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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 the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$ | 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 observation for solving for the Tree $4$ 's height. Tree $4$ can't be $11$ meters for the sum of the five tree heights to still end with $1$ . Therefore, the Tree $4$ is $44$ meters tall.
Now, Tree $5$ can either be $22$ or $88$ . Find the average height for both cases of Tree $5$ . Doing this, we realize the Tree $5$ must be $22$ for the average height to end with $0.2$ and that the average height is $\boxed{24.2}$ | 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 the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$ | 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 $5$ . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both $22$ meters. At this point, our table looks as follows: \[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\]
If Tree 4 now has a height of $11$ , then Tree 5 would need to have height $22$ , but in that case $S$ would equal $88$ , which is not $1$ more than a multiple of $5$ . So we instead take Tree 4 to have height $44$ . Then the sum of the heights of the first 4 trees is $22+11+22+44 = 99$ , so using a height of $22$ for Tree 5 gives $S=121$ , which is $1$ more than a multiple of $5$ (whereas $88$ gives $S = 187$ , which is not). Thus the average height of the trees is $\frac{121}{5} = \boxed{24.2}$ meters. | 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 the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$ | 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 $6$ , because those are the only numbers from $0$ to $9$ that are congruent to $1$ after taking modulo $5$ . The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are $66$ and $121$ . Since the sum of the heights of Tree 1, 2, and 3 is already $55$ , we know that $66$ is impossible to obtain. Then, we can decide with relative confidence that the answer should be $\frac{121}{5} = \boxed{24.2}$ | 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 the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$ | 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 get $c+d \equiv 1 \pmod {5}$ .
The only possible combination of $c$ and $d$ that abides the condition are 44 and 22, respectively.
Adding these up, we get 121, and $\frac{121}{5}$ is $\boxed{24.2}$ | 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 there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$ | 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 $P$ to each of the white squares immediately beneath it(also called the Water Fall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated).
[asy] int N = 7; for (int i = 0; i < 8; ++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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("$1$", (5.5, .5)); label("$1$", (4.5, 1.5)); label("$1$", (6.5, 1.5)); label("$1$", (3.5, 2.5)); label("$1$", (7.5, 2.5)); label("$2$", (5.5, 2.5)); label("$1$", (2.5, 3.5)); label("$3$", (6.5, 3.5)); label("$3$", (4.5, 3.5)); label("$4$", (3.5, 4.5)); label("$3$", (7.5, 4.5)); label("$6$", (5.5, 4.5)); label("$10$", (4.5, 5.5)); label("$9$", (6.5, 5.5)); label("$19$", (5.5, 6.5)); label("$9$", (7.5, 6.5)); label("$28$", (6.5, 7.5)); [/asy]
The answer is therefore $\boxed{28}$ | 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 there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$ | 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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("$P$", (5.5,.5)); label("$Q$", (6.5,7.5)); label("$X$", (8.5,3.5)); label("$Y$", (8.5,5.5)); [/asy]
The total number of paths from $P$ to $Q$ , including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\binom{4+3}{3} = \binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$ . To get to $X$ , the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through $Y$ is $\binom{4+1}{1}\cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both $X$ and $Y$ ; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$ , and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{28}$ | 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 there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$ | 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 ${7 \choose 3}=35$ . However, we originally promised that we will not go over the border, and now we have to subtract the paths that do go over the border. The paths that go over the border are any paths that start with RRR(1 path), RR(5 paths) and LRRR(1 path) so our final number of paths is $35-7=\boxed{28}.$ ~PEKKA | 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 there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++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) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$ | 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. Notice how, at the first step, we figured that the answer was even, so choice D and E are eliminated, and after the second row, we realized it must be a multiple of 4, so choice B is eliminated. When we get to the fourth row, we do not multiply by 2 again, since we have limited possibilities rather than multiplying by 2 again. Choice C implies that there are two possibilities per row; however, we know that if you go to the farthest possible, you only have one possibility, so it is not $2^5 = 32$ so we know that the answer is choice $\boxed{28}$ .
