2015 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

How many square yards of red carpet are required to cover a rectangular floor that is 1212 feet long and 99 feet wide? (There are 3 feet in a yard.)

12 12

36 36

108 108

324 324

972 972

Answer: A
Solution(s):

Since one side is 1212 feet, it would be 123=4\dfrac{12}{3} = 4 yards.

Since another side is 99 feet, it would be 93=3\dfrac{9}{3} = 3 yards.

Since the dimensions are 4 yards×3 yards,4 \text{ yards} \times 3 \text{ yards}, the area is equal to 43=12.4\cdot3 = 12.

Thus, the correct answer is A .

2.

Point OO is the center of the regular octagon ABCDEFGH,ABCDEFGH, and XX is the midpoint of the side AB.\overline{AB}. What fraction of the area of the octagon is shaded?

1132\dfrac{11}{32}

38\dfrac{3}{8}

1332\dfrac{13}{32}

716\dfrac{7}{16}

1532\dfrac{15}{32}

Answer: D
Solution(s):

First notice that there are 88 equally sized triangles that can be created with OO and any two consecutive points. Therefore, they each take up 18\dfrac {1}{8} of the total area of the octagon.

The shaded area has three complete triangles and half of the triangle ABO.ABO. Therefore, the shaded area is 3.58=716\dfrac{3.5}{8} = \dfrac{7}{16} of the total area of the octagon.

Thus, the correct answer is D.

3.

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of 1010 miles per hour. Jack walks to the pool at a constant speed of 44 miles per hour. How many minutes before Jack does Jill arrive?

5 5

6 6

8 8

9 9

10 10

Answer: D
Solution(s):

Jack travels at a rate of 44 miles per 6060 minutes. Therefore, it takes him 604=15\dfrac{60}{4} = 15 minutes to get to the pool.

Jill travels at a rate of 1010 miles per 6060 minutes. Therefore it takes her 6010=6\dfrac{60}{10} = 6 minutes to get to the pool.

Therefore, the difference in their times is 156=915-6 = 9 minutes.

Thus, the correct answer is D.

4.

The Centerville High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?

2 2

4 4

5 5

6 6

12 12

Answer: E
Solution(s):

Since there are 22 boys and two places to put them, there are 2!=22! = 2 ways to place them.

Since there are 33 boys and three places to put them, there are 3!=63! = 6 ways to place them.

Therefore, the total number of places to put them is 62=12.6\cdot2=12.

Thus, the correct answer is E.

5.

Billy's basketball team scored the following points over the course of the first 1111 games of the season: 42,47,53,53,58,58,42, 47, 53, 53, 58, 58, 58,61,64,65,73.58, 61, 64, 65, 73. If his team scores 4040 in the 12th12^{\text{th}} game, which of the following statistics will show an increase?

range \text{range}

median \text{median}

mean \text{mean}

mode \text{mode}

mid-range \text{mid-range}

Answer: A
Solution(s):

When considering all 1212 games, 4040 -- from the 12th12^{\text{th}} game -- will be the lowest score. Therefore, compared to the range of just the first 1111 games, the range of all 1212 games would increase from 7342=3173-42 = 31 to 7340=33.73-40=33.

Thus, the correct answer is A .

6.

In ABC,\bigtriangleup ABC, AB=BC=29,AB=BC=29, and AC=42.AC=42. What is the area of ABC?\bigtriangleup ABC?

100 100

420 420

500 500

609 609

701 701

Answer: B
Solution(s):

Let's begin by sketching the triangle:

Define XX to be the midpoint of AC.AC. Therefore, AX=21.AX=21.

Then, since AB=BC,AB=BC, we know that AXBAXB is a right triangle. By the Pythagorean Theorem, we know AX2+BX2=AB2,AX^2+BX^2 = AB^2, so we get BX=AB2BX2=292212=400=20.\begin{align*}BX &= \sqrt{AB^2-BX^2} \\&= \sqrt{29^2-21^2} \\&= \sqrt{400} = 20.\end{align*}

Then, as the area of a triangle is equal to 12bh\dfrac12 bh formula for area, we can see that the area is 42202=420.\dfrac{42\cdot20}2 = 420.

Thus, the correct answer is B.

7.

