2016 AMC 8 Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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1.

The longest professional tennis match ever played lasted a total of 1111 hours and 55 minutes. How many minutes was this?

605 605

655 655

665 665

1005 1005

1105 1105

Answer: C
Solution:

There are 6060 minutes in an hour, so the total time is 6011+5=66560 \cdot 11 + 5 = 665 minutes.

Thus, C is the correct answer.

2.

In rectangle ABCD,ABCD, AB=6AB=6 and AD=8.AD=8. Point MM is the midpoint of AD.\overline{AD}. What is the area of AMC?\triangle AMC?

12 12

15 15

18 18

20 20

24 24

Answer: A
Solution:

From the diagram, we can see that the base of AMC\triangle AMC is 44 and the altitude is 4.4. The area is therefore 1246=12.\dfrac{1}{2} \cdot 4 \cdot 6 = 12.

Thus, A is the correct answer.

3.

Four students take an exam. Three of their scores are 70,80,70, 80, and 90.90. If the average of their four scores is 70,70, then what is the remaining score?

40 40

50 50

55 55

60 60

70 70

Answer: A
Solution:

From the average, we can calculate the sum of the scores to be 470=280.4 \cdot 70 = 280. This means that the remaining score is 280708090=40.280 - 70 - 80 - 90 = 40.

Thus, A is the correct answer.

4.

When Cheenu was a boy he could run 1515 miles in 33 hours and 3030 minutes. As an old man he can now walk 1010 miles in 44 hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?

6 6

10 10

15 15

18 18

30 30

Answer: B
Solution:

To better compare the rates, we can change his speed into minutes per mile.

As a boy he ran 1515 miles in 360+30=2103 \cdot 60 + 30 = 210 minutes, which means that he ran at a pace of 210/15=14210 / 15 = 14 minutes per mile.

As an adult, he can walk 1010 miles in 460=2404 \cdot 60 = 240 minutes, which means he walks at a pace of 240/10=24240 / 10 = 24 minutes per mile.

Subtracting the two, we get that he takes 1010 more minutes to walk a mile as an adult.

Thus, B is the correct answer.

5.

The number NN is a two-digit number with the following properties:

\quad• When NN is divided by 9,9, the remainder is 1.1.

\quad• When NN is divided by 10,10, the remainder is 3.3.

What is the remainder when NN is divided by 11?11?

0 0

2 2

4 4

5 5

7 7

Answer: E
Solution:

The two-digit numbers that leave a remainder of 11 when divided by 99 are: 10,19,28,37,46,55,64,73,82,91.\begin{align*}&10, 19, 28, 37, 46,\\& 55, 64, 73, 82, 91.\end{align*} The two-digit numbers that leave a remainder of 33 when divided by 1010 are: 13,23,33,43,53,63,73,83,93.\begin{align*}&13, 23, 33, 43, 53,\\& 63, 73, 83, 93.\end{align*} Among these numbers, 7373 is the only common number. The remainder of 7373 when divided by 1111 is 7.7.

Thus, E is the correct answer.

6.

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

3 3

4 4

5 5

6 6

7 7

Answer: B
Solution:

Since there are 1919 people, each with one corresponding name length, the middle length will be the tenth one. Counting from the left side, the tenth value that we arrive upon is 4.4.

Thus, B is the correct answer.

7.

Which of the following numbers is not a perfect square?

12016 1^{2016}

22017 2^{2017}

32018 3^{2018}

42019 4^{2019}

52020 5^{2020}

Answer: B
Solution:

Since any number with an even exponent is a perfect square, we can eliminate A, C, and E. Also, a square number to any power remains a square number, so that rules out D.

Thus, B is the correct answer.

8.

Find the value of the expression 10098+9694+9290+86+42. \begin{align*} 100 - 98 + 96 - 94 + 92 - 90 \\ + \cdots 8 - 6 + 4 - 2. \end{align*}

20 20

40 40

50 50

80 80

100 100

Answer: C
Solution:

We can group the sum as follows: (10098)+(9694)++(42). \begin{gather*} (100 - 98) + (96 - 94) \\ + \cdots + (4 - 2). \end{gather*} Note that each pair evaluates to 22 and there are 2525 pairs. Therefore, the total sum is 225=50.2 \cdot 25 = 50.

