2010 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.

At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are 1111 students in Mrs. Germain's class, 88 students in Mr. Newton's class, and 99 students in Mrs. Young's class taking the AMC 88 this year. How many mathematics students at Euclid Middle School are taking the contest?

 26 \ 26

 27 \ 27

 28 \ 28

 29 \ 29

 30 \ 30

Solution(s):

There are 11+8+9=2811+8+9 = 28 students.

Therefore, the answer is C.

2.

If ab=a×ba+ba * b = \dfrac{a\times b}{a+b} for a,ba,b positive integers, then what is 510?5 *10?

 310 \ \dfrac{3}{10}

 1 \ 1

 2 \ 2

 103 \ \dfrac{10}{3}

 50 \ 50

Solution(s):

Given our definition of ab,a*b, we have 510=5×105+10=5015=103.5 *10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}.

Thus, the answer is D.

3.

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?

 50 \ 50

 62 \ 62

 70 \ 70

 89 \ 89

 100 \ 100

Solution(s):

The highest price is 1717 and the lowest price is 10.10. This means the percent is (17101)100=70. (\dfrac{17}{10}-1) * 100 = 70.

Thus, the answer is C.

4.

What is the sum of the mean, median, and mode of the numbers 2,3,0,3,1,4,0,3?2,3,0,3,1,4,0,3?

 6.5 \ 6.5

 7 \ 7

 7.5 \ 7.5

 8.5 \ 8.5

 9 \ 9

Solution(s):

The list reordered is 0,0,1,2,3,3,3,4.0,0,1,2,3,3,3,4.

The median is the mean of the middle two numbers, which would be 2+32=2.5.\dfrac{2+3}2 = 2.5. The mode is 33 since 33 appears the most. The mean is 0+0+1+2+3+3+3+48 \dfrac{0+0+1+2+3+3+3+4}{8}=168=2. = \dfrac{16}{8} = 2 .

Their sum is 2.5+3+2=7.5.2.5+3+2 = 7.5.

Thus, their answer is C.

5.

Alice needs to replace a light bulb located 1010 centimeters below the ceiling in her kitchen. The ceiling is 2.42.4 meters above the floor. Alice is 1.51.5 meters tall and can reach 4646 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

 32 \ 32

 34 \ 34

 36 \ 36

 38 \ 38

 40 \ 40

Solution(s):

We know the height of the ceiling is 2.4100=2402.4 \cdot 100 = 240 cm. Subtracting out all the given values, we get 2401504610=34. 240 - 150 - 46 - 10 = 34. Therefore, the height of the stool is 34 cm. Thus, B is the correct answer.

6.

Which of the following figures has the greatest number of lines of symmetry?

equilateral triangle

non-square rhombus

non-square rectangle

isosceles trapezoid

square

Solution(s):

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn't on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, lets look at each of the shapes. An equilateral triangle has 33 lines that intersect a corner or a midpoint, so it has at most 33 symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangles only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn't bisect the angle. An isosceles trapezoid only has 11 symmetry line, that goes through the midpoints of opposite sides. A square has 44 symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E.

7.

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?

 6 \ 6

 10 \ 10

 15 \ 15

 25 \ 25

 99 \ 99

Solution(s):

Freddie needs 44 pennies as this is the only way to pay 44 cents. Next, he needs 11 nickel or 55 pennies to make a 99 cents after having the 44 starting pennies. Here, we have 55 coins. Using 11 nickel minimizes the coins needed. Now, to make 2424 cents, we need 1515 more cents. This can be done with a dime and nickel. Now, we have 77 coints Now, for the last 7575 cents, we can use 33 quarters to make the rest. This leaves us with 1010 coins.

Therefore, the answer is B.

8.

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/21/2 mile in front of her. After she passes him, she can see him in her rear mirror until he is 1/21/2 mile behind her. Emily rides at a constant rate of 1212 miles per hour, and Emerson skates at a constant rate of 88 miles per hour. For how many minutes can Emily see Emerson?

 6 \ 6

 8 \ 8

 12 \ 12

 15 \ 15

 16 \ 16

Solution(s):

Let dd be how far Emily is ahead of Emerson. Emily sees Emerson if 12d12.-\dfrac 12 \leq d \leq \frac 12. Suppose at t=0,t= 0, where tt is in hours, that d=12.d = -\frac 12. Then, d=(128)t12.d = (12-8)t - \frac 12 . Since we must find where d=12,d = \frac 12, we find the time where 12=4t12    1=4t\frac 12 = 4t - \frac 12 \implies 1 = 4t     t=0.25.\implies t = 0.25. Since 0.250.25 hours passed, we know that 600.25=1560*0.25 = 15 minutes passed.

