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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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... | Notice the average height of the trees ends with $0.2$ ; therefore, the sum of all five heights of the trees must end with $1$ or $6$ . ( $0.2$ $\cdot$ $5$ $1$ )
We already know Tree $2$ is $11$ meters tall. Both Tree $1$ and Tree $3$ must $22$ meters tall - since neither can be $5.5$ .
Once again, apply our observat... | B | 24.2 |
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... | As in Solution 1, we shall show that the heights of the trees are $22$ $11$ $22$ $44$ , and $22$ meters. Let $S$ be the sum of the heights, so that the average height will be $\frac{S}{5}$ meters. We note that $0.2 = \frac{1}{5}$ , so in order for $\frac{S}{5}$ to end in $.2$ $S$ must be one more than a multiple of $5$... | B | 24.2 |
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... | Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of $22$ . Now using the information given by the last column (that the average of the heights of the trees ends in $.2$ ), we can tell that the sum of all the heights of the trees ends in either $1$ or $6$ , be... | B | 24.2 |
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... | Since her average ends with 0.2, the sum of her tree heights must be $\equiv 1 \pmod {5}$ .
If Tree 1's height is $a$ , Tree 3's is $b$ , Tree 4's is $c$ , and Tree 5's is $d$ , then we $a$ and $b$ are both 22, since all of the tree heights are integers.
Now we have $22+11+22+c+d \equiv 1 \pmod {5}$ . Simplifying, we g... | B | 24.2 |
ec3790c1839d1dd100f3d31fc1f976ce | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are... | Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $... | A | 28 |
ec3790c1839d1dd100f3d31fc1f976ce | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are... | Suppose we "extend" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.
[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filld... | A | 28 |
ec3790c1839d1dd100f3d31fc1f976ce | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are... | On any white square, we may choose to go left or right, as long as we do not cross over the border of the board. Call the moves $L$ and $R$ respectively. Every single legal path consists of $4$ $R's$ and $3$ $L's$ , so now all we have to find is the number of ways to order $4 R's$ and $3 L's$ in any way, which is ${7 \... | A | 28 |
ec3790c1839d1dd100f3d31fc1f976ce | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are... | We label the rows starting from the bottom. At row 1, there is $1$ way: at P. We draw all the possible ways to get to Q. There are two ways to choose for row 2, and another two ways to choose for row 3. However, you can go to the "edge" or the farthest possible square westward of Q, so you can't multiply by 2 again. No... | A | 28 |
0c61235c3941cfe4c30980c467223e57 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22 | When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.2... | We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,... | E | 83 |
0c61235c3941cfe4c30980c467223e57 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22 | When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.2... | As in Solution 1, we work backwards from $1$ , this time showing the possible cases in a tree diagram:
[asy] // Upper branches draw((-6, 1.5)--(-5, 1)--(-3, 1)--(-2,0)--(0, 0)); draw((-6, 0.5)--(-5, 1)); // Lower branches draw((-6, -1.5)--(-5, -1.5)--(-4, -1)--(-3, -1)--(-2, 0)); draw((-6, -0.5)--(-5, -0.5)--(-4, -1));... | E | 83 |
0c61235c3941cfe4c30980c467223e57 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22 | When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.2... | We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$ . If $I$ is even, $O=\frac{I}{2}$ , and if $I$ is odd, $O=3I+1$ . We can therefore see that $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$ , if $I$ is even, $I=2$ , a... | E | 83 |
617e810cecf970a2a3b5c7feb5a1340a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23 | Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$ | Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award;... | B | 150 |
617e810cecf970a2a3b5c7feb5a1340a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23 | Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$ | Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.
In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 ... | B | 150 |
617e810cecf970a2a3b5c7feb5a1340a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23 | Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$ | Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn'... | B | 150 |
43e8027d92dff63978a8a8ae52f14c53 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)... | Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_... | A | 651 |
43e8027d92dff63978a8a8ae52f14c53 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)... | Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$ . Let the sum of the side lengths of $S_1$ and $S_3$ be $x$ , and let the length of square $S_2$ be $y$ . We then have the system \[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\] which we solve ... | A | 651 |
43e8027d92dff63978a8a8ae52f14c53 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)... | Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$ , the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{651}$ | A | 651 |
43e8027d92dff63978a8a8ae52f14c53 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)... | Let the side length of $S_2$ be s, and the shorter side length of $R_1$ and $R_2$ be $r$ . We have
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$... | A | 651 |
6ad9edf661be5bfc54af9e2bd6686167 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1 | Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textb... | We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they... | D | 9 |
6ad9edf661be5bfc54af9e2bd6686167 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1 | Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textb... | Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of \[4.50s+d=30\] In the question, it states that Ike and Mike buys as many sandwiches as possible.