Thus, C is the correct answer.

9.

What is the sum of the distinct prime integer divisors of 2016?2016?

9 9

12 12

16 16

49 49

63 63

Answer: B
Solution:

We can prime factorize 20162016 as 25327.2^5 \cdot 3^2 \cdot 7. This shows that the prime divisors of 20162016 are 2,3,2, 3, and 7.7. The sum of these is 12,12, so B is the correct answer.

10.

Suppose that aba * b means 3ab.3a - b. What is the value of xx if 2(5x)=1?2 * (5 * x) = 1?

110 \dfrac{1}{10}

2 2

103 \dfrac{10}{3}

10 10

14 14

Answer: D
Solution:

We can simplify the equation as follows: 1=2(5x)=2(35x)=2(15x)=32(15x)=x9. \begin{align*} 1 &= 2 * (5 * x) \\ &= 2 * (3 \cdot 5 - x) \\ &= 2 * (15 - x) \\ &= 3 \cdot 2 - (15 - x) \\ &= x - 9. \end{align*} Solving yields x=10.x = 10.

Thus, D is the correct answer.

11.

Determine how many two-digit numbers satisfy the following property:

When the number is added to the number obtained by reversing its digits, the sum is 132.132.

5 5

7 7

9 9

11 11

12 12

Answer: B
Solution:

Let abab be the two-digit number in question. Then, it follows that the number obtained by reversing its digits is ba.ba. Therefore, in order for abab to satisfy the property in the question: 10(a+b)+a+b=13211(a+b)=132a+b=12.\begin{align*}10(a + b) + a + b &=132\\11(a + b)&=132\\a+b&=12.\end{align*} The only possible solutions (a,b(a,b to this equation, where a,ba,b are both one digit, are: (3,9),(4,8),(5,7),(6,6),(3,9), (4,8), (5,7), (6,6),(7,5),(8,4),(9,3). (7,5), (8,4),(9,3). As such, there are 77 solutions.

Thus, B is the correct answer.

12.

Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students on the field trip were girls?

12 \dfrac{1}{2}

917 \dfrac{9}{17}

713 \dfrac{7}{13}

23 \dfrac{2}{3}

1415 \dfrac{14}{15}

Answer: B
Solution:

To more easily compare, we can convert the fractions to have the same denominator:34=912\dfrac{3}{4} = \dfrac{9}{12} 23=812\dfrac{2}{3} = \dfrac{8}{12} This shows that the ratio of girls to boys is 9:8,9 : 8, which means that the fraction of girls on the field trip is 917.\dfrac{9}{17}.

Thus, B is the correct answer.

13.

Two different numbers are randomly selected from the set {2,1,0,3,4,5}\{ - 2, -1, 0, 3, 4, 5\} and multiplied together. What is the probability that the product is 0?0?

16 \dfrac{1}{6}

15 \dfrac{1}{5}

14 \dfrac{1}{4}

13 \dfrac{1}{3}

12 \dfrac{1}{2}

Answer: D
Solution:

The only way for the product to be 00 is if one of the number chosen is 0.0. If the first number chosen is 0,0, then there are 55 options for the second number.

Similarly, there are 55 combinations if 00 was chosen second.

Therefore, there are 1010 total pairs where the product is 0.0. The total number of pairs is 65=30,6 \cdot 5 = 30, so the probability is 1030=13.\dfrac{10}{30} = \dfrac{1}{3}.

Thus, D is the correct answer.

14.

Karl's car uses a gallon of gas every 3535 miles, and his gas tank holds 1414 gallons when it is full.

One day, Karl started with a full tank of gas, drove 350350 miles, bought 88 gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?

525 525

560 560

595 595

665 665

735 735

Answer: A
Solution:

If Karl drove 350350 miles, then he used 350/35350 / 35 gallons of gas.

When he bought more gas, he added 88 gallons to 1410=414 - 10 = 4 gallons, attaining a total of 1212 gallons.

If his tank was half full when he arrived, he used 127=512 - 7 = 5 gallons, which equates to 535=1755 \cdot 35 = 175 miles.

