2008 AMC 8 考试题目

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1.

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

$12$

$14$

$26$

$38$

$50$

Answer: B

Solution:

Susan spent $2\cdot 12=24$ dollars on rides, so she spent $12+24=36$ dollars total.

She had $50-36=14$ dollars left.

Thus, B is the correct answer.

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 $\text{CLUE}?$

$\ 8671$

$\ 8672$

$\ 9781$

$\ 9782$

$\ 9872$

Answer: A

Solution:

The code BEST OF LUCK represents the digits $0$ through $9$ in order, so $C=8$ , $L=6$ , $U=7$ , and $E=1$ .

Therefore, $\text{CLUE}=8671$ .

Thus, A is the correct answer.

3.

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

$\ \text{Sunday}$

$\ \text{Monday}$

$\ \text{Wednesday}$

$\ \text{Thursday}$

$\ \text{Saturday}$

Answer: A

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?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

Answer: C

Solution:

The outer triangle has area $16$ , and the inner triangle has area $1$ .

So the three congruent trapezoids have total area $16-1=15$ , and each trapezoid has area $15\div3=5$ .

Thus, C is the correct answer.

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?

$\ 15$

$\ 16$

$\ 18$

$\ 20$

$\ 22$

Answer: E

Solution:

The odometer increased by $1661-1441=220$ miles.

Barney rode for $4+6=10$ hours, so his average speed was $220\div10=22$ miles per hour.

Thus, E is the correct answer.

6.

In the figure, what is the ratio of the area of the shaded squares to the area of the unshaded squares?

$\ 3:10$

$\ 3:8$

$\ 3:7$

$\ 3:5$

$\ 1:1$

Answer: D

Solution:

After subdividing the central shaded square, the figure has $16$ equal small squares.

Of these, $6$ are shaded and $10$ are unshaded, so the desired ratio is $6:10=3:5$ .

Thus, D is the correct answer.

7.

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

$\ 27$

$\ 29$

$\ 45$

$\ 105$

$\ 127$

Answer: E

Solution:

From $\dfrac{3}{5}=\dfrac{M}{45}$ , we get $M=\dfrac{3\cdot45}{5}=27$ .

From $\dfrac{3}{5}=\dfrac{60}{N}$ , we get $3N=300$ , so $N=100$ .

Therefore, $M+N=27+100=127$ .

Thus, E is the correct answer.

8.

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

$\ 60$

$\ 70$

$\ 75$

$\ 80$

$\ 85$

Answer: D

Solution:

The four monthly sales are $100,60,40,$ and $120$ dollars.

Their average is $\dfrac{100+60+40+120}{4}=\dfrac{320}{4}=80$ dollars.

Thus, D is the correct answer.

9.

In $2005$ Tycoon Tammy invested $\$100$ 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?

$\ 5\%\text{ loss}$

$\ 2\%\text{ loss}$

$\ 1\%\text{ gain}$

$\ 2\% \text{ gain}$

$\ 5\%\text{ gain}$

Answer: D

Solution:

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

The $20\%$ gain then changes it to $1.2\cdot\$85=\$102$ .

The investment went from $\$100$ to $\$102$ , which is a $2\%$ gain.

Thus, D is the correct answer.

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?

$\ 32.5$

$\ 33$

$\ 33.5$

$\ 34$

$\ 35$

Answer: D

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?

$\ 7$

$\ 13$

$\ 19$

$\ 39$

$\ 46$

Answer: A

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?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

Answer: C

Solution:

The first bounce rises to $2$ meters. Each later bounce is $\dfrac{2}{3}$ of the previous bounce.

The fourth bounce rises to $2\left(\dfrac{2}{3}\right)^3=\dfrac{16}{27}$ , which is greater than $\dfrac{1}{2}$ .

The fifth bounce rises to $2\left(\dfrac{2}{3}\right)^4=\dfrac{32}{81}$ , which is less than $\dfrac{1}{2}$ .

Thus, C is the correct answer.

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?

$\ 160$

$\ 170$

$\ 187$

$\ 195$

$\ 354$

Answer: C

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?

$\ 2$

$\ 3$

$\ 4$

$\ 5$

$\ 6$

Answer: C

Solution:

The other two A's must occupy one square in each of the remaining two rows and columns. There are two possible diagonal patterns for the three A's.

For either A-pattern, the two remaining positions in the top row can be filled as B,C or C,B, and then the rest of the grid is forced.

So there are $2\cdot2=4$ arrangements.

Thus, C is the correct answer.

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?

$\ 35$

$\ 40$

$\ 48$

$\ 56$

$\ 72$

Answer: B

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?

