problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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c9614b6913980d73802c6d63b24149b8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$
... | First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$ . Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\l... | A | 14.6 |
c9614b6913980d73802c6d63b24149b8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$
... | We can call the length of AC as $x$ . Therefore, the length of the base of the triangle with height $11$ is $11/h = a/x$ . Therefore, the base of the smaller triangle is $11x/h$ . We find that the area of the trapezoid is $(hx)/2 - 11^2x/2h$
Using similar triangles once again, we find that the base of the shaded triang... | A | 14.6 |
c9614b6913980d73802c6d63b24149b8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$
... | Since the length of AC does not matter, we can assume the base of triangle ABC is $h$ . Therefore, the area of the trapezoid in the first diagram is $h^2/2 - \frac{11^2}{2}$
The area of the triangle in the second diagram is now $\frac{(h-5)^2}{2}$
Therefore, $\frac{h^2 - 11^2}{2} = \frac{(h-5)^2}{2}$ . Multiplying both... | A | 14.6 |
c9614b6913980d73802c6d63b24149b8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_24 | Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$
... | The answers are there on the bottom, so start with the middle one, ${\textbf{(C)}15}$ . After calculating, we find that we need a shorter length, so try ${\textbf{(B)}14.8}$ . Still, we need a shorter answer, so we simply choose $\boxed{14.6}$ without trying it out. | A | 14.6 |
aad3ac2bba915811d335c7315c6013a2 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25 | Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$
$\textbf{(A)}\ 8 \qqu... | We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$ , and the maximum– $250-13=237$ . There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$ , so $17$ satisfies $22... | A | 8 |
aad3ac2bba915811d335c7315c6013a2 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25 | Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$
$\textbf{(A)}\ 8 \qqu... | Let the common difference between consecutive $a_i$ be $d$ .
Since $a_{15} - a_1 = 14d$ , we find from the first and last inequalities that $231 \le 14d \le 249$ . As $d$ must be an integer, this means $d = 17$ . Substituting this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \... | A | 8 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Draw the following four lines as shown: [asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)... | A | 10 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | There are $5$ lattice points in the interior of the logo and $12$ lattice points on the boundary of the logo. Because of Pick's Theorem, the area of the logo is $5+\frac{12}{2}-1=\boxed{10}$ | A | 10 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Notice that the area of the figure is equal to the area of the $4 \times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \cdot \frac{1}{2} = 6$ , so the answer is $16-6 = \boxed{10}$ | A | 10 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Draw the following four lines as shown:
[asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)... | A | 10 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\boxed{10}$ | A | 10 |
dcaa034b0bf2f88fb2af30742e8c32cc | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | If the triangles are rearranged such that the gaps are filled, there would be a $4$ by $2$ rectangle, and two $1$ by $1$ squares are present. Thus, the answer is $\boxed{10}$ | A | 10 |
f5a2930556b34ef57bc7dfef1024f1b3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_2 | Consider these two operations: \begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*} What is the output of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$
$\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220$ | We can substitute $5$ $3$ , and $6$ into the functions' definitions: \begin{align*} (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\ (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(25-9\right) \, \bigstar \, 6 \\ &= 16 \, \bigstar \, 6 \\ &= (16-6)^2 \\ &= \boxed{100} ~pog ~Mat... | D | 100 |
443d8834e113d46d25f334097ce0931d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3 | When three positive integers $a$ $b$ , and $c$ are multiplied together, their product is $100$ . Suppose $a < b < c$ . In how many ways can the numbers be chosen?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$ | The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$
Together, the numbers $a,b,$ and $c$ can be chosen in $\boxed{4}$ ways. | E | 4 |
443d8834e113d46d25f334097ce0931d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3 | When three positive integers $a$ $b$ , and $c$ are multiplied together, their product is $100$ . Suppose $a < b < c$ . In how many ways can the numbers be chosen?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$ | The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] We apply casework to $a$
If $a=1$ , then there are $3$ cases:
If $a=2$ , then there is only $1$ case:
In total, there are $3+1=\boxed{4}$ ways to choose distinct positive integer values of $a,b,c$ | E | 4 |
caf502d71847dfc7453032584c0a7882 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_5 | Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qqu... | Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.
Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.
Therefore, Anna is $14-11=\boxed{3}$ years older than Bella. | C | 3 |
c4fc086c5d6207dc123bd84de929c42b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the smallest number be $x.$ It follows that the largest number is $4x.$
Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{6} ~MRENTHUSIASM | C | 6 |
c4fc086c5d6207dc123bd84de929c42b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the common difference of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\implies d=9$ . Finally, we find that the smallest number is $15-9=\boxed{6}$ | C | 6 |
c4fc086c5d6207dc123bd84de929c42b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the smallest number be $x$ . Since the integers are equally spaced, and there are three of them, the middle number ( $15$ ) is the arithmetic mean of the other two numbers ( $x$ and $4x$ ). Thus, we set up the equation $(4x + x)/3 = 15$ , and, solving for $x$ , get $x = 6$ . Since $6$ is the smallest number out of ... | C | 6 |
c4fc086c5d6207dc123bd84de929c42b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the smallest number be $x$ . Because $x$ and $4x$ are equally spaced from $15$ $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$ , we get that the mean is also $2.5x$ . We get that $2.5x=15$ , and solving gets $x=\boxed{6}$ | C | 6 |
2569d8831452af98ac921c7165d8298f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{... | Notice that the number of kilobits in this song is $4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.$
We must divide this by $56$ in order to find out how many seconds this song would take to download: $\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.$
Finally, we divide this number by $60$ because this... | B | 10 |
2569d8831452af98ac921c7165d8298f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{... | We seek a value of $x$ that makes the following equation true, since every other quantity equals $1$
\[\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.\] Solving yields $x=\boxed{10}$ | B | 10 |
7f5b2ceee58b2c69e3bbf03b54cfaf68 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_9 | A cup of boiling water ( $212^{\circ}\text{F}$ ) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$ . Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?
$... | Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.
After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.
After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.
After $15$ ... | B | 86 |
6c94adfb918975f0c18cd8a9036e544b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_11 | Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?
$\textbf{(A) } 34\qquad\textbf{... | If there are $10$ pieces of pasta, Henry took $10-1=9$ bites. Each of these $9$ bites took $3$ inches of pasta out, and thus his bites in total took away $9\cdot 3 = 27$ inches of pasta. Thus, the original piece of pasta was $27+17=\boxed{44}$ inches long. | D | 44 |
e44247bc94190fdaa650b0e77e875dec | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_13 | How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$ .”
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$ | Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$
By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there ... | D | 9 |
c9c1bfde64dd08019d5ead2e3834e536 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_14 | In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$ | All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\] The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which ther... | D | 24 |
849d4729175498880d5f4f368246cc39 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | [asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*"$1$", (1,-0.3), black); label(scale(0.7)*"$2$", (2,-0.3), black); label(scale(0.7)*"$3$", (3,-0.3), black); label(scale(0.7)*"$4$", (4,-0.3), black); label(scale(0.7)*"$5$",... | C | 3 |
849d4729175498880d5f4f368246cc39 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram:
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*"$1$", (1,-0.3), black); label(scale(0.7)*"... | C | 3 |
849d4729175498880d5f4f368246cc39 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | We can find the lowest point in each line ( $1$ $2$ $3$ $4$ , or $5$ ) and find the price per pound. (Note that we don't need to find the points higher than the points below since we are finding the lowest price per pound.)
