### 2008 AMC 8 Exam Solutions

Problems used with permission of the Mathematical Association of America. Scroll down to view solutions, print PDF solutions, view answer key, or:

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1.

Susan had $\50$ 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?

$12$

$14$

$26$

$38$

$50$

###### Solution(s):

Susan spent 2 \cdot $12 =$24 on rides. This means she spent a total of $12 +$24 = $36 at the carnival. This means that she has$50 - $36 =$14 left to spend.

2.

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 $\text{CLUE}?$

$\ 8671$

$\ 8672$

$\ 9781$

$\ 9782$

$\ 9872$

###### Solution(s):

The letter $C$ is in the $9$th position, so it would be the letter $8.$ The letter $L$ is in the $7$th position, so it would be the letter $6.$ The letter $U$ is in the $8$th position, so it would be the letter $7.$ The letter $E$ is in the $2$nd position, so it would be the letter $1.$

Therefore, when putting together the word $\text{CLUE},$ we get $8671.$

3.

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

$\ \text{Sunday}$

$\ \text{Monday}$

$\ \text{Wednesday}$

$\ \text{Thursday}$

$\ \text{Saturday}$

###### Solution(s):

Since the $13$th is a Friday, we know that the $6$th is also a Friday. The $1$st is $5$ days before the Friday, making it a Sunday.

4.

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?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

###### Solution(s):

Since the area of the larger triangle is $16$ and the area of the smaller triangle is $1.$ Thus, the area of the other $3$ trapezoids is $15.$ Since the $3$ trapezoids have a combined area of $15,$ each of their areas is $\dfrac{15}{3} =5.$

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?

$\ 15$

$\ 16$

$\ 18$

$\ 20$

$\ 22$

###### Solution(s):

The total distance traveled is $1661-1441 = 220$ miles. He also travelled $10$ hours. Thus, the average speed is $\dfrac{220}{10} = 22$ miles per hour.

6.

In the figure, what is the ratio of the area of the colored squares to the area of the uncolored squares?

$\ 3:10$

$\ 3:8$

$\ 3:7$

$\ 3:5$

$\ 1:1$

###### Solution(s):

The total area of the entire square is $16$ since it is a $4 \times 4$ square. The area of the colored region is $6,$ making the remaining area $16-6 = 10.$ Thus, the ratio is $6:10 = 3:5.$

7.

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

$\ 27$

$\ 29$

$\ 45$

$\ 105$

$\ 127$

###### Solution(s):

$\dfrac{3}{5}=\dfrac{M}{45}$ means $M = \dfrac{3\cdot 45}{5}=27.$

$\dfrac{3}{5}=\dfrac{60}{N}$ means $N = \dfrac{5\cdot 60}{3}=100.$

Therefore, $M +N = 127.$

8.

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

$\ 60$

$\ 70$

$\ 75$

$\ 80$

$\ 85$

###### Solution(s):

The total sales are $100+60+40+120 = 320.$

The average sales is then $\dfrac{320}{4} =80.$

9.

In $2005$ Tycoon Tammy invested $100$ dollars for two years. During 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?

$\ 5\%\text{ loss}$

$\ 2\%\text{ loss}$

$\ 1\%\text{ gain}$

$\ 2\% \text{ gain}$

$\ 5\%\text{ gain}$

###### Solution(s):

The $15\%$ loss means the investment went from $\100$ to $0.85\cdot \100 = \85.$

The $20\%$ gain means the investment went from $\85$ to \$85 \cdot 1.2 =$102. The total investment went from $\100$ to \$102, making a gain of $2\%.$

10.

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?

$\ 32.5$

$\ 33$

$\ 33.5$

$\ 34$

$\ 35$

###### Solution(s):

The sum of the ages in Room A is $6\cdot 40 = 240.$

The sum of the ages in Room B is $4\cdot 25 = 100.$

The total sum is $240+100 = 340.$

The average is therefore $\dfrac{340}{10} =34.$

11.

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?

$\ 7$

$\ 13$

$\ 19$

$\ 39$

$\ 46$

###### Solution(s):

The number of people that have both animals is equal to the number of people that own a cat plus the number of people that own a dog minus the number of people that own either.

Therefore, the number of people who own both is $20+26-39 = 7.$

12.

