed evenly among all ten data points, increasing each by 1.5"} +{"id": "GMAT QUANT_p13_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 13, "page_end": 13, "topic_guess": "statistics", "text": "y 71 - 56 = 15. When you add the new term to the set, those 15 \"extra points\" are distributed evenly among all ten data points, increasing each by 1.5\n\nRESIDUALS\nResiduals = Data point – Mean \nFor any set, the residuals sum to zero. Alternatively, the positive residuals (\"overs\") and negative residuals (\"unders\") for any set will cancel out\nIf the mean of the set {97, 100, 85, 90, 94, 80, 92, x} is 91, what is the value of x?\nYou could certainly solve this problem with the traditional formula for averages, but you would waste a lot of time on arithmetic. Instead, just use the given mean of 91 to compute the residuals for all the terms except x: +6, +9, -6, -1, +3, -11, + 1. These residuals sum to +1.Therefore, x must leave a residual of -1, since all the residuals sum to zero. As a result, x is one less than the mean, or 90"} +{"id": "GMAT QUANT_p14_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 14, "page_end": 14, "topic_guess": "statistics", "text": "MEDIAN – THE MIDDLE TERM\n\nMEAN = MEDIAN in a set of consecutive integers. However, just because the mean and the median are equal does not mean that the set comprises of evenly spaced/consecutive integers \nUltimately, if your mean and median are the same, what it tells you is that your data is arranged symmetrically around the median. For every bit over the mean on one side, there's a corresponding bit under the mean on the other side, balancing it out. When they're different, that tells you that this symmetry has been broken\n\nDETERMINING THE MEDIAN\nTo find the median, list the numbers in ascending or descending order, and find the middle term"} +{"id": "GMAT QUANT_p14_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 14, "page_end": 14, "topic_guess": "statistics", "text": "this symmetry has been broken\n\nDETERMINING THE MEDIAN\nTo find the median, list the numbers in ascending or descending order, and find the middle term\n\nWhen number of terms N is large, then the following two formulae are useful for finding the middle term:\nWhen N is odd, the middle term is (N+1)/2th term\nWhen N is even, the median will be the average of the middle two terms, i.e., average of the (N/2)th term and the (N/2)th + 1 term\nVALUE OF THE MEDIAN\nWhen the number of items in a set of integers is odd, the median is an integer\nWhen the number of terms in a set of integers is even, the median can only end with 0.5 as it is the average of the middle two terms. However, it is important to note that the median does not ALWAYS end in 0.5 when the number of terms in the set is even. For example, the median for 2,4,6,8,10,12 is 7 \nADVANCED PROPERTIES OF MEDIANS \nMedian of combined set will always lie between the two medians or could be equal to one or both the medians of the two sets"} +{"id": "GMAT QUANT_p14_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 14, "page_end": 14, "topic_guess": "statistics", "text": "CED PROPERTIES OF MEDIANS \nMedian of combined set will always lie between the two medians or could be equal to one or both the medians of the two sets\n\nFor example: a,b,c are integers and a= 0\nRange of a set has to be equal or greater than difference between any two numbers in the set"} +{"id": "GMAT QUANT_p16_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 16, "page_end": 16, "topic_guess": "statistics", "text": "ge = 0 when all values are the same \nRange is always >= 0\nRange of a set has to be equal or greater than difference between any two numbers in the set\n\nSTANDARD DEVIATION \nFormula: √ (term1 – average)2 + (term2 – average)2 + (term3 – average)2 / √number of terms\nRules\nThe set in which numbers are farther away from each other than the numbers in some other set will have greater value of standard deviation\nA = {2,4,6,8,10}\nB = {3,6,9,12,15}\nIn set B, the numbers are wider away from each other and hence will have greater value of standard deviation\nStandard deviation does not change when a constant value is either added to or subtracted from every term in a set\nStandard deviation is multiplied or divided by that constant value which is used to multiply or divide every term in a set\nTaking the absolute value of the elements of a set will not change the standard deviation if all elements are positive or negative\nIf y=ax + b, and if the standard deviation of x series is ‘S’, then the standard deviation of y series will be a*s \nThe SD of any list is not dependent on the average, but on the deviation of the numbers from the average. So just by knowing that two lists having different averages doesn't say anything about their standard deviation - different averages can have the same SD\nIf mean = maximum value it means that all values are equal and SD is 0"} +{"id": "GMAT QUANT_p16_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 16, "page_end": 16, "topic_guess": "statistics", "text": "ything about their standard deviation - different averages can have the same SD\nIf mean = maximum value it means that all values are equal and SD is 0\n\nVARIANCE = STANDARD DEVIATION2"} +{"id": "GMAT QUANT_p17_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 17, "page_end": 17, "topic_guess": "fractions_decimals_percents", "text": "FRACTIONS,DIGITS AND DECIMALS AND PERCENTAGES"} +{"id": "GMAT QUANT_p18_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 18, "page_end": 18, "topic_guess": "fractions_decimals_percents", "text": "FRACTIONS\nPROPER FRACTIONS\nProper fractions fall between 0 and 1. In proper fractions, the numerator is always smaller than the denominator\n\nIMPROPER FRACTIONS\nIn improper fractions, the numerator is always greater than the denominator. Improper fractions are therefore always greater than 1. For example, 3/2 = 1.5, 5/4 = 1.25\nImproper fractions can be rewritten as mixed numbers, i.e. as an integer and a proper fraction\n\nCOMPARING FRACTIONS\nThe traditional method of comparing fractions entails finding a common denominator and comparing the numerators\nThe shortcut method entails multiplying the numerator of one fraction with the denominator of the other fraction"} +{"id": "GMAT QUANT_p18_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 18, "page_end": 18, "topic_guess": "fractions_decimals_percents", "text": "ominator and comparing the numerators\nThe shortcut method entails multiplying the numerator of one fraction with the denominator of the other fraction\n\nFRACTION RULES FOR POSITIVE, PROPER FRACTIONS\nAs the numerator increases, the value of the fraction increases as it approaches 1, assuming that the denominator is held constant\nAs the denominator increases, the value of the fraction decreases as it approaches 1, assuming that the numerator is held constant\nIncreasing both the numerator and denominator by the same value brings the fraction closer to 1, regardless of the original value of the fraction. Hence, a proper fraction will increase in value, while an improper fraction will decrease in value \nExample: 3/2 > 4/3 > 13/12 > 1,013/1,012"} +{"id": "GMAT QUANT_p18_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 18, "page_end": 18, "topic_guess": "fractions_decimals_percents", "text": "fraction. Hence, a proper fraction will increase in value, while an improper fraction will decrease in value \nExample: 3/2 > 4/3 > 13/12 > 1,013/1,012\n\nIMPACT OF ARITHMETIC OPERATIONS ON PROPER FRACTIONS\nAdding fractions increase their value \nSubtracting fractions decrease their value\nMultiplying fractions decrease their value\nSquaring a positive fraction reduces its value. For example, (1/2)2 = (1/4), where 1/ 2 > 1/ 4\nSquaring a negative fraction increases its value. For example, (-1/2)2 = (1/4), where 1/ 4 > - 1/ 2\nCubing a positive fraction reduces its value\nCubing a negative fraction increases its value (i.e., it becomes a smaller NEGATIVE number)\nMultiplying the numerator of a positive, proper fraction increases the value of the fraction as it approaches 1\nThere will be no change in the value of the fraction if both the numerator and the denominator are multiplied by the same constant\nDividing fractions increase their value\nTaking the square root of a fraction increases its value\n\nIf there is an addition or subtraction operation in the denominator, the fraction CANNOT be simplified further"} +{"id": "GMAT QUANT_p19_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 19, "page_end": 19, "topic_guess": "fractions_decimals_percents", "text": "DIGITS AND DECIMALS"} +{"id": "GMAT QUANT_p20_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 20, "page_end": 20, "topic_guess": "fractions_decimals_percents", "text": "DECIMALS – ADVANCED CONCEPTS"} +{"id": "GMAT QUANT_p21_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 21, "page_end": 21, "topic_guess": "sequences_patterns", "text": "DIGITS – LAST DIGIT\nLast digit of (xyz)n is the same as that of Zn\n\nDetermine the cyclicity number of Z \nInteger ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base\nIntegers ending with 2, 3, 7 and 8 have a cyclicity of 4 \n2 – 2, 4, 8, 6 \n3 – 3, 9, 7, 1\n7 – 7, 9, 3, 1 \n8 – 8, 4, 2, 6\n - Integers ending with 4 have a cyclicity of 2. When n is odd, last digit will be 4 and when n is even, last digit will be 6\n- Integers ending with 9 have a cyclicity of 2. When n is odd, last digit will be 9 and when n is even, last digit will be 1\nFind the remainder when n is divided by the cyclicity\nWhen the remainder > 0, the last digit will be the same as Zr \nWhen remainder = 0, the last digit will be the ZC, where c is the cyclicity number\n\nExample # 1:\nWhat is the last digit of 12739?\n\nLast digit of 12739 is the same as that of 739. First we need to determine the cyclicity of 7, which is 4.\n\nNow divide 39 (power) by 4 (cyclicity), remainder is 3. Therefore, the last digit of 12739 is the same as the last digit of 73 which is 3\n\nExample # 2:\nGiven that x is a positive integer, what is the units digit of 24(2x+1)* 33 (x+1)* 17(x+2)* 9(2x)"} +{"id": "GMAT QUANT_p21_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 21, "page_end": 21, "topic_guess": "sequences_patterns", "text": "ame as the last digit of 73 which is 3\n\nExample # 2:\nGiven that x is a positive integer, what is the units digit of 24(2x+1)* 33 (x+1)* 17(x+2)* 9(2x)\n\nx is a positive integer, so 2x is always even, while 2x + 1 is always odd\n\nThus,\n4(2x + 1) = (4)(odd), which always has a units digit of 4\n9(2x) = 9(even), which always has a units digit of 1\n\nThat leaves us to find the units digit of (3)(x + 1)*(7)(x + 2). Rewriting, and dropping all but the units digit at each intermediate step,\n= (3)(x + 1)(7)(x + 2)\n= (3)(x + 1)(7)(x + 1)(7)\n= (3 × 7)(x + 1)(7)\n= (21)(x + 1)(7)\n= (1)(x + 1)(7) = 7, for any value of x\n\nSo, the units digit of is (4)(7)(1) = 28, then once again drop all but the units digit to get \nSolve the problem step by step. However, pay attention only to the last digit of the intermediate product\n\nNote 1: you just need a 5 and a 2 (or a multiple of 2) to decide the units digit, as it will be always be 0 if you find these two numbers in a product\n\nExample: \nWhat is the units digit of 5*5*9*9*4*4*4? \n5*5 = 25; 9*9 = 81; 4*4*4= 64; Multiply the last digit of each of the intermediary products 5*1*4 = 20; Units digit will therefore be 0"} +{"id": "GMAT QUANT_p21_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 21, "page_end": 21, "topic_guess": "sequences_patterns", "text": "*5*9*9*4*4*4? \n5*5 = 25; 9*9 = 81; 4*4*4= 64; Multiply the last digit of each of the intermediary products 5*1*4 = 20; Units digit will therefore be 0\n\nLast digit of a product\nLast digit of a power\nWhat is the units digit of 1727 ?\n1727 will end in the same units digit as 727\n27 / 4 results in a remainder of 3. Therefore, the units digit of 1727 will be the same as the units digit of 73 , which is 3\n\nIf x is a positive integer, what is the units digit of 24(2x+1) 33(x+1) 17(x+2) 9 (2x)\nNote that x is a positive integer, so 2x is always even, while 2x + 1 is always odd.  Thus,\u000b(4)(2x + 1) = (4)(odd), which always has a units digit of 4\u000b(9)(2x) = (9)(even), which always has a units digit of 1\u000bThat leaves us to find the units digit of (3)(x + 1)(7)(x + 2).  Rewriting, and dropping all but the units digit at each intermediate step,\u000b(3)(x + 1)(7)(x + 2)\u000b= (3)(x + 1)(7)(x + 1)(7)\u000b= (3 × 7)(x + 1)(7)\u000b= (21)(x + 1)(7)\u000b= (1)(x + 1)(7) = 7, for any value of x.\nSo, the units digit of (4)(2x + 1)(3)(x + 1)(7)(x + 2)(9)(2x) is (4)(7)(1) = 28, then once again drop all but the units digit to get 8"} +{"id": "GMAT QUANT_p22_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 22, "page_end": 22, "topic_guess": "fractions_decimals_percents", "text": "PERCENTAGES – 1"} +{"id": "GMAT QUANT_p23_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 23, "page_end": 23, "topic_guess": "word_problems", "text": "PERCENTAGES – 2 \nMIXTURE PROBLEMS \nThe first type of problem will have a mixture of two components and will ask you to alter the make-up of the mixture by adding or subtracting one of the components\n\nFor example, a 50-ounce mixture of sugar and water is made up of 40% sugar and 60% water. How much sugar should you add so that the mixture is 60% sugar and 40% water \nAlways remember: the key to these problems is the component that does not change.  In this case, we are adding sugar, while water remains constant.  