Each of two boxes contains three chips numbered 1,1, 2,2, 3.3. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

19 \dfrac{1}{9}

29 \dfrac{2}{9}

49 \dfrac{4}{9}

12 \dfrac{1}{2}

59 \dfrac{5}{9}

Answer: E
Solution(s):

First, let us try to find the probability that the product is odd. We know that the product will be odd if both boxes yield odd numbers. As there is a 23\dfrac 23 chance of a box being odd, there is a (23)2=49\left(\dfrac 23\right)^2 = \dfrac 49 chance of the product being odd.

The probability that a number is even is 11 minus the probability that it is odd (as they are complements), so the odds of it being even is 149=59.1 - \dfrac 49 = \dfrac 59.

Thus, the correct answer is E.

8.

What is the smallest whole number larger than the perimeter of any triangle with a side of length 5 5 and a side of length 19?19?

24 24

29 29

43 43

48 48

57 57

Answer: D
Solution(s):

Let the third side be s.s. By the triangle inequality, we know s<5+19.s < 5+19. Therefore, the perimeter of the triangle: s+5+19<2(5+19)=48.s+5 +19 < 2(5+19) = 48.

As such, the largest triangle we can make under these constraints is a triangle with sides 5,19,23,5,19,23, and therefore a perimeter of 47,47, as required. Therefore, the smallest possible number greater than any perimeter is 48.48.

Thus, the correct answer is D.

9.

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 2020 days?

39 39

40 40

210 210

400 400

401 401

Answer: D
Solution(s):

We want to find 1+3++39.1+3 + \cdots + 39. This sum can be rewritten as (1+39)+(3+37)++(19+21),\begin{align*}&(1+39) + (3+ 37) \\&+\cdots +(19+21),\end{align*} which we can see has 1010 terms. Further notice that each term is equal to 40.40. Therefore, the sum is 1040=400.10\cdot40=400.

Thus, the correct answer is D .

10.

How many integers between 10001000 and 99999999 have four distinct digits?

3024 3024

4536 4536

5040 5040

6480 6480

6561 6561

Answer: B
Solution(s):

First, there are 99 digits to choose for the thousands digit since 00 can't be chosen.

Then, after that, there are 99 ways to choose the hundreds digit, 88 ways to choose the tens digit, and 77 ways to choose the ones digit. Therefore, we get 9987=45369\cdot9\cdot8\cdot7 = 4536 ways to choose such an integer.

Thus, the correct answer is B .

11.

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

122,050 \dfrac{1}{22,050}

121,000 \dfrac{1}{21,000}

110,500 \dfrac{1}{10,500}

12,100 \dfrac{1}{2,100}

11,050 \dfrac{1}{1,050}

Answer: B
Solution(s):

First, we find how many license plates are possible.

For the first letter, we have 55 choices. For the second letter, we have 2121 choices. For the third letter, we have 211=2021-1=20 choices since we must exlude the letter chosen for the second letter. For the final letter, we have 1010 choices. Overall, there are 5212010=21,0005\cdot21\cdot20\cdot10 = 21,000 choices.

The choice AMC8 is one possible choice of these 21,00021,000 possibilites, so the probability of it being our choice is 121,000.\dfrac{1}{21,000}.

Thus, the correct answer is B.

12.

How many pairs of parallel edges, such as AB\overline{AB} and GH\overline{GH} or EH\overline{EH} and FG,\overline{FG}, does a cube have?

6 6

1212

18 18

2424

3636

Answer: C
Solution(s):

We have 1212 edges, and each edge has 33 parallel lines to it. However, this would double count it, so we divide by 2.2. Therefore, there are 3122=18\dfrac{3\cdot12}{2} = 18 set of parallel edges.

Thus, the correct answer is C.

13.

How many subsets of two elements can be removed from the set {1,2,3,4,5,6,7,8,9,10,11}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} so that the mean (average) of the remaining numbers is 6?

 1 \text{ 1}

 2 \text{ 2}

 3 \text{ 3}

 5 \text{ 5}

 6 \text{ 6}

Answer: D
Solution(s):

Currently, the sum is 66.66. If we remove 22 elements, then there are 99 elements remaining. Since their mean is 6,6, the sum of the remaining elements is 54.54. This means the sum of the two elements we take out are 6654=12.66-54=12. There are 55 two elements sets that work, namely: {1,11}{2,10}{3,9}{4,8}{5,7}\begin{align*} &\{1,11\}\\ &\{2,10\}\\&\{3,9\}\\&\{4,8\}\\&\{5,7\} \end{align*}

Thus, the correct answer is D.

14.

