## AMC 8 2015

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 1 point for each correct answer. There is no penalty for wrong answers.
- No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
- Figures are not necessarily drawn to scale.
- You will have
**40 minutes**working time to complete the test.

How many square yards of carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are $3$ feet in a yard.)

$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$

$\textbf{A}$

The side lengths of the rectangular floor are $12/3=4$ yards and $9/3=3$ yards, respectively. So the area of the carpet is $3\times4=12$ square yards.

Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?

$\textbf{(A) }\dfrac{11}{32} \quad\textbf{(B) }\dfrac{3}{8} \quad\textbf{(C) }\dfrac{13}{32} \quad\textbf{(D) }\dfrac{7}{16}\quad \textbf{(E) }\dfrac{15}{32}$

$\textbf{D}$

$\triangle OBC$, $\triangle OCD$, $\triangle ODE$ each takes up $\dfrac18$ of the area of the octagon, while $\triangle OXB$ takes up $\dfrac1{16}$. So the fraction is $\dfrac18\times3+\dfrac1{16}=\dfrac7{16}$.

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

$\textbf{D}$

It takes Jill $\dfrac1{10}$ hour to the pool, which is 6 minutes. However, Jack spent $\dfrac14$ hour, or 15 minutes, to walk to the pool. Hence, Jill arrived $15-6=9$ minutes before Jack.

The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?

$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$

$\textbf{E}$

There are $2!=2$ ways to arrange boys, and $3!=6$ ways to arrange girls. So the answer is $2\times6=12$.

Billy's basketball team scored the following points over the course of the first $11$ games of the season:\[42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73\]If his team scores 40 in the 12th game, which of the following statistics will show an increase?

$\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}$

$\textbf{A}$

Since 40 is the lowest score in the set, the range is extended.

In $\triangle ABC$, $AB=BC=29$, and $AC=42$. What is the area of $\triangle ABC$?

$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$

$\textbf{B}$

If we take $AC$ as the base of the isosceles triangle, the height should be $\sqrt{29^2-\left(\dfrac{42}{2}\right)^2}=20$. Therefore, the area of the triangle is $\dfrac12\times42\times20=420$.

Each of two boxes contains three chips numbered $1$, $2$, $3$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

$\textbf{(A) }\dfrac{1}{9}\qquad\textbf{(B) }\dfrac{2}{9}\qquad\textbf{(C) }\dfrac{4}{9}\qquad\textbf{(D) }\dfrac{1}{2}\qquad \textbf{(E) }\dfrac{5}{9}$

$\textbf{E}$

Among all $3\times3=9$ results, the product is odd only happened when both numbers are odd. So the probability that the product is odd is $\dfrac{2\times2}{9}=\dfrac49$. Hence, the probability that the product is even is $1-\dfrac49=\dfrac59$.

What is the smallest whole number larger than the perimeter of any triangle with a side of length 5 and a side of length 19?

$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$

$\textbf{D}$

The length of the third side can not be greater than the sum of the other two sides, which is $5+19=24$. So the perimeter of the triangle must be less than $5+19+24=48$. 48 is the smallest whole number for this question.

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?

$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$

$\textbf{D}$

The number of widgets Janabel sold on day 20 is $1+2(20-1)=39$. The total number of sales is

\begin{align*}

1+3+5+\dots+37+39&=(1+39)+(3+37)+\dots\\&=40+40+\dots\\&=40\times10\\&=400

\end{align*}

How many integers between $1000$ and $9999$ have four distinct digits?

$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$

$\textbf{B}$

There are $9$ choices for the first number, since it cannot be $0$. Then there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This means there are $9\times9\times8\times7=4536$ integers between $1000$ and $9999$ with four distinct digits.

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel ($A, E, I, O,$ or $U$), the second and third must be two different letters among the $21$ non-vowels, and the fourth must be a digit ($0$ through $9$). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "$AMC8$"?