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 $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

###### Solution(s):

Since the height of a bounce decreases by $\frac 23$ each bounce, the height of each bounce is $3\cdot \left(\dfrac 23\right)^n.$

After $4$ bounces, the ball bounces $3\cdot \left(\dfrac 23\right)^4 = \dfrac{16}{27},$ which is greater than $\frac 12.$

After $5$ bounces, the ball bounces $3\cdot \left(\dfrac 23\right)^5 = \dfrac{32}{81},$ which is less than $\frac 12.$

Therefore, we are under $\frac 12$ after $5$ bounces.

13.

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?

$\ 160$

$\ 170$

$\ 187$

$\ 195$

$\ 354$

###### Solution(s):

Let the weights be $a,b,c.$ We know $\begin{cases}a+b = 122\\a+c = 125\\b+c = 127.\end{cases}$ Adding all of this yields $2(a+b+c) =$$122+125+127=374.$ This makes $a+b+c = 187.$

14.

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?

$\ 2$

$\ 3$

$\ 4$

$\ 5$

$\ 6$

###### Solution(s):

There are $2$ rows and $2$ columns to put B in, so there are $4$ places for it. After that, there is $1$ column and $1$ row available for C, so the number of combinations is $2\cdot 2\cdot 1\cdot 1 = 4.$

15.

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?

$\ 35$

$\ 40$

$\ 48$

$\ 56$

$\ 72$

###### Solution(s):

The sum of her first $8$ scores is $37.$ Since the average of the first $9$ scores is an integer, the sum of the first $9$ scores is a multiple of $9.$

Since the score is less than $10,$ the sum of the scores after $9$ games is between $37$ and $47,$ and is a multiple of $9,$ making the sum $45.$ Thus, the score of the $9$th game is $45-37=8.$

The sum of Theresa's first $9$ scores is $45.$ Since the average of the first $10$ scores is an integer, the sum of the first $10$ scores is a multiple of $10.$

Since the score is less than $10,$ the sum of the scores after $10$ games is between $45$ and $55,$ and is a multiple of $10,$ making the sum $50.$ Thus, the score of the $10$th game is $50-45=5.$

Therefore, their product is $8\cdot 5=40.$

16.

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?

$\:1 : 6$

$\: 7 : 36$

$\: 1 : 5$

$\: 7 : 30$

$\: 6 : 25$

###### Solution(s):

There are $7$ unit cubes, so the volume is $7.$

If we look at the perspective of the inner cube, we can see that there is one cube connected to each side. Furthermore, there are $5$ exposed on each of the outer cubes. This makes the surface area $5\cdot 6=30.$

This makes the ratio of volume to surface area $7:30.$

17.

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?

$\ 76$

$\ 120$

$\ 128$

$\ 132$

$\ 136$

###### Solution(s):

Let the length and width be $l,w$ respectively. The rectangle would have an area of $l+r+l+r = 2(l+r)=50.$ Thus, $l+r =25,$ so $r=25-l.$ The area is $lr = l(25-l)$$= 156.25 - (l-12.5)^2.$ This makes the area largest when $l$ is as close as possible to $12.5$ and the area is the smallest when $l$ is as far from $12.5.$ Therefore, the largest area is when $l=12,13$ and the smallest area is when $l=1,24.$

This makes the larger area equal to $156.25-(13-12.5)^2$$= 156.25-0.25 =156$ and the smaller area equal to $156.25-(24-12.5)^2$$= 156.25-132.25 =24.$

Thus, the difference is $156-24=132.$

18.

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?

$\ 10\pi+20$

$\ 10\pi+30$

$\ 10\pi+40$

$\ 20\pi+20$

$\ 20\pi+40$

###### Solution(s):

The circumference of a circle is $\pi d = 2r\pi ,$ so going a quarter of the way around is $\dfrac {\pi r }2.$

He goes a quarter of the way around the large circle, so this part is $10 \pi$ meters. He then goes from the larger circle to the smaller circle, which is $20-10=10$ meters.

He goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He then goes through the diameter of the smaller circle, which is $10\cdot 2=20$ meters.

He then goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He finally goes from the smaller circle to the larger circle, which is $20-10=10$ meters.

The total length is $10\pi + 10+5\pi + 20 + 5\pi + 10$$=20\pi + 40.$

19.

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?

$\ \dfrac{1}{4}$

$\ \dfrac{2}{7}$

$\ \dfrac{4}{11}$

$\ \dfrac{1}{2}$

$\ \dfrac{4}{7}$

###### Solution(s):

Each dot has $2$ dots that are one unit away from it.

Therefore, regardless of the choice of the first dot, $2$ of the other $7$ dots would be within one unit, so the probability is $\dfrac 27.$

20.

