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2013 AMC 8 真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2017年7月25日下午12:59

2013 AMC 8 真题

答案详细解析请参考文末

Problem 1

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Problem 2

A sign at the fish market says, "50% off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Problem 3

What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$ ?

$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$

Problem 4

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

Problem 5

Hammie is in the $6^\text{th}$ grade and weighs 106 pounds. Her quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?

$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5} \qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$

Problem 6

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?

$[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]$

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Problem 7

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?

$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Problem 8

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34$

Problem 9

The Inedible Bulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?

$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$

Problem 10

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Problem 11

Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Problem 12

At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?

$\textbf{(A)}\ 25\% \qquad \textbf{(B)}\ 30\% \qquad \textbf{(C)}\ 33\% \qquad \textbf{(D)}\ 40\% \qquad \textbf{(E)}\ 45\%$

Problem 13

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$

Problem 14

Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?

$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$

Problem 15

If $3^p + 3^4 = 90$ , $2^r + 44 = 76$ , and $5^3 + 6^s = 1421$ , what is the product of $p$ , $r$ , and $s$ ?

$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$

Problem 16

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$

Problem 17

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Problem 18

Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?

$[asy] import three; size(3inch); currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]$

$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$

Problem 19

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?

$\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \qquad \textbf{(B)}\ \text{Hannah, Bridget, Cassie} \\ \qquad \textbf{(C)}\ \text{Cassie, Bridget, Hannah} \qquad \textbf{(D)}\ \text{Cassie, Hannah, Bridget} \\ \qquad \textbf{(E)}\ \text{Bridget, Cassie, Hannah}$

Problem 20

A $1\times 2$ rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?

$\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3$

Problem 21

Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Problem 22

Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?

$[asy] picture corner; draw(corner,(5,0)--(35,0)); draw(corner,(0,-5)--(0,-35)); for (int i=0; i<3; ++i) { for (int j=0; j>-2; --j) { if ((i-j)<3) { add(corner,(50i,50j)); } } } draw((5,-100)--(45,-100)); draw((155,0)--(185,0),dotted); draw((105,-50)--(135,-50),dotted); draw((100,-55)--(100,-85),dotted); draw((55,-100)--(85,-100),dotted); draw((50,-105)--(50,-135),dotted); draw((0,-105)--(0,-135),dotted); [/asy]$

$\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932$

Problem 23

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$ , and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ . What is the radius of the semicircle on $\overline{BC}$ ?

$[asy] import graph; pair A,B,C; A=(0,8); B=(0,0); C=(15,0); draw((0,8)..(-4,4)..(0,0)--(0,8)); draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); real theta = aTan(8/15); draw(arc((15/2,4),17/2,-theta,180-theta)); draw((0,8)--(15,0)); dot(A); dot(B); dot(C); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE);[/asy]$

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$

Problem 24

Squares $ABCD$ , $EFGH$ , and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE$ , respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?

$[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, SW); label("$G$", G, S); label("$H$", H, N); label("$I$", I, NE); label("$J$", J, SE); [/asy]$

$\textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}$

Problem 25

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

$[asy] pair A,B; size(8cm); A=(0,0); B=(480,0); draw((0,0)--(480,0),linetype("3 4")); filldraw(circle((8,0),8),black); draw((0,0)..(100,-100)..(200,0)); draw((200,0)..(260,60)..(320,0)); draw((320,0)..(400,-80)..(480,0)); draw((100,0)--(150,-50sqrt(3)),Arrow(size=4)); draw((260,0)--(290,30sqrt(3)),Arrow(size=4)); draw((400,0)--(440,-40sqrt(3)),Arrow(size=4)); label("$A$", A, SW); label("$B$", B, SE); label("$R_1$", (100,-40), W); label("$R_2$", (260,40), SW); label("$R_3$", (400,-40), W);[/asy]$

$\textbf{(A)}\ 238\pi\qquad\textbf{(B)}\ 240\pi\qquad\textbf{(C)}\ 260\pi\qquad\textbf{(D)}\ 280\pi\qquad\textbf{(E)}\ 500\pi$

In order to have her model cars in perfect, complete rows of 6, Danica must have a number of cars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy $\boxed{\textbf{(A)}\ 1}$ more model car.

The 50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is $6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is $\boxed{\textbf{(D)}\ 12}$ dollars.

Notice that we can pair up every two numbers to make a sum of 1: $\begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\ &=& 500\end{eqnarray*}$ Therefore, the answer is $4 \cdot 500= \boxed{\textbf{(E)}\ 2000}$ .

