2008 AMC 8 Exam Solutions

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:

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.

Thus, the answer is B.

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:

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 $8671.$

Thus, the answer is A.

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:

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.

Thus, the answer is A.

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:

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 $\dfrac{15}{3} =5.$

Thus, the answer is C.

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:

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.

Thus, the answer is E.

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:

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 $6:10 = 3:5.$

Thus, the answer is D.

7.

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

$\ 27$

$\ 29$

$\ 45$

$\ 105$

$\ 127$

Solution:

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

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

Therefore, $M +N = 127.$

Thus, the answer is E.

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:

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

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

Thus, the answer is D.

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:

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 $2\%.$

Thus, the answer is D.

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:

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 $\dfrac{340}{10} =34.$

Thus, the answer is D.

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:

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 $20+26-39 = 7.$

Thus, the answer is A.

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:

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.

Thus, the answer is C.

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:

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 $a+b+c = 187.$

Thus, the answer is C.

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:

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 $2\cdot 2\cdot 1\cdot 1 = 4.$

Thus, the answer is C.

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:

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 $8\cdot 5=40.$

Thus, the answer is B.

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:

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 $7:30.$

Thus, the answer is D.

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:

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 $156-24=132.$

Thus, the answer is D.

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:

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 $10\pi + 10+5\pi + 20 + 5\pi + 10$ $=20\pi + 40.$

Thus, the answer is E.

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:

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 $\dfrac 27.$

Thus, the answer is B.

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:

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 $17.$

Thus, the answer is B.

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:

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 $48\pi \approx 154 -\pi \approx 154-3 = 151.$

Thus, the answer is C.

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:

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.

Thus, the answer is A.

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:

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 $BFD = s^2 - \dfrac {s^2}3-\dfrac {s^2}3-\dfrac {s^2}{18}$ $= \dfrac {5s^2}{18}.$

Thus, the answer is C.

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:

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 $\dfrac{11}{60} .$

Thus, the answer is C.

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:

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 $42 \%.$

Thus, the answer is A.