~hh99754539 | 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((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$ | 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: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\boxed{83}$ | 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((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$ | 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)); label("$1$", (0, 0), UnFill(0.1mm)); label("$2$", (-1, 0), UnFill(0.1mm)); label("$4$", (-2, 0), UnFill(0.1mm)); // Upper branches label("$1$", (-3, 1), UnFill(0.1mm)); label("$2$", (-4, 1), UnFill(0.1mm)); label("$4$", (-5, 1), UnFill(0.1mm)); label("$\textbf{8}$", (-6, 1.5), UnFill(0.1mm)); label("$\textbf{1}$", (-6, 0.5), UnFill(0.1mm)); // Lower branches label("$8$", (-3, -1), UnFill(0.1mm)); label("$16$",(-4, -1), UnFill(0.1mm)); label("$5$", (-5, -0.5), UnFill(0.1mm)); label("$32$", (-5, -1.5), UnFill(0.1mm)); label("$\textbf{10}$", (-6, -0.5), UnFill(0.1mm)); label("$\textbf{64}$", (-6, -1.5), UnFill(0.1mm)); [/asy]
The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are $1$ $8$ $64$ , and $10$ . Thus the answer is $1+8+64+10=\boxed{83}$ | 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((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$ | 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, $I=2$ , and if $I$ is odd, $I=0$ , but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$ . Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\frac{1}{3}$ , which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$ . Substituting in $4$ , if $I$ is even, $I=8$ , but if $I$ is odd, $I=1$ . Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$ . If it is $1$ , then by the argument we have just made, the 5th-to-last number has to be $2$ , the 6th-to-last number has to be $4$ , and the 7th-to-last number, which is the first number, must be either $1$ or $8$ . In this way, we have ultimately found two solutions: $N=1$ and $N=8$
On the other hand, if the 4th-to-last number is $8$ , substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\frac{7}{3}$ , but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$ , but if I is odd, $I=5$ , and both of these are valid. If the 6th-to-last number is $32$ , then the first number must be $64$ , since $\frac{31}{3}$ is not an integer; if the 6th-to-last number is $5,$ then the first number has to be $10$ , as $\frac{4}{3}$ is not an integer. This means that, in total, there are $4$ solutions for $N$ , specifically, $1$ $8$ $10$ , and $64$ , which sum to $\boxed{83}$ | 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; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn.
If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.
In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{150}$ | 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 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$
In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1 = 90$ ways for this case.
Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{150}$ | 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't receive an award, there are $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$
However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{150}$ | 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)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$ | 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 $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{651}$ | 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)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$ | 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 to determine $y=\boxed{651}$ | 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)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$ | 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)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$ | 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("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); label("$r$",(5.2,5/2)); label("$r$",(3.2,1/2)); label("$s$",(3.2,3/2)); [/asy]
From this diagram, it is evident that $r+s+r=2020$ . Also, the side length of $S_1$ and $S_3$ is $r+s$ . Then, $r+s+s+r+s=3322$ . Now, we have 2 systems of equations.
\begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*}
We can see an $r+s+r$ in the 2nd equation, so substituting that in gives us $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{651}$ | 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 $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$ | We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{9}$ | D | 9 |
6ad9edf661be5bfc54af9e2bd6686167 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1 | Ike and Mike go into a sandwich shop with a total of $$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?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$ | 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{align*} We don't want a remainder so the maximum number of sandwiches is $6$ .
The total money spent is $6\cdot 4.50=27$ .
The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items.
Hence, the answer is $\boxed{9}$ | 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)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$ | 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 $\boxed{150}$ . ~avamarora | 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)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$ | 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 rectangle is $50 \cdot 3 = \boxed{150}$ . ~~fath2012 | 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)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$ | 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 $15$ . The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\cdot10$ which is $\boxed{150}$ .
~~mathboy282 | 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)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$ | 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 length of the rectangle is $15$ . Multiplying $10$ and $15$ gives us $15\cdot10$ which is $\boxed{150}$ .
~~awesomepony566 | 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)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$ | 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 rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer $\boxed{150}$ .