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ 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?

$\ 12$

$\ 17$

$\ 24$

$\ 27$

$\ 36$

###### Solution(s):

Since the number of boys who passed and the number of girls who passed are the same, we can assign them the same variable. Let this number be $p.$

The number of boys in the class is $\dfrac 32 p,$ and the number of girls in the class is $\dfrac 43 p.$ Thus, the total number of people is $\dfrac 43p + \dfrac 32p = \dfrac {17}6p.$

The total number of people must be a multiple of the numerator of this fraction, so the number of people must be a multiple of $17,$ making the number of people $17.$

21.

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?

$\ 48$

$\ 75$

$\ 151$

$\ 192$

$\ 603$

###### Solution(s):

The total volume is $Ah$ where $A$ is the area of the base and $h$ is the height.

To find $A,$ we take the area of the circle. It has a diameter of $8$cm, so it has a radius of area $4$cm. Thus, the area is $r^2\pi = 16\pi.$ Since $h=6$cm, the volume of the whole thing is $16\pi \cdot 6 = 96\pi.$

Since the wedge is half of the volume of the cylinder, the volume of the wedge is $48\pi.$

To estimate this, we can find that $49\pi \approx 49 \cdot \dfrac{22}{7} =154,$ so $48\pi \approx 154 -\pi \approx 154-3 = 151.$

22.

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

$\ 12$

$\ 21$

$\ 27$

$\ 33$

$\ 34$

###### Solution(s):

Let $x = \dfrac n3.$ We know $x$ and $9\dfrac n3 = 9x$ are both three digit numbers.

Thus, $100 \leq x, 9x \leq 999,$ so $100 \leq x \leq 111.$ Every integer in this range has $x$ that creates a valid $n,$ so there are $12$ valid numbers.

23.

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

$\ \dfrac{1}{6}$

$\ \dfrac{2}{9}$

$\ \dfrac{5}{18}$

$\ \dfrac{1}{3}$

$\ \dfrac{7}{20}$

###### Solution(s):

Let the side length be $s.$ Note that the total area is $s^2.$

Since $AF = 2FE,$ we know $FE = \dfrac{AF}3 = \dfrac s3,$$AF = 2 \dfrac{AE}3 = \dfrac{2s}3 .$

Since $CD = 2DE,$ we know $CD = \dfrac{CE}3 = \dfrac s3,$$AF = 2 \dfrac{AE}3 = \dfrac {2s}{3}.$

This makes the area of $ABF = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3,$ the area of $BCD = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3,$ and the area of $FED = \dfrac {\dfrac s3 \cdot \dfrac s3 }2 = \dfrac {s^2} {18}.$ Thus, the area of $BFD = s^2 - \dfrac {s^2}3-\dfrac {s^2}3-\dfrac {s^2}{18}$$= \dfrac {5s^2}{18}.$

24.

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?

$\ \dfrac{1}{10}$

$\ \dfrac{1}{6}$

$\ \dfrac{11}{60}$

$\ \dfrac{1}{5}$

$\ \dfrac{7}{30}$

###### Solution(s):

If the rolled number was $1,$ then the tile can be $1,4,9,$ yielding $3$ combinations.

If the rolled number was $2,$ then the tile can be $2,8,$ yielding $2$ combinations.

If the rolled number was $3,$ then the tile can be $3,$ yielding $1$ combination.

If the rolled number was $4,$ then the tile can be $1,4,9,$ yielding $3$ combinations.

If the rolled number was $5,$ then the tile can be $5,$ yielding $1$ combination.

If the rolled number was $6,$ then the tile can be $6,$ yielding $1$ combination.

The total number of combinations is $3+2+1+3+1+1=11.$ There are $60$ combinations each with equal likelihood, so the probability is $\dfrac{11}{60} .$

25.

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 dark-colored?

$\ 42$

$\ 44$

$\ 45$

$\ 46$

$\ 48$

###### Solution(s):

The largest circle is of radius $12,$ so the entire design has area of $12^2 \pi = 144 \pi.$

Each dark region is the area of its circle minus the area of the previous circle. The largest dark area would be of area $(10^2-8^2) \pi = 36 \pi,$ the next area would be $(6^2-4^2) \pi = 20 \pi ,$ and the smallest area would be $(2^2-0^2) \pi = 4 \pi .$ Therefore, the combined dark area would be $60 \pi.$

This fraction would be $\dfrac{60 \pi }{144 \pi} = \dfrac 5{12} ,$ which is approximately $42 \%.$