Therefore, he travelled a total distance of 350+175=525 miles.350 + 175 = 525\text{ miles.}

Thus, A is the correct answer.

15.

What is the largest power of 22 that is a divisor of 134114?13^4 - 11^4?

8 8

16 16

32 32

64 64

128 128

Answer: C
Solution:

We can factor this expression using difference of squares.

(132+112)(132112)=29048=321453 \begin{align*} (13^2 + 11^2)&(13^2 - 11^2) \\ &= 290 \cdot 48 \\ &= 32 \cdot 145 \cdot 3 \end{align*}

This shows that 3232 is the largest power of 22 that divides the expression.

Thus, C is the correct answer.

16.

Annie and Bonnie are running laps around a 400400-meter oval track. They started together, but Annie has pulled ahead, because she runs 25%25\% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

114 1\dfrac{1}{4}

313 3\dfrac{1}{3}

4 4

5 5

25 25

Answer: D
Solution:

Since Annie is 25%25\% faster than Bonnie, for every lap Bonnie finishes, Annie completes 1141 \dfrac{1}{4} laps. Therefore, Annie gains a quarter lap every time Bonnie finished a lap.

With this in mind, for Annie to completely lap Bonnie, Bonnie must finish 44 laps, which means that Annie finished 55 laps.

Thus, D is the correct answer.

17.

An ATM password at Fred's Bank is composed of four digits from 00 to 9,9, with repeated digits allowable. If no password may begin with the sequence 9,1,1,9,1,1, then how many passwords are possible?

30 30

7290 7290

9000 9000

9990 9990

9999 9999

Answer: D
Solution:

The total number of passwords with no conditions is 104.10^4. The condition removes 1010 possible passwords since the first 33 are determined, and the last one can be anything. Therefore, the number of acceptable passwords is 10,00010=9990.10,000 - 10 = 9990.

Thus, D is the correct answer.

18.

In an All-Area track meet, 216216 sprinters enter a 100100-meter dash competition. The track has 66 lanes, so only 66 sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race.

How many races are needed to determine the champion sprinter?

36 36

42 42

43 43

60 60

72 72

Answer: C
Solution:

Note that each race eliminates 55 people. For there to be a winner, 215215 must be eliminated. Therefore, 215/15=43215 / 15 = 43 races are required to eliminate this number of people.

Thus, C is the correct answer.

19.

The sum of 2525 consecutive even integers is 10,000.10,000. What is the largest of these 2525 consecutive integers?

360 360

388 388

412 412

416 416

424 424

Answer: E
Solution:

The average of these numbers is 10,000/25=400.10,000 / 25 = 400. The largest number is 1212 even numbers away, which means that it equals 400+122=424.400 + 12 \cdot 2 = 424.

Thus, E is the correct answer.

20.

The least common multiple of aa and bb is 12,12, and the least common multiple of bb and cc is 15.15. What is the least possible value of the least common multiple of aa and c?c?

20 20

30 30

60 60

120 120

180 180

Answer: A
Solution:

We know that bb has to divide both 1212 and 15,15, so it must equal either 11 or 3.3.

If b=1,b = 1, then a=12a = 12 and c=15,c = 15, making their least common multiple 60.60. If b=3,b = 3, then the smallest value of aa is 1212 and cc is 5.5. The least common multiple in this scenario is 20.20.

Thus, A is the correct answer.

21.

A top hat contains 33 red chips and 22 green chips. Chips are drawn randomly, one at a time without replacement, until all 33 of the reds are drawn or until both green chips are drawn. What is the probability that the 33 reds are drawn?

310 \dfrac{3}{10}

25 \dfrac{2}{5}

12 \dfrac{1}{2}

35 \dfrac{3}{5}

23 \dfrac{2}{3}

Answer: B
Solution:

The only way for the 33 reds to be drawn first is if there is only one green drawn in the first 44 draws. The green can be in any of the first 44 spots, yielding 44 possibilities.

We can find the total number of possibilities to be 1010 by listing them out. Therefore, the desired probability is 410=25.\dfrac{4}{10} = \dfrac{2}{5}.