Each of her seven friends paid $\text{\textdollar}2.50$ to cover Judi's portion. Therefore, Judi's portion must be $\text{\textdollar}2.50 \cdot 7 = {\textdollar}17.50$ . Since each friend was supposed to pay the same amount, each friend was supposed to pay $\frac{1}{8}$ of the total bill; the total bill must be $8 \cdot 7 \cdot \text{\textdollar}2.50 = \boxed{\textbf{(C)} \ \text{\textdollar}140}$ .

The median here is obviously less than the mean, so options $(A)$ and $(B)$ are out.Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.The average weight of the five kids is $\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26$ .Therefore, the average weight is bigger, by $26-6 = 20$ pounds, making the answer $\boxed{\textbf{(E)}\ \text{average, by 20}}$ .

Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ . $y$ is the answer we're looking for. $[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]$ We see that $600 = 30x$ , making $x = 20$ . $[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("20",(1,1)); label("$y$",(2,3)); [/asy]$ It follows that $20 = 5y$ , so $y = \boxed{\textbf{(C)}\ 4}$ .

If Trey saw $\frac{6\text{ cars}}{10\text{ seconds}}$ , then he saw $\frac{3\text{ cars}}{5\text{ seconds}}$ .2 minutes and 45 seconds can also be expressed as $2\cdot60 + 35 = 155$ seconds.Trey's rate of seeing cars, $\frac{3\text{ cars}}{5\text{ seconds}}$ , can be multiplied by $165\div5 = 33$ on the top and bottom (and preserve the same rate): $\frac{3\cdot 33\text{ cars}}{5\cdot 33\text{ seconds}} = \frac{3\text{ cars}}{5\text{ seconds}}$ . It follows that the most likely number of cars is $\textbf{\boxed{(C)}\ 100}$ .

First, there are $2^3 = 8$ ways to flip the coins, in order.The ways to get two consecutive heads are HHT and THH.The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\ \frac38}$ .

This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$ .However, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}$

To find either the LCM or the GCF of two numbers, always prime factorize first.The prime factorization of $180 = 3^2 \times 5 \times 2^2$ .The prime factorization of $594 = 3^3 \times 11 \times 2$ .Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.Thus the answer = $\frac{5940}{18}$ = $\boxed{\textbf{(C)}\ 330}$ .

We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.On Monday, he was at a rate of $5 \text{ m.p.h}$ . So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$ .For Wednesday, he walked at a rate of $3 \text{ m.p.h}$ . Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3} \text { hours}$ .On Friday, he walked at a rate of $4 \text{ m.p.h}$ . So, $4x = 2 \text{ miles}\implies x=\frac{2}{4} \text {hours}$ .Adding up the hours yields $\frac{2}{5} \text { hours}$ + $\frac{2}{3} \text { hours}$ + $\frac{2}{4} \text { hours}$ = $\frac{94}{60} \text { hours}$ .We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times 3 \text{ days}\implies x = \frac{3}{2} \text { hours}$ .To find the amount of time saved, subtract the two amounts: $\frac{94}{60} \text { hours}$ - $\frac{3}{2} \text { hours}$ = $\frac{4}{60} \text { hours}$ . To convert this to minutes, we multiply by $60$ .Thus, the solution to this problem is $\dfrac{4}{60}\times 60=\boxed{\textbf{(D)}\ 4}$

First, find the amount of money one will pay for three sandals without the discount. We have $\textdollar 50\times 3 \text{ sandals} = \textdollar 150$ .Then, find the amount of money using the discount: $50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105$ .Finding the percentage yields $\frac{105}{150} = 70 \%$ .To find the percent saved, we have $100 \% -70 \%= \boxed{\textbf{(B)}\ 30}$

Let the two digits be $a$ and $b$ .The correct score was $10a+b$ . Clara misinterpreted it as $10b+a$ . The difference between the two is $|9a-9b|$ which factors into $|9(a-b)|$ . Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is $\boxed{\textbf{(A)}\ 45}$ .

The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$ . The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$ . Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

First, we're going to solve for $p$ . Start with $3^p+3^4=90$ . Then, change $3^4$ to $81$ . Subtract $81$ from both sides to get $3^p=9$ and see that $p$ is $2$ . Now, solve for $r$ . Since $2^r+44=76$ , $2^r$ must equal $32$ , so $r=5$ . Now, solve for $s$ . $5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$ . Therefore, $s=4$ . $prs$ equals $2*5*4$ which equals $40$ . So, the answer is $\boxed{\textbf{(B)}\ 40}$ .