Which of the following integers cannot be written as the sum of four consecutive odd integers?

16 16

40 40

72 72

100100

200 200

Answer: D
Solution(s):

Let 2k+12k+1 be the lowest of the odd integers in question. Then, the next three consecutive odd integers are 2k+3,2k+5,2k+3,2k+5, and 2k+7.2k+7.

It follows that their sum is 8k+16=8(k+2).8k+16 = 8(k+2). This means the integer must a multiple of 8.8. The only answer choice that is not divisible by 88 is 100.100.

Thus, the correct answer is D.

15.

At Euler Middle School, 198198 students voted on two issues in a school referendum with the following results: 149149 voted in favor of the first issue and 119119 voted in favor of the second issue. If there were exactly 2929 students who voted against both issues, how many students voted in favor of both issues?

49 49

70 70

79 79

99 99

149 149

Answer: D
Solution(s):

Since 2929 students voted against both, we know that 19829=169198-29 = 169 people voted for at least one.

As we know that 149149 students voted for the first issue, and 119119 students voted for the second issue, and 169169 students that voted for at least one issue, we conclude that the number of students that voted for both is 149+119169=99.149+119-169 = 99.

Thus, the correct answer is D.

16.

In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 13\dfrac{1}{3} of all the ninth graders are paired with 25\dfrac{2}{5} of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

215 \dfrac{2}{15}

411 \dfrac{4}{11}

1130 \dfrac{11}{30}

38 \dfrac{3}{8}

1115 \dfrac{11}{15}

Answer: B
Solution(s):

First, let the number of sixth-grader be ss and let the number of ninth-graders be n.n. Since they are paired in a way that everyone has one buddy, we know 13n=25s.\dfrac{1}{3}n = \dfrac{2}{5}s . This means 5n=6s.5n = 6s. Using algebra, we get 11n=6(s+n)11n = 6(s+n)11s=5(s+n),11s = 5(s+n), so we get n=(s+n)611n = (s+n)\dfrac{6}{11}s=(s+n)511. s = (s+n)\dfrac{5}{11}. Now, we must determine how many people have a buddy divided by the total number of people. There are 13n+25s\dfrac{1}{3}n + \dfrac{2}{5}s people who have buddies and s+ns+n total people. Therefore: 13n+25ss+n=13(611(s+n))s+n+25(511(s+n))s+n=211+211=411.\begin{align*}\dfrac{\dfrac{1}{3}n + \dfrac{2}{5}s}{s+n} &=\dfrac{\dfrac{1}{3}\left(\dfrac{6}{11}(s+n)\right)}{s+n} \\&+ \dfrac{\dfrac{2}{5}\left(\dfrac{5}{11}(s+n)\right)}{s+n} \\ &= \dfrac{2}{11} + \dfrac{2}{11} \\&= \dfrac{4}{11}. \end{align*}

Thus, the correct answer is B.

17.

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?

4 4

6 6

8 8

9 9

12 12

Answer: D
Solution(s):

Let the speed in rush hour be ss miles per hour. Since 2020 minutes is 13\dfrac 13 hours, Jeremy's father drives s(13)s(\dfrac 13) miles during rush hour. When it's not rush hour, Jeremy's father drives at a speed of s+18s+18 miles per hour. Since 1212 minutes is 15\dfrac 15 hours, he drives (s+18)(15)(s+18)(\dfrac 15) miles when its not rush hour.

As the distance to school is the same, we know s13=(s+18)15s53=s+18s=27.\begin{align*}s\dfrac{1}{3} &= (s+18) \dfrac {1}{5}\\ s\dfrac{5}{3} &= s+ 18\\ s &= 27.\end{align*} Since the distance to school is s(13),s(\dfrac 13 ), we can deduce that the distance to school is (27)(13)=9(27)(\dfrac 13) = 9 miles.

Thus, the correct answer is D.

18.

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, 2,5,8,11,142,5,8,11,14 is an arithmetic sequence with five terms, in which the first term is 22 and the constant 33 is added. Each row and each column in this 5×55\times5 array is an arithmetic sequence with five terms. What is the value of X?\text{X}?

