## AMC 8 2008

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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.

Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 38 \qquad \textbf{(E)}\ 50$

$\textbf{B}$

She had $50-12-12\times2=14$ dollars left.

The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$, in order. What 4-digit number is represented by the code word $\text{CLUE}$?

$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$

$\textbf{A}$

We can derive that $C=8$, $L=6$, $U=7$, and $E=1$. Therefore, the answer is $8671$.

If February is a month that contains Friday the $13^{\text{th}}$, what day of the week is February 1?

$\textbf{(A)}\ \text{Sunday} \qquad \textbf{(B)}\ \text{Monday} \qquad \textbf{(C)}\ \text{Wednesday} \qquad \textbf{(D)}\ \text{Thursday}\qquad \textbf{(E)}\ \text{Saturday}$

$\textbf{A}$

Since February 13 is a Friday, February 6 is also a Friday. Then we go back by days and find February 1 is a Sunday.

In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

$\textbf{C}$

The sum of the areas of the three trapezoids is $16-1=15$. Since the three trapezoids are congruent, the area of each of them is $15/3=5$.

Barney Schwinn notices that the odometer on his bicycle reads $1441$, a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661$. What was his average speed in miles per hour?

$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 22$

$\textbf{E}$

It takes him $4+6=10$ hours to ride $1661-1441=220$ miles. So his average speed is $220/10=22$ mph.

In the figure, what is the ratio of the area of the gray squares to the area of the white squares?

$\textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1$

$\textbf{D}$

The gray squares are made up of of 6 little squares while the white squares are made up of 10. So the ratio of the area of the gray squares to the area of the white squares is $6:10=3:5$.

If $\dfrac{3}{5}=\dfrac{M}{45}=\dfrac{60}{N}$, what is $M+N$?

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$

$\textbf{E}$

We can separate the given equation into two equations $\dfrac35 = \dfrac{M}{45}$ and $\dfrac35 = \dfrac{60}{N}$. Hence, we get $M=27$ and $N=100$. So $M+N=127$.

Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?

$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85$

$\textbf{D}$

There are a total of $100+60+40+120=320$ dollars of sales spread through $4$ months, for an average of $320/4 =80$ dollars.

In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the the first year her investment suffered a $15\%$ loss, but during the second year the remaining investment showed a $20\%$ gain. Over the two-year period, what was the change in Tammy's investment?

$\textbf{(A)}\ 5\%\text{ loss}\qquad \textbf{(B)}\ 2\%\text{ loss}\qquad \textbf{(C)}\ 1\%\text{ gain}\qquad \textbf{(D)}\ 2\% \text{ gain} \qquad \textbf{(E)}\ 5\%\text{ gain} \qquad$

$\textbf{D}$

Now Tammy has $100\times(1-15\%)\times(1+20\%)=102$ dollars. So finally she gets a $2\%$ gain.

The average age of the $6$ people in Room A is $40$. The average age of the $4$ people in Room B is $25$. If the two groups are combined, what is the average age of all the people?

$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$

$\textbf{D}$

The total age of the first group is $6\times40=240$. The total age of the second group is $4\times25=100$. So the average age of all people is $(240+100)/(6+4)=34$.

Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

$\textbf{A}$

The number of students that have both a dog and a cat is $20+26-39 =7$.

A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\dfrac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it rise to a height less than $0.5$ meters?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

$\textbf{C}$

For the $n\text{th}$ bounce, the height is $3\left(\dfrac23\right)^n$ meters. The first bounce reaches $2$ meters, the second $4/3$, the third $8/9$, the fourth $16/27$, and the fifth $32/81$. Half of $81$ is $40.5$, so the ball does not reach the required height on the 5th bounce.

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$, $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?

$\textbf{(A)}\ 160\qquad \textbf{(B)}\ 170\qquad \textbf{(C)}\ 187\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 354$

$\textbf{C}$

The sum of each results $122+125+127=374$ is twice the combined weight of the three boxes because each boxes is weighted twice. So the combined weight of the three boxes is $374/2=187$ pounds.

Three $\text{A's}$, three $\text{B's}$, and three $\text{C's}$ are placed in the nine spaces so that each row and column contain one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?

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

$\textbf{C}$

For the first row, we have two ways to arrange B and C. For the first column, we also have two ways to arrange B and C. After the positions of two B's and two C's are fixed, the rest letters are fixed either. Hence, we have $2\times2=4$ arrangements.

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$

$\textbf{B}$

The total number of points for the first $8$ games is $7+4+3+6+8+3+1+5=37$. We have to make this a multiple of $9$ by scoring less than $10$ points in the ninth game. The closest multiple of $9$ is $45$. So the score of the ninth game is $45-37=8$. Now we have to add a number to get a multiple of 10. The next multiple is $50$. So the score of the tenth game is $50-45=5$. So the answer is $8\times5=40$.

A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?

$\textbf{(A)} \:1 : 6 \qquad\textbf{ (B)}\: 7 : 36 \qquad\textbf{(C)}\: 1 : 5 \qquad\textbf{(D)}\: 7 : 30\qquad\textbf{ (E)}\: 6 : 25$

$\textbf{D}$

The volume of the seven unit cubes is $7$. The surface areas from the top, the bottom, the front, the back, the left side and the right side are all 5. So the surface of the shape is $5\times 6=30$. The ratio of the volume to the surface area is $7 : 30$.

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 128\qquad \textbf{(D)}\ 132\qquad \textbf{(E)}\ 136$

$\textbf{D}$

A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \times 13 = 156$. Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \times 24 = 24$. The difference in areas is $156-24=132$.

Two circles that share the same center have radii $10$ meters and $20$ meters. An aardvark runs along the path shown, starting at $A$ and ending at $K$. How many meters does the aardvark run?

$\textbf{(A)}\ 10\pi+20\qquad\textbf{(B)}\ 10\pi+30\qquad\textbf{(C)}\ 10\pi+40\qquad\textbf{(D)}\ 20\pi+20\qquad\textbf{(E)}\ 20\pi+40$

$\textbf{E}$

The trace of the aardvark is composed of $\dfrac14$ of the perimeter of the large circle, $\dfrac12$ of the perimeter of the small circle, and twice the radius of the large circle. So the total distance is $$\dfrac14\times2\pi\times20+\dfrac12\times2\pi\times10+2\times20=20\pi+40$$

Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart?

$\textbf{(A)}\ \dfrac{1}{4}\qquad\textbf{(B)}\ \dfrac{2}{7}\qquad\textbf{(C)}\ \dfrac{4}{11}\qquad\textbf{(D)}\ \dfrac{1}{2}\qquad\textbf{(E)}\ \dfrac{4}{7}$

$\textbf{B}$

We choose the first point randomly. No matter which point it is, it has two adjacent points who are located 1 unit away. We have 7 choices for the second point, but only 2 of them are 1 unit away from the first point. So the probability is $\dfrac27$.

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\dfrac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

$\textbf{B}$

Let $b$ be the number of boys and $g$ be the number of girls. Here we have \[\dfrac23 b = \dfrac34 g \Rightarrow b = \dfrac98 g\] For $g$ and $b$ to be positive integers, $g$ must be a multiple of 8, and the smallest possible value is $8$. This yields $9$ boys. The minimum number of students is $8+9=17$.

Jerry cuts a wedge from a $6$-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?

$\textbf{(A)} 48 \qquad \textbf{(B)} 75 \qquad \textbf{(C)}151\qquad \textbf{(D)}192 \qquad \textbf{(E)}603$

$\textbf{C}$

The slice is cutting the cylinder into two equal wedges. The volume of the cylinder is $\pi r^2 h = \pi (4^2)(6) = 96\pi$. The volume of the wedge is half of the volume of the cylinder. So the answer is $48\pi \approx 151$.

For how many positive integer values of $n$ are both $\dfrac{n}{3}$ and $3n$ three-digit whole numbers?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

$\textbf{A}$

Since both $\dfrac{n}3$ and $3n$ are three-digit whole numbers, we have $300\le n\le333$, and $n$ is a multiple of 3. $\dfrac{333-300}3+1=12$ numbers between 300 and 333 are multiples of 3. So the answer is 12.

In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$?

$\textbf{(A)}\ \dfrac{1}{6}\qquad\textbf{(B)}\ \dfrac{2}{9}\qquad\textbf{(C)}\ \dfrac{5}{18}\qquad\textbf{(D)}\ \dfrac{1}{3}\qquad\textbf{(E)}\ \dfrac{7}{20}$

$\textbf{C}$

Let the side length of the square $ABCE$ be 1. The area of $\triangle ABF$ is $\dfrac12\times1\times\dfrac23=\dfrac13$. The area of $\triangle BCD$ is also $\dfrac13$. The area of $\triangle DEF$ is $\dfrac12\times\dfrac13\times\dfrac13=\dfrac1{18}$. So the area of $\triangle BFD$ is $1-\dfrac13-\dfrac13-\dfrac1{18}=\dfrac5{18}$. The ratio of the area of $\triangle BFD$ to the area of square $ABCE$ is $\dfrac5{18}$.

Ten tiles numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?

$\textbf{(A)}\ \dfrac{1}{10}\qquad \textbf{(B)}\ \dfrac{1}{6}\qquad \textbf{(C)}\ \dfrac{11}{60}\qquad \textbf{(D)}\ \dfrac{1}{5}\qquad \textbf{(E)}\ \dfrac{7}{30}$

$\textbf{C}$

Among the $10\times6=60$ possible results, the pairs that have products of squares are $(1,1)$, $(1,4)$, $(2,2)$, $(4,1)$, $(3,3)$, $(9,1)$, $(4,4)$, $(8,2)$, $(5,5)$, $(6,6),$ and $(9,4)$. So the probability is $\dfrac{11}{60}$.

Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Which of the following is closest to the percent of the design that is black?

$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 46\qquad \textbf{(D)}\ 47\qquad \textbf{(E)}\ 49\qquad$

$\textbf{A}$

Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc.

The entire circle's area is $144\pi$. The area of the black regions is $(100-64)\pi + (36-16)\pi + 4\pi = 60\pi$. The percentage of the design that is black is $\dfrac{60\pi}{144\pi} = \dfrac{5}{12} \approx 42\%$.