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+l... | C | 3 |
533d6a318769e6044f336615a28da965 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16 | Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) ... | Note that the sum of the first two numbers is $21\cdot2=42,$ the sum of the middle two numbers is $26\cdot2=52,$ and the sum of the last two numbers is $30\cdot2=60.$
It follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of the first ... | B | 25 |
533d6a318769e6044f336615a28da965 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16 | Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) ... | Let $a,b,c,$ and $d$ be the four numbers in that order. We are given that \begin{align*} \frac{a+b}{2} &= 21, &(1) \\ \frac{b+c}{2} &= 26, &(2) \\ \frac{c+d}{2} &= 30, &(3) \end{align*} and we wish to find $\frac{a+d}{2}.$
We add $(1)$ and $(3),$ then subtract $(2)$ from the result: \[\frac{a+d}{2}=21+30-26=\boxed{25}.... | B | 25 |
533d6a318769e6044f336615a28da965 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16 | Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) ... | We can just assume some of the numbers. For example, let the first two numbers both be $21.$ It follows that the third number is $31,$ and the fourth number is $29.$ Therefore, the average of the first and last numbers is $\dfrac{21+29}2=\dfrac{50}2=\boxed{25}.$ | B | 25 |
3e4b464dd3f10c44ed34a1ac38483e57 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_17 | If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]
$\textbf{(A) } 0\qquad... | Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{2... | B | 2 |
ca4dc60e852a9eff0d8c494765f6c938 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18 | The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the
area of the rectangle?
$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$ | The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.
Let $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rho... | C | 40 |
ca4dc60e852a9eff0d8c494765f6c938 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18 | The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the
area of the rectangle?
$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$ | If a rectangle has area $K,$ then the area of the quadrilateral formed by its midpoints is $\frac{K}{2}.$
Define points $A,B,C,$ and $D$ as Solution 1 does. Since $A,B,C,$ and $D$ are the midpoints of the rectangle, the rectangle's area is $2[ABCD].$ Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $\overline... | C | 40 |
3411ecc20c66c7104fe881ee2fbc011a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_19 | Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores. [asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((... | We set up our cases as solution 1 showed, realizing that only the second case is possible.
We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$ . We find the median of these two numbers, getting:
\[13-5=8\] \[\frac{8}{2}=4\] \[13-4=9\]
Thus, we realize that $4$ students must ... | C | 4 |
9d96716d08bb4871032feccc9df4bf79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ [asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3));... | D | 8 |
9d96716d08bb4871032feccc9df4bf79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$
Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$
We express the other two missing numbers in terms of $x$ and $y,$ as shown below: [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); ... | D | 8 |
9d96716d08bb4871032feccc9df4bf79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | This is based on the Solution 2 above and it is perhaps a little simpler than that.
Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following.
[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3));... | D | 8 |
9d96716d08bb4871032feccc9df4bf79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | Note that the sum of the rows and columns must be $8+5-1=12$ . We proceed to test the answer choices.
Testing $\textbf{(A)}$ , when $x = -1$ , the number above $x$ must be $15$ , which contradicts the precondition that the numbers surrounding $x$ is less than $x$
Testing $\textbf{(B)}$ , the number above $x$ is $9$ , w... | D | 8 |
9d96716d08bb4871032feccc9df4bf79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | The sum of the numbers in each column and row should be $5+(-1)+8=12$ . If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$
[asy] draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3... | D | 8 |
2aa7758f7737912b480a34eb0bcbe2d6 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21 | Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the sam... | Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x... | C | 9 |
2aa7758f7737912b480a34eb0bcbe2d6 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21 | Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the sam... | Clearly, Steph made $15 + 10 = 25$ shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is $\boxed{9}.$ | C | 9 |
2aa7758f7737912b480a34eb0bcbe2d6 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21 | Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the sam... | Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is 8 baskets. The most she could have made in the second half is 17 baskets. Steph makes 25 and misses 5 baskets and the only way for Candace to make 25 shots is... | C | 9 |
0c727a7b347f9092e5ff13fed3444cea | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22 | A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwi... | Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$
We construct the following table of $5$ -minute intervals: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ [0.5ex] \hline & & \\ [-2ex] \boldsymbol{5} \ \textbf{Minutes} &... | A | 17 |
0c727a7b347f9092e5ff13fed3444cea | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22 | A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwi... | Since Zia will wait for the bus if the bus is at the previous stop, we can create an equation to solve for when the bus is at the previous stop. The bus travels $\frac{1}{3}$ of a stop per minute, and Zia travels $\frac{1}{5}$ of a stop per minute. Now we create the equation, $\frac{1}{3}m = \frac{1}{5}m + 3 - 1$ (the ... | A | 17 |
565b6de5d24de6c9bafd9a37feecd8e4 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23 | $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,... | Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.