21 21

31 31

36 36

40 40

42 42

Answer: B
Solution(s):

We can write an arbitrary 5-term arithmetic sequence as: a,a+b,a+2b,a+3b,a+4b.a,a+b,a+2b,a+3b,a+4b. In this case, this representation of sequences implies that a+2b=(a)+(a+4b)2,a+2b = \dfrac{(a)+(a+4b)}{2}, so the third team is the average of the first and last terms. Therefore, the middle number on the top row is the average of the top two corners 1+252=13,\dfrac{1+25}{2} = 13, and the middle number on the bottom row is the average of the bottom two corners, which would be 17+812=49.\dfrac{17+81}{2} = 49. Finally, the middle entry is the average of the third entry in the first and last row, which is 13+492=31.\dfrac{13 + 49}{2} = 31.

Thus, the correct answer is B.

19.

A triangle with vertices as A=(1,3),A=(1,3), B=(5,1),B=(5,1), and C=(4,4)C=(4,4) is plotted on a 6×56\times5 grid. What fraction of the grid is covered by the triangle?

16 \dfrac{1}{6}

15 \dfrac{1}{5}

14 \dfrac{1}{4}

13 \dfrac{1}{3}

12 \dfrac{1}{2}

Answer: A
Solution(s):

The total area of the grid is 65=30.6\cdot5=30. In order to find the fraction of this grid that the triangle covers, we must now find the area of the triangle. To do this, we will use the following diagram:

Thus, the area of the triangle A(ABC)A(\triangle ABC) is equal to: A(PQRB)A(PAB)A(BCR)A(CAQ)=(3)(4)12(4)(2)2(32)=1243=5. \begin{align*} &A(PQRB)-A(\triangle PAB)\\&-A(\triangle BCR)-A(\triangle CAQ)\\ &=(3)(4)-\dfrac12 (4)(2)-2 \left(\dfrac32\right)\\ &=12 - 4 - 3\\ &=5. \end{align*}

Therefore, the fraction of the area is 530=16.\dfrac{5}{30} = \dfrac{1}{6} .

Thus, the correct answer is A.

20.

Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?

4 4

5 5

6 6

7 7

8 8

Answer: D
Solution(s):

Let aa be the number of $1\$1 pairs, let bb be the number of $3\$3 pairs, and let cc be the number of $4\$4 pairs of socks Ralph bought.

We know a+b+c=12a+b+c = 12a+3b+4c=24.a+3b+4c=24. Therefore, we can subtract the two equations to get that 2b+3c=12.2b+3c=12. Since b>0,b > 0, we know 2b>02b > 0 and therefore 123c>0.12-3c > 0. Therefore, c<4.c < 4.

Also, we know c>0.c > 0. Taking (mod2)\pmod 2 of 2b+3c=122b+3c=12 yields c0(mod2).c \equiv 0 \pmod 2. Since c0(mod2)c \equiv 0 \pmod 2 and 0<c<4,0 < c < 4, we know c=2.c=2. Substituting, we deduce b=3.b=3. Finally, with a+b+c=12,a+b+c = 12, we get a+2+3=12,a+2+3 = 12, so a=7.a=7.

This means the number of $1 pairs is 7.7.

Thus, the correct answer is D.

21.

In the given figure hexagon ABCDEFABCDEF is equiangular, ABJIABJI and FEHGFEHG are squares with areas 1818 and 3232 respectively, JBK\triangle JBK is equilateral and FE=BC.FE=BC. What is the area of KBC?\triangle KBC?

62 6\sqrt{2}

99

1212

929\sqrt{2}

3232

Answer: C
Solution(s):

Since ABJIABJI is a square, each side has a length equal to the square root of the area. This means JB=18=32.JB = \sqrt{18} = 3 \sqrt{2}. Since JBKJBK is equiangular, K=J.\angle K = \angle J. This means BK=JB=32.BK = JB = 3\sqrt{2}.

Similarly, we get EF=32=42.EF = \sqrt{32} = 4\sqrt{2} .

Since ABCDEFABCDEF is equiangular, we can get that BC=EF=42.BC=EF = 4\sqrt{2}.

Therefore, the area of KBCKBC is equal to (BC)(BK)2=(42)(32)2=242=12.\begin{align*}\dfrac{(BC)(BK)}{2} &=\dfrac{(4\sqrt{2})(3\sqrt{2})}{2} \\&= \dfrac{24}{2} \\&= 12.\end{align*}

Thus, the correct answer is C.

22.

On June 1, a group of students is standing in rows, with 15 students in each row.

On June 2, the same group is standing with all of the students in one long row.

On June 3, the same group is standing with just one student in each row.

On June 4, the same group is standing with 6 students in each row.