So, we drop the number of sodas for a while.
We have: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{alig... | D | 9 |
04fa84157a48ae907460ce977e966a64 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{150}$ . ~avam... | E | 150 |
04fa84157a48ae907460ce977e966a64 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | Using the diagram we find that the larger side of the small rectangle is $2$ times the length of the smaller side. Therefore, the longer side is $5 \cdot 2 = 10$ . So the area of the identical rectangles is $5 \cdot 10 = 50$ . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large... | E | 150 |
04fa84157a48ae907460ce977e966a64 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | We see that if the short sides are 5, the long side has to be $5\cdot2=10$ because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle $ABCD$ )
is $10+5=15$ because long side + short side of the small rectangle is $15$... | E | 150 |
04fa84157a48ae907460ce977e966a64 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | We see that the $2$ rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the $2$ rectangles put together is a square. So, we can infer that the length of the rectangles is $10$ . Adding that to the width of the third rectangle which is $5$ , we get that the... | E | 150 |
04fa84157a48ae907460ce977e966a64 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the re... | E | 150 |
807b327a91d7f9b39f16db6f4074255a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4 | Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$
[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label... | [asy] draw((-12,0)--(0,5)); draw((0,5)--(12,0)); draw((12,0)--(0,-5)); draw((0,-5)--(-12,0)); draw((0,0)--(12,0)); draw((0,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,-5)); dot((-12,0)); dot((0,5)); dot((12,0)); dot((0,-5)); label("A",(-12,0),W); label("B",(0,5),N); label("C",(12,0),E); label("D",(0,-5),S); label(... | D | 120 |
807b327a91d7f9b39f16db6f4074255a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4 | Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$
[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label... | Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted ... | D | 120 |
4f920158d12a12c0f92c9e068a227c6a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257... | null | 48 |
4f920158d12a12c0f92c9e068a227c6a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ test... | A | 48 |
4f920158d12a12c0f92c9e068a227c6a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | We can compare each of the scores with the average of $81$ $76$ $\rightarrow$ $-5$ $94$ $\rightarrow$ $+13$ $87$ $\rightarrow$ $+6$ $100$ $\rightarrow$ $+19$
So the last one has to be $-33$ (since all the differences have to sum to $0$ ), which corresponds to $81-33 = \boxed{48}$ | null | 48 |
4f920158d12a12c0f92c9e068a227c6a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | We know that she scored $76, 94,$ and $87$ points on her first $3$ tests for a total of $257$ points and that she wants her average to be $81$ for her $5$ total tests. Therefore, she needs to score a total of $405$ points. In addition, one of the final $2$ tests needs to be the maximum of $100$ points, to make the fina... | A | 48 |
3fbc19723e11bd84ebda8802cd496c04 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8 | Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\q... | After Gilda gives $20$ % of the marbles to Pedro, she has $80$ % of the marbles left. If she then gives $10$ % of what's left to Ebony, she has $(0.8*0.9)$ $72$ % of what she had at the beginning. Finally, she gives $25$ % of what's left to her brother, so she has $(0.75*0.72)$ $\boxed{54}$ of what she had in the begin... | E | 54 |
3fbc19723e11bd84ebda8802cd496c04 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8 | Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\q... | Suppose Gilda has 100 marbles.
Then, she gives Pedro 20% of 100 = 20, she remains with 80 marbles.
Out of 80 marbles, she gives 10% of 80 = 8 to Ebony.
Thus, she remains with 72 marbles.
Then, she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.
And, $\frac{54}{100}$ =54%= $\boxed{54}$ | E | 54 |
3fbc19723e11bd84ebda8802cd496c04 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8 | Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\q... | (Only if you have lots of time do it this way)
Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So, the only option that is greater than... | E | 54 |
60fc5e9801861e2017a5d8ae36544b67 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | Let $x$ be the number of students taking both a math and a foreign language class.