Therefore, we will focus on the water for most of the problem\nFirst, determine how much water is in the mixture; 60% of 50 is 30 ounces\nBecause we are not adding or subtracting any water, we will still have 30 ounces of it after we add more sugar. However, we now want that 30 ounces to represent 40% of the total, rather than 60%. Therefore, 40% * x = 30; x = 75\nSince the increase in total volume is only made up of additional sugar, and we went from 50 total ounces to 75 total ounces, we must add 25 ounces of sugar"} +{"id": "GMAT QUANT_p23_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 23, "page_end": 23, "topic_guess": "word_problems", "text": "the increase in total volume is only made up of additional sugar, and we went from 50 total ounces to 75 total ounces, we must add 25 ounces of sugar\n\nIn the second type of mixture problem, substances with different characteristics are combined, and it is necessary to determine the characteristics of the resulting mixture\nThis is similar to a weighted average problem\n\nFor example, mixture X is 10% acid and mixture Y is 30% acid. If mixture X and mixture Y are combined, the new mixture is 15% acid. What percentage of the new mixture is mixture X? \n(0.1X + 0.30Y)/ (X+Y) = 0.15 \nSolving for X and Y, we get 0.05 X = 0.15 Y \nX/Y = 3 / 1, which means that for every 3 parts of X, there is 1 part of Y \nThe ratio of mixture X to the total ratio is 3/4. Therefore, mixture X constitutes 75% of the new mixture"} +{"id": "GMAT QUANT_p23_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 23, "page_end": 23, "topic_guess": "fractions_decimals_percents", "text": "or every 3 parts of X, there is 1 part of Y \nThe ratio of mixture X to the total ratio is 3/4. Therefore, mixture X constitutes 75% of the new mixture\n\nHow many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so that the resulting solution is 10 percent salt?\nLet n represent the number of liters of the 15% solution. The amount of salt in the 15% solution [0.15n] plus the amount of salt in the 8% solution [(0.08)(5)] must be equal to the amount of salt in the 10% mixture [0.10 (n + 5)]. Therefore, \n0.15n + 0.08(5) = 0.10(n + 5)\n15n + 40 = 10n + 50\n5n = 10\nn = 2 liters"} +{"id": "GMAT QUANT_p24_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 24, "page_end": 24, "topic_guess": "fractions_decimals_percents", "text": "PERCENTAGES – 3 \nDISCOUNT PROBLEMS \nIf a price is discounted by n percent, then the price becomes (100 – n) percent of the original price\nDiscount is calculated on Marked price and NOT on Cost price\n\nExample # 1\nA certain customer paid $24 for a dress. If that price represented a 25 percent discount on the original price of the dress, what was the original price of the dress? \nSolution # 1\nIf p is the original price of the dress, then 0.75p is the discounted price and 0.75p = $24, or p = $32. The original price of the dress was $32.\n\nExample #2 \nThe price of an item is discounted by 20 percent and then this reduced price is discounted by an additional 30 percent. These two discounts are equal to an overall discount of what percent? \nSolution # 2\nIf p is the original price of the item, then 0.8p is the price after the first discount. The price after the second discount is (0.7)(0.8)p = 0.56p. This represents an overall discount of 44 percent (100% – 56%)"} +{"id": "GMAT QUANT_p24_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 24, "page_end": 24, "topic_guess": "word_problems", "text": "e after the first discount. The price after the second discount is (0.7)(0.8)p = 0.56p. This represents an overall discount of 44 percent (100% – 56%)\n\nSIMPLE INTEREST\nSI = Simple Interest, P = Principal, T = Time period in years, R = Rate of interest per annum\nS. I. = (PTR)/100\nIf simple interest for 2 years is 200, then S.I for 4 years is 400, for 1 year it is 100, for 10 years it is 1000. If S.I. for any length of time is known, then it can be calculated for any other length of time\n\nA man invests $1000 in a bank which pays him 10% p.a. rate of simple interest for 2 years. How much does the man get after 2 years?\n\nS.I. = (1000 ×2 ×10)/100 = 200\nTherefore, amount after 2 years = P + S.I. = 1000 + 200 = 1200\n\nCOMPOUND INTEREST \nA = P × [1 + (R/100)] ^n\n\nIf interest is compounded monthly, divide the annual rate by 12, and multiply n by 12\nIf interest is compounded quarterly, divide the annual rate by 4, and multiply n by 4\nIf interest is compounded semi-annually, divide the annual rate by 2, and multiply n by 2"} +{"id": "GMAT QUANT_p25_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 25, "page_end": 25, "topic_guess": "fractions_decimals_percents", "text": "PERCENTAGES – 4"} +{"id": "GMAT QUANT_p26_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 26, "page_end": 26, "topic_guess": "fractions_decimals_percents", "text": "PERCENTAGES – 5\nAll of the furniture for sale at Al’s Discount Furniture is offered for less than the manufacturer’s suggested retail price. Once a year, Al’s holds a clearance sale. If jamie purchased a certain desk during the sale, did she get a discount of more than 50% of Al’s regular price for the desk? \nAl’s regular price for the desk is 60%, rounded to the nearest percent, of the MSRP of $2000\nThe sale price was $601 less than Al’s regular price for the desk"} +{"id": "GMAT QUANT_p26_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 26, "page_end": 26, "topic_guess": "fractions_decimals_percents", "text": "ular price for the desk is 60%, rounded to the nearest percent, of the MSRP of $2000\nThe sale price was $601 less than Al’s regular price for the desk\n\nIn order to determine the percent discount received by Jamie, we need to know two things: the regular price of the desk and the sale price of the desk. Alternatively, we could calculate the percent discount from the price reduction and either the regular price or the sale price\n\u000b(1) INSUFFICIENT: This statement tells us the regular price of the desk at Al’s, but provides no information about how much Jamie actually paid for the desk during the annual sale.\u000b(2) INSUFFICIENT: This statement tells us how much the price of the desk was reduced during the sale, but provides no information about the regular price. For example, if the regular price was $6010, then the discount was only 10%. On the other hand, if the regular price was $602, then the discount was nearly 100%. \u000b\u000b(1) AND (2) INSUFFICIENT: At first glance, it seems that the statements together provide enough information"} +{"id": "GMAT QUANT_p26_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 26, "page_end": 26, "topic_guess": "statistics", "text": "$602, then the discount was nearly 100%. \u000b\u000b(1) AND (2) INSUFFICIENT: At first glance, it seems that the statements together provide enough information\n\nStatement (1) seems to provide the regular price of the desk, while statement (2) provides the discount in dollars. However, pay attention to the words “rounded to the nearest percent” in statement (1). This indicates that the regular price of the desk at Al’s is 60% of the MSRP, plus or minus 0.5% of the MSRP. Rather than clearly stating that the regular price is (0.60)($2000) = $1200, this statement gives a range of values for the regular price: $1200 plus or minus $10 (0.5% of 2000), or between $1190 and $1210. If the regular price was $1190, then the discount was ($601/$1190) × 100% = 50.5% (you can actually see that this is greater than 50% without calculating). If the regular price was $1210, then the discount was ($601/$1210) × 100% = 49.7% (you can actually see that this is less than 50% without calculating).\u000bThe uncertainty about the regular price means that we cannot answer with certainty whether the discount was more than 50% of the regular price. \u000bThe correct answer is E"} +{"id": "GMAT QUANT_p27_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 27, "page_end": 27, "topic_guess": "fractions_decimals_percents", "text": "COMMON FRACTIONS, DECIMALS AND PERCENTAGE EQUIVALENTS"} +{"id": "GMAT QUANT_p28_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 28, "page_end": 28, "topic_guess": "fractions_decimals_percents", "text": "RATIO AND PROPORTION"} +{"id": "GMAT QUANT_p29_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 29, "page_end": 29, "topic_guess": "fractions_decimals_percents", "text": "RATIO AND PROPORTION \nA ratio expresses a particular relationship between two or more quantities. Here are some examples of ratios: \nThe two partners spend time working in the ratio of 1 to 3. For every 1 hour the first partner works, the second partner works 3 hours.\nThree sisters invest in a certain stock in the ratio of 2 to 3 to 8. For every $2 the first sister invests, the second sister invests $3, and the third sister invests $8\nThe ratio of men to women in the room is 3 to 4. For every 3 men, there are 4 women\nRatios can express a part-part relationship or a part-whole relationship\nA part-part relationship: The ratio of men to women in the office is 3:4\nA part-whole relationship: There are 3 men for every 7 employees\nNotice that if there are only two parts in the whole, you can derive a part-whole ratio from a part-part ratio, and vice versa\nRemember that ratios only express a relationship between two or more items; they do not provide enough information, on their own, to determine the exact quantity for each item. For example, knowing that the ratio of men to women in an office is 3 to 4 does NOT tell us exactly how many men and how many women are in the office. All we know is that the number of men is less than the number of women. However, ratios are surprisingly powerful on Data Sufficiency - they often provide enough information to answer the question\nMultiple ratios: Make a common Term\nIf C:A = 3:2, and C:L = 5:4, what is A:L?\nC : A : L C : A : L\n3 : 2 :  Multiply by 5  15 : 10 : \n5 : : 4  Multiply by 3  15 : : 12 \nOnce you have converted C to make both ratios equal (15), you can combine the ratios\nCombined Ratio C:A:L = 15:10:12, and ratio A:L = 10:12 or 5:6\nIf a/b = c/d, then: \n(a+b)/b = (c+d)/d\n(a – b)/b = (c – d)/d\n(a + b)/(a – b) = (c + d)/(c – d)"} +{"id": "GMAT QUANT_p30_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 30, "page_end": 30, "topic_guess": "general", "text": "EXPONENTS & ROOTS"} +{"id": "GMAT QUANT_p31_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 31, "page_end": 31, "topic_guess": "general", "text": "EXPONENTS"} +{"id": "GMAT QUANT_p32_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 32, "page_end": 32, "topic_guess": "algebra", "text": "EXPONENTS\nSimplifying exponential expressions\nAlways try to simplify exponential expressions when they have the same base or the same exponent. You can only simplify exponential expressions that are linked by multiplication or division. When expressions with the same base are linked by a sum, you cannot simplify but you can factor the expression:\n7² + 7³ = 7².(7 + 1) = 49.8\nExample: what is the largest prime factor of 4²¹ + 4²2 + 4²³?\nAnswer: 4²¹ + 42² + 4²³ = 4²¹(1 + 4 + 16) = 4²¹.(21) = 4²¹(3.7). The largest prime factor of 4²¹ + 422 + 4²³ is 7\n\n*** When both sides of the equation are broken down to the product of prime bases, the respective exponents of like bases must be equal"} +{"id": "GMAT QUANT_p33_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 33, "page_end": 33, "topic_guess": "algebra", "text": "ROOTS\nA ROOT is the inverse operation of raising a number to an exponent, and answers the question: Which number do I multiply by itself n times in order to get a product of b ?\n\nEVEN ROOTS ONLY HAVE ONE SOLUTION – A POSITIVE VALUE\nUnlike even exponents, which yield both a positive and a negative solution, even roots have only one solution. √4 can only be 2, and not -2\nA root can only have a negative value if it is an odd root, and the base is negative\n\nSIMPLIFYING A ROOT\nGMAT very often tries to trick you by giving a root linked by addition where it is tempting to simplify the terms, for example: √(25 + 16). It is tempting to think that this will result into 5 + 4, but you can only simplify roots when the terms inside/outside are linked by multiplication or division\nSimplifying radicals in the denominator with conjugate radical expressions is very useful on challenging GMAT radical questions\nExamples of roots simplification:\n√25.16 = √25 . √16 = 5.4\n√50 . √18 = √50.18 = √900 = 30\n√144:16 = √144 : √16 = 12/4 = 3\n√25+16 = √41  you cannot simplify this one\n2√7 + 3√7 = 5√7"} +{"id": "GMAT QUANT_p33_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 33, "page_end": 33, "topic_guess": "algebra", "text": "n:\n√25.16 = √25 . √16 = 5.4\n√50 . √18 = √50.18 = √900 = 30\n√144:16 = √144 : √16 = 12/4 = 3\n√25+16 = √41  you cannot simplify this one\n2√7 + 3√7 = 5√7\n\nKNOWING COMMON ROOTS\nGMAT requires you to know all the perfect square roots from 1 to 30 and also the following imperfect rules:\n√2 ≈ 1.4\n√3 ≈ 1.7\n√5 ≈ 2.25\n\nSUM OF THE ROOTS OF AN EQUATION ax2+bx+c=0 IS (-B/A)"} +{"id": "GMAT QUANT_p34_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 34, "page_end": 34, "topic_guess": "general", "text": "COMMON EXPONENTS AND ROOTS"} +{"id": "GMAT QUANT_p35_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 35, "page_end": 35, "topic_guess": "word_problems", "text": "RATE AND WORK\nBasic motion problems (speed, distance, time)\nAverage rate problems\nSimultaneous motion problems\nWork problems\nPopulation problems"} +{"id": "GMAT QUANT_p36_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 36, "page_end": 36, "topic_guess": "word_problems", "text": "RATE, DISTANCE TIME \nRATE\nRate is expressed as a ratio of distance and time, with two corresponding units. Some examples of rates include: 30 miles per hour, 10 meters/second etc.\n\nDISTANCE \nDistance is expressed using a unit of distance. Some examples of distances include: 18 miles, 20 meters, 100 kilometres\n\nTIME\nTime is expressed using a unit of time. Some examples of times include: 6 hours, 23 seconds, 5 months, etc.