~elenafan | 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("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$ | [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("E",(0,0),SW); [/asy]
A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ $12$ $\overline{EC}$
Consider one of the right triangles:
[asy] draw((-12,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,5)); dot((-12,0)); dot((0,5)); label("A",(-12,0),W); label("B",(0,5),N); label("E",(0,0),SE); [/asy]
$\overline{AB}$ $13$ , and $\overline{AE}$ $12$ . Using the Pythagorean theorem, we find that $\overline{BE}$ $5$ .
You know the Pythagorean triple, (5, 12, 13).
Thus the values of the two diagonals are $\overline{AC}$ $24$ and $\overline{BD}$ $10$ .
The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ $\frac{24\cdot{10}}{2}$ $120$
$\boxed{120}$ | 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("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$ | 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 side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get $\boxed{120}$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$ | 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. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{48}$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$ | 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 $2$ tests is $100$ , the least possible score she can get is $\boxed{48}$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$ | 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 final test score—the one that we are looking for—the lowest score possible for her to earn. We can now see here that the sum of $76+94+100$ has a units digit of $0$ and that the final test score must have a units digit ending with a $5$ . Now, $87$ needs to be added to a number that makes the sum divisible by $5$ . Among the answer choices of $48, 52, 66, 70,$ and $74$ , only $\boxed{48}$ has a units digit that works. ( $48+87=135$ , giving us a units digit of $5$ .) | 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?
$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$ | 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 beginning left. | 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?
$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$ | 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?
$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$ | (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 100% - 55% is $\boxed{54}$ | 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 language class?
$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$ | 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 language class?
$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$ | 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 only one math class, so we get $70 - 31 = \boxed{39}$ | 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 language class?
$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$ | [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 students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\boxed{39}$ | 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 language class?
$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$ | 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 necessary to know that $70$ students take math.)
~hailstone | 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 $N$
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$ | 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 written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$ . Then, $N = 110$ , and the sum of the digits of $N$ is $1+1+0 = \boxed{2}$ | 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 $N$
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$ | 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 palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$ ’s digits is $1+1+0 = \boxed{2}$ | 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 $N$
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$ | 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-digit multiple of 11 which is $110$ . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \boxed{2}$ | 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 $\boxed{110}.$ | 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: \[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\] Simplifying the equation, we get \[15+x=25+\frac{10x}{11}.\] Solving the equation yields $x=110,$ so our answer is $\boxed{110}$ | 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 $55$ miles per hour for some time, so $d=55t$ . Plug that in for d to get $55t = 10+50t$ and $t=2$ and since $d=55t$ $d = 2\cdot55 =110$ , the answer is $\boxed{110}$ | 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 traveled in the first part and $55h$ miles in the second. This gives the following equation:
\[\dfrac{15+55h}{\frac{1}{2}+h} = 50.\]
Cross multiplying, we get that $15 + 55h = 50h + 25$ , and simple algebra gives $h=2$ . In 2 hours traveling at 55 mph, the distance traveled is $\frac{2 \hspace{0.05 in} \text{hours}}{1} \cdot \frac{55 \hspace{0.05 in} \text{miles}}{1 \hspace{0.05 in} \text{hour}} = 2 \cdot 55 \hspace{0.05 in} \text{miles} = 110 \hspace{0.05 in} \text{miles}$ , which is choice $\boxed{110}$ | 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 each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$ | 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 points, or $\boxed{24}$ | 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 each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$ | 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 maximum score, we can say that in match $AB$ $A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$ . This tells us that each team $A$ $B$ , and $C$ win and lose twice. This gives each team a total of $3 + 3 + 0 + 0 = 6$ points. Now, we need to include the other three teams. We can label these teams as $D$ $E$ , and $F$ . We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD$ $AD$ $AE$ $AE$ $AF$ $AF$ $BD$ $BD$ $BE$ $BE$ $BF$ $BF$ $CD$ $CD$ $CE$ $CE$ $CF$ , and $CF$ . We can say $A$ $B$ , and $C$ win each of these in order to obtain the maximum score that $A$ $B$ , and $C$ can have. If $A$ $B$ , and $C$ win all six of their matches, $A$ $B$ , and $C$ will have a score of $18$ $18 + 6$ results in a maximum score of $\boxed{24}$ | 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 each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$ | 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 the top three teams be $A,B$ , and $C.$ $A$ plays $B$ and $C$ twice so in order to maximize the games being played, we can split it $50-50$ between the $4$ games $A$ plays against $B$ or $C$ . We find that we just subtract $2$ games or $6$ points. Therefore the answer is $30-6$ $24$ or $\boxed{24}$ | 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 $x$ , so by the Fundamental Theorem of Algebra , there can be at most four real 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 \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$ | 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 can be both positive and negative, we have $4$ solutions: $-3,3,-1,1$ (we can find these solutions by setting $x^2-9$ and $x^2-1$ equal to $0$ and solving for $x$ ). So, the answer is $\boxed{4}$ | 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 $y=1-x$ intersect at $(-4,5)$ , and we call this line y.