By P-I-E, we get $70 + 54 - x$ $93$
Solving gives us $x = 31$
But we want the number of students taking only a math class,
which is $70 - 31 = 39$
$\boxed{39}$ | D | 39 |
60fc5e9801861e2017a5d8ae36544b67 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | We have $70 + 54 = 124$ people taking classes. However, we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class, we subtract the people who didn't take on... | D | 39 |
60fc5e9801861e2017a5d8ae36544b67 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | [asy] draw(circle((-0.5,0),1)); draw(circle((0.5,0),1)); label("$\huge{x}$", (0, 0)); label("$70-x$", (-1, 0)); label("$54-x$", (1, 0)); [/asy]
We know that the sum of all three areas is $93$ So, we have: \[93 = 70-x+x+54-x\] \[93 = 70+54-x\] \[93 = 124 - x\] \[-31=-x\] \[x=31\]
We are looking for the number of student... | null | 39 |
60fc5e9801861e2017a5d8ae36544b67 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | We are looking for students in math only, which is the complement (exactly the rest of the students) compared to those taking a language class. Since $54$ students take a language (with or without math), we subtract that from the total number of students. Since $93-54= 39,$ our answer is $\boxed{39}.$ (It's not necessa... | D | 39 |
c835ee23fd691d1a813e846a06f122dc | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13 | palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$
$\textbf{(A) }2\q... | Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be wri... | A | 2 |
c835ee23fd691d1a813e846a06f122dc | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13 | palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$
$\textbf{(A) }2\q... | We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88,$ and $99$ . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. $110 \ldots$ Aha! This multiple of 11, 110, not only isn’t a... | A | 2 |
c835ee23fd691d1a813e846a06f122dc | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13 | palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$
$\textbf{(A) }2\q... | As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$ . Since we know this fact and we are finding the smallest value possible, we can start with the first three-dig... | A | 2 |
8db56dc8ff3d7b0101248301d5ae5a23 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16 | Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | The only option that is easily divisible by $55$ is $110$ , which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$
And, $\text{Average Speed}$ $\frac{\text{Total Distance}}{\text{Total Time}}$
Thus, $\frac{125}{50} = \frac{5}{2}$
Both are equal and thus our answer is $\boxed{1... | D | 110 |
8db56dc8ff3d7b0101248301d5ae5a23 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16 | Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | To calculate the average speed, simply evaluate the total distance over the total time. Let the number of additional miles he has to drive be $x.$ Therefore, the total distance is $15+x$ and the total time (in hours) is \[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\] We can set up the following equation: \[\fr... | D | 110 |
8db56dc8ff3d7b0101248301d5ae5a23 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16 | Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | If he travels $15$ miles at a speed of $30$ miles per hour, he travels for 30 min. Average rate is total distance over total time so $(15+d)/(0.5 + t) = 50$ , where d is the distance left to travel and t is the time to travel that distance. Solve for $d$ to get $d = 10+50t$ . You also know that he has to travel $55$ mi... | D | 110 |
8db56dc8ff3d7b0101248301d5ae5a23 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16 | Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | Let $h$ be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is $50$ mph, means total distance over total time. For 15 miles at 30 mph, the time taken is $\frac{1}{2}$ hour, so the total time for this trip would be $\frac{1}{2} + h$ hours. For the total distance, 15 miles are trave... | D | 110 |
9fa79832a2b1c62aa46a0a6f59fe4b37 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19 | In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for ... | This isn't finished
to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.