\n\nRATE = DISTANCE / TIME \nRATE * TIME = DISTANCE\n\nSpeed and time are inversely proportional to each other. Therefore, if a man increases his speed by 3/2, then the time taken will become 2/3*original time\nIf three men cover the same distance with speeds in the ratio a : b : c, the times taken by these three will be in the ratio 1/a : 1/b : 1/c respectively"} +{"id": "GMAT QUANT_p37_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 37, "page_end": 37, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS – SPEED RELATIONS"} +{"id": "GMAT QUANT_p38_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 38, "page_end": 38, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS"} +{"id": "GMAT QUANT_p39_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 39, "page_end": 39, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS – THE KISS/CRASH"} +{"id": "GMAT QUANT_p40_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 40, "page_end": 40, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS – CATCHING UP/OVERTAKING"} +{"id": "GMAT QUANT_p41_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 41, "page_end": 41, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS – ROUND TRIP"} +{"id": "GMAT QUANT_p42_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 42, "page_end": 42, "topic_guess": "word_problems", "text": "MULTIPLE RATE, DISTANCE TIME PROBLEMS – SAMPLE SITUATIONS"} +{"id": "GMAT QUANT_p43_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 43, "page_end": 43, "topic_guess": "general", "text": "MISCELLANEOUS PROBLEMS \nBOAT PROBLEMS\n\nIf the speed of a boat (or man) in still water is X km/hour, and the speed of the stream (or current) is Y km/hour, then:\nSpeed of boat with the stream (or Downstream or D/S) = (X + Y) km/hour\nSpeed of boat against the stream (or upstream or U/S) = (X – Y) km/hour\n\nX = [(X + Y) + (X – Y)] / 2 and Y = [(X + Y) – (X – Y)] / 2\nBoat’s speed in still water = [Speed downstream + Speed upstream] / 2\nSpeed of current = [Speed downstream – Speed upstream] / 2"} +{"id": "GMAT QUANT_p44_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 44, "page_end": 44, "topic_guess": "word_problems", "text": "AVERAGE RATE / RELATIVE SPEED \nIf an object moves the same distance twice, but at different rates, then the average rate will NEVER be the average of the two rates given for the two legs of the journey. In fact, because the object spends more time traveling at the slower rate, the average rate will be closer to the slower of the two rates than to the faster\nIt is not necessary to know the total distance travelled or the total time taken while computing the average speed\nWhen two objects travel in the same direction, the time to catch up/overtake will be = lead distance/ difference of speeds (note: in order for A to overtake B, A needs to be traveling at a higher speed than B otherwise A will never catch up) \nWhen two objects travel in different directions, time to meet will be = lead distance/ sum of speeds \nAVERAGE RATE = TOTAL DISTANCE/ TOTAL TIME\nOR\nAVERAGE SPEED = (2*SPEED OF A*SPEED OF B) / (SPEED A + B)  only to be used when the distance travelled is the same"} +{"id": "GMAT QUANT_p45_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 45, "page_end": 45, "topic_guess": "word_problems", "text": "SPEED , DISTANCE, TIME PROBLEMS"} +{"id": "GMAT QUANT_p46_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 46, "page_end": 46, "topic_guess": "word_problems", "text": "WORK\n\nIn distance problems, if the rate (speed) is known, it will normally be given to you as a ready-to-use number. In work problems, though, you will usually have to figure out the rate from some given information about how many jobs the agent can complete in a given amount of time\nFor example, if Oscar can perform one surgery in 1.5 hours, the work rate = # of given jobs/ given amount of time i.e., 1/1.5 hours, which is equal to 2/3 surgeries per hour"} +{"id": "GMAT QUANT_p46_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 46, "page_end": 46, "topic_guess": "word_problems", "text": "an perform one surgery in 1.5 hours, the work rate = # of given jobs/ given amount of time i.e., 1/1.5 hours, which is equal to 2/3 surgeries per hour\n\nWORKING TOGETHER – ADD WORK RATES \nIf A can finish a job alone in x days and B can finish the same job in y days alone, then\nOne-day work of both A and B is (x + y)/x*y\nNumber of days A and B take to finish the same job working together is: (x*y) / (x + y)\nException: one agent undoes the other’s work, like a pump putting water into a tank and another drawing water out. Rate will be x – y\nExample: Working alone, A finish a job in 2 days and B can finish a job in 3 days. If A and B work together, in how many days will they finish the work?\nA’s rate = 1/ 2 \nB’s rate = 1/ 3\nIn one day, A and B finish 5 / 6 of the job; therefore, they take 6/ 5 days to complete the job\nExample: Four men working together all day, can finish a piece of work in 11 days; but two of them having other engagements can work only one half–time and quarter time respectively. How long will it take them to complete the work ?\nEach man will take 11 x 4 = 44 days to complete the work. If one man works half day/day he will take 44 x 2 = 88 days to finish the work. Similarly, a man working quarter day/day will take 44 x 4 = 176 days to finish the work. When these work together they will require 1 / [ (1/44) + (1/44) + (1/88) + (1/176) ] = 16 days\nMen * hours * days \nExample: If 2 men can finish a job in 3 days working 4 hours per day, then how many days will 3 men working 2 hours per day take to finish the same job?\nSolution: In the first set the product is: 2 men ×3 days× 4 hours. This must remain constant in the second set in which we need to find number of days: 2 men × 3 days × 4 hours = 3 men × N days × 2 hours\n2 × 3 × 4 = 3 × N ×2\nTherefore, N = 4 days"} +{"id": "GMAT QUANT_p46_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 46, "page_end": 46, "topic_guess": "word_problems", "text": "the second set in which we need to find number of days: 2 men × 3 days × 4 hours = 3 men × N days × 2 hours\n2 × 3 × 4 = 3 × N ×2\nTherefore, N = 4 days\n\nIf a man can do a piece of work in N days (or hours or any other unit of time), then the work done by him in one day will be 1/N of the total work \nIf A is twice as good a workman as B, then A will take half the time B takes to finish a piece of work"} +{"id": "GMAT QUANT_p47_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 47, "page_end": 47, "topic_guess": "general", "text": "WORK – 2"} +{"id": "GMAT QUANT_p48_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 48, "page_end": 48, "topic_guess": "number_theory", "text": "NUMBER PROPERTIES OF INTEGERS\nIMPORTANT CONCEPTS TO ALWAYS KEEP IN MIND \nAn integer is any number in the following set – {..-3,-2,-1,0,1,2,3..}\nAn integer can be either positive or negative, even or odd, a prime or a consonant (0 and 1 are neither prime nor consonant) \nNever assume a number to be an integer unless EXPLICITLY stated in the problem\nIf n2 is an integer, it is not necessary that n is also an integer. For example, the square of √2 is 2.\n\nImpact of arithmetic operations on integers \nWhen integers are added to, subtracted from, or multiplied with each other, the result is ALWAYS an integer. However, during division, it is not necessary that the result will be an integer – when an even number is divided by an odd number, it may result in a non – integer (Example: 10/4). Similarly, an odd integer divided by an odd integer may result in a non – integer (Example: 11/3). An odd integer divided by an even integer ALWAYS results in a non – integer"} +{"id": "GMAT QUANT_p48_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 48, "page_end": 48, "topic_guess": "number_theory", "text": "r divided by an odd integer may result in a non – integer (Example: 11/3). An odd integer divided by an even integer ALWAYS results in a non – integer\n\nDivisibility \nThe product of 2 even numbers will always be divisible by 4\nThe product of x consecutive integers will always be divisible by x and x!. For example, x*(x +1)* (x+2) will be divisible by 3\nThe sum of x consecutive integers \nIf x is odd, the sum of the integers is always divisible by x. Example, 2+3+4 = 9, which is divisible by 3 \nIf x is even, the sum of the integers is never divisible by x. Example, 5+6+7+8 = 26, which is not divisible by 4\n\nPrime factorization – Lowest Common Multiple (LCM), Greatest Common Factor (GCF)\nLCM: The smallest number that has all given numbers as a factor\nGCF: The largest number which is a factor of all given numbers\nProduct of two numbers = LCM * GCF"} +{"id": "GMAT QUANT_p49_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 49, "page_end": 49, "topic_guess": "general", "text": "PROPERTIES OF ZERO"} +{"id": "GMAT QUANT_p50_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 50, "page_end": 50, "topic_guess": "number_theory", "text": "PRIME NUMBERS \nPOSITIVE INTEGER greater than 1 that has EXACTLY TWO different positive factors – ONE AND ITSELF\n\nNote: If a positive integer greater than or equal to 2 has only two factors, then the integer must be a prime number \nTWO is the smallest and only EVEN prime number\n\nIf the product of two primes is even or if the sum of two primes is odd, one of the primes has to be 2\nConversely, if you know that 2 CANNOT be one of the primes in the sum, then the sum of the two primes must be even\nMethod to determine a prime number\nFind approximate square root of the number. Then check if all the prime numbers below the square root are factors of the given number. If none are then the number is prime else not\nFor example: the approximate square root of 91 is 10, and the prime numbers below 10 are 2,3,5,7. 91 is not divisible by 2,3 or 5, but it is divisible by 7. Hence, it is not a prime number\nFirst 100 prime numbers: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97\nONE is NOT a prime number\nEXAMPLES\n\nWhat is the value of integer x?\n\n(1) x has exactly 2 factors.\n(2) When x is divided by 2, the remainder is 0"} +{"id": "GMAT QUANT_p50_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 50, "page_end": 50, "topic_guess": "number_theory", "text": "89,97\nONE is NOT a prime number\nEXAMPLES\n\nWhat is the value of integer x?\n\n(1) x has exactly 2 factors.\n(2) When x is divided by 2, the remainder is 0\n\nStatement (1) indicates that x is prime, because it has only 2 factors. This statement is insufficient by itself, since there are infinitely many prime numbers\nStatement (2) indicates that 2 divides evenly into x, meaning that x is even; that is also insufficient by itself\nTaken together, however, the two statements reveal that x must be an even prime-and the only even prime number is 2. The answer is (C): BOTH statements TOGETHER are sufficient"} +{"id": "GMAT QUANT_p51_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 51, "page_end": 51, "topic_guess": "number_theory", "text": "DIVISIBILITY RULES \nFor any integer n, n3 – n is divisible by 3, n5 – n is divisible by 5, n11 – n is divisible by 11, n13 – n is divisible by 13. In general, if p is a prime number then for any whole number a, ap – a is divisible by p. For example, 53 – 5 is divisible by 3: (53 – 5) = 120, which is divisible by 3\n\nFACTORIALS: The factorial of N, symbolized by N!, is the product of all integers from 1 up to and including N. For instance, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. Because it is the product of all the integers from 1 to N, any factorial N! must be divisible by all integers from 1 to N. Another way of saying this is that N! is a multiple of all the integers from 1 to N"} +{"id": "GMAT QUANT_p52_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 52, "page_end": 52, "topic_guess": "number_theory", "text": "PRIME FACTORIZATION\nCOMMON FACTOR \nBreak down both numbers to their prime factors to see what factors they have in common. Multiply shared prime factors to find all common factors\n\nWhat factors greater than 1 do 135 and 225 have in common?\n135 = 3 x 3 x 3 x 5\n225 = 3 x 3 x 5 x 5\nBoth share 3 x 3 x 5 in common—find all combinations of these numbers: 3 x 3 = 9; 3 x 5 = 15; 3 x 3 x 5 = 45\n\nGREATEST COMMON FACTOR (GCF)\nThe number that contains all the factors common to both numbers (i.e., biggest number that will divide into both the numbers)\n\nNote: If the numbers do not share any common factors, then the GCF is 1\n\nThe GCF for two consecutive numbers is always 1. However, just because the GCF is 1 doesn’t guarantee that the numbers are consecutive\n\nExample: \nThe prime factorizations for 2940 & 3150 are:\n\n2940 = 2 x 2 x 3 x 5 x 7 x 7\n3150 = 2 x 3 x 3 x 5 x 5 x 7\n\nTherefore, GCF is 2 x 3 x 5 x 7 = 210\n\nLOWEST COMMON MULTIPLE (LCM)\nThe LCM is the smallest number that contains both numbers as factors i.e. the smallest number that is a multiple of both these values\n\nL.C.M of two numbers x and y is maximum power of the prime factors in x and y"} +{"id": "GMAT QUANT_p52_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 52, "page_end": 52, "topic_guess": "number_theory", "text": "actors i.e. the smallest number that is a multiple of both these values\n\nL.C.M of two numbers x and y is maximum power of the prime factors in x and y\n\nExample:\nThe prime factorizations for 2940 & 3150 are:\n\n2940 = 2 x 2 x 3 x 5 x 7 x 7\n3150 = 2 x 3 x 3 x 5 x 5 x 7\n\nTherefore, LCM is 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 = 44,100\n\nLCM * GCF = PRODUCT OF TWO NUMBERS (DOES NOT WORK FOR MORE THAN TWO NUMBERS)\n\nGCF of fractions = GCF of numerators ÷ LCM of denominators\n\nLCM of fractions = LCM of numerators ÷ GCF of denominators"} +{"id": "GMAT QUANT_p53_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 53, "page_end": 53, "topic_guess": "number_theory", "text": "PRIME FACTORIZATION AND DIVISIBILITY – EXAMPLES"} +{"id": "GMAT QUANT_p54_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 54, "page_end": 54, "topic_guess": "number_theory", "text": "FACTORS AND MULTIPLES \nFACTORS/DIVISORS\nX IS A FACTOR OF Y\n\nEssentially, a factor is a POSITIVE integer that divides evenly into an integer. Ex: 1,2,4 and 8 are all the factors of 8\n\nOne is a factor of all numbers\n\nThe largest factor of a number is the number itself\n\nMULTIPLES\nY IS A MULTIPLE OF X\n\nEssentially, a multiple of an integer is formed by multiplying that integer by any integer, so 8, 16,24, and 32 are some of the multiples of 8\n\nFor the purpose of the GMAT, multiples are always POSITIVE\n\n** Divisibility can be stated in multiple ways. For example, 12 items can be shared among 3 people so that each person has the same number of items. This just means that 12 (y) is divisible by 3 (x), and that each of the individuals has 4 items each (n)\n\nThe smallest multiple of a number is the number itself\n\nY = X*INTEGER\nWhere x and y are both integers, and x is not equal to zero"} +{"id": "GMAT QUANT_p55_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 55, "page_end": 55, "topic_guess": "number_theory", "text": "FACTORS AND MULTIPLES – IMPORTANT CONCEPTS \nPERFECT SQUARES\nAlways have an odd number of factors. Therefore, if an integer has an odd number of factors, it has to be a perfect square\nThe prime factorization of a perfect square contains only even powers of primes. This is because perfect squares are formed from the product of two copies of the same prime factors. For example, 902 = (2*32*5) * (2*32*5). By contrast, if a number's prime