2008 AMC 8 考试题目

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Used with permission of the Mathematical Association of America.

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1.

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

12 12

14 14

26 26

38 38

50 50

Answer: B
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.

Thus, the answer is B.

2.

The ten-letter code BEST OF LUCK\text{BEST OF LUCK} represents the ten digits 09,0-9, in order. What 4-digit number is represented by the code word CLUE?\text{CLUE}?

 8671 \ 8671

 8672 \ 8672

 9781 \ 9781

 9782 \ 9782

 9872 \ 9872

Answer: A
Solution(s):

The letter CC is in the 99th position, so it would be the letter 8.8. The letter LL is in the 77th position, so it would be the letter 6.6. The letter UU is in the 88th position, so it would be the letter 7.7. The letter EE is in the 22nd position, so it would be the letter 1.1.

Therefore, when putting together the word CLUE,\text{CLUE}, we get 8671.8671.

Thus, the answer is A.

3.

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

 Sunday \ \text{Sunday}

 Monday \ \text{Monday}

 Wednesday \ \text{Wednesday}

 Thursday \ \text{Thursday}

 Saturday \ \text{Saturday}

Answer: A
Solution(s):

Since the 1313th is a Friday, we know that the 66th is also a Friday. The 11st is 55 days before the Friday, making it a Sunday.

Thus, the answer is A.

4.

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

 3 \ 3

 4 \ 4

 5 \ 5

 6 \ 6

 7 \ 7

Answer: C
Solution(s):

Since the area of the larger triangle is 1616 and the area of the smaller triangle is 1.1. Thus, the area of the other 33 trapezoids is 15.15. Since the 33 trapezoids have a combined area of 15,15, each of their areas is 153=5.\dfrac{15}{3} =5.

Thus, the answer is C.

5.

Barney Schwinn notices that the odometer on his bicycle reads 1441,1441, a palindrome, because it reads the same forward and backward. After riding 44 more hours that day and 66 the next, he notices that the odometer shows another palindrome, 1661.1661. What was his average speed in miles per hour?

 15 \ 15

 16 \ 16

 18 \ 18

 20 \ 20

 22 \ 22

Answer: E
Solution(s):

The total distance traveled is 16611441=2201661-1441 = 220 miles. He also travelled 1010 hours. Thus, the average speed is 22010=22\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 \ 3:10

 3:8 \ 3:8

 3:7 \ 3:7

 3:5 \ 3:5

 1:1 \ 1:1

Answer: D
Solution(s):

The total area of the entire square is 1616 since it is a 4×44 \times 4 square. The area of the colored region is 6,6, making the remaining area 166=10.16-6 = 10. Thus, the ratio is 6:10=3:5.6:10 = 3:5.

Thus, the answer is D.

7.

If 35=M45=60N,\dfrac{3}{5}=\dfrac{M}{45}=\dfrac{60}{N}, what is M+N?M+N?

 27 \ 27

 29 \ 29

 45 \ 45

 105 \ 105

 127 \ 127

Answer: E
Solution(s):

35=M45\dfrac{3}{5}=\dfrac{M}{45} means M=3455=27.M = \dfrac{3\cdot 45}{5}=27.

35=60N\dfrac{3}{5}=\dfrac{60}{N} means N=5603=100.N = \dfrac{5\cdot 60}{3}=100.

Therefore, M+N=127.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 \ 60

 70 \ 70

 75 \ 75

 80 \ 80

 85 \ 85

Answer: D
Solution(s):

The total sales are 100+60+40+120=320.100+60+40+120 = 320.

The average sales is then 3204=80.\dfrac{320}{4} =80.

Thus, the answer is D.

9.

In 20052005 Tycoon Tammy invested 100100 dollars for two years. During the first year her investment suffered a 15%15\% loss, but during the second year the remaining investment showed a 20%20\% gain. Over the two-year period, what was the change in Tammy's investment?