Therefore, we use Case 2 since it brings the greater amount of poi... | C | 24 |
9fa79832a2b1c62aa46a0a6f59fe4b37 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19 | In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for ... | We can name the top three teams as $A$ $B$ , and $C$ . We can see that (respective scores of) $A=B=C$ because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: $AB$ $BC$ , and $AC$ come twice. In order to even out the scores and get the ... | C | 24 |
9fa79832a2b1c62aa46a0a6f59fe4b37 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19 | In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for ... | To start, we calculate how many games each team plays. Each team can play against $5$ people twice, so there are $10$ games that each team plays. So the answer is $10\cdot 3$ which is $30!$ But wait... if we want $3$ teams to have the same amount of points, there can't possibly be a player who wins all their games. Let... | C | 24 |
b8f4ef8926aaf3100e44a06f702f6035 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20 | How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$ | We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$ . If $x^2-5 = 4$ , then $x^2 = 9 \implies x = \pm 3$ , giving 2 solutions. If $x^2-5 = -4$ , then $x^2 = 1 \implies x = \pm 1$ , giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{4}$ . Further, the equation is a quartic in $x... | D | 4 |
b8f4ef8926aaf3100e44a06f702f6035 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20 | How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$ | We can expand $(x^2-5)^2$ to get $x^4-10x^2+25$ , so now our equation is $x^4-10x^2+25=16$ . Subtracting $16$ from both sides gives us $x^4-10x^2+9=0$ . Now, we can factor the left hand side to get $(x^2-9)(x^2-1)=0$ . If $x^2-9$ and/or $x^2-1$ equals $0$ , then the whole left side will equal $0$ . Since the solutions ... | D | 4 |
b8f4ef8926aaf3100e44a06f702f6035 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20 | How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$ | Subtract 16 from both sides and factor using difference of squares:
\[(x^2 - 5)^2 = 16\] \[(x^2 - 5)^2 - 16 =0\] \[(x^2 - 5)^2 - 4^2 = 0\] \[[(x^2 - 5)-4][(x^2 - 5) + 4] = 0\] \[(x^2 - 9)(x^2 - 1) =0\] \[(x+3)(x-3)(x+1)(x-1) = 0\]
Quite obviously, this equation has $\boxed{4}$ solutions. | D | 4 |
f807402c8b794087eddf48dfc2700f7d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21 | What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ | First, we need to find the coordinates where the graphs intersect.
We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$ , and $y=1+x$ intersect at $(4,5)$ , we call this line x.
Doing the same thing, we get $x=-4.$ Thus, $y=5$ . Also, $y=5$ and ... | E | 16. |
f807402c8b794087eddf48dfc2700f7d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21 | What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ | Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{16}$ | E | 16 |
f807402c8b794087eddf48dfc2700f7d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21 | What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ | $y = x + 1$ and $y = -x + 1$ have $y$ -intercepts at $(0, 1)$ and slopes of $1$ and $-1$ , respectively. Since the product of these slopes is $-1$ , the two lines are perpendicular. From $y = 5$ , we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this t... | E | 16 |
51e56ddf93bce8450ef0fd68c6eb40c4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }... | Suppose the fraction of discount is $x$ . That means $(1-x)(1+x)=0.84$ ; so, $1-x^{2}=0.84$ , and $(x^{2})=0.16$ , obtaining $x=0.4$ . Therefore, the price was increased and decreased by $40$ %, or $\boxed{40}$ | E | 40 |
51e56ddf93bce8450ef0fd68c6eb40c4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }... | After the first increase by $p$ percent, the shirt price became $(1+p)$ times greater than the original. Upon the decrease in p percent on this price, the shirt price became $(1-p)$ times less than $(1+p)$ , or $(1-p)(1+p)$ . We know that this price is $84$ percent of the original, so $(1-p)(1+p) = 0.84$
From here, we ... | E | 40 |
51e56ddf93bce8450ef0fd68c6eb40c4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }... | We can try out every option and see which one works. By this method, we get $\boxed{40}$ | E | 40 |
51e56ddf93bce8450ef0fd68c6eb40c4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }... | Let our original cost be $$ 100.$ We are looking for a result of $$ 84,$ then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try ... | null | 40 |
51e56ddf93bce8450ef0fd68c6eb40c4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22 | A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }... | Let our original cost be $$ 100$ , so we are looking for a whole number of $$ 84$ . Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) $$100$ increased by 20% = $$120$ , and $$120$ decreased by 20% = $... | E | 40 |
fc470b3a431554ea1e0809bff9a88b59 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23 | After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $... | Given the information above, we start with the equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$ ,where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$ , or $28x+28\cdot 15=13... | B | 11 |
fc470b3a431554ea1e0809bff9a88b59 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23 | After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $... | We first start by setting the total number of points as $28$ , since $\text{LCM}(4,7) = 28$ . However, we see that this does not work since we surpass the number of points just with the information given ( $28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30$ $(> 28)$ ). Next, we can see that the total number of points sco... | B | 11 |
fc470b3a431554ea1e0809bff9a88b59 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23 | After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $... | Adding together Alexa's and Brittany's fractions, we get $\frac{15}{28}$ as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: $\frac{15x}{28x}$ where $x$ is the common ratio. Let $y$ and $z$ and $w$ be the number of people who scored 1, 2, an... | B | 11 |
fc470b3a431554ea1e0809bff9a88b59 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23 | After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $... | We can rewrite the question as an algebraic equation: $\frac{1}{4} x + \frac{2}{7} x + 15 + y$ , where $x$ represents the total amount of points and $y$ the amount of points the $7$ other players scored. From there, we add the two fractions to get $\frac{15}{28} x + 15 = x$ . Subtracting $\frac{15}{28} x$ from both sid... | B | 11 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | We use the line-segment ratios to infer area ratios and height ratios.