factorization contains any odd powers of primes, then the number is not a perfect square. For instance, 132,300 = 22 x 33 X 52 X 72 is not a perfect square, because the 3 is raised to an odd power. If this number is multiplied by 3, then the result, 396,900, is a perfect square: 396,900 = 22 x 34 X 52x 72\nThe sum of all distinct factors of a perfect square is always odd\n\nPERFECT CUBES\n- If a number is a perfect cube, then it is formed from three identical sets of primes, so all the powers of primes are multiples of 3 in the factorization of a perfect cube. For instance, 903 = (2 X 32 x 5) * (2 X 32 x 5) * (2 X 32 x 5) = 23 x 36 x 53"} +{"id": "GMAT QUANT_p55_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 55, "page_end": 55, "topic_guess": "number_theory", "text": "ers of primes are multiples of 3 in the factorization of a perfect cube. For instance, 903 = (2 X 32 x 5) * (2 X 32 x 5) * (2 X 32 x 5) = 23 x 36 x 53\n\nTHE SUM OR DIFFERENCE OF TWO MULTIPLES OF A NUMBER IS ALSO A MULTIPLE OF THAT NUMBER\nAlgebraically, if N is a divisor of X and of Y, then N is a divisor of (X+Y) \nExample: 55 and 155 are both multiples of 5. Thus, 55 + 155 = 210 and 155 – 55 = 100 are also multiples of 5\nIf you add two non-multiples of N, the result could be either a multiple of N or a non-multiple of N\n19 + 13 = 32 (Non-multiple of 3) + (Non-multiple of 3) = (Non-multiple of 3)\n19 + 14 = 33 (Non-multiple of 3) + (Non-multiple of 3) = (Multiple of 3)\nIf you add a multiple of N to a non-multiple of N, the result is a non-multiple of N\n\nIF A NUMBER IS A FACTOR OF TWO OTHER NUMBERS, IT IS A FACTOR OF THEIR SUM OR DIFFERENCE\nExample: 4 is a factor of 64 and 44. Thus, 4 is a factor of 64 + 44 = 108 and 64 – 44 = 20"} +{"id": "GMAT QUANT_p55_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 55, "page_end": 55, "topic_guess": "number_theory", "text": "WO OTHER NUMBERS, IT IS A FACTOR OF THEIR SUM OR DIFFERENCE\nExample: 4 is a factor of 64 and 44. Thus, 4 is a factor of 64 + 44 = 108 and 64 – 44 = 20\n\nFACTOR FOUNDATION RULE \nIf a is a factor of b, and b is a factor of c, then a is actor of c\nAll the factors of both x and y must be factors of the product, x*y\nExample:\nGiven that the integer n is divisible by 3, 7, and 11, what other numbers must be divisors (factors) of n? \nSince we know that 3, 7, and 11 are prime factors of n, we know that n must also be divisible by all the possible products of the primes (3, 7 and 11): 21, 33, 77, and 231 (given that n > 231)\n\nIN ANY LIST OF N CONSECUTIVE INTEGERS, EXACTLY ONE OF THE INTEGERS IS A MULTIPLE OF N. AS A RESULT, THE PRODUCT OF N CONSECUTIVE INTEGERS IS ALWAYS A MULTIPLE OF N (OR, IT IS DIVISIBLE BY N)\nIn any list of three consecutive integers, one number (33) is a multiple of 3  33, 34, 35\nProducts of 113 × 114 × 115 and (y + 5)(y + 6)(y + 7) are both divisible by 3"} +{"id": "GMAT QUANT_p55_c4", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 55, "page_end": 55, "topic_guess": "number_theory", "text": "ee consecutive integers, one number (33) is a multiple of 3  33, 34, 35\nProducts of 113 × 114 × 115 and (y + 5)(y + 6)(y + 7) are both divisible by 3\n\nTHE NUMBER OF FACTORS A NUMBER HAS (INCLUDING 1 AND THE NUMBER ITSELF) CAN BE COMPUTED BY FOLLOWING THE STEPS DETAILED BELOW: \nFirst, determine all the prime factors of the number through LCM\nSecond, determine the number of copies of each of the prime factors and express them in the following format - p^a * q^b * .... *r^k \nThird, the number of factors would be (a+1)*(b+1)*...(k+1) \n Example\nIf your number is 360, the factors are 8, 9 and 5\n8*9*5 = 2^3 * 3^2 * 5^1 \nThe number of factors is (3+1)*(2+1)*(1+1) = 24"} +{"id": "GMAT QUANT_p56_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 56, "page_end": 56, "topic_guess": "number_theory", "text": "QUOTIENTS AND REMAINDERS\nY = (QUOTIENT * DIVISOR) + REMAINDER"} +{"id": "GMAT QUANT_p57_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 57, "page_end": 57, "topic_guess": "number_theory", "text": "QUOTIENTS AND REMAINDERS"} +{"id": "GMAT QUANT_p58_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 58, "page_end": 58, "topic_guess": "number_theory", "text": "QUOTIENTS AND REMAINDERS"} +{"id": "GMAT QUANT_p59_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 59, "page_end": 59, "topic_guess": "number_theory", "text": "QUOTIENTS AND REMAINDERS"} +{"id": "GMAT QUANT_p60_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 60, "page_end": 60, "topic_guess": "number_theory", "text": "EVEN AND ODD INTEGERS \nONLY integers can be even/odd\nAn odd integer can be expressed as 2n + 1 and an even integer can be expressed as 2n \nSum of two different prime numbers will ALWAYS be EVEN, unless one of the numbers is 2 \nIf two even integers are multiplied, the result will be divisible by 4, and if three even integers are multiplied, the result will be divisible by 8 etc. \nExample: If x is even, and N = (x)(x+1)(x+2), is N divisible by 24?\nYes. The terms x and x+2 must each have a factor of 2, and the product (x)(x+1)(x+2) must have factors 1, 2, 3 and 4 Therefore, the prime factorization of N must be a multiple of 1*2*3*2*2 = 24"} +{"id": "GMAT QUANT_p61_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 61, "page_end": 61, "topic_guess": "number_theory", "text": "POSITIVE AND NEGATIVE INTEGERS\nZERO is neither positive nor negative – be very careful !"} +{"id": "GMAT QUANT_p62_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 62, "page_end": 62, "topic_guess": "number_theory", "text": "CONSECUTIVE INTEGERS \nSUM OF INTEGERS IN A SET = AVERAGE * # OF ITEMS\n\nExample\n\nWhat is the sum of all integers from 20 to 100, inclusive? \nAverage the first and last term to find the precise \"middle\" of the set  (100+20)/2 = 60\nCount the number of terms: 100 - 20 = 80, plus 1 yields 81\nMultiply the \"middle\" number by the number of terms to find the sum: 60 x 81 = 4,860"} +{"id": "GMAT QUANT_p63_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 63, "page_end": 63, "topic_guess": "algebra", "text": "EQUATIONS"} +{"id": "GMAT QUANT_p64_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 64, "page_end": 64, "topic_guess": "algebra", "text": "BASIC EQUATIONS (THOSE WITHOUT EXPONENTS)\nSIMULTANEOUS EQUATIONS – THREE VARIABLES \nLook for ways to simplify the work. Look especially for shortcuts or symmetries in the form of the equations to reduce the number of steps needed to solve the system\nExample: What is the sum of x, y and z?\nx + y=8; x + z=1; y + z=7\nIn this case, DO NOT try to solve for x, y, and z individually. Instead, notice the symmetry of the equations-each one adds exactly two of the variables-and add them all together. Therefore, 2x + 2y + 2z = 26; and x + y+ z = 13"} +{"id": "GMAT QUANT_p64_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 64, "page_end": 64, "topic_guess": "algebra", "text": "ce the symmetry of the equations-each one adds exactly two of the variables-and add them all together. Therefore, 2x + 2y + 2z = 26; and x + y+ z = 13\n\nMISMATCH PROBLEMS \nMISMATCH problems, which are particularly common on the Data Sufficiency portion of the test, are those in which the number of unknown variables does NOT correspond to the number of given equations\nA MASTER RULE for determining whether 2 equations involving 2 variables (say, x and y) will be sufficient to solve for the variables is this: (1) If both of the equations are linear-that is, if there are no squared terms (such as x2 or y2) and no xy terms-the equations will be sufficient UNLESS the two equations are mathematically identical (e.g., x +y = 10 is identical to 2x + 2y = 20) and (2) If there are ANY non linear terms in either of the equations (such as x2, y2, x/y, or xy), there will USUALLY be two (or more) different solutions for each of the variables and the equations will not be sufficient.\nExample:\nWhat is X?\n3x/ (3y + 5z) = 8 ; 6y + 10 z = 18\nIt is tempting to say that these two equations are not sufficient to solve for x, since there are 3 variables and only 2 equations. However, the question does NOT ask you to solve for all three variables. It only asks you to solve for x, which IS possible. First, get the x term on one side of the equation: 3x = 8(3y + 5z). Then, notice that the second equation gives us a value for 3y + 5z, which we can substitute into the first equation in order to solve for x. Thus, BOTH statements TOGETHER are sufficient"} +{"id": "GMAT QUANT_p64_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 64, "page_end": 64, "topic_guess": "algebra", "text": "n gives us a value for 3y + 5z, which we can substitute into the first equation in order to solve for x. Thus, BOTH statements TOGETHER are sufficient\n\nCOMBO PROBLEMS – MANIPULATION \nThe GMAT often asks you to solve for a combination of variables, called COMBO problems. For example, a question might ask, what is the value of x +y?\nIn these cases, since you are not asked to solve for one specific variable, you should generally NOT try to solve for the individual variables right away. Instead, you should try to manipulate the given equation(s) so that the COMBO is isolated on one side of the equation. There are four easy manipulations that are the key to solving most COMBO problems:\nM: Multiply or divide the whole equation by a certain number\nA: Add or subtract a number on both sides of the equation\nD: Distribute or factor an expression on ONE side of the equation.\nS: Square or un square both sides of the equation\nCombo problems occur most frequently in Data Sufficiency. Whenever you detect this is a Data Sufficiency question that may involve a combo, you should try to manipulate the given equation(s) in either the question or the statement, so that the combo is isolated on one side of the equation. Then, if the other side of an equation from a statement contains a VALUE, that equation is SUFFICIENT. If the other side of the equation contains a VARIABLE EXPRESSION, that equation is NOT SUFFICIENT"} +{"id": "GMAT QUANT_p65_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 65, "page_end": 65, "topic_guess": "algebra", "text": "ABSOLUTE VALUE EQUATIONS \nAbsolute value refers to the POSITIVE value of the expression within the absolute value brackets. Equations that involve absolute value generally have TWO SOLUTIONS. In other words, there are TWO numbers that the variable could equal in order to make the equation true. The reason is that the value of the expression inside the absolute value brackets could be POSITIVE OR NEGATIVE. For instance, if we know |x| = 5, then x could be either 5 or -5, and the equation would still be true\nIf |x| > x then x must be negative\nThe following three-step method should be used when solving for a variable expression inside absolute value brackets. Consider the following example: Solve for w, given that 12 + |w – 4| = 30.\nIsolate the expression within the absolute value brackets: |w – 4| = 18 \nRemove the absolute value brackets and solve the equation for 2 different cases (positive and negative)\nw – 4 = 18; w = 22\n- (w – 4 ) = - 18; - w + 4 = 18; - w = 14; w = -14 \n3. Check to see whether each solution is valid by putting each one back into the original equation and verifying that the two sides of the equation are in fact equal. The possibility of a failed solution is a peculiarity of absolute value equations. For most other types of equations, it is good to check your solutions, but doing so is less critical"} +{"id": "GMAT QUANT_p66_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 66, "page_end": 66, "topic_guess": "algebra", "text": "EXPONENTIAL AND QUADRATIC EQUATIONS"} +{"id": "GMAT QUANT_p67_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 67, "page_end": 67, "topic_guess": "general", "text": "INEQUALITIES"} +{"id": "GMAT QUANT_p68_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 68, "page_end": 68, "topic_guess": "general", "text": "INEQUALITIES"} +{"id": "GMAT QUANT_p69_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 69, "page_end": 69, "topic_guess": "statistics", "text": "INEQUALITIES\nCOMBINING INEQUALITIES \nMany GMAT inequality problems involve more than one inequality. To solve such problems, you may need to convert several inequalities to a compound inequality, which is a series of inequalities strung together\nFor example, If x > 8, x < 17, and x + 5 < 19, what is the range of possible values for x?\nFirst, solve any inequalities that need to be solved. In this example, only the last inequality needs to be solved: x < 14 \nSecond, simplify the inequalities so that all the inequality symbols point in the same direction, preferably to the left (less than): 8 < x, x <17, x < 14\nThird, combine the inequalities by taking the more limiting upper and lower extremes: 8 < x < 14"} +{"id": "GMAT QUANT_p69_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 69, "page_end": 69, "topic_guess": "algebra", "text": "ferably to the left (less than): 8 < x, x <17, x < 14\nThird, combine the inequalities by taking the more limiting upper and lower extremes: 8 < x < 14\n\nAnother helpful approach is to combine inequalities by adding the inequalities together. However, note that we should never subtract or divide two inequalities, and can only multiply inequalities together under certain circumstances \nFor example, is mn < 10 given that (1) m < 2 (2) n < 5\nIt is tempting to multiply these two statements together and conclude that mn < 10. That would be a mistake, however, because both m and n could be negative numbers that yield a number larger than 10 when multiplied together. For example, if m = -2 and n = -6, then mn = 12, which is greater than 10. Since you can find cases with mn < 10 and cases with mn > 10, the correct answer is (E): The two statements together are INSUFFICIENT\nConsider the following variation: if both m and n are positive, is mn < 10 given that (1) m < 2 (2) n < 5 . Since the variables are positive, we can multiply these inequalities together and conclude that mn < 10. The correct answer is (C)"} +{"id": "GMAT QUANT_p69_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 69, "page_end": 69, "topic_guess": "algebra", "text": ") m < 2 (2) n < 5 . Since the variables are positive, we can multiply these inequalities together and conclude that mn < 10. The correct answer is (C)\n\nMANIPULATING COMPOUND INEQUALITIES\nSometimes a problem with compound inequalities will require you to manipulate the inequalities in order to solve the problem. You can perform operations on a compound inequality as long as you remember to perform those operations on every term in the inequality, not just the outside terms.