### 2008 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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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? $$\ 42$$

$$\ 44$$

$$\ 45$$

$$\ 46$$

$$\ 48$$

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 $$42 \%.$$