Therefore, the answer is D.

9.

Ryan got 80%80\% of the problems correct on a 2525-problem test, 90%90\% on a 4040-problem test, and 70%70\% on a 1010-problem test. What percent of all the problems did Ryan answer correctly?

 64 \ 64

 75 \ 75

 80 \ 80

 84 \ 84

 86 \ 86

Solution(s):

There were a total of 25+40+10=7525+40+10 = 75 problems.

On the first test, he solved 0.825=200.8*25 = 20 problems.

On the second test, he solved 0.940=360.9*40 = 36 problems.

On the third test, he solved 0.710=70.7*10 = 7 problems.

Therefore, he solved a total of 20+36+7=63.20+36+7=63. This means the fraction he solved is 6375=84%. \dfrac{63}{75} = 84\%.

Therefore, the answer is D.

10.

Six pepperoni circles will exactly fit across the diameter of a 1212-inch pizza when placed. If a total of 2424 circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?

 12 \ \dfrac 12

 23 \ \dfrac 23

 34 \ \dfrac 34

 56 \ \dfrac 56

 78 \ \dfrac 78

Solution(s):

Each circle has 16\dfrac{1}{6} the diameter of the large circle, so it has (16)2=136(\frac 16)^2 = \frac 1{36} of the total area.

Since there are 2424 pepperoni, they take up 24136=2324 * \dfrac 1{36} = \frac 23 of the area.

Therefore, the answer is B.

11.

The top of one tree is 1616 feet higher than the top of another tree. The heights of the two trees are in the ratio 3:4.3:4. In feet, how tall is the taller tree?

 48 \ 48

 64 \ 64

 80 \ 80

 96 \ 96

 112 \ 112

Solution(s):

Let b,sb,s be the heights of bigger and smaller trees respectively. Then, b=s+16b= s+16 and s=0.75b. s = 0.75b. If we substitute, we get b=0.75b+16    0.25b=16b = 0.75b + 16 \implies 0.25b = 16     b=64.\implies b = 64.

Thus, the answer is B.

12.

Of the 500500 balls in a large bag, 80%80\% are red and the rest are blue. How many of the red balls must be removed so that 75%75\% of the remaining balls are red?

 25 \ 25

 50 \ 50

 75 \ 75

 100 \ 100

 150 \ 150

Solution(s):

If rr is the number of red balls, then r=0.8500=400.r = 0.8*500 = 400. Therefore, if bb is the number of blue balls, then b=500400=100.b = 500-400 = 100.

If there are 7575% red balls after removing balls, then there are 2525% blue balls. This means the total number of balls is 100.25=400.\dfrac{100}{.25} = 400. This means the total number of balls decreased by 500400=100.500-400 = 100.

Thus, the answer is D.

13.

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30%30\% of the perimeter. What is the length of the longest side?

 7 \ 7

 8 \ 8

 9 \ 9

 10 \ 10

 11 \ 11

Solution(s):

Let ss be the smallest length. Then, all the side lengths are s,s+1,s+2.s,s+1,s+2. This would make the perimeter equal to 3s+3.3s+3. Since s=0.3(3s+3),s = 0.3(3s+3), then s=0.9s+0.9.s = 0.9s + 0.9 . This makes 0.1s=0.9,0.1s = 0.9, so s=9s = 9 which makes the longest side length s+2=11.s + 2 = 11.

Thus, the answer is E.

14.

What is the sum of the prime factors of 2010?2010?

 67 \ 67

 75 \ 75

 77 \ 77

 201 \ 201

 210 \ 210

Solution(s):

First, the primes 2,52,5 are factors of 20102010 since it is a multiple of 10.10. Dividing 20102010 by 1010 is 201.201. Then, 33 is a factor of 201201 since the digit sum of 201201 is a multiple of 3.3. Dividing this by 33 yields the prime number 67.67. This means the prime factors are 2,3,5,67,2,3,5,67, which makes their sum 77.77.

Thus, the correct answer is C.

15.

A jar contains 55 different colors of gumdrops. 30%30\% are blue, 20%20\% are brown, 15%15\% are red, 10%10\% are yellow, and the other 3030 gum drops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

 35 \ 35

 36 \ 36

 42 \ 42

 48 \ 48

 64 \ 64

Solution(s):

Since we have percentages for every color except green, the percent of green is 100%100\% minus the sum of the other colors. This would make the percent of green equal to 100%30%20%100\%-30\%-20\%-15%10%=25%15\%-10\% = 25\%

Since we know 3030 jelly beans is 25%,25\%, we know that the total number of jelly beans is 30.25=120.\dfrac { 30}{.25} = 120.