We take casework:
Case 1: 3 l... | D | 84 |
565b6de5d24de6c9bafd9a37feecd8e4 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23 | $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,... | We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$ 's and $2$ ways to choose a column with all $\triangle$ 's. The third column can be filled in $2^3=8$ wa... | D | 84 |
dd741c514d8ee0ccbf0bbd4dc0dfe360 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24 | The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?
[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); ... | While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$
Since $GH=14,$ then $JG=14-8=6.$ So, the area ... | C | 192 |
b911f4f6218991b2b9b5b808d6d6029a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1 | Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E... | We have $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{24}$ cups. | E | 24 |
b911f4f6218991b2b9b5b808d6d6029a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1 | Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E... | Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\cdot2=6$ cups of sugar.
Since the amount of water is $4$ times the amount of sugar, he uses $6\cdot4=\boxed{24}$ cups of water. | E | 24 |
b911f4f6218991b2b9b5b808d6d6029a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1 | Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E... | The ratio is $\text{Water}:\text{Sugar}:\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\cdot4=\boxed{24}$ cups of water. | E | 24 |
79e4ea64a269320b08715b6e5eb4751f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_2 | Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others?
$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \... | The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$ . Since the friend who earned $$40$ will need to leave with $$25$ , he will have to give $$\left(40-25\right)=\boxed{15}$ to the others. | C | 15 |
4888e09eb151f913d36b00881364aa3d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3 | Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \t... | Note that the unit of the answer is strawberries , which is the product of
By conversion factors, we have \[\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\rig... | D | 1920 |
4888e09eb151f913d36b00881364aa3d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3 | Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \t... | The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{1920}$ strawberries. | D | 1920 |
acb25a3031a50594a799e4fda6aba6e4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/25... | Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{37}$ dots. | B | 37 |
acb25a3031a50594a799e4fda6aba6e4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/25... | The dots in the next hexagon have four bands. From innermost to outermost:
Together, the answer is $1+6+12+18=\boxed{37}.$ | B | 37 |
acb25a3031a50594a799e4fda6aba6e4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/25... | The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon has $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus, the num... | B | 37 |
acb25a3031a50594a799e4fda6aba6e4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/25... | From the full diagram below, the answer is $\boxed{37} draw(shift(R)*scale(7.25)*hex,black+linewidth(1.25)); [/asy] ~MRENTHUSIASM | B | 37 |
a5ae8fde3653142a8d1c34e82ba0cb28 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 | Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25... | Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{15}$ | C | 15 |
a5ae8fde3653142a8d1c34e82ba0cb28 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 | Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25... | The pitcher is $\frac{3}{4}$ full, i.e. $75\%$ full. Therefore each cup receives $\frac{75}{5}=\boxed{15}$ percent of the total capacity. | C | 15 |
a5ae8fde3653142a8d1c34e82ba0cb28 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 | Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25... | Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that e... | C | 15 |
dbbe134836f8714e604c81204030fa7d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7 | How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)
$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$ | Firstly, observe that the second digit of such a number cannot be $1$ or $2$ , because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$ , so it must be $3$ . It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4... | C | 15 |
dbbe134836f8714e604c81204030fa7d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7 | How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)
$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$ | As in Solution 1, we find that the first two digits must be $23$ , and the third digit must be at least $4$ . If it is $4$ , then there are $5$ choices for the last digit, namely $5$ $6$ $7$ $8$ , or $9$ . Similarly, if the third digit is $5$ , there are $4$ choices for the last digit, namely $6$ $7$ $8$ , and $9$ ; if... | C | 15 |
8811840ae549f42110db27dc8236b6de | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8 | Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?