\nFor example, consider the equation x + 3 d and (2) ab2 – b > b2c – d \nThis is a multiple variable inequality problem, so you must solve it by doing algebraic manipulations on the inequalities\n (1) INSUFFICIENT: Statement (1) relates b to d, while giving us no knowledge about a and c\n(2) INSUFFICIENT: Statement (2) does give a relationship between a and c, but it still depends on the values of b and d. One way to see this clearly is by realizing that only the right side of the equation contains the variable d. Perhaps ab2 – b is greater than b2c – d simply because of the magnitude of d. Therefore there is no way to draw any conclusions about the relationship between a and c.\n(1) AND (2) SUFFICIENT: By adding the two inequalities from statements (1) and (2) together, we can come to the conclusion that a > c. Two inequalities can always be added together as long as the direction of the inequality signs is the same:\u000b\u000bab2 – b > b2c – d\u000b(+)     b >  d           \u000b-----------------------\u000b      ab2 > b2c\u000b\u000bNow divide both sides by b2. Since b2 is always positive, you don't have to worry about reversing the direction of the inequality. The final result: a > c"} +{"id": "GMAT QUANT_p70_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 70, "page_end": 70, "topic_guess": "algebra", "text": "INEQUALITIES – USING EXTREME VALUES \nOne effective technique for solving GMAT inequality problems is to focus on the EXTREME VALUES of a given inequality. This is particularly helpful when solving the following types of inequality problems: \nProblems with multiple inequalities where the question involves the potential range of values for variables in the problem \nProblems involving both equations and inequalities"} +{"id": "GMAT QUANT_p71_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 71, "page_end": 71, "topic_guess": "algebra", "text": "INEQUALITIES AND ABSOLUTE VALUES \nSolving inequalities \n√x2 = |x|\nIt is often helpful to try to visualize the problem with a number line. For example, take the inequality |x| = 5. One way to understand this inequality is to say \"x must be less than 5 units from zero on the number line.\" Indeed, one interpretation of absolute value is simply distance on the number line. For a simple absolute value expression such as lx|, we are evaluating distance from zero. The range therefore, would be -5 < x < 5 \nEquations involving absolute value require you to consider two scenarios: one where the expression inside the absolute value brackets is positive, and one where the expression is negative. The same is true for inequalities. Note that you should never change I x – 5 |to x + 5. Remember, when you drop the absolute value signs, you either leave the expression alone or enclose the ENTIRE expression in parentheses and put a negative sign in front. \nExample: Given that | x – 2 | < 5, what is the range of possible values of x? \nTo work out the FIRST scenario, we simply remove the absolute value brackets and solve: x – 2 < 5 ; x < 7\nTo work out the SECOND scenario, we reverse the signs of the terms inside the absolute value brackets, remove the brackets, and solve again: - (x - 2) < 5 ; - x + 2 < 5 ; - x < 3 ; x > - 3 \nTherefore, range is -3 < x < 7 \niii. Inequalities with two absolute value expressions \nBecause there are two absolute value expressions, each of which yields two algebraic cases, it seems that we need to test four cases overall: positive/positive, positive/negative, negative/positive, and negative/negative\nExample: Given that | x – 2 | = |2x – 3 |, what is the range of possible values of x? \nThe positive/positive case: (x- 2) = (2x- 3)\nThe positive/negative case: (x - 2) = -(2x - 3)\nThe negative/positive case: - (x - 2) = (2x- 3)\nThe negative/negative case: - (x- 2) = - (2x- 3)"} +{"id": "GMAT QUANT_p72_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 72, "page_end": 72, "topic_guess": "general", "text": "INEQUALITIES AND ABSOLUTE VALUES – EXAMPLES"} +{"id": "GMAT QUANT_p73_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 73, "page_end": 73, "topic_guess": "general", "text": "INEQUALITIES AND ABSOLUTE VALUES – EXAMPLES"} +{"id": "GMAT QUANT_p74_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 74, "page_end": 74, "topic_guess": "sequences_patterns", "text": "SEQUENCES AND FUNCTIONS\nSequences may be of two major types – arithmetic or geometric \nArithmetic progression is a series of numbers in which every term after the first can be derived from the immediately preceding term by adding to it a fixed quantity called common difference (d) \nGeometric progression is a series of numbers in which each term is formed from the preceding by multiplying it by a constant factor ( r ). The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it\nTo answer a question about a sequence, we not only need to know the rule the sequence follows, but also at least one term in the sequence"} +{"id": "GMAT QUANT_p75_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 75, "page_end": 75, "topic_guess": "sequences_patterns", "text": "PROGRESSIONS AND SEQUENCES – ARITHMETIC PROGRESSION\n\nArithmetic progression is a series of numbers in which every term after the first can be derived from the immediately preceding term by adding to it a fixed quantity called common difference (d) \na, a + d, a + 2d, a + 3d are in Arithmetic Progression \nThree numbers in an A.P should be taken as a – d, a, a+ d \nFour numbers in an A.P. should be taken as a – 3d, a – d, a + d, a + 3d\nA series of quantities is said to be in harmonic progression when their reciprocals are in A.P"} +{"id": "GMAT QUANT_p76_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 76, "page_end": 76, "topic_guess": "sequences_patterns", "text": "PROGRESSIONS AND SEQUENCES – GEOMETRIC PROGRESSION \nGeometric progression is a series of numbers in which each term is formed from the preceding by multiplying it by a constant factor ( r ). The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it. For example, in the series 3,9,27,81 and 243, the factor is 3\nThree numbers in a G.P. should be taken as a/r, a, ar\nFour numbers in a G.P. should be taken as a/r3, a/r, ar, ar3"} +{"id": "GMAT QUANT_p77_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 77, "page_end": 77, "topic_guess": "coordinate_geometry", "text": "FUNCTIONS\nThe \"domain\" of a function indicates the possible inputs and the \"range\" of a function indicates the possible outputs. For instance, the function f(x) = x2 can take any input but never produces a negative number. So the domain is all numbers, but the range is f(x) >= O\n\nCOMPOUND FUNCTIONS – work from the INSIDE out \nIf f(x) = x³ + √x and g(x) = 4x – 3, what is f(g(3))?\nFirst, solve g(3): 4.3 – 3 = 9. Now, plug-in 9 on f(x): 9³ + √9 = 729 + 3"} +{"id": "GMAT QUANT_p77_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 77, "page_end": 77, "topic_guess": "coordinate_geometry", "text": "k from the INSIDE out \nIf f(x) = x³ + √x and g(x) = 4x – 3, what is f(g(3))?\nFirst, solve g(3): 4.3 – 3 = 9. Now, plug-in 9 on f(x): 9³ + √9 = 729 + 3\n\nDETERMINING WHETHER A FUNCTION IS EVEN/ODD - To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f(–x) = f(x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f(–x) = –f(x), so all of the \"plus\" signs become \"minus\" signs, and vice versa), then the function is odd. So I'll plug –x in for x, and simplify: In all other cases, the function is \"neither even nor odd“\nExample # 1 : Determine algebraically whether f(x) = –3x2 + 4 is even, odd, or neither\nf(–x) = –3(–x)2 + 4 \u000b= –3(x2) + 4 \u000b= –3x2 + 4 \u000bSince the new expression is the same as the initial expression, the function is even"} +{"id": "GMAT QUANT_p77_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 77, "page_end": 77, "topic_guess": "fractions_decimals_percents", "text": "n, odd, or neither\nf(–x) = –3(–x)2 + 4 \u000b= –3(x2) + 4 \u000b= –3x2 + 4 \u000bSince the new expression is the same as the initial expression, the function is even\n\nCOMMON FUNCTION TYPES\ni) Direct proportionality means that the two quantities always change by the same factor and in the same direction. For instance, tripling the input will cause the output to triple as well. Direct proportionality relationships are of the form Y = kx, where x is the input value, y is the output value and k is the proportionality constant. This equation can also be written as y/x = k\nExample: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?\nSolution: Typically with direct proportion problems, you will be given \"before\" and \"after\" values. Simply set up ratios to solve the problem for example, y1 and x1 can be used for the \"before” values and y2 and x2 can be used for the \"after\" values. We then write y1/x1 = y2/x2 , since both ratios are equal to the same constant k. In the problem given above, note that the direct proportion is between the height and the square of the velocity, not the velocity itself. Therefore, write the proportion as (4/162) = 9/v2 ; v2 = 576; v = 24\nii) Inverse proportionality means that the two quantities change by RECIPROCAL factors. Cutting the input in half will actually double the output. Inverse proportionality relationships are of the form yx = k. In these cases, set up the ratio as a product, and then solve\niii) Linear growth - Many GMAT problems, especially word problems, feature quantities with linear growth (or decay), i.e., they grow at a constant rate. Such quantities are determined by the linear function: Y = mx + b. In this equation, the slope m is the constant rate at which the quantity grows. The y-intercept b is the value of the quantity at time zero, and the variable (in this case, x) stands for time. For instance, if a baby weighs 9 pounds at birth and gains 1.2 pounds per month, the function of his weight is: w = 1.2t + 9 (t = nº of months)"} +{"id": "GMAT QUANT_p78_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 78, "page_end": 78, "topic_guess": "sets_probability_counting", "text": "COMBINATORICS AND PROBABILITY"} +{"id": "GMAT QUANT_p79_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 79, "page_end": 79, "topic_guess": "sets_probability_counting", "text": "COMBINATORICS \nPERMUTATION\nORDER MATTERS - a permutation is nothing but an ordered combination. Essentially, permutation questions are about taking a group of objects and totaling the number of ways in which they can be arranged in specific/pre-defined ways\nThe number of permutations of r items, chosen from a pool of n items, is n! / (n-r)! \nThe number of permutations when things are not all different : If there be n things, p of them of one kind, q of another kind, r of still another kind and so on, then the total number of permutations is given by n! / (p! q! r!...)\nAs a special case, the number of permutations of all n items in a pool of n items is just n! , and the number of ways to arrange n distinct objects along a fixed circle is: (n – 1)!\nCOMBINATION\nORDER DOES NOT MATTER (i.e., whether the events are taking place simultaneously or sequentially). Since the order does not matter, we need to remove all the “duplicate” cases \nThe number of combinations of r items, chosen from a pool of n items, is n! / (n-r)! r!"} +{"id": "GMAT QUANT_p79_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 79, "page_end": 79, "topic_guess": "sets_probability_counting", "text": "does not matter, we need to remove all the “duplicate” cases \nThe number of combinations of r items, chosen from a pool of n items, is n! / (n-r)! r!\n\n“Does it matter if the two items (or however many items you have) positions are changed?” If the answer is yes, it is a permutations problem, if not, it is a combinations problem\nReplacement / Repetition\n\nPotential outcomes that are replaced or constant\nExamples\nHow many possibilities for a combination lock with 40 numbers that requires 3 selections\n\nNon – replacement \nPotential outcomes that decrease with each selection\nExamples \nHow many possibilities for a combination lock with 40 numbers that requires 3 selections and cannot have the same number twice\nCombinations from multiple groups\nIn this situation, combinations are being drawn from several groups to form a complete set. Figure out the combinations from each group and then multiply them together"} +{"id": "GMAT QUANT_p79_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 79, "page_end": 79, "topic_guess": "sets_probability_counting", "text": ", combinations are being drawn from several groups to form a complete set. Figure out the combinations from each group and then multiply them together\n\nCombinations from pairings\nThese are pairing questions – so they must take place in two \nExample: 7 basketball teams with five players each are at a tournament. If each player shakes hands with every other player NOT on his own team, how many handshakes take place?\nHow many people take part in a handshake? Two. There are 35 people who could be in the first spot, but that person cannot shake hands with anyone on his own team. So that person has only 30 people who’s hands he can shake. Therefore, the number of combinations is (35*30)/2"} +{"id": "GMAT QUANT_p80_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 80, "page_end": 80, "topic_guess": "sets_probability_counting", "text": "PERMUTATIONS VS. COMBINATIONS \nOne concept that you need to know for the exam is that when dealing with combinations and permutations, each result corresponds to a unique set of circumstances. For example, if you have z people and know that choosing two of them would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would yield exactly 15 different groups of two. So if you know how many subgroups of a certain size you can choose from an unknown original larger group, you can deduce the size of the larger group\nSeating 5 people in 3 chairs, is the same as seating 3 people in 5 chairs"} +{"id": "GMAT QUANT_p81_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 81, "page_end": 81, "topic_guess": "word_problems", "text": "SOLVING COMBINATORICS PROBLEMS – BASIC STRATEGIES \nFUNDAMENTAL COUNTING PRINCIPLE\nIf you must make a number of separate decisions, then multiply the number of ways to make each individual decision to find the number of ways to make all decisions together\nExample: If you have 4 types of bread, 3 types of cheese and 2 types of ham and wish to make a sandwich, you can make it in 4*3*2 = 24 different ways"} +{"id": "GMAT QUANT_p81_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 81, "page_end": 81, "topic_guess": "sequences_patterns", "text": "Example: If you have 4 types of bread, 3 types of cheese and 2 types of ham and wish to make a sandwich, you can make it in 4*3*2 = 24 different ways\n\nSLOT METHOD\nFor problems where some choices are restricted and/or affect other choices, choose most restricted options first (slot method)\nFor example, you must insert a 5-digit lock code, but the first and last numbers need to be odd, and no repetition is allowed. How many codes are possible?