It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$ , we call this line z.
Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{16.}$ | 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 $y=1-x$
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ | 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 triangle is a $45-45-90$ triangle, so the legs are $4\sqrt{2}$ each and the area is $\frac{(4\sqrt{2})^2}{2} = \boxed{16}$ | 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 $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 can list the factors of $0.84$ and see which are equidistant from $1$ . We see that $0.6$ and $1.4$ are both $0.4$ from $1$ , so $p = 0.4 = 40 \%$ , or choice $\boxed{40}$ | 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 $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $\boxed{40}$ , and we have the answer; it worked.
(OR: try (C) first to eliminate 2 answer choices) | 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 $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $$120$ decreased by 20% = $$96$ , a whole number, and (E) $$100$ increased by 40% = $$140$ , and $$140$ decreased by 40% = $$84$ , a whole number.
Thus, $40$ % or $\boxed{40}$ is the answer. | 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 $7$ team members?
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$ | 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 $28x+28\cdot 15=13t$ . Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$ . Then, it is easy to see $u=2$ $t=56$ ) is the only candidate remaining, giving $x=\boxed{11}$ | 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 $7$ team members?
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$ | 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 scored is $56$ as, if it is more than or equal to $84$ , at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: $56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45$ , and thus, the other seven players would have scored a total of $56-45 = \boxed{11}$ . (We see that this works since we could have $4$ of them score $2$ points, and the other $3$ of them score $1$ point.) | 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 $7$ team members?
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$ | 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, and 0 points, respectively. Writing an equation, we have $\frac{13x}{28x} = 15 + y + 2z + 0w.$ We want all of our variables to be integers. Thus, we want $15 + y + 2z = 0 \pmod {13}.$ Simplifying, $y+2z = 11 \pmod {13}.$ The only possible value, as this integer sum has to be less than $7 \cdot 2 + 1 = 15,$ must be 11. Therefore, $y+2z = 11,$ and the answer is $\boxed{11}$ | 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 $7$ team members?
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$ | 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 sides, we get $\frac{13}{28} x = y + 15$ . We multiply each side by $28$ to get rid of the denominator, in which we get $13x = 420 + 28y$ . Now let’s think of this logically. This equation is telling us that if you add $420$ and $28$ times the amount of points scored by the extra $7$ players, you get $13$ times the amount of points total. And since we have to have a whole number of points total, this means that $420 + 28y$ must be divisible by $13$ . Plugging in all the answer choices for $y$ , we find that the only answer that makes $420 + 28y$ divisible by $13$ is $\boxed{11}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $A$
$AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$
$BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$
Conclusion:
$\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$ , and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$
So, $\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$ , and thus, $[EBF] = \boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $XE=EF$ $AX = XE = EF$ , and since $AX + XE + EF = AF$ , all of these are equal to $\frac{AF}{3}$ , and so the altitude of triangle $BEF$ is equal to $\frac{1}{3}$ of the altitude of $ABC$
The area of $ABC$ is $360$ , so the area of $\triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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; /* point style */ real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.28,2.39),dotstyle); label("$A$", (0.36,2.59), NE * labelscalefactor); dot((-2.8,-1.17),dotstyle); label("$B$", (-2.72,-0.97), NE * labelscalefactor); dot((3.78,-1.05),dotstyle); label("$C$", (3.86,-0.85), NE * labelscalefactor); dot((1.2887445398528459,1.3985482236874887),dotstyle); label("$D$", (1.36,1.59), NE * labelscalefactor); dot((-0.7199623188673492,-1.1320661821070033),dotstyle); label("$F$", (-0.64,-0.93), NE * labelscalefactor); dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); label("$E$", (-0.2,0.57), NE * labelscalefactor); label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.