This means there are .2120=24.2*120 = 24 brown jelly beans to start and .3120=36.3*120 = 36 blue jelly beans. If half of the blue jelly beans are turned to brown, then 362=18 \dfrac{36}{2} = 18 more brown jelly beans are added. Therefore, we have 24+18=4224+18= 42 brown jelly beans.

Thus, the answer is C.

16.

A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

 π2 \ \dfrac{\sqrt{\pi}}{2}

 π \ \sqrt{\pi}

 π \ \pi

 2π \ 2\pi

 π2 \ \pi^{2}

Solution(s):

Let ss be the side length of the square and let rr be the radius of the circle. Then, since they have the same areas, s2=πr2.s^2 = \pi r^2 . This means (sr)2=π,(\dfrac sr)^2 = \pi, so sr=π.\frac sr = \sqrt{ \pi} .

Thus, the answer is B.

17.

The diagram shows an octagon consisting of 1010 unit squares. The portion below PQ\overline{PQ} is a unit square and a triangle with base 5.5. If PQ\overline{PQ} bisects the area of the octagon, what is the ratio XQQY?\dfrac{XQ}{QY}?

25 \dfrac{2}{5}

12 \dfrac{1}{2}

35 \dfrac{3}{5}

23 \dfrac{2}{3}

34 \dfrac{3}{4}

Solution(s):

Since PQPQ bisects the area, the area under the line is 5.5. Removing the square on the right makes the bottom a triangle of base 55 with area 4.4. Let the base of this triangle be PZ.PZ.

The area being 44 means (PZ)(ZQ)2=5(ZQ)2=4\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4     ZQ=1.6.\implies ZQ = 1.6.

Therefore, QY=QZ1=0.6QY = QZ-1 = 0.6 and XQ=2QZ=0.4. XQ = 2-QZ = 0.4. This would make XQQY=0.40.6=23.\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.

Thus, the answer is D.

18.

A decorative window is made up of a rectangle with semicircles at either end. The ratio of ADAD to ABAB is 3:2.3:2. And ABAB is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?

 2:3 \ 2:3

 3:2 \ 3:2

 6:π \ 6:\pi

 9:π \ 9:\pi

 30:π \ 30 :\pi

Solution(s):

Combining the semicircles would make a circle of diameter d=30.d=30. This would make the radius equal to d2. \dfrac{d}{2}. Therefore, the combined area of the semicircles is (d2)2π=πd24.(\dfrac{d}{2})^2 * \pi = \dfrac{\pi d^2}{4}.

Side AD=32AD = \dfrac 32 and AB=32d.AB = \dfrac 32 d. The area of the rectangle is therefore 32dd=32d2. \dfrac 32 d*d = \dfrac 32 d^2. The ratio of the area of the rectangle to the area of the semicircles is 32d2πd24=6π.\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}.

Thus, the answer is C.

19.

The two circles pictured have the same center C.C. Chord AD\overline{AD} is tangent to the inner circle at B,B, ACAC is 10,10, and chord AD\overline{AD} has length 16.16. What is the area between the two circles?

 36π \ 36 \pi

 49π \ 49 \pi

 64π \ 64 \pi

 81π \ 81 \pi

 100π \ 100 \pi

Solution(s):

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be (AC)2π(CB)2π(AC)^2\pi - (CB)^2 \pi =π(AC2CB2).= \pi(AC^2 - CB^2). By the Pythagorean Theorem, we can get AC2CB2=AB2.AC^2 - CB^2 = AB^2. Therefore, we need to find AB2π.AB^2 \pi.

SInce ABAB is half of AD,AD, we get AB=8.AB = 8. This makes AB2π=64π.AB^2 \pi = 64 \pi.

Thus, the area is C.

20.

In a room, 2/52/5 of the people are wearing gloves, and 3/43/4 of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?

 3 \ 3

 5 \ 5

 8 \ 8

 15 \ 15

 20 \ 20

Solution(s):

Since our room has 25\dfrac 25 of the people wearing gloves, the number of people must be a multiple of 5.5. Since our room has 34\dfrac 34 of the people wearing hats, the number of people must be a multiple of 4.4. Therefore, the people in the room must be a multiple of 20.20.

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to 25+34 \dfrac{2}{5} + \dfrac 34 - Fraction of people who wear either. If we wish to minimize the number of who wear both, we maximize the number of people who wear both, to make it 1.1. Therefore, the fraction of people that wear both is 25+341=320.\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.