\nIn this problem, you do not want to choose the five digits in order. Instead, you should start by picking the first and last digits (which must be odd), because these digits are the most restricted. Because there are 5 different odd digits (1, 3, 5, 7, and 9), there are 5 ways of picking the first digit. Since no repetition is allowed, there are only 4 odd digits left for the last digit. You can then pick the other three digits in any order, but make sure you account for the lack of repetition. For those three choices, you have only 8, 7, and 6 digits available. Therefore, the total number of lock codes is 4 x 5 x 8 x 7 x 6 = 6,720"} +{"id": "GMAT QUANT_p81_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 81, "page_end": 81, "topic_guess": "statistics", "text": "epetition. For those three choices, you have only 8, 7, and 6 digits available. Therefore, the total number of lock codes is 4 x 5 x 8 x 7 x 6 = 6,720\n\nSIMPLE FACTORIALS\nThe number of ways in which n different things can be arranged in a straight line, such that no one thing is allowed to appear more than once in any of the arrangements (i.e. when repetitions are not allowed) is n!. n! counts the rearrangements of n distinct objects as a special (but very common) application of the Slot Method\nFor example, In staging a house, a real-estate agent must place six different books on a bookshelf. In how many different orders can she arrange the books? Using the Fundamental Counting Principle, we see that we have 6 choices for the book that goes first, 5 choices for the book that goes next, and so forth. Ultimately, we have this total: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders"} +{"id": "GMAT QUANT_p81_c4", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 81, "page_end": 81, "topic_guess": "statistics", "text": "hat goes first, 5 choices for the book that goes next, and so forth. Ultimately, we have this total: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders\n\nANAGRAMS\nAn anagram is a rearrangement of the letters in a word or phrase. For instance, the word DEDUCTIONS is an anagram of DISCOUNTED, and so is the gibberish \"word“ CDDEINOSTU\nAnagrams with repeated words \nThe number of anagrams of a word is the factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters"} +{"id": "GMAT QUANT_p81_c5", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 81, "page_end": 81, "topic_guess": "statistics", "text": "umber of anagrams of a word is the factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters\n\nExample # 1\nHow many different anagrams (meaningful or nonsense) are possible for the word PIZZAZZ?\nWe might expect 7! possible combinations from the 7 letters in the word PIZZAZZ. However, there are only 210. Why are there, relatively speaking, so few anagrams? The answer lies in repetition: the four Z's are indistinguishable from each other. If the four Z's were all different letters, then we would have 7! = 5,040 different anagrams. To picture that scenario, imagine labeling the Z's with subscripts: Z1, Z2, Z3 and Z4. We could then list the 5,040 anagram. Now erase the subscripts from those 5,040 anagrams. You will notice that many \"different“ arrangements-like AIPZ1Z2Z3Z4, AIPZ4Z2Z3Z1' and so on-are now the same. In fact, for any genuinely unique anagram-like AIPZZZZ-there are now 4! = 24 identical copies in the list of 5,040 anagrams, because there are 4! = 24 ways to rearrange the four Z's in the word PIZZAZZ without changing anything. Because this 24-fold repetition occurs for every unique anagram of PIZZAZZ, we take 7! (which counts the arrangements as if the letters were all distinct) and divide by 4! (= 24) to account for the 4 repeated Z's"} +{"id": "GMAT QUANT_p82_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 82, "page_end": 82, "topic_guess": "word_problems", "text": "SOLVING COMBINATORICS PROBLEMS – ADVANCED STRATEGIES"} +{"id": "GMAT QUANT_p83_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 83, "page_end": 83, "topic_guess": "word_problems", "text": "SOLVING COMBINATORICS PROBLEMS – ADVANCED STRATEGIES"} +{"id": "GMAT QUANT_p84_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 84, "page_end": 84, "topic_guess": "word_problems", "text": "SOLVING COMBINATORICS PROBLEMS – ADVANCED STRATEGIES"} +{"id": "GMAT QUANT_p85_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 85, "page_end": 85, "topic_guess": "sets_probability_counting", "text": "PROBABILITY\nPROBABILITY = NUMBER OF FAVORABLE OUTCOMES TOTAL NUMBER OF OUTCOMES \nPROBABILITY OF MULTIPLE EVENTS \nAND\nIf two events have to occur together, generally an \"and\" is used. For example: \"I will only be happy today if I get email and win the lottery.\" The \"and\" means that both events are expected to happen together\n\nIn the case of “and,” we multiply probabilities together to get a lower overall probability\n\nExample:\nIf a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails?\nThe probability that the coin will land on heads is 1/2. The probability that the coin will land on Tails is also 1/2. Since we want them to happen together, we multiply individual probabilities 1/2 × 1/2 = 1/4\n\nOR\nIf both events do not necessarily have to occur together, an “or” may be used. For example, I will be happy today if I win the lottery OR have email\n\n“OR” means that we add probabilities together to get a higher overall probability"} +{"id": "GMAT QUANT_p85_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 85, "page_end": 85, "topic_guess": "sets_probability_counting", "text": "r example, I will be happy today if I win the lottery OR have email\n\n“OR” means that we add probabilities together to get a higher overall probability\n\nExample: \nJohn will win $100 if, from a deck of 52 standard playing cards, he chooses either a 7 or a 9 when pulling a single card from the deck. What is the probability that John will win $100?\nHow can John win? He can win by pulling out either a 7 or a 9. His chances of doing that are higher than if he could win only by pulling out a 7. In that case, he’d only have 4 cards that would make him win $100 (because there are 4 7's in a standard deck), now he has 8 cards. To find the total probability, we need to figure out the probability of each event and then add them together. Therefore probability = (4/52) + (4/52) = 2/13\n\nINDEPENDENT EVENTS\n\nEvents where the outcome of first event does not affect the probability of the second event \nExample: coin toss \nDEPENDENT EVENTS\n\nEvents where the probability of the second event is affected by the outcome of the first event \nExample: Drawing from a pack of cards without replacing the card drawn \nMUTUALLY EXCLUSIVE EVENTS"} +{"id": "GMAT QUANT_p85_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 85, "page_end": 85, "topic_guess": "sets_probability_counting", "text": "event is affected by the outcome of the first event \nExample: Drawing from a pack of cards without replacing the card drawn \nMUTUALLY EXCLUSIVE EVENTS\n\nTwo events cannot occur together  Add the probabilities of individual events\n\nEVENTS CAN OCCUR TOGETHER\n\nP (A or B) = P(A) + P(B) – P (A&B)\n\nIf you simply add P(A) and P(B) in situations when A and B can occur together, then you are overestimating the probability of either A or B (or both) occurring"} +{"id": "GMAT QUANT_p86_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 86, "page_end": 86, "topic_guess": "sets_probability_counting", "text": "PROBABILITY – 2 \nThe greatest probability- the certainty that an event will occur-is 1. Thus, a probability of 1 means that the event must occur\nThe lowest probability-the impossibility that an event will occur is 0. Thus, a probability of 0 means that an event will NOT occur. For example, the probability that you roll a fair die once and it lands on the number 9 is impossible, or 0\nIf on a GMAT problem, \"success“ contains multiple possibilities –especially if the wording contains phrases such as “at least\" and “at most“ then consider finding the probability that success does not happen. If you can find this “failure\" probability more easily (call it x) the probability you really want to find will be 1 – x\nExample: What is the probability that, on three rolls of a single fair die, at least of the rolls will be a six?\nFailure: What is the probability that NONE of the rolls will yield a 6? On each roll, there is a 5/6 probability that the die will not yield a 6. Thus, the probability that on all 3 rolls the die will not yield a 6 is (5/6)*(5/6)*(5*6) = 125/216\nNow, we originally defined success as rolling at least one six. Since we have found the probability of failure, we answer the original question by subtracting this probability from 1: 1 – (125/216) = 91/256"} +{"id": "GMAT QUANT_p87_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 87, "page_end": 87, "topic_guess": "geometry", "text": "GEOMETRY"} +{"id": "GMAT QUANT_p88_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 88, "page_end": 88, "topic_guess": "geometry", "text": "LINES AND ANGLES \nTwo straight lines which meet at a point form an angle between them\nAcute angle is between 0 and 90 degrees \nRight angle is 90 degrees\nObtuse angle is between 90 degrees and 180 degrees\nReflex angle is between 180 degrees and 360 degrees \nA straight line is the shortest distance between two points and measures 180 degrees \nParallel lines are lines that never intersect, no matter how far they are stretched, and are equidistant \nPerpendicular lines intersect at 90 degrees (may or may not bisect the line segment) \nThe sum of angles at a point is 360 degrees\n\nINTERSECTING LINES\n\nEXTERIOR ANGLES OF A TRIANGLE"} +{"id": "GMAT QUANT_p89_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 89, "page_end": 89, "topic_guess": "geometry", "text": "LINES AND ANGLES \nPARALLEL LINES CUT BY A TRANSVERSAL\n\nGMAT can disguise the lines, by not drawing them completely. Extending them can help you to find out whether they are transversal or parallel. GMAT uses the symbol II to indicate 2 lines are parallel. MN II OP means those segments are parallel\n\nProperties\nAlternate interior angles are equal \nAngles 3 and 6 are equal; angles 4 and 5 are equal\n\nAlternate exterior angles are equal \nAngles 1 and 8 are equal; angles 2 and 7 are equal\n\nCorresponding angles are equal (corresponding angles are one interior and one exterior angle that are on the same side of the transversal)\nAngles 1 and 5 \nAngles 2 and 6 \nAngles 3 and 7\nAngles 4 and 8\n\nIt is important to note that if two lines cut by a transversal have any of the above properties, then the two lines must be parallel. For example, if alternate interior angles are equal, then the two lines cut by a transversal must be parallel"} +{"id": "GMAT QUANT_p90_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 90, "page_end": 90, "topic_guess": "geometry", "text": "TWO – DIMENSIAL POLYGONS\n\nTWO DIMENSIONAL POLYGONS\n\nTRIANGLES – 3 SIDED\n\nQUADRILATERALS – 4 SIDED\n\nOTHERS – N SIDED\n\nEquilateral \nIsosceles\nScalene\n\nPentagon – 5 sides\n\nHexagon – 6 sides \nComprised of 6 equilateral triangles. Therefore, the area of a hexagon will be 6 * area of an equilateral triangle\n\nOctagon – 8 sides\n\nIrregular quadrilaterals\n\nParallelogram \nRectangle \nRhombus\nSquare\n\nAll quadrilaterals are not necessarily parallelograms. For a quadrilateral to be a parallelogram opposite sides and opposite angles have to be equal and parallel\n\nTrapezium\n\nSUM OF THE INTERIOR ANGLES OF A POLYGON – This depends on the number of sides (n) a polygon has: (n-2)*180\nSUM OF THE EXTERIOR ANGLES OF A POLYGON = 360"} +{"id": "GMAT QUANT_p91_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 91, "page_end": 91, "topic_guess": "geometry", "text": "TRIANGLES \nEQUILATERAL \nISOSCELES \nSCALENE\nTYPES OF TRIANGLES – SIDES \nTWO SIDES ARE EQUAL AND THE THIRD SIDE IS LESS THAN THE SUM OF THE OTHER TWO SIDES\nAll THREE SIDES ARE EQUAL \nALL THREE SIDES ARE DIFFERENT\nACUTE \nOBTUSE \nRIGHT – ANGLED \nTYPES OF TRIANGLES – ANGLES \nONE ANGLE IS GREATER THAN 90 DEGREES\n\nIn a triangle ABC with the vertex at B, AB2 + BC2 < AC2 \nALL ANGLES ARE LESS THAN 90 DEGREES\n\nIn a triangle ABC with the vertex at B, AB2 + BC2 > AC2 \nONE ANGLE IS 90 DEGREES\n\nThe 45-45-90 Triangle is known as a right-angled isosceles triangle\n\nIn a triangle ABC with the vertex at B, AB2 + BC2 = AC2"} +{"id": "GMAT QUANT_p92_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 92, "page_end": 92, "topic_guess": "geometry", "text": "BASIC PROPERTIES OF TRIANGLES – A"} +{"id": "GMAT QUANT_p93_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 93, "page_end": 93, "topic_guess": "geometry", "text": "BASIC PROPERTIES OF TRIANGLES – C"} +{"id": "GMAT QUANT_p94_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 94, "page_end": 94, "topic_guess": "geometry", "text": "BASIC PROPERTIES OF TRIANGLES – B\nSUM OF THE ANGLES OF A TRIANGLE = 180 DEGREES\n\nWhen one side is extended in any direction, an angle is formed with another side. This is called the exterior angle. There are six exterior angles in a triangle\n\nANGLES CORRESPOND TO THEIR OPPOSITE SIDE\n\nThis means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side. Additionally, if two angles are equal, their sides are also equal\n\nSUM OF ANY TWO SIDES OF A TRIANGLE MUST BE GREATER THAN THE THIRD SIDE\n\nDIFFERENCE BETWEEN ANY TWO SIDES IS LESS THAN THE THIRD SIDE\n\nTHE LENGTH OF THE THIRD SIDE MUST LIE BETWEEN THE SUM AND THE DIFFERENCE OF THE TWO GIVEN SIDES"} +{"id": "GMAT QUANT_p95_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 95, "page_end": 95, "topic_guess": "geometry", "text": "SIMILAR AND CONGRUENT TRIANGLES \nSimilarity – if any of the conditions are met – \nAngle, angle, angle – if the corresponding angles of each triangle have the same measurement. In other words, the above triangles are similar if:\u000bAngle L = Angle O; Angle N = Angle Q; Angle M = Angle P\nSide, angle, side - An angle in one triangle is the same measurement as an angle in the other triangle and the two sides containing these angles have the same ratio. In other words, the above triangles are similar if:\u000bAngle L = Angle O; Side LM/Side OP = Side LN/Side OQ\nSide, side, side - Each pair of corresponding sides have the same ratio. In other words, the above triangles are similar if:\u000bSide LM/Side OP = Side LN/Side OQ = Side MN/Side PQ"} +{"id": "GMAT QUANT_p96_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 96, "page_end": 96, "topic_guess": "geometry", "text": "RIGHT- ANGLED TRIANGLE AND THE PYTHAGOREAN THEOREM \nPythagorean Theorem: c2 = a2 + b2\na and b are the legs, while c is the hypotenuse\n\nTHE 45-45-90 ISOSCELES TRIANGLE\n\nThe 45-45-90 triangle is a special triangle with 2 equal sides and a relation between each side. If you are given one dimension on a 45-45-90 triangle, you can find the others.