Areas:
$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$
$BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$
Heights:
Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$
$AD:DC = 1... | null | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | Draw $X$ on $\overline{AF}$ such that $\overline{XD}$ is parallel to $\overline{BC}$
Triangles $BEF$ and $EXD$ are similar, and since $BE = ED$ , they are also congruent, and so $XE=EF$ and $XD=BF$
$AC:AD = 3$ implies $\frac{AF}{AX} = 3 = \frac{FC}{XD} = \frac{FC}{BF}$ , so $BC=BF + 3BF = 4BF$ $BF=\frac{BC}{4}$
Since $... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* p... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | $\frac{BF}{FC}$ is equal to $\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}$ . The area of triangle $ABE$ is equal to $60$ because it is equal to on half of the area of triangle $ABD$ , which is equal to one-third of the area of triangle $ABC$ , which is $360$ . The area of triangle $ACE$ i... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | Extend $\overline{BD}$ to $G$ such that $\overline{AG} \parallel \overline{BC}$ as shown: [asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ES... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | Let $ADB$ be a right triangle, and $BD=CD$
Let $A=(-2\sqrt{30}, 0)$
$B=(0, 4\sqrt{30})$
$C=(4\sqrt{30}, 0)$
$D=(0, 0)$
$E=(0, 2\sqrt{30})$
$F=(\sqrt{30}, 3\sqrt{30})$
The line $\overleftrightarrow{AE}$ can be described with the equation $y=x-2\sqrt{30}$
The line $\overleftrightarrow{BC}$ can be described with $x+y=4\sq... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /*... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$A$",A,N); dot(B); label("$B$", B,SW);dot(C); label("$C$",C,... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$A$",A,N); dot(B); label("$B$", B,SW);dot(C); label("$C$",C,SE); dot(DD); label("$D$",DD,NE)... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed)... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label("$A$",A,N); label("$B$", B,SW); label("$C$",C,SE); label("$D$",DD,NE); la... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D... | B | 30 |
24bfde7529dba29c74dc7996166640d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24 | In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] un... | [asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D... | B | 30 |
27316f96e067163949d5c3f51279a4b7 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25 | Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ | Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can ... | C | 190 |
27316f96e067163949d5c3f51279a4b7 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25 | Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ | Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ $3$ , or $6$ distinguishable ways to assign $a$ $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1... | C | 190 |
27316f96e067163949d5c3f51279a4b7 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25 | Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ | Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we ... | C | 190 |
2163c16fc3cb8abde1d473c6bf50a6f2 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1 | An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\... | You can see that since the ratio of real building's heights to the model building's height is $1:20$ . We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\b... | A | 14 |
2163c16fc3cb8abde1d473c6bf50a6f2 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1 | An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\... | We can compute $\frac{289}{20}$ and round our answer to get $\boxed{14}$ .