$\textbf{(A) } \dfrac{1}{22,050} \qquad \textbf{(B) } \dfrac{1}{21,000}\qquad \textbf{(C) } \dfrac{1}{10,500}\qquad \textbf{(D) } \dfrac{1}{2,100} \qquad \textbf{(E) } \dfrac{1}{1,050}$

$\textbf{B}$

There are 5 choices for the first letter, 21 choices for the second letter, 20 choices for the third letter since it must differ from the second letter, and 10 choices for the last digit. So there are $5\times21\times20\times10=21,000$ possible license plates. Hence, the probability that the plate is chosen as AMC8 is $\dfrac{1}{21,000}$.

How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH},$ or $\overline{EH}$ and $\overline{FG},$ does a cube have?

$\textbf{(A) }6 \qquad\textbf{(B) }12 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 24 \qquad \textbf{(E) } 36$

$\textbf{C}$

12 edges can be divided into 3 groups. Each group has 4 edges which are in parallel. Now we choose 2 edges from a group into a pair, which makes $\dbinom{4}{2}=6$ pairs from 1 group. So there are $3\times6=18$ pairs from 3 groups.

How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is $6$?

$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$

$\textbf{D}$

The average of the original set is 6. To keep the average unchanged, the two-element subset must have an average of 6 either. Hence, we have 5 subsets $\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}$.

Which of the following integers cannot be written as the sum of four consecutive odd integers?

$\textbf{(A)}\text{ 16}\qquad\textbf{(B)}\text{ 40}\qquad\textbf{(C)}\text{ 72}\qquad\textbf{(D)}\text{ 100}\qquad\textbf{(E)}\text{ 200}$

$\textbf{D}$

Let the four consecutive odd integers be $2n-3$, $2n-1$, $2n+1$ and $2n+3$. The sum of the four integers is $8n$, which means the sum must be a multiple of 8. Apparently 100 is not a multiple of 8.

At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?

$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$

$\textbf{D}$

Given that $149$ voted in favor of the first issue and $119$ voted in favor of the second issue, we know that $198-149=49$ voted against the first issue and $198-119=79$ voted against the second issue. Since there are 29 students who voted against both issue, we know that $49-29=20$ students voted against the first but in favor of the second, and $79-29=50$ students voted against the second but in favor of the first. Hence, the number of students who voted in favor of both issues is $198-29-20-50=99$.

In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\dfrac{1}{3}$ of all the ninth graders are paired with $\dfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

$\textbf{(A) } \dfrac{2}{15} \qquad \textbf{(B) } \dfrac{4}{11} \qquad \textbf{(C) } \dfrac{11}{30} \qquad \textbf{(D) } \dfrac{3}{8} \qquad \textbf{(E) } \dfrac{11}{15}$

$\textbf{B}$

Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\dfrac{n}{3}=\dfrac{2s}{5}$, which simplifies to $n=\dfrac{6s}{5}$. So the fraction of the total number of sixth and ninth graders have a buddy is $$\dfrac{\dfrac{n}{3}+\dfrac{2s}{5}}{n+s}=\dfrac4{11}$$

Jeremy's father drives him to school in rush hour traffic in $20$ minutes. One day, there is no traffic, so his father can drive him $18$ miles per hour faster and gets him to school in $12$ minutes. How far in miles is it to school?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

$\textbf{D}$

Let the distance be $d$ miles. His father drives $d$ miles in 20 minutes in rush hour, so the speed is $3d$ miles per hour. When there is no traffic, his father can drive $d$ miles in 12 minutes, so the speed is $5d$ miles per hour. Hence, we get $$3d+18=5d\rightarrow d=9$$

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$. Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$?

$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$

$\textbf{B}$

Since the top row is a arithmetic sequence, we get the middle of the top row as $(25+1)/2=13$. Similarly, we get the middle of the bottom row as $(17+81)/2=49$. Now look at the middle column, $X$ should be the average of the top and the bottom, which is $X=(13+49)/2=31$.

A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?

$\textbf{(A) }\dfrac{1}{6} \qquad \textbf{(B) }\dfrac{1}{5} \qquad \textbf{(C) }\dfrac{1}{4} \qquad \textbf{(D) }\dfrac{1}{3} \qquad \textbf{(E) }\dfrac{1}{2}$

$\textbf{A}$

The area of the $6\times5$ grid is 30. The area of the triangle is the area of the red rectangle subtracts three little triangles. Hence, the area of $\triangle ABC$ is $12-1.5-1.5-4=5$. The fraction of the grid covered by $\triangle ABC$ is $\dfrac{5}{30}=\dfrac16$.