Thus, B is the correct answer.

22.

Rectangle DEFADEFA below is a 3×43 \times 4 rectangle with DC=CB=BA=1.DC=CB=BA=1. The area of the "bat wings" (shaded area) is

2 2

212 2 \dfrac{1}{2}

3 3

312 3 \dfrac{1}{2}

5 5

Answer: C
Solution:

Define II to be the midpoint of AD\overline{AD} and GG to be the midpoint of EF.\overline{EF}. Also define HH to be the intersection of CF\overline{CF} and BE.\overline{BE}.

The area of BCE\triangle BCE equals 1214=2.\dfrac{1}{2} \cdot 1 \cdot 4 = 2. By symmetry, we can see that BCH\triangle BCH and EFH\triangle EFH are similar. Since their bases are in a 1:31 : 3 ratio, so are their altitudes. This means that 3IH=HG,3IH = HG, which implies that IH=1.IH = 1.

Therefore, the area of BCH=1211=12.\begin{align*}\triangle BCH &= \dfrac{1}{2} \cdot 1 \cdot 1\\ &= \dfrac{1}{2}.\end{align*} This implies that the area of ECH=212=32.\begin{align*}\triangle ECH &= 2 - \dfrac{1}{2} \\&= \dfrac{3}{2}.\end{align*} Since the figure is symmetric, the total area of the bat wings is 232=3.2 \cdot \dfrac{3}{2} = 3.

Thus, C is the correct answer.

23.

Two congruent circles centered at points AA and BB each pass through the other circle's center. The line containing both AA and BB is extended to intersect the circles at points CC and D.D.

The circles intersect at two points, one of which is E.E. What is the degree measure of CED?\angle CED?

90 90

105 105

120 120

135 135

150 150

Answer: C
Solution:

We know that AE=EB=ABAE = EB = AB since they are all radii of congruent circles, so they form an equilateral triangle, which means that AEB=60.\angle AEB = 60^{\circ}.

Also, since DB\overline{DB} and AC\overline{AC} are diameters, DEB=AEC=90.\angle DEB = \angle AEC = 90^{\circ}. Therefore, CED=DEB+AECAEB,\begin{align*}\angle CED = \angle DEB &+ \angle AEC\\ &- \angle AEB,\end{align*} which equals 120.120^{\circ}.

Thus, C is the correct answer.

24.

The digits 1,1, 2,2, 3,3, 4,4, and 55 are each used once to write a five-digit number PQRST.PQRST. The three-digit number PQRPQR is divisible by 4,4, the three-digit number QRSQRS is divisible by 5,5, and the three-digit number RSTRST is divisible by 3.3. What is P?P?

1 1

2 2

3 3

4 4

5 5

Answer: A
Solution:

Since QRSQRS is divisible by 5,5, we know that S=5.S = 5.

Since PQRPQR is divisible by 4,4, QRQR equals either 12,24,12, 24, or 32.32.

This means that RSTRST will equal either 25T25T or 45T.45T. Note that RST=25TRST = 25T would require TT to be 22 or 5,5, each of which has solutions. Therefore the remaining digits when considering RST=45T,RST = 45T, 453453 is the only number divisible by \(3.\

Therefore, PQRST=12435.PQRST = 12435.

Thus, A is the correct answer.

25.

A semicircle is inscribed in an isosceles triangle with base 1616 and height 1515 so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

43 4 \sqrt{3}

12017 \dfrac{120}{17}

10 10

1722 \dfrac{17\sqrt{2}}{2}

1732 \dfrac{17\sqrt{3}}{2}

Answer: B
Solution:

Let OO be the center of the circle, which is the midpoint of AB.\overline{AB}.

We then get that BC=17BC = 17 via the Pythagorean theorem.

In addition, we can also calculate the area of BOC\triangle BOC as: BOC=12815=60.\begin{align*}\triangle BOC &= \dfrac{1}{2} \cdot 8 \cdot 15 \\&= 60.\end{align*} As the area of BOC=60=12OECB\triangle BOC=60=\dfrac{1}{2} OE \cdot CB we can see that OE=12017.OE = \dfrac{120}{17}.

Thus, B is the correct answer.