The triangle we will consider is $\triangle ABC$ (obviously), and we will let $E$ be the center of mass, so that $D$ balances $A$ and $C$ (this is true since $E$ balances $B$ and $D$ , but $E$ also balances $A$ and $B$ and $C$ so $D$ balances $A$ and $C$ ), and $F$ balances $B$ and $C$
We know that $AD:CD=1:2$ and $D$ balances $A$ and $C$ so we assign $2$ to $A$ and $1$ to $C$ . Then, since $D$ balances $A$ and $C$ , we get $D = A + C = 2 + 1 = 3$ (by mass points addition).
Next, since $E$ balances $B$ and $D$ in a ratio of $BE:DE=1:1$ , we know that $B=D=3$ . Similarly, by mass points addition, $E=B+D=3+3=6$
Finally, $F$ balances $B$ and $C$ so $F=B+C=3+1=4$ . We can confirm we have done everything right by noting that $E$ balances $A$ and $F$ , so $E$ should equal $A+F$ , which it does.
Now that our points have weights, we can solve the problem. $BF:FC=1:3$ so $BF:BC=1:4$ so $[ABF]=\frac{1}{4}[ABC]=90$ . Also, $EF:EA=2:4=1:2$ so $EF:AF=1:3$ so $[EBF]=\frac{1}{3}[ABF]=\boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | $\frac{BF}{FC}$ 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 $ACE$ is the sum of the areas of triangles $AED$ and $CED$ , which is respectively $60$ and $120$ . So, $\frac{BF}{FC}$ is equal to $\frac{60}{180}$ $\frac{1}{3}$ , so the area of triangle $ABF$ is $90$ . That minus the area of triangle $ABE$ is $\boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SE); label("$F$", F, S); label("$G$", G, ENE); [/asy] Then, $\triangle ADG \sim \triangle CDB$ and $\triangle AEG \sim \triangle FEB$ . Since $CD = 2AD$ , triangle $CDB$ has four times the area of triangle $ADG$ . Since $[CDB] = 240$ , we get $[ADG] = 60$
Since $[AED]$ is also $60$ , we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus, triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$ , giving $[BEF] = (60+60)/4 = \boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); label("$x$", (B+E+F)/3); label("$120-x$", (F+E+C)/3); [/asy] As before, we figure out the areas labeled in the diagram. Then, we note that \[\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}\] Even simpler: \[\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120}{240}\] Solving gives $x = \boxed{30}$ .
(Credit to scrabbler94 for the idea) | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | 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 $x+y=4\sqrt{30}$
Solving, we get $x=3\sqrt{30}$ and $y=\sqrt{30}$
Now we can find $EF=BF=2\sqrt{15}$
$[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{30}\blacksquare$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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; /* point style */ real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */ /* draw figures */ draw(circle((0,0), 5), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); draw((-4,-3)--(0,5), linewidth(2)); draw((0,5)--(4,3), linewidth(2)); draw((12,-1)--(-4,-3), linewidth(2)); draw((0,5)--(0,-5), linewidth(2)); draw((-4,-3)--(0,-5), linewidth(2)); draw((4,3)--(0,2.48), linewidth(2)); draw((4,3)--(12,-1), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); /* dots and labels */ dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((-5,0),dotstyle); dot((-4,-3),dotstyle); label("B", (-4.45,-3.38), NE * labelscalefactor); dot((4,3),dotstyle); label("$D$", (4.15,3.2), NE * labelscalefactor); dot((0,5),dotstyle); label("A", (-0.09,5.26), NE * labelscalefactor); dot((12,-1),dotstyle); label("C", (12.23,-1.24), NE * labelscalefactor); dot((0,-5),dotstyle); label("$G$", (0.19,-4.82), NE * labelscalefactor); dot((0,2.48),dotstyle); label("I", (-0.33,2.2), NE * labelscalefactor); dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((0,-2.5),dotstyle); label("F", (0.23,-2.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
Let $A[\Delta XYZ]$ $\text{Area of Triangle XYZ}$
$A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240$
$A[\Delta ABE] = A[\Delta AED] = 60$ (the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around $\Delta ABD$ with $BD$ as is diameter. Extend $AF$ to $G$ such that it meets the circle at $G$ . Draw line $BG$
$A[\Delta ABD] = A[\Delta ABG] = 120$ (Since $\square ABGD$ is cyclic)
But $A[\Delta ABE]$ is common in both with an area of 60. So, $A[\Delta AED] = A[\Delta BEG]$
Therefore $A[\Delta AED] \cong A[\Delta BEG]$ (SAS Congruency Theorem).