Since our number is a (positive) multiple of 20,20, we have the number of people wearing both as 33 if we choose to have just 2020 people.

Therefore, A is the correct answer.

21.

Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read 1/51 / 5 of the pages plus 1212 more, and on the second day she read 1/41 / 4 of the remaining pages plus 1515 pages. On the third day she read 1/31 / 3 of the remaining pages plus 1818 pages. She then realized that there were only 6262 pages left to read, which she read the next day. How many pages are in this book?

 120 \ 120

 180 \ 180

 240 \ 240

 300 \ 300

 360 \ 360

Solution(s):

The pages left after the third day is 62.62. Before reading the last 1818 pages, she had 8080 pages left. This is 23\dfrac 23 of the pages remaining, so she had 120120 pages left before the third day.

Before reading the 1515 pages, she had 120+15=135120+15=135 pages left. This is 34\dfrac 34 of the pages remaining, so she had 180180 pages left before the second day.

Before reading the 1212 pages, she had 180+12=192180+12=192 pages left. This is 45\dfrac 45 of the pages remaining, so she had 240240 pages left before the first day, making the book 240240 pages.

Thus, the answer is C.

22.

The hundreds digit of a three-digit number is 22 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

0 0

2 2

4 4

6 6

8 8

Solution(s):

Let the units digit be u,u, and let the tens digit be t.t. This makes the hundreds digit be u+2.u+2. This makes the number equal to 100(u+2)+10t+u100(u+2) +10t+u=201u+200+10t=201u+200+10t and the reversed number is 100u+10t+(u+2)100u +10t+(u+2)=201u+2+10t.=201u+2+10t. This makes the difference equal to (201u+10t+200)(201u+10t+200)-(201u+10t+2)=198.(201u+10t+2)=198. This makes the units digit 8.8.

Therefore, the units digit is E.

23.

Semicircles POQPOQ and ROSROS pass through the center O.O. What is the ratio of the combined areas of the two semicircles to the area of circle O?O?

24 \dfrac{\sqrt 2}{4}

12 \dfrac{1}{2}

2π \dfrac{2}{\pi}

23 \dfrac{2}{3}

22 \dfrac{\sqrt 2}{2}

Solution(s):

The area of each of the smaller circles have an area of πr22.\pi \dfrac {r^2}{2} . Each of them have a radius of 1,1, so their combined area is π12+π12=π.\pi \frac{1}{2} + \pi \frac 12 = \pi.

Next, the radius of the larger circle is equal to the length of OQ,OQ, which is equal to 12+12=2. \sqrt{1^2+1^2} = \sqrt 2. Its area is πr2=π(2)2=2π.\pi r^2 = \pi(\sqrt{2})^2 = 2\pi.

This means the ratio is π2π=12.\dfrac{\pi}{2\pi} = \frac 12.

Thus, the answer is B.

24.

What is the correct ordering of the three numbers, 108,10^8, 512,5^{12}, and 224?2^{24}?

2^{24} < 10^8 < 5^{12}

2^{24} < 5^{12} < 10^8

5^{12} < 2^{24} < 10^8

10^8 < 5^{12} < 2^{24}

10^8 < 2^{24} < 5^{12}

Solution(s):

First, we get 224=(28)(48)<(28)(58)=108.2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.

Next, we get 108=(28)(58)=10^8 = (2^8)(5^8) = (44)(58)<(54)(58)=512.(4^4)(5^8) < (5^4)(5^8) = 5^{12}. This means 224<108<512.2^{24} < 10^8 < 5^{12} .

Thus, the answer is A.

25.

Everyday at school, Jo climbs a flight of 66 stairs. Jo can take the stairs 1,1, 2,2, or 33 at a time. For example, Jo could climb 3,3, then 1,1, then 2.2. In how many ways can Jo climb the stairs?

 13 \ 13

 18 \ 18

 20 \ 20

 22 \ 22

 24 \ 24

Solution(s):

Lets count first the number of ways to climb a flight of stairs. She must land on the last stair, and for each of the other stairs, she can either step on it or not step on it. This means there are 25=322^5 = 32 total combinations. Now, we must exclude the ways that include stepping more than 33 steps. This would mean we climb 4,54,5 or 66 stairs in one step. To do this, we can do it in the following ways: 6,15,51,42,24,6,1-5,5-1,4-2,2-4,114,141,411.1-1-4,1-4-1,4-1-1. This means we have 88 cases to exlude, so we have 328=2432-8 = 24 total valid cases.

Thus, the answer is E.