This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

21 21

30 30

60 60

90 90

1080 1080

Answer: C
Solution(s):

Let nn be the number of students. Since we consider only 1212 days, and each day's row and column pair represents a unique factor pair of n,n, we have 1212 factors of n.n.

Since 1515 students were in one row, nn is a multiple of 15.15.

Since 66 students were in one row on a different day, nn is a multiple of 6.6.

Since nn is a multiple of 66 and 15,15, it must be a multiple of lcm(6,15)=30.lcm(6,15) =30. This means nn is a multiple of 235.2\cdot3\cdot5. The number of factors is the product of the exponents plus 1,1, so this would have 88 factors. However, multiplying it by 22 yields 2235=60.2^2\cdot3\cdot5=60. This yields 1212 total factors as required.

Thus, the correct answer is C.

23.

Tom has twelve slips of paper which he wants to put into five cups labeled A,A, B,B, C,C, D,D, E.E.

He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from AA to E.E. The numbers on the papers are: 2,2,2,2.5,2.5,3,3,3,3,3.5,4,4.5.\begin{align*} &2, 2, 2, 2.5, 2.5, 3,\\& 3, 3, 3, 3.5, 4, 4.5.\end{align*} If a slip with 22 goes into cup EE and a slip with 33 goes into cup B,B, then the slip with 3.53.5 must go into what cup?

A A

B B

C C

D D

E E

Answer: D
Solution(s):

The sum of every slip combined is 35.35. Therefore, the average of the cups would be 7.7. This would make A=5A=5B=6B=6C=7C=7D=8D=8E=9.E=9. Since BB has one 3,3, the sum of the remaining slips is 3,3, which can only be done with another 3.3.

As such, what remains after filling BB is: 2,2,2,2.5,2.5,3,3,3.5,4,4.5.\begin{align*}&2, 2, 2, 2.5, 2.5, 3,\\& 3, 3.5, 4, 4.5.\end{align*} Let EE' be the sum of EE after putting the 22 slip. This equals 92=7.9-2=7.

Therefore, we need to fill C,EC,E' each to be 77 and have AA have 55 and DD having 8.8. Now, lets try fitting the 3.53.5 slip into each of the cups.

When putting it in A,A, we have 1.51.5 left which is invalid. Similarly, we can't put it in BB as it is full. When putting it in C,C, we have 3.53.5 left, which cannot work as no other values sum to 3.5.3.5. Similar to C,C, putting it in EE would have 3.5,3.5, which we just showed to be invalid. Finally, we can make cup DD work by making the cups be {2.5,2.5},{3,3},{3,4},{3.5,4.5},{2,2,2,3}.\begin{align*}&\{2.5,2.5\},\{3,3\},\{3,4\},\\& \{3.5,4.5\},\{2,2,2,3\}. \end{align*}

Thus, the correct answer is D.

24.

A baseball league consists of two four-team divisions. Each team plays every other team in its division NN games. Each team plays every team in the other division MM games with N>2MN > 2M and M>4.M > 4. Each team plays a 7676 game schedule.

How many games does a team play within its own division?

36 36

48 48

54 54

60 60

72 72

Answer: B
Solution(s):

Since there are 33 other teams in the division and 44 teams not in the division, a team plays 3N+4M=763N+4M=76 teams. Also notice that 76=3N+4M>10M,76=3N+4M > 10M, since N>2M.N > 2M. This implies that 76>10M,76 > 10M, and as such, 7M.7 \geq M.

Under the given conditions, we know 7M>4.7 \geq M > 4. Also, taking 3N+4N=763N+4N=76 under modulus 33 yields M1(mod3).M \equiv 1 \pmod 3. The only possible MM in the range is 7,7, suggesting that he plays 47=284\cdot7 = 28 games against non-divisional opponents.

As such, he plays 7628=4876-28 = 48 games against divisional opponents.

Thus, the correct answer is B.

25.

One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?

9 9

1212 12\dfrac{1}{2}

15 15

1512 15\dfrac{1}{2}

17 17

Answer: C
Solution(s):

The desired square can be drawn like this. It consists a smaller square of size 3×33 \times 3 in it, with area 33=9,3\cdot 3=9, as well as 44 triangles. As each square has base 33 and height 1,1, the combined area of the triangles is 4(312)=6.4(\dfrac{3\cdot 1}{2}) = 6.

This makes the total area of this square equal to 15.15.

Thus, the correct answer is C.