 5% loss \ 5\%\text{ loss}

 2% loss \ 2\%\text{ loss}

 1% gain \ 1\%\text{ gain}

 2% gain \ 2\% \text{ gain}

 5% gain \ 5\%\text{ gain}

Answer: D
Solution(s):

The 15%15\% loss means the investment went from $100\$100 to 0.85$100=$85.0.85\cdot \$100 = \$85.

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

Thus, the answer is D.

10.

The average age of the 66 people in Room A is 40.40. The average age of the 44 people in Room B is 25.25. If the two groups are combined, what is the average age of all the people?

 32.5 \ 32.5

 33 \ 33

 33.5 \ 33.5

 34 \ 34

 35 \ 35

Answer: D
Solution(s):

The sum of the ages in Room A is 640=240.6\cdot 40 = 240.

The sum of the ages in Room B is 425=100.4\cdot 25 = 100.

The total sum is 240+100=340.240+100 = 340.

The average is therefore 34010=34.\dfrac{340}{10} =34.

Thus, the answer is D.

11.

Each of the 3939 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 2626 students have a cat. How many students have both a dog and a cat?

 7 \ 7

 13 \ 13

 19 \ 19

 39 \ 39

 46 \ 46

Answer: A
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+2639=7.20+26-39 = 7.

Thus, the answer is A.

12.

A ball is dropped from a height of 33 meters. On its first bounce it rises to a height of 22 meters. It keeps falling and bouncing to 23\frac{2}{3} of the height it reached in the previous bounce. On which bounce will it not rise to a height of 0.50.5 meters?

 3 \ 3

 4 \ 4

 5 \ 5

 6 \ 6

 7 \ 7

Answer: C
Solution(s):

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

After 44 bounces, the ball bounces 3(23)4=1627,3\cdot \left(\dfrac 23\right)^4 = \dfrac{16}{27}, which is greater than 12.\frac 12.

After 55 bounces, the ball bounces 3(23)5=3281,3\cdot \left(\dfrac 23\right)^5 = \dfrac{32}{81}, which is less than 12.\frac 12.

Therefore, we are under 12\frac 12 after 55 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 100100 pounds or more than 150150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122,122, 125125 and 127127 pounds. What is the combined weight in pounds of the three boxes?

 160 \ 160

 170 \ 170

 187 \ 187

 195 \ 195

 354 \ 354

Answer: C
Solution(s):

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

Thus, the answer is C.

14.

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

 2 \ 2

 3 \ 3

 4 \ 4

 5 \ 5

 6 \ 6

Answer: C
Solution(s):

There are 22 rows and 22 columns to put B in, so there are 44 places for it. After that, there is 11 column and 11 row available for C, so the number of combinations is 2211=4.2\cdot 2\cdot 1\cdot 1 = 4.

Thus, the answer is C.

15.

In Theresa's first 88 basketball games, she scored 7,4,3,6,8,3,17, 4, 3, 6, 8, 3, 1 and 55 points. In her ninth game, she scored fewer than 1010 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 1010 points and her points-per-game average for the 1010 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

 35 \ 35

 40 \ 40

 48 \ 48

 56 \ 56

 72 \ 72

Answer: B
Solution(s):

The sum of her first 88 scores is 37.37. Since the average of the first 99 scores is an integer, the sum of the first 99 scores is a multiple of 9.9.

Since the score is less than 10,10, the sum of the scores after 99 games is between 3737 and 47,47, and is a multiple of 9,9, making the sum 45.45. Thus, the score of the 99th game is 4537=8.45-37=8.

The sum of Theresa's first 99 scores is 45.45. Since the average of the first 1010 scores is an integer, the sum of the first 1010 scores is a multiple of 10.10.

Since the score is less than 10,10, the sum of the scores after 1010 games is between 4545 and 55,55, and is a multiple of 10,10, making the sum 50.50. Thus, the score of the 1010th game is 5045=5.50-45=5.

Therefore, their product is 85=40.8\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 \:1 : 6

7:36 \: 7 : 36

1:5 \: 1 : 5

7:30 \: 7 : 30

6:25 \: 6 : 25

Answer: D
Solution(s):

There are 77 unit cubes, so the volume is 7.7.