$\textbf{(A) }\text{806} \qquad ... | Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$ , giving a total of $(2019\cdot 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the greatest ... | C | 8072 |
8811840ae549f42110db27dc8236b6de | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8 | Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?
$\textbf{(A) }\text{806} \qquad ... | Suppose Ricardo has $p$ pennies, so then he has $(2020-p)$ nickels. In order to have at least one penny and at least one nickel, we require $p \geq 1$ and $2020 - p \geq 1$ , i.e. $1 \leq p \leq 2019$ . The number of cents he has is $p+5(2020-p) = 10100-4p$ , so the maximum is $10100-4 \cdot 1$ and the minimum is $1010... | C | 8072 |
f550ac11126d7e92b03135b958094a19 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each ... | C | 12 |
f550ac11126d7e92b03135b958094a19 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{12}\] | C | 12 |
f550ac11126d7e92b03135b958094a19 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are $2!$ ways to arrange Steelie and Tiger within this "super marble." Then there are $3!$ ways to arrange the "super m... | C | 12 |
f550ac11126d7e92b03135b958094a19 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | We will use the following
$\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.
$\textit{Proof. (Created by AoPS user RedFireTruck)}$
Let ou... | C | 12 |
0198b8fb1a229b2c172b52650826695c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11 | After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
[asy] unitsize(1.25cm); dotfactor = 10; p... | Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$ , her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{24}$ | E | 24 |
0198b8fb1a229b2c172b52650826695c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11 | After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
[asy] unitsize(1.25cm); dotfactor = 10; p... | Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \cdot 2 = 12$ miles per hour. The difference is consequently $36-12=\boxed{24}$ | E | 24 |
64b4e5991b0af939fc08f77bfed9ae2d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12 | For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$ | We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{10}$ | A | 10 |
64b4e5991b0af939fc08f77bfed9ae2d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12 | For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$ | Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{10}$ | A | 10 |
64b4e5991b0af939fc08f77bfed9ae2d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12 | For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$ | We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$
We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$
Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{10}.$ | A | 10 |
cfe7516ac1250a0d2de8319415859a1e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13 | Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) ... | After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{9}$ | B | 9 |
cfe7516ac1250a0d2de8319415859a1e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13 | Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) ... | As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$ . Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$ . Thus, Jamal added $\boxed{9}$ purple socks. | B | 9 |
cfe7516ac1250a0d2de8319415859a1e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13 | Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) ... | $6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$ . Therefore, $45-6-18-12=\boxed{9}$ purple socks were added. | B | 9 |
78319c29f0e646bee09e421742aa9127 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14 | There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
[asy] // made by SirCalcsALot size(300); pen shortdashed... | We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \cdot 20 = \boxed{95000}$ | D | 95000 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | Since $20\% = \frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{75}$ percent of $x$ | C | 75 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$ . Clearly, it follows that $y$ is $75\%$ of $x$ , so the answer is $\boxed{75}$ | C | 75 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$ , so $\frac{3}{20}x=\frac{1}{5}y$ . Solving for $y$ , we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$ | C | 75 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | We are given $0.15x = 0.20y$ , so we may assume without loss of generality that $x=20$ and $y=15$ . This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$ , and thus the answer is $\boxed{75}$ | C | 75 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | $15\%$ of $x$ is $0.15x$ , and $20\%$ of $y$ is $0.20y$ . We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$ . Solving for $y$ , dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$ | C | 75 |