Ralph went to the store and bought $12$ pairs of socks for a total of $\$24$. Some of the socks he bought cost $\$1$ a pair, some of the socks he bought cost $\$3$ a pair, and some of the socks he bought cost $\$4$ a pair. If he bought at least one pair of each type, how many pairs of $\$1$ socks did Ralph buy?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

$\textbf{D}$

Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ dollars. This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\times1=9$ dollars spent, which leaves $7$ dollars. If we replace one pair with a $3$ dollar pair, then we would spend an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would spend an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$s and $3$s is $2+2+3$. If we change $3$ pairs, we would have $6$ pairs of $\$1$ socks left. Adding the one pair from previously, we have $7$ pairs.

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?

$\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32$

$\textbf{C}$

Since the hexagon $ABCDEF$ is equiangular, we have $\angle ABC=120^\circ$. Given that $\triangle JBK$ is equilateral, we have $\angle JBK=60^\circ$. So $\angle CBK=360^\circ-120^\circ-90^\circ-60^\circ=90^\circ$, which makes $\triangle CBK$ a right triangle. $BC=EF=\sqrt{32}=4\sqrt2$, $BK=BJ=\sqrt{18}=3\sqrt2$, so the area of $\triangle CBK$ is $\dfrac12\times4\sqrt2\times3\sqrt2=12$.

On June $1$, a group of students are standing in rows, with $15$ students in each row. On June $2$, the same group is standing with all of the students in one long row. On June $3$, the same group is standing with just one student in each row. On June $4$, the same group is standing with $6$ students in each row. This process continues through June $12$ with a different number of students per row each day. However, on June $13$, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

$\textbf{C}$

The number has 12 factors, including 15 and 6. The least common multiple of 15 and 6 is 30. So the number must be a multiple of $30$. The smallest possible number is $60=2^2\times3^1\times5^1$ with $(2+1)(1+1)(1+1)=12$ factors.

Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup?

$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$

$\textbf{D}$

The sum of the 12 numbers is 35. Since the 5 integers in cups are consecutive, the 5 integers should be 5, 6, 7, 8, 9. If the slip with 3.5 goes into cup A, we need a slip of $5-3.5=1.5$ in cup A at the same time, which is impossible. Given that a slip with 3 goes into cup B, we need another slip with 3 in cup B to get a sum $3+3=6$. So the slip with 3.5 can not go into cup B. If the slip with 3.5 goes into cup C, the rest slips in cup C should have a sum of $7-3.5=3.5$, which is impossible. Similarly, the slip of 3.5 can not go into cup E because there is a slip of 2 inside, which makes the sum of the rest slips be $9-2=7$. This is exactly the same situation we met in cup C. Finally, the only possible choice for the slip with 3.5 is in cup D.

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$. Each team plays a $76$-game schedule. How many games does a team play within its own division?

$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

$\textbf{B}$

A team plays $3N$ games in its division and $4M$ games in the other division. This gives $3N+4M=76$. Given that $M>4$ and $N>2M$, we have $$76=3N+4M>3(2M)+4M=10M$$ So $4<M<7.6$, $M$ could be 5, 6, or 7. Trying each value of $M$ into the equation $3N+4M=76$, we found that only $M=7$ is possible. So the number of games a team play within its own division is $3N=76-4M=48$.

One-inch squares are cut from the corners of this $5$ inch square. What is the area in square inches of the largest square that can fit into the remaining space?

$\textbf{(A) } 9\qquad \textbf{(B) } 12\dfrac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\dfrac{1}{2}\qquad \textbf{(E) } 17$

$\textbf{C}$

The largest square is the white square in the center plus 4 red triangles. The area of the white square in the center is $3\times3=9$. The area of the red triangle is $\dfrac12(3)(1)=\dfrac32$. So the area of the largest square is $9+4\times\dfrac32=15$.