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 55 exposed on each of the outer cubes. This makes the surface area 56=30.5\cdot 6=30.

This makes the ratio of volume to surface area 7:30.7:30.

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 5050 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 \ 76

 120 \ 120

 128 \ 128

 132 \ 132

 136 \ 136

Answer: D
Solution(s):

Let the length and width be l,wl,w respectively. The rectangle would have an area of l+r+l+r=2(l+r)=50.l+r+l+r = 2(l+r)=50. Thus, l+r=25,l+r =25, so r=25l.r=25-l. The area is lr=l(25l)lr = l(25-l) =156.25(l12.5)2.= 156.25 - (l-12.5)^2. This makes the area largest when ll is as close as possible to 12.512.5 and the area is the smallest when ll is as far from 12.5.12.5. Therefore, the largest area is when l=12,13l=12,13 and the smallest area is when l=1,24.l=1,24.

This makes the larger area equal to 156.25(1312.5)2156.25-(13-12.5)^2 =156.250.25=156= 156.25-0.25 =156 and the smaller area equal to 156.25(2412.5)2156.25-(24-12.5)^2 =156.25132.25=24.= 156.25-132.25 =24.

Thus, the difference is 15624=132.156-24=132.

Thus, the answer is D.

18.

Two circles that share the same center have radii 1010 meters and 2020 meters. An aardvark runs along the path shown, starting at AA and ending at K.K. How many meters does the aardvark run?

 10π+20 \ 10\pi+20

 10π+30 \ 10\pi+30

 10π+40 \ 10\pi+40

 20π+20 \ 20\pi+20

 20π+40 \ 20\pi+40

Answer: E
Solution(s):

The circumference of a circle is πd=2rπ,\pi d = 2r\pi , so going a quarter of the way around is πr2.\dfrac {\pi r }2.

He goes a quarter of the way around the large circle, so this part is 10π10 \pi meters. He then goes from the larger circle to the smaller circle, which is 2010=1020-10=10 meters.

He goes a quarter of the way around the smaller circle, so this part is 5π5 \pi meters. He then goes through the diameter of the smaller circle, which is 102=2010\cdot 2=20 meters.

He then goes a quarter of the way around the smaller circle, so this part is 5π5 \pi meters. He finally goes from the smaller circle to the larger circle, which is 2010=1020-10=10 meters.

The total length is 10π+10+5π+20+5π+1010\pi + 10+5\pi + 20 + 5\pi + 10 =20π+40.=20\pi + 40.

Thus, the answer is E.

19.

Eight points are spaced around at intervals of one unit around a 2×22 \times 2 square, as shown. Two of the 88 points are chosen at random. What is the probability that the two points are one unit apart?

 14 \ \dfrac{1}{4}

 27 \ \dfrac{2}{7}

 411 \ \dfrac{4}{11}

 12 \ \dfrac{1}{2}

 47 \ \dfrac{4}{7}

Answer: B
Solution(s):

Each dot has 22 dots that are one unit away from it.

Therefore, regardless of the choice of the first dot, 22 of the other 77 dots would be within one unit, so the probability is 27.\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 34\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 \ 12

 17 \ 17

 24 \ 24

 27 \ 27

 36 \ 36

Answer: B
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.p.

The number of boys in the class is 32p,\dfrac 32 p, and the number of girls in the class is 43p.\dfrac 43 p. Thus, the total number of people is 43p+32p=176p.\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 17,17, making the number of people 17.17.

Thus, the answer is B.

21.

Jerry cuts a wedge from a 66-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 \ 48

 75 \ 75

 151 \ 151

 192 \ 192

 603 \ 603

Answer: C
Solution(s):

The total volume is AhAh where AA is the area of the base and hh is the height.

To find A,A, we take the area of the circle. It has a diameter of 88cm, so it has a radius of area 44cm. Thus, the area is r2π=16π.r^2\pi = 16\pi. Since h=6h=6cm, the volume of the whole thing is 16π6=96π.16\pi \cdot 6 = 96\pi.