\n\nRelationship between sides\n45° - 45° - 90°\nLeg - leg - hypotenuse\n: 1 : √2\nx : x : x√2\n\nTherefore, in a 45-45-90 degree triangle, you only need to know the value of one side to determine the others\n\nA 45-45-90 is exactly half of a square. Two 45-45-90 triangles put together make up a square. So, if you are given the diagonal of a square, you can find the side by using the relation above. For example, if the diagonal of a square is 6√2, find the length of each of the two sides. \n(6√2)2 = x2 + x2\n36*2 = 2x2\n36 = x2\n X = 6"} +{"id": "GMAT QUANT_p97_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 97, "page_end": 97, "topic_guess": "geometry", "text": "HYPOTENUSE\nHYPOTENUSE\nEQUILATERAL TRIANGLES AND THE 30-60-90 TRIANGLE \nAn equilateral triangle is one in which all three sides and all three angles are equal\nEquilateral Triangles and the 30-60-90 Triangle:\n\n30\n30\n60\n60\nSHORT\nL\nONG\nL\nONG\nSHORT"} +{"id": "GMAT QUANT_p98_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 98, "page_end": 98, "topic_guess": "geometry", "text": "INSCRIBED AND CIRCUMSCRIBED TRIANGLES AND CIRCLES\nA right triangle inscribed in a semi- circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle"} +{"id": "GMAT QUANT_p99_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 99, "page_end": 99, "topic_guess": "general", "text": "EXAMPLES"} +{"id": "GMAT QUANT_p100_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 100, "page_end": 100, "topic_guess": "geometry", "text": "QUADRILATERALS \nHow to prove a quadrilateral is a parallelogram \nIf both pairs of opposite sides of a quadrilateral are equal\nIf both pairs of opposite angles of a quadrilateral are equal\nIf all pairs of consecutive angles of a quadrilateral are supplementary\nIf one pair of opposite sides of a quadrilateral is both equal and parallel\nIf the diagonals of a quadrilateral bisect each other\n\nThe area of a parallelogram is twice the area of a triangle created by one of its diagonals\nOf all quadrilaterals with a given perimeter, the square has the largest area. Conversely, of all the quadrilaterals with a given area, the square is the one with the smaller perimeter\nThe diagonal formed by joining vertices of two smaller angles will be greater than the diagonal formed by joining vertices of two greater angles in a rhombus\nThe diagonals of a kite are perpendicular bisectors of each other. The kite also has two pairs of adjacent, congruent sides (i.e., all sides are not equal)– THE KITE IS NOT A PARALLELOGRAM"} +{"id": "GMAT QUANT_p101_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 101, "page_end": 101, "topic_guess": "general", "text": "QUADRILATERALS"} +{"id": "GMAT QUANT_p102_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 102, "page_end": 102, "topic_guess": "geometry", "text": "QUADRILATERALS\n\nTRAPEZIOD\n\nOne pair of opposite sides is parallel, but not equal \nArea = 1/2 * (a+b)*h\n\nIsosceles Trapezoid\nWhat is the area of the trapezoid as show? \nAngle A = 120 degrees\nThe perimeter of the trapezoid ABCD = 36\n\nThe area of a trapezoid is equal to the average of the bases multiplied by the height. In this problem, you are given the top base (AB = 6), but not the bottom base (CD) or the height. (Note: 8 is NOT the height!) In order to find the area, you will need a way to figure out this missing data. Drop 2 perpendicular lines from points A and B to the horizontal base CD, and label the points at which the lines meet the base E and F."} +{"id": "GMAT QUANT_p102_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 102, "page_end": 102, "topic_guess": "geometry", "text": "missing data. Drop 2 perpendicular lines from points A and B to the horizontal base CD, and label the points at which the lines meet the base E and F.\n\nEF = AB = 6 cm. The congruent symbols in the drawing tell you that Angle A and Angle B are congruent, and that Angle C and Angle D are congruent. This tells you that AC = BD and CE = FD. \nStatement (1) tells us that Angle A = 120. Therefore, since the sum of all 4 angles must yield 360 (which is the total number of degrees in any four-sided polygon), we know that Angle B = 120, Angle C = 60, and Angle D = 60. This means that triangle ACE and triangle BDF are both 30-60-90 triangles. The relationship among the sides of a 30-60-90 triangle is in the ratio of x: x √3: 2x , where x is the shortest side. For triangle ACE, since the longest side AC = 8 , CE = 4 and AE = 4 √3 . The same measurements hold for triangle BFD. Thus we have the length of the bottom base (4 + 6 + 4) and the height and we can calculate the area of the trapezoid"} +{"id": "GMAT QUANT_p102_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 102, "page_end": 102, "topic_guess": "geometry", "text": "asurements hold for triangle BFD. Thus we have the length of the bottom base (4 + 6 + 4) and the height and we can calculate the area of the trapezoid\n\nStatement (2) tells us that the perimeter of trapezoid ABCD is 36. We already know that the lengths of sides AB (6), AC (8), and BD (8) sum to 22. We can deduce that CD = 14. Further, since EF = 6, we can determine that CE = FD = 4. From this information, we can work with either Triangle ACE or Triangle BDF, and use the Pythagorean theorem to figure out the height of the trapezoid. Now, knowing the lengths of both bases, and the height, we can calculate the area of the trapezoid\n\nEACH STATEMENT ALONE IS SUFFICIENT"} +{"id": "GMAT QUANT_p103_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 103, "page_end": 103, "topic_guess": "general", "text": "POLYGONS – 3 DIMENSIONAL"} +{"id": "GMAT QUANT_p104_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 104, "page_end": 104, "topic_guess": "general", "text": "POLYGONS – 3 DIMENSIONAL"} +{"id": "GMAT QUANT_p105_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 105, "page_end": 105, "topic_guess": "geometry", "text": "CIRCLES\n\nCircumference of a circle= 2πr\nCircumference of a semi-circle= πr\nPerimeter of a semi-circle = πr + D (Diameter of the circle)\nArea of a circle = πr² \nArea of a semi-circle = πr² / 2\nLength of the arc = (θ/360) × (2 π r)\nArea of the sector = (θ/360) × πr² \nDistance travelled by a wheel in n revolutions = n x circumference\nPerimeter of a sector = (θ/360) × (2 π r) + 2r \nCircles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin.  When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2"} +{"id": "GMAT QUANT_p106_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 106, "page_end": 106, "topic_guess": "geometry", "text": "CIRCLES, ARCS AND SECTORS \nA circle is a set of points that are equidistant from a fixed point called the centre. It comprises 360 degrees. \nRADIUS: distance from the centre of the circle to any point on the circle; can be shown as a line segment connecting the centre to a point on the circle – OC \nDIAMETER: is a line segment that connects two points on the circle and goes through the centre of the circle - AE\nDiameter = 2r \nCHORD: is any line segment whose endpoints are any two points on the circle – BD \nThe tangent line and the radius of the circle that has an endpoint at the point of tangency are perpendicular to each other – AO IS PERPENDICULAR TO XY \nARC – a portion/distance on the circle\n\nA\nC\nB\nE\nO\nD\nY\nX\nA\nB\nO\nC"} +{"id": "GMAT QUANT_p107_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 107, "page_end": 107, "topic_guess": "geometry", "text": "INSCRIBED CIRCLES \nWhen a circle is inscribed inside a square, the side equals the diameter"} +{"id": "GMAT QUANT_p108_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 108, "page_end": 108, "topic_guess": "geometry", "text": "CO – ORDINATE GEOMETRY"} +{"id": "GMAT QUANT_p109_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 109, "page_end": 109, "topic_guess": "geometry", "text": "CO-ORDINATE GEOMETRY – GENERAL CONCEPTS \nCO-ORDINATE POINTS\n\nPoints in the plane are identified by using an ordered pair of numbers, (x , y), where x is the x co-ordinate and y is the y co-ordinate. For example, in (4, -2), the x co-ordinate is 4 and the y co-ordinate is -2 \nAny point on the X axis can be taken as (a,0)\nAny point on the Y axis can be taken as (0,b)\nThe point (0, 0), where the axes cross, is called the origin\nAlong the X axis, the y co-ordinate is 0 and along the Y axis, the x co-ordinate is 0\n\nINTERCEPTS: The point where a line intersects a co-ordinate axis is called an intercept \nX intercept: the point at which the line intersects the x axis\nThe x-intercept is expressed using the ordered pair (x, 0), where x is the point where the line intersects the x-axis. The x-intercept is the point on the line at which y = O\nY intercept: the point at which the line intersects the y axis\nThe y-intercept is expressed using the ordered pair (0, y), where y is the point where the line intersects the y-axis. The y-intercept is the point on the line at which x = O"} +{"id": "GMAT QUANT_p109_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 109, "page_end": 109, "topic_guess": "algebra", "text": "ssed using the ordered pair (0, y), where y is the point where the line intersects the y-axis. The y-intercept is the point on the line at which x = O\n\nLINE \nA line can be formed in any of the three ways \nBy knowing the co-ordinates of any two points on the line \nBy knowing the co-ordinates of one point on the line and the slope of the line \nBy knowing the equation of the line\n\nSLOPE OF A LINE: Y2 – Y1 / X2 – X1 \nA line in the plane is formed by the connection of two or more points. Therefore, the slope is defined as the “rise” over the “run” – that is, how much the line rises vertically divided by how much the line runs horizontally\nFor example, what is the slope of the line for the following co-ordinates: (2,-3) and (6, 5) \nSlope = 5 – (-3) / (6 – 2) = 8/4, which is equal to 2\n\nThe slope is always the co-efficient of x in the equation (point-intercept) form y = mx + b. Therefore, in the equation y = 4x + 8; the slope is 8 \nIf the equation of the line is given in the general form ax + by + c = 0; then the slope is –a/b and the intercept is –c/b"} +{"id": "GMAT QUANT_p109_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 109, "page_end": 109, "topic_guess": "algebra", "text": "y = 4x + 8; the slope is 8 \nIf the equation of the line is given in the general form ax + by + c = 0; then the slope is –a/b and the intercept is –c/b\n\nThe slope of all points/ co-ordinates on a line will ALWAYS be the same. However, just because two points have the same slope, doesn’t necessarily mean that the lie on the same line and vice versa. For instance, parallel lines have the same slope; conversely, if two lines have the same slope they could either be parallel or lie on each other\nIf the slope is 1, the angle formed by the slope is 45 degrees \nParallel lines have equal slopes\nPerpendicular lines have negative reciprocal slopes\nSLOPE DIRECTION - There are four types of slope an equation can have\nPositive – rises upward from left to right \nNegative – falls downward from left to right \nZero – parallel to the x axis \nUndefined – parallel to the y axis \nEvery line (but the one that crosses the origin OR the one that is parallel to X or Y axis OR the X and Y axis themselves) crosses three quadrants"} +{"id": "GMAT QUANT_p110_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 110, "page_end": 110, "topic_guess": "algebra", "text": "CO-ORDINATE GEOMETRY – EQUATION OF A LINE/ SLOPE-INTERCEPT LINE\nWhat is the slope-intercept form for a line with the equation 6x + 3y = 18? \nRewrite the equation by solving for y as follows: \n6x + 3y = 18 \n2x + y = 6 \ny = – 2x + 6\nThus, the slope is -2 and the y-intercept (b) is 6\n\nHorizontal and vertical lines are not expressed in the y = mx +b form. Instead, they are expressed as simple, one-variable equations. Horizontal lines are expressed in the form: y = some number, such as y = 3 or y = 5 (i.e., equation of the line parallel to the x axis is 3 or 5). Vertical lines are expressed in the form: x = some number, such as x = 4 or x = 7 (i.e., equation of the line parallel to the y axis is 4 or 7)\nEquation of the X axis is Y= 0 \nEquation of the Y axis is X= 0"} +{"id": "GMAT QUANT_p111_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 111, "page_end": 111, "topic_guess": "word_problems", "text": "CO-ORDINATE GEOMETRY – 3\nDISTANCE BETWEEN TWO POINTS \nThe distance between any two points in the coordinate plane can be calculated by using the Pythagorean Theorem\nExample: what is the distance between the points (1, 3) and (7, -5)?