3b27f4e5862f67c58bfa53a3b8b90a63 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | $15\%$ of $x$ can be written as $\frac{15}{100}x$ , or $\frac{15x}{100}$ $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$ , or $\frac{20y}{100}$ . So now, $\frac{15x}{100} = \frac{20y}{100}$ . Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$ . Dividing both sides by $500$ , we get... | C | 75 |
c5664966ee8689960c48494c7e756ab6 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16 | Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $... | We can form the following expressions for the sum along each line: \[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (w... | E | 5 |
c5664966ee8689960c48494c7e756ab6 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16 | Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $... | Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testin... | E | 5 |
33a3110a13b7b1100384be28275b57ff | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17 | How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{7}$ factors have more than $3$ factors. | B | 7 |
33a3110a13b7b1100384be28275b57ff | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17 | How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number whic... | B | 7 |
33a3110a13b7b1100384be28275b57ff | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17 | How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ wit... | B | 7 |
dc5b7f0aa5f81ceaad824ac9900595cb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); ... | A | 240 |
dc5b7f0aa5f81ceaad824ac9900595cb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | Let the midpoint of segment $FE$ be the origin. Evidently, point $D=(-8,0)$ and $A=(8,0)$ . Since points $C$ and $B$ share $x$ -coordinates with $D$ and $A$ respectively, it suffices to find the $y$ -coordinate of $B$ (which will be the height of the rectangle) and multiply this by $DA$ (which we know is $16$ ). The ra... | A | 240 |
dc5b7f0aa5f81ceaad824ac9900595cb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | We can use a result from the Art of Problem Solving Introduction to Algebra book Sidenote: for a semicircle with diameter $(1+n)$ , such that the $1$ part is on one side and the $n$ part is on the other side, the height from the end of the $1$ side (or the start of the $n$ side) is $\sqrt{n}$ . To use this formula, we ... | A | 240 |
dc5b7f0aa5f81ceaad824ac9900595cb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | Draw the other half of the circle as follows: [asy] draw(arc((0,0),17,360,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E... | A | 240 |
dc5b7f0aa5f81ceaad824ac9900595cb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); ... | A | 240 |
2e06cde8177c0e20f737f75264c2361a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | A number is divisible by $15$ precisely if it is divisible by $3$ and $5$ . The latter means the last digit must be either $5$ or $0$ , and the former means the sum of the digits must be divisible by $3$ . If the last digit is $0$ , the first digit would be $0$ (because the digits alternate), which is not possible. Hen... | B | 4 |
2e06cde8177c0e20f737f75264c2361a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | As in Solution 1, we find that such numbers must start with $5$ and alternate with $5$ (i.e. must be of the form $5\square 5\square 5$ ), where the two digits between the $5$ s need to be the same. Call that digit $x$ . For the number to be divisible by $3$ , the sum of the digits must be divisible by $3$ ; since the s... | B | 4 |
2e06cde8177c0e20f737f75264c2361a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | After finding out that the last digit must be $5$ , the number is of the form $5\square 5\square 5$ . If the unknown digit is $x$ , we can find that one of the solutions to $x$ is $0$ , since $5+5+5$ is equal to $15$ , which is divisible by $3$ . After trying every one digit number, you'll notice that $x$ must be a mul... | B | 4 |
2e06cde8177c0e20f737f75264c2361a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | assume the number is $ababa$ $10101a+1010b=0 (mod 15)\newline$ $6a+5b=0 (mod 15)\newline$ $a=0 (mod 5)\newline$ $5b=0 (mod 15)\newline$ $b=0 (mod 3)\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\newline$ $\boxed{4}$ | null | 4 |
7c7d03eba309acae3722263567dc3012 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_20 | A scientist walking through a forest recorded as integers the heights of $5$ trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating th... | We will show that $22$ $11$ $22$ $44$ , and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Tree... | B | 24.2 |
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