It is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem. | A | 14 |
2163c16fc3cb8abde1d473c6bf50a6f2 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1 | An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\... | We know that $20 \cdot 14 = 280 ,$ and that $20 \cdot 15 = 300 .$ These are the multiples of $20$ around $289 ,$ and the closest one of those is $280.$ Therefore, the answer is $\dfrac {280} {20} = \boxed{14} .$ | A | 14 |
ac4ead35d69de5c71100b4e772ba0132 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_2 | What is the value of the product
\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(... | By adding up the numbers in each of the $6$ parentheses, we get:
$\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$
Using telescoping, most of the terms cancel out diagonally. We are left with $\frac{7}{1}$ which is equivalent to $7$ . Thus, the answer would be $\box... | D | 7 |
a8030b076ff05a2e8c53be8f1dd18d2b | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4... | We count $3 \cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\boxed{13}$ | C | 13 |
a8030b076ff05a2e8c53be8f1dd18d2b | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4... | We can see here that there are $9$ total squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then, we can easily find that each corner has an area of one square and there are $4$ corners so we add that to the original 9 squares to get $9+4=\boxed... | C | 13 |
a8030b076ff05a2e8c53be8f1dd18d2b | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4... | We can apply Pick's Theorem here. There are $8$ lattice points, and $12$ lattice points on the boundary. Then,
\[8 + 12 \div 2 - 1 = \boxed{13}.\] | C | 13 |
22b5d061dc065cb2287a8ae6fa232a49 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$ , and our answer is $-1009+2019=\boxed{1010}$ | E | 1010 |
22b5d061dc065cb2287a8ae6fa232a49 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get $\boxed{1010}$ | E | 1010 |
22b5d061dc065cb2287a8ae6fa232a49 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | It is similar to the Solution 1:
Rearranging the terms, we get $1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)$ , and our answer is $1+1009=\boxed{1010}$ | E | 1010 |
22b5d061dc065cb2287a8ae6fa232a49 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | Note that the sum of consecutive odd numbers can be expressed as a square, namely $1+3+5+7+...+2017+2019 = 1010^2$ . We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have $1010^2-1009^2-1009$ . ... | null | 1010 |
256dd39a0bcc17e2d1e6145633d54067 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6 | On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\text... | Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is [mathjax]r=\dfrac dt=\dfrac{10}{0.5}=20[/mathjax] mph. His speed on the highway then is [mathjax]60[/mathjax] mph. He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]5... | C | 80 |
256dd39a0bcc17e2d1e6145633d54067 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6 | On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\text... | Since Anh drives [mathjax]3[/mathjax] times as fast on the highway, it takes him [mathjax]\dfrac{1}{3}[/mathjax] of the time to drive [mathjax]10[/mathjax] miles on the highway than on the coastal road. [mathjax]\dfrac{1}{3}[/mathjax] of [mathjax]30[/mathjax] is [mathjax]10[/mathjax], and since he drives [mathjax]50[/m... | C | 80 |
5ae0e5f2a55483523dd12add31424d16 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_7 | The $5$ -digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$ . What is the remainder when this number is divided by $8$
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$ . This means $2+0+1+8+U$ must be divisible by $9$ . The only possible value for $U$ then must be $7$ . Since we are looking for the remainder when divided by $8$ , we can ignore the thousands. The remainder when $18... | B | 3 |
34f26b2cea34e49c9d3cb59812cea591 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_8 | Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.
[asy] size(8cm); void drawbar(real x, real h) { fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0... | The mean, or average number of days is the total number of days divided by the total number of students. The total number of days is $1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109$ . The total number of students is $1+3+2+6+8+3+2=25$ . Hence, $\frac{109}{25}=\boxed{4.36}$ | C | 4.36 |
b10f85a3ef592eb40a89ee804700fa7e | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_9 | Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?
$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf... | She will place $(12\cdot2)+(14\cdot2)=52$ tiles around the border. For the inner part of the room, we have $10\cdot14=140$ square feet. Each tile takes up $4$ square feet, so he will use $\frac{140}{4}=35$ tiles for the inner part of the room. Thus, the answer is $52+35= \boxed{87}$ | B | 87 |
b10f85a3ef592eb40a89ee804700fa7e | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_9 | Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?
$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf... | The area around the border: $(12 \cdot 2) + (14 \cdot 2) = 52$ . The area of tiles around the border: $1 \cdot 1 = 1$ . Therefore, $\frac{52}{1} = 52$ is the number of tiles around the border.
The inner part will have $(12 - 2)(16 - 2) = 140$ . The area of those tiles are $2 \cdot 2 = 4$ $\frac{140}{4} = 35$ is the amo... | null | 87 |
7be54399bfbdd9ae6e5aec44c54d589b | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_13 | Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for La... | Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$
The value $y$ has to be greater than $82$ , because otherwise she would receive the same score on her last test. Additionally, the greatest value for $y$ is $98$ (as $y=100$ would make $x$ as a decimal), so theref... | A | 4 |
7be54399bfbdd9ae6e5aec44c54d589b | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_13 | Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for La... | The average point is $82$ leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests is $81$ points, then the last tests should be $86$ points to keep the average point is fixed. The... | A | 4 |
901ecba54db1e3832a2fae79df10cd17 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_14 | Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$ . The next pla... | D | 18 |
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