$\:1 : 6$

$\: 7 : 36$

$\: 1 : 5$

$\: 7 : 30$

$\: 6 : 25$

Answer: D

Solution:

The shape uses $7$ unit cubes, so its volume is $7$ cubic units.

The center cube has no exposed faces. Each of the six outer cubes has $5$ exposed faces, for a total surface area of $6\cdot5=30$ square units.

The ratio of volume to surface area is $7:30$ .

Thus, D is the correct answer.

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?

$\ 76$

$\ 120$

$\ 128$

$\ 132$

$\ 136$

Answer: D

Solution:

If the side lengths are $l$ and $w$ , then $2l+2w=50$ , so $l+w=25$ .

The smallest possible area comes from side lengths $1$ and $24$ , giving area $24$ .

The largest possible area comes from the closest integer pair with sum $25$ , namely $12$ and $13$ , giving area $156$ .

The difference is $156-24=132$ .

Thus, D is the correct answer.

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?

$\ 10\pi+20$

$\ 10\pi+30$

$\ 10\pi+40$

$\ 20\pi+20$

$\ 20\pi+40$

Answer: E

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 at intervals of one unit around a $2\times2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the points are one unit apart?

$\ \dfrac{1}{4}$

$\ \dfrac{2}{7}$

$\ \dfrac{4}{11}$

$\ \dfrac{1}{2}$

$\ \dfrac{4}{7}$

Answer: B

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?

$\ 12$

$\ 17$

$\ 24$

$\ 27$

$\ 36$

Answer: B

Solution:

Let $p$ be the common number of boys and girls who passed.

Since $\dfrac{2}{3}$ of the boys passed, the number of boys is $\dfrac{3p}{2}$ . Since $\dfrac{3}{4}$ of the girls passed, the number of girls is $\dfrac{4p}{3}$ .

The smallest positive $p$ that makes both counts whole is $6$ . Then there are $9$ boys and $8$ girls, for a minimum total of $17$ students.

Thus, B is the correct answer.

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?

$\ 48$

$\ 75$

$\ 151$

$\ 192$

$\ 603$

Answer: C

Solution:

The cylinder has diameter $8$ cm, so its radius is $4$ cm. Its length is $6$ cm.

The whole cylinder has volume $\pi r^2h=\pi\cdot4^2\cdot6=96\pi$ .

The dashed cut takes half of the cylinder, so the wedge has volume $48\pi$ , which is about $151$ cubic centimeters.

Thus, C is the correct answer.

22.

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

$\ 12$

$\ 21$

$\ 27$

$\ 33$

$\ 34$

Answer: A

Solution:

Let $x=\dfrac{n}{3}$ . Then $n=3x$ , so $3n=9x$ .

Both $x$ and $9x$ must be three-digit whole numbers. Thus $100\le x$ and $9x\le999$ , so $100\le x\le111$ .

There are $111-100+1=12$ integer values of $x$ , and each gives one positive integer value of $n$ .

Thus, A is the correct answer.

23.

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

$\ \dfrac{1}{6}$

$\ \dfrac{2}{9}$

$\ \dfrac{5}{18}$

$\ \dfrac{1}{3}$

$\ \dfrac{7}{20}$

Answer: C

Solution:

Because the answer is a ratio, choose the side length of the square to be $3$ . Then $AF=2$ , $FE=1$ , $CD=2$ , and $DE=1$ .

The square has area $9$ . The areas of triangles $ABF$ and $BCD$ are each $\dfrac{1}{2}\cdot3\cdot2=3$ .

Triangle $DEF$ has area $\dfrac{1}{2}\cdot1\cdot1=\dfrac{1}{2}$ .

So $[BFD]=9-3-3-\dfrac{1}{2}=\dfrac{5}{2}$ , and the desired ratio is $\dfrac{5/2}{9}=\dfrac{5}{18}$ .

Thus, C is the correct answer.

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?

$\ \dfrac{1}{10}$

$\ \dfrac{1}{6}$

$\ \dfrac{11}{60}$

$\ \dfrac{1}{5}$

$\ \dfrac{7}{30}$

Answer: C

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. Approximately what percent of the design is shaded?

$\ 42$

$\ 44$

$\ 45$

$\ 46$

$\ 48$

Answer: A

Solution:

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

The shaded parts have areas $2^2\pi=4\pi$ , $(6^2-4^2)\pi=20\pi$ , and $(10^2-8^2)\pi=36\pi$ .

The total shaded area is $4\pi+20\pi+36\pi=60\pi$ , so the shaded fraction is $\dfrac{60\pi}{144\pi}=\dfrac{5}{12}$ , which is about $42\%$ .

Thus, A is the correct answer.