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

To estimate this, we can find that 49π49227=154,49\pi \approx 49 \cdot \dfrac{22}{7} =154, so 48π154π1543=151.48\pi \approx 154 -\pi \approx 154-3 = 151.

Thus, the answer is C.

22.

For how many positive integer values of nn are both n3\dfrac{n}{3} and 3n3n three-digit whole numbers?

 12 \ 12

 21 \ 21

 27 \ 27

 33 \ 33

 34 \ 34

Answer: A
Solution(s):

Let x=n3.x = \dfrac n3. We know xx and 9n3=9x9\dfrac n3 = 9x are both three digit numbers.

Thus, 100x,9x999,100 \leq x, 9x \leq 999, so 100x111.100 \leq x \leq 111. Every integer in this range has xx that creates a valid n,n, so there are 1212 valid numbers.

Thus, the answer is A.

23.

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

 16 \ \dfrac{1}{6}

 29 \ \dfrac{2}{9}

 518 \ \dfrac{5}{18}

 13 \ \dfrac{1}{3}

 720 \ \dfrac{7}{20}

Answer: C
Solution(s):

Let the side length be s.s. Note that the total area is s2.s^2.

Since AF=2FE,AF = 2FE, we know FE=AF3=s3,FE = \dfrac{AF}3 = \dfrac s3,AF=2AE3=2s3. AF = 2 \dfrac{AE}3 = \dfrac{2s}3 .

Since CD=2DE,CD = 2DE, we know CD=CE3=s3,CD = \dfrac{CE}3 = \dfrac s3,AF=2AE3=2s3. AF = 2 \dfrac{AE}3 = \dfrac {2s}{3}.

This makes the area of ABF=s2s32=s23,ABF = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3, the area of BCD=s2s32=s23,BCD = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3, and the area of FED=s3s32=s218.FED = \dfrac {\dfrac s3 \cdot \dfrac s3 }2 = \dfrac {s^2} {18}. Thus, the area of BFD=s2s23s23s218BFD = s^2 - \dfrac {s^2}3-\dfrac {s^2}3-\dfrac {s^2}{18}=5s218. = \dfrac {5s^2}{18}.

Thus, the answer is C.

24.

Ten tiles numbered 11 through 1010 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?

 110 \ \dfrac{1}{10}

 16 \ \dfrac{1}{6}

 1160 \ \dfrac{11}{60}

 15 \ \dfrac{1}{5}

 730 \ \dfrac{7}{30}

Answer: C
Solution(s):

If the rolled number was 1,1, then the tile can be 1,4,9,1,4,9, yielding 33 combinations.

If the rolled number was 2,2, then the tile can be 2,8,2,8, yielding 22 combinations.

If the rolled number was 3,3, then the tile can be 3,3, yielding 11 combination.

If the rolled number was 4,4, then the tile can be 1,4,9,1,4,9, yielding 33 combinations.

If the rolled number was 5,5, then the tile can be 5,5, yielding 11 combination.

If the rolled number was 6,6, then the tile can be 6,6, yielding 11 combination.

The total number of combinations is 3+2+1+3+1+1=11.3+2+1+3+1+1=11. There are 6060 combinations each with equal likelihood, so the probability is 1160.\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 \ 42

 44 \ 44

 45 \ 45

 46 \ 46

 48 \ 48

Answer: A
Solution(s):

The largest circle is of radius 12,12, so the entire design has area of 122π=144π.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 (10282)π=36π,(10^2-8^2) \pi = 36 \pi, the next area would be (6242)π=20π,(6^2-4^2) \pi = 20 \pi , and the smallest area would be (2202)π=4π.(2^2-0^2) \pi = 4 \pi . Therefore, the combined dark area would be 60π.60 \pi.

This fraction would be 60π144π=512, \dfrac{60 \pi }{144 \pi} = \dfrac 5{12} , which is approximately 42%.42 \%.

Thus, the answer is A.