In $\Delta AED$ , let $DI$ be the median of $\Delta AED$
which means $A[\Delta AID] = 30 = A[\Delta EID]$
Rotate $\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$ $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\frac{AE}{EF} = 2:1$
$AE$ is a radius and $EF$ is half of it implies $EF$ $\frac{radius}{2}$
which means $A[\Delta BEF] \cong A[\Delta DEI]$
Thus, $A[\Delta BEF] = \boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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,SE); dot(DD); label("$D$",DD,NE); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label("$M$",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy] Using the ratio of $\overline{AD}$ and $\overline{CD}$ , we find the area of $\triangle ADB$ is $120$ and the area of $\triangle BDC$ is $240$ . Also using the fact that $E$ is the midpoint of $\overline{BD}$ , we know $\triangle ADE = \triangle ABE = 60$ .
Let $M$ be a point such $\overline{EM}$ is parellel to $\overline{CD}$ . We immediatley know that $\triangle BEM \sim BDC$ by $2$ . Using that we can conclude $EM$ has ratio $1$ . Using $\triangle EFM \sim \triangle AFC$ , we get $EF:AE = 1:2$ . Therefore using the fact that $\triangle EBF$ is in $\triangle ABF$ , the area has ratio $\triangle BEF : \triangle ABE=1:2$ and we know $\triangle ABE$ has area $60$ so $\triangle BEF$ is $\boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy] By Menelaus's Theorem on triangle $BCD$ , we have \[\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.\] Therefore, \[[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{30}.\] | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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); draw(A--B--C--cycle); draw(A--F); draw(B--D); dot(A); label("$A$",A,NE); dot(B); label("$B$",B,dir(180)); dot(C); label("$C$",C,SE); dot(D); label("$D$",D,dir(0)); dot(E); label("$E$",E,SE); dot(F); label("$F$",F,SW); [/asy] Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\overline{AC}$ in a $1:2$ ratio, we draw $\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$ . By definition, Point $E$ splits line segment $\overline{BD}$ in a $1:1$ ratio, so we draw $\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$ $1$ unit away from both. We then draw line segments $\overline{AB}$ and $\overline{BC}$ . We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$ , we can easily see that triangle $EBF$ forms a right triangle occupying $\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus $\frac{1}{4}\div3=\frac{1}{12}$ , and since the area of triangle $ABC$ is $360$ , this means that the area of triangle $EBF$ is $\frac{1}{12}\times360=\boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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); label("$E$",EE,NW); label("$F$",FF,S); label("$G$",G,S); [/asy] We know that $AD = \dfrac{1}{3} AC$ , so $[ABD] = \dfrac{1}{3} [ABC] = 120$ . Using the same method, since $BE = \dfrac{1}{2} BD$ $[ABE] = \dfrac{1}{2} [ABD] = 60$ . Next, we draw $G$ on $\overline{BC}$ such that $\overline{DG}$ is parallel to $\overline{AF}$ and create segment $DG$ . We then observe that $\triangle AFC \sim \triangle DGC$ , and since $AD:DC = 1:2$ $FG:GC$ is also equal to $1:2$ . Similarly (no pun intended), $\triangle DBG \sim \triangle EBF$ , and since $BE:ED = 1:1$ $BF:FG$ is also equal to $1:1$ . Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$ , or equivalently, $BF = \dfrac{1}{4} BC$ . Thus, $[BFA] = \dfrac{1}{4} [BCA] = 90$ . We already know that $[ABE] = 60$ , so the area of $\triangle EBF$ is $[BFA] - [ABE] = \boxed{30}$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); label("$x$", (B+E+F)/3); label("$120-x$", (F+E+C)/3); [/asy] Since $AD:DC=1:2$ thus $\triangle ABD=\frac{1}{3} \cdot 360 = 120.$
Similarly, $\triangle DBC = \frac{2}{3} \cdot 360 = 240.$
Now, since $E$ is a midpoint of $BD$ $\triangle ABE = \triangle AED = 120 \div 2 = 60.$
We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\triangle BEC$ and $\triangle DEC$ share 2 sides.