\nDraw a right triangle connecting the points\nFind the lengths of the two legs of the triangle by calculating the rise and the run\nThe y-coordinate changes from 3 to -5, a difference of 8 (the vertical leg)\nThe x-coordinate changes from 1 to 7, a difference of 6 (the horizontal leg)\nUse the Pythagorean Theorem to calculate the length of the diagonal, which is the distance between the points\n62 + 82 = c2\n36 + 64 = c2\n100 = c2 \nc = 10\n\nPOSITIVE AND NEGATIVE QUADRANTS \nQuadrant I: x and y are both positive \nQuadrant II: x is negative, y is positive\nQuadrant III: x is negative, y is negative\nQuadrant IV: x is positive, y is negative"} +{"id": "GMAT QUANT_p111_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 111, "page_end": 111, "topic_guess": "word_problems", "text": "nd y are both positive \nQuadrant II: x is negative, y is positive\nQuadrant III: x is negative, y is negative\nQuadrant IV: x is positive, y is negative\n\nThe GMAT sometimes asks you to determine which quadrants a given line passes through. For example: Which quadrants does the line 2x + Y = 5 pass through?\n2x + y = 5\ny = 5 – 2x \nSince b = 5, the y-intercept is the point (0, 5). The slope is -2, so the line slopes downward steeply to the right from the y-intercept. Although we do not know exactly where the line intersects the x-axis, we can now see that the line passes through quadrants I, II, and IV\n\nPARALLEL LINES\nTwo lines are said to be parallel when their slopes are equal and the y – intercept is different \nFor example: If Point A is (5,20) and Point B is (30,7), define a line through point C (12,10) that is parallel to the line that passes through both A and B\nSlope of AB = (7-20)/(30-5)  -13/25  -0.52\nSubstituting in equation y-y1 = m(x-x1)  y = -0.52(x – 12) + 10 \ny = -0.52x + 16.24\nDistance between two parallel lines y = mx + b and y= mx + c can be found by the formula D = | b – c |/ √m2 + 1"} +{"id": "GMAT QUANT_p112_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 112, "page_end": 112, "topic_guess": "coordinate_geometry", "text": "CO-ORDINATE GEOMETRY – 4\nPERPENDICULAR BISECTORS \nThe perpendicular bisector of a line segment forms a 90° angle with the segment and divides the segment exactly in HALF. The key to solving perpendicular bisector problems is remembering this property: the perpendicular bisector has the negative reciprocal slope of the line segment it bisects. That is, the product of the two slopes is -1 (the only exception occurs when one line is horizontal and the other line is vertical, since vertical lines have undefined slopes)\nExample: If the coordinates of point A are (2, 2) and the coordinates of point B are (0, -2), what is the equation of the perpendicular bisector of line segment AB?\nSlope of AB is 2 \nSlope of the perpendicular bisector is – (1/2)\nNow we know that the equation of the perpendicular bisector has the following form: y = (-1/2)x + b. However, we still need to find the value of b To do this, we will need to find one point on the perpendicular bisector\nThe perpendicular bisector passes through the midpoint of AB. Thus, if we find the midpoint of AB, we will have found a point on the perpendicular bisector. The x co-ordinate of the perpendicular bisector will be the average of the x co-ordinates of A and B. Similarly, the y co-ordinate of the perpendicular bisector of AB will be the average of the y co-ordinates of A and B. \nPlug in the values of x (1) and y (0) to get b = 1/2"} +{"id": "GMAT QUANT_p112_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 112, "page_end": 112, "topic_guess": "algebra", "text": "dinate of the perpendicular bisector of AB will be the average of the y co-ordinates of A and B. \nPlug in the values of x (1) and y (0) to get b = 1/2\n\nINTERSECTION OF TWO LINES\nRecall that a line in the coordinate plane is defined by a linear equation relating x and y. That is, if a point (x, y) lies on the line, then those values of x and y satisfy the equation. For instance, the point (3, 2) lies on the line defined by the equation y = 4x - 10, since the equation is true when we plug in x = 3 and y = 2. On the other hand, the point (7, 5) does not lie on that line, because the equation is false when we plug in x = 7 and y = 5\nSo, what does it mean when two lines intersect in the coordinate plane? It means that at the point of intersection, BOTH equations representing the lines are true. That is, the pair of numbers (x, y) that represents the point of intersection solves BOTH equations. Finding this point of intersection is equivalent to solving a system of two linear equations\nExample: At what point does the line represented by y = 4x - 10 intersect the line represented by 2x + 3y = 26?\nSolving the equations simultaneously, we get x=4 and y=6 \nTherefore, the point of intersection is (4,6)\nIf two lines in a plane do not intersect, then the lines are parallel. If this is the case, there is NO pair of numbers (x, y) that satisfies both equations at the same time\nThe two equations/lines in a plane may also represent the same line. In this case, infinitely many points (x, y) along the line satisfy the two equations (which must actually be the same equation in two disguises)"} +{"id": "GMAT QUANT_p112_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 112, "page_end": 112, "topic_guess": "algebra", "text": "line. In this case, infinitely many points (x, y) along the line satisfy the two equations (which must actually be the same equation in two disguises)\n\nThe coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m:n are (nx1 + mx2)/m+n and (ny1 + my2)/m+n"} +{"id": "GMAT QUANT_p113_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 113, "page_end": 113, "topic_guess": "geometry", "text": "ADVANCED CONCEPTS \nEQUATION OF A CIRCLE \nThe equation of a circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2\nThe equation of a circle with centre (0, 0) and radius r is x2 + y2 = r2\n\nIn the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?\n(5 – 2)2 + (0 – (-3))2 = r2\n(3)2 + (3)2 = r2\n18 = r2\nTherefore, area = 18 π\n\nFind the equation of a circle whose center is at (4, 2) and is tangent to y-axis\nSince the circle is tangent to y-axis, the radius of the circle is perpendicular to y-axis. It also means that the length of the radius is also the length of the perpendicular segment from the center of the circle to y –axis. The point of tangency is at (0, 2). To find the length of the radius, use the distance formula:\n(0 – 4)2 + (2 – 2)2 = r2\nr = 4"} +{"id": "GMAT QUANT_p113_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 113, "page_end": 113, "topic_guess": "geometry", "text": "the circle to y –axis. The point of tangency is at (0, 2). To find the length of the radius, use the distance formula:\n(0 – 4)2 + (2 – 2)2 = r2\nr = 4\n\nWrite the equation of the circle with the given condition: (10, 8) and (4, -2) are the endpoints of the diameter\nIn a circle, the radius is one-half of the diameter. Since the given are the endpoints of the diameter, the center of the circle is the midpoint of the diameter\n(h,k) = [(10+4)/2, (8-2)/2)]  (7,3)\nThe next step is to get the length of the radius. Since radius is one-half of the circle, it will equal the distance from the center to any one end point of the diameter: (10 – 7)2 + (8 – 3)2 = r2  r = root 34 \nEquation = (x – 7)2 + (y – 3)2 = 34"} +{"id": "GMAT QUANT_p113_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 113, "page_end": 113, "topic_guess": "word_problems", "text": "equal the distance from the center to any one end point of the diameter: (10 – 7)2 + (8 – 3)2 = r2  r = root 34 \nEquation = (x – 7)2 + (y – 3)2 = 34\n\nWhat is the radius for the following – Circle tangent to the line 3x – 4y = 24 with center at (1, 0) \nThe radius of the circle is equal to the distance of the center (1, 0) from the line 3x – 4y = 24\nTo find the distance/radius from a point to a line, we use the formula: d = Ax + By + C / - √(A2 + B2 ), where A and B are the coefficients of x and y, and C is the constant in the equation of line. X and y are co-ordinates of the point \nTherefore, A = 3, B = -4 and C = -24, x = 1, y = 0\nRadius = 21/5"} +{"id": "GMAT QUANT_p114_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 114, "page_end": 114, "topic_guess": "algebra", "text": "ADVANCED CONCEPTS \nPARABOLAS\n\nThe equation y = Ax2 + Bx + C is the equation of the quadratic graph which is a parabola with axis parallel to y axis\nC is the y intercept (i.e., the value of y when x is 0)\n\nIf A > 0, the parabola opens upwards. If A < 0, the parabola opens downwards\n\nIf B2 > 4AC, the parabola cuts the x-axis at 2 different points\nIf B2 = 4AC, the parabola touches the x-axis at one point (the two points become co-incident)\nIf B2 < 4AC, the parabola does not cut the x-axis at all\n\nThe vertex of a parabola represents the maximum or minimum value of the function. The vertex is located at point (-b/2a, c – (b2/4a))"} +{"id": "GMAT QUANT_p116_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 116, "page_end": 116, "topic_guess": "general", "text": "MINOR PROBLEM TYPES"} +{"id": "GMAT QUANT_p117_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 117, "page_end": 117, "topic_guess": "fractions_decimals_percents", "text": "MINOR PROBLEM TYPES \nOPTIMIZATION\nSCHEDULING\nGROUPING\nScheduling problems, which require you to determine possible schedules satisfying a variety of constraints, can usually be tackled by careful consideration of extreme possibilities, usually the earliest and latest possible time slots for the events to be scheduled\n\nExample\nHow many days after the purchase of Product X does its standard warranty expire? (1997 is not a leap year\nWhen Mark purchased Product X in January 1997, the warranty did not expire until March 1997\nWhen Santos purchased Product X in May 1997, the warranty expired in May 1997\n\nRephrase the two statements in terms of extreme possibilities:\nStatement 1 \nShortest possible warranty period: Jan. 31 to Mar. 1 (29 days later)\nLongest possible warranty period: Jan. 1 to Mar. 31 (89 days later)\nStatement 2 \nShortest possible warranty period: May 1 to May 2, or similar (1 day later)\nLongest possible warranty period: May 1 to May 31 (30 days later)"} +{"id": "GMAT QUANT_p117_c2", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 117, "page_end": 117, "topic_guess": "algebra", "text": "ement 2 \nShortest possible warranty period: May 1 to May 2, or similar (1 day later)\nLongest possible warranty period: May 1 to May 31 (30 days later)\n\nEven taking both statements together, there are still two possibilities-29 days and 30 days -so both statements together are still insufficient\nNote that, had the given year been a leap year, the two statements together would have become sufficient!\nIn general optimization problems, the objective is to maximize or minimize some quantity, given constraints on other quantities. These quantities are all related through some equation"} +{"id": "GMAT QUANT_p117_c3", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 117, "page_end": 117, "topic_guess": "algebra", "text": "the objective is to maximize or minimize some quantity, given constraints on other quantities. These quantities are all related through some equation\n\nExample\nThe guests at a football banquet consumed a total of 401 pounds of food. If no individual guest consumed more than 2.5 pounds of food, what is the minimum number of guests that could have attended the banquet?\nBegin by considering the extreme case in which each guest eats as much food as possible, or 2.5 pounds apiece. The corresponding number of guests at the banquet works out to 401/2.5 = 160.4 people\nYou obviously cannot have a fractional number of guests at the banquet. Thus the answer must be rounded. To determine whether to round up or down, consider the explicit constraint: the amount of food per guest is a maximum of 2.5 pounds per guest. Therefore, the minimum number of guests is 160.4 (if guests could be fractional), and we must round up to make the number of guests an integer: 161\nNote the careful reasoning required! Although the phrase \"minimum number of guests\" may tempt you to round down, you will get an incorrect answer if you do so. In general, as you solve this sort of problem, put the extreme case into the underlying equation, and solve. Then round appropriately"} +{"id": "GMAT QUANT_p117_c4", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 117, "page_end": 117, "topic_guess": "algebra", "text": "er if you do so. In general, as you solve this sort of problem, put the extreme case into the underlying equation, and solve. Then round appropriately\n\nIn grouping problems, you make complete groups of items, drawing these items out of a larger pool. The goal is to maximize or minimize some quantity, such as the number of complete groups or the number of leftover items that do not fit into complete groups. As such, these problems are really a special case of optimization. One approach is to determine the limiting factor on the number of complete groups"} +{"id": "GMAT QUANT_p117_c5", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 117, "page_end": 117, "topic_guess": "algebra", "text": ". As such, these problems are really a special case of optimization. One approach is to determine the limiting factor on the number of complete groups\n\nExample\nOrange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?\nThe first step is to find out how many groups you can make with the people from each division separately, ignoring the other divisions. There are enough Division A people for 20 groups, but only enough Division B people for 15 groups and only enough Division C people for 13 groups. So the limiting factor is Division C. Therefore the total number of people left over will be 7 from A + 4 from B + 1 from C = 12 \nCOMPUTATION\nVery occasionally, the GMAT features problems centered on computation problems that contain no variables at all. \nExample\nFive identical pieces of wire are soldered together to form a longer wire, with the pieces overlapping by 4 cm at each join. If the wire thus made is exactly 1 meter long, how long is each of the identical pieces? (1 meter = 100 cm)\nIt is easy to assume that, because there are 5 pieces, there must be 5 joins. Instead, there are only 4 joins. Each join includes 4 cm of both wires joined, but is only counted once in the total length of 100 cm. Therefore, the total length of all the original wires is 100 + 4(4) = 116 cm. Because there are 5 wires, each wire is (116/5) = 23.2 cms long"} +{"id": "GMAT QUANT_p118_c1", "source_name": "GMAT Quantitative Review PPT", "source_file": "GMAT QUANT.pptx", "source_type": "pptx", "page_start": 118, "page_end": 118, "topic_guess": "general", "text": "MENTAL MATH / SHORT CUTS"}