We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together: $5:3 = 5(8):3(8) = 40:24$ $8:5 = 8(5):5(5) = 40:25$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{\textbf{(E)}\ 89}$ .

The mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$ . Therefore the numbers are $333, 334, 335, 336, 337, 338$ , so the answer is $\boxed{\textbf{(B)}\ 338}$

There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}$ .

If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is $\boxed{\textbf{(D) Cassie, Hannah, Bridget}}$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = 2.392515856236789, xmax = 4.844947145877386, ymin = 6.070697674418619, ymax = 8.062241014799170; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)--cycle, zzttqq); /* draw figures */ draw((2.912600422832983,6.903678646934476)--(4.326813985206080,6.903678646934476)); draw(shift((3.619707204019532,6.903678646934476))*xscale(0.7071067811865487)*yscale(0.7071067811865487)*arc((0,0),1,0.000000000000000,180.0000000000000)); draw((3.619707204019532,6.903678646934476)--(4.119707204019532,6.903678646934476)); draw((3.619707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)); draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476), zzttqq); draw((4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476), zzttqq); draw((4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476), zzttqq); draw((3.119707204019531,6.903678646934476)--(3.119707204019531,7.403678646934482), zzttqq); label("$1$",(3.847061310782247,6.924820295983102),SE*labelscalefactor); label("$1$",(4.155729386892184,7.208118393234687),SE*labelscalefactor); draw((3.619707204019532,6.903678646934476)--(4.119707204019532,7.403678646934476)); label("$\sqrt{2}$",(3.711754756871041,7.288456659619466),SE*labelscalefactor); label("$2$",(3.563763213530660,7.563298097251601),SE*labelscalefactor); /* dots and labels */ dot((2.912600422832983,6.903678646934476)); dot((4.326813985206080,6.903678646934476)); dot((3.619707204019532,6.903678646934476)); dot((4.119707204019532,6.903678646934476),blue); dot((3.619707204019532,6.903678646934476)); dot((3.119707204019531,6.903678646934476),blue); dot((3.119707204019531,7.403678646934482),blue); dot((4.119707204019532,7.403678646934476),blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, $\sqrt{1^2+1^2}=\sqrt{2}$ . The area is $\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}$ .

$[asy] unitsize(8mm); for(int i=0; i<=8; ++i) { draw((0,i)--(8,i)); draw((i,0)--(i,8)); } fill((2,1)--(6,1)--(6,6)--(2,6)--cycle); for(int j=0; j<= 38; ++j) { draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0), black+linewidth(3)); draw((6,6)--(6,8)--(8,8)--(8,6)--(6,6), black+linewidth(3)); }[/asy]$ We see that Samantha only has one way to go through in the city park, 6 ways to the city park to school, and 3 ways from her house to the city park. After this, we can see that there are $6 \cdot 3 = 18$ ways for Samantha to go to school. Hence, $\boxed{\textbf{E}}$

There are $61$ vertical columns with a length of $32$ toothpicks, and there are $33$ horizontal rows with a length of $60$ toothpicks. An effective way to verify this is to try a small case, i.e. a $2 \times 3$ grid of toothpicks. Thus, our answer is $61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}$ .

If the semicircle on AB were a full circle, the area would be $16\pi$ . Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is $\boxed{\textbf{(B)}\ 7.5}$ .

$[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= extension(I,J,A,B); dot(X,red); draw(I--X--B,red); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); label("$X$", X,SE); dot("$J$", J, SE);[/asy]$ First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\overline{IJ}$ until it hits the extension of $\overline{AB}$ . Call this point $X$ . The area of triangle $AXJ$ then is $\dfrac{3 \cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \cdot 1 = 2$ . Thus, our desired area is $6-2 = 4$ . Now, the ratio of the shaded area to the combined area of the three squares is $\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$ .

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses $2\pi*2/2=2\pi$ inches, and it gains $2\pi$ inches on B. $[asy] unitsize(0.04cm); import graph; draw(circle(96*dir(0),4),linewidth(1.3)); draw(circle(96*dir(-45),4),linetype("4 4")); draw(circle(96*dir(-90),4),linetype("4 4")); draw(circle(96*dir(-135),4),linetype("4 4")); draw(circle(96*dir(180),4),linetype("4 4")); draw((-100,0)..(0,-100)..(100,0)); draw((-96,0)..(0,-96)..(96,0),dotted); label("1",(-87,0)); label("2",(-60,-60)); label("3",(0,-87)); label("4",(60,-60)); label("5",(87,0)); [/asy]$ So, the departure from the length of the track means that the answer is $\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$ .