We know that $\triangle BEC=\triangle DEC=240 \div 2 = 120$ since $E$ is a midpoint of $BD.$
Let's label $\triangle BEF$ $x$ . We know that $\triangle EFC$ is $120-x$ since $\triangle BEC = 120.$
Note that with this information now, we can deduct more things that are needed to finish the solution.
Note that $\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$
We want to find $x.$
This is a simple equation, and solving we get $x=\boxed{30}.$ | 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] 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); dot(EE); label("$E$",EE,NW); dot(FF); label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$ | [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)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); [/asy]
Because $AD:DC=1:2$ and $E$ is the midpoint of $BD$ , we know that the areas of $ABE$ and $AED$ are $60$ and the areas of $DEC$ and $EBC$ are $120$ \[\frac{[EBF]}{[EFC]} = \frac{[ABF]}{[AFC]} = \frac{ [ABE]}{[AEC]} = \frac{60}{180}\] $[EBF] = \frac{120}{4} = \boxed{30}$ | 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 remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$ . In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$ . This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$ . We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$ , therefore the answer is $\boxed{190}$ | 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 $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\boxed{190}$ | 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 can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{190}$ possible sets of values in total. | 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?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | 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 $\boxed{14}$ | 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?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | 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?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | 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
\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | 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 $\boxed{7}$ | 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)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$ | 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)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$ | 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 $9+4=\boxed{13}$ . That is how I did it. | 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)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$ | 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 $1010^2-1009^2-1009$ . Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{1010}$ | 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\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax] | 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]50[/mathjax] minutes (Note that [mathjax]60[/mathjax] miles per hour is the same as [mathjax]1[/mathjax] mile per minute). The total amount of minutes spent on his trip is [mathjax]30+50\implies \boxed{80}[/mathjax]. | 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\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax] | 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[/mathjax] miles on the highway, we multiply [mathjax]10[/mathjax] by [mathjax]5[/mathjax] to get [mathjax]50[/mathjax]. This means it took him [mathjax]50[/mathjax] minutes to drive on the highway, and if we add the [mathjax]30[/mathjax] minutes it took for him to drive on the coastal road, we would get [mathjax]\boxed{80}[/mathjax]. | 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 $187$ is divided by $8$ is $\boxed{3}$ | 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+0.15,h)--(x-0.15,h)--cycle,gray); } draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle); for (real i=1; i<5; i=i+0.5) { draw((0.5,i)--(7.5,i),gray); } drawbar(1.0,1.0); drawbar(2.0,2.0); drawbar(3.0,1.5); drawbar(4.0,3.5); drawbar(5.0,4.5); drawbar(6.0,2.0); drawbar(7.0,1.5); for (int i=1; i<8; ++i) { label("$"+string(i)+"$",(i,0.25)); } for (int i=1; i<9; ++i) { label("$"+string(i)+"$",(0.5,0.5*(i+1)),W); } label("Number of Days of Exercise",(4,-0.1)); label(rotate(90)*"Number of Students",(-0.1,2.75)); [/asy] What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?
$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$ | 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?
$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$ | 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 amount of tiles for the inner part. So, $52 + 35 = \boxed{87}$ | 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 Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$ | 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 therefore, the greatest value $x$ can be is $98$ . As a result, only $4$ numbers work, $86, 90, 94$ and $98$ . Thus, the answer is $\boxed{4}$ | 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 Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$ | 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 possible points are $86$ $90$ $94$ $98$ . The answer is $\boxed{4}$ | 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 place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18\implies \boxed{18}$ | D | 18 |