2004 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

On a map, a 1212-centimeter length represents 7272 kilometers. How many kilometers does a 1717-centimeter length represent?

66

102102

204204

864864

12241224

Answer: B
Solution(s):

Note that 11 cm represents 72÷12=672 \div 12 = 6 kilometers. This means that 1717 cm represents 617=1026 \cdot 17 = 102 kilometers.

Thus, B is the correct answer.

2.

How many different four-digit numbers can be formed be rearranging the four digits in 2004?2004?

44

66

1616

2424

8181

Answer: B
Solution(s):

Note that there are 22 non-zero digits that could be the thousands digit.

After choosing that, we need to arrange the other 33 digits. There are 33 spots for the other non-zero digit.

This gives us 23=62 \cdot 3 = 6 possible numbers.

Thus, B is the correct answer.

3.

Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for 1818 people. If they shared, how many meals should they have ordered to have just enough food for the 1212 of them?

88

99

1010

1515

1818

Answer: A
Solution(s):

Note that 1212 meals feed 1818 people. This means that one meal feeds 1812=32\dfrac{18}{12} = \dfrac{3}{2} people.

This means that they need 12÷32=812 \div \dfrac{3}{2} = 8 meals for 1212 people.

Thus, A is the correct answer.

4.

Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?

22

44

66

88

1010

Answer: B
Solution(s):

If there are 33 starters, then one person must not be starting. Choosing the person who doesn't start determines the starters.

There are 44 choices for the person who doesn't start.

Thus, B is the correct answer.

5.

Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?

44

77

88

1515

1616

Answer: D
Solution(s):

Note that after every game, one team gets eliminated. For there to be one team remaining, 1515 teams must have been eliminated.

This means that 1515 games had to have been played.

Thus, D is the correct answer.

6.

After Sally takes 2020 shots, she has made 55%55 \% of her shots. After she takes 55 more shots, she raises her percentage to 56%.56 \%. How many of the last 55 shots did she make?

11

22

33

44

55

Answer: C
Solution(s):

Sally made 20×.55=1120 \times .55 = 11 of her first 2020 shots. Then we get that 11+x25=.56, \dfrac{11 + x}{25} = .56, which tells us that 11+x=14 11 + x = 14 and x=3.x = 3.

Thus, C is the correct answer.

7.

An athlete's target heart rate, in beats per minute, is 80%80 \% of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from 220.220. To the nearest whole number, what is the target heart rate of an athlete who is 2626 years old?

134134

155155

176176

194194

243243

Answer: B
Solution(s):

The maximum heart rate for this athlete would be 22026=194.220 - 26 = 194. Then the target heart rate would be 194×.8155.194 \times .8 \approx 155.

Thus, B is the correct answer.

8.

Find the number of two-digit positive integers whose digits total 7.7.

66

77

88

99

1010

Answer: B
Solution(s):

Note that the tens digit can range from 11 to 7,7, and this digit determines the units digit.

Therefore, there are 77 numbers.

Thus, B is the correct answer.

9.

The average of the five numbers in a list is 54.54. The average of the first two numbers is 48.48. What is the average of the last three numbers?

5555

5656

5757

5858

5959

Answer: D
Solution(s):

The sum of all 55 numbers is 545=270.54 \cdot 5 = 270. The sum of the first 22 numbers is 482=96.48 \cdot 2 = 96.

The sum of the last 33 numbers is 27096=174.270 - 96 = 174. The average is therefore 174÷3=58.174 \div 3 = 58.

Thus, D is the correct answer.

10.

Handy Aaron helped a neighbor 1141 \frac{1}{4} hours on Monday, 5050 minutes on Tuesday, from 8:208:20 to 10:4510:45 on Wednesday morning, and a half-hour on Friday. He is paid $ 3 per hour. How much did he earn for the week?

$ 8

$ 9

$ 10

$ 12

$ 15

Answer: E
Solution(s):

Aaron worked 5060=56\dfrac{50}{60} = \dfrac{5}{6} hours on Tuesday. He worked 22 hours and 2525 minutes on Wednesday, which equals 2+2560=29122 + \dfrac{25}{60} = \dfrac{29}{12} hours.

This means that he would have earned 3(54+56+2912+12) 3\left(\dfrac{5}{4} + \dfrac{5}{6} + \dfrac{29}{12} + \dfrac{1}{2}\right) =35= 3 \cdot 5 = $ 15.

Thus, E is the correct answer.

11.

The numbers 2,4,6,9-2, 4, 6, 9 and 1212 are rearranged according to these rules:

1. The largest isn’t first, but it is in one of the first three places.

2. The smallest isn’t last, but it is in one of the last three places.

3. The median isn’t first or last.

What is the average of the first and last numbers?

3.53.5

55

6.56.5

7.57.5

88

Answer: C
Solution(s):

Note that the largest, smallest, and median numbers cannot be the first or last number.

This means that the first and last numbers are 44 and 99 in some order. The average is (4+9)÷2 (4 + 9) \div 2 =13÷2= 13 \div 2 =6.5.= 6.5.

Thus, C is the correct answer.

12.

Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for 2424 hours. If she is using it constantly, the battery will last for only 33 hours. Since the last recharge, her phone has been on 99 hours, and during that time she has used it for 6060 minutes. If she doesn’t use it any more but leaves the phone on, how many more hours will the battery last?

77

88

1111

1414

1515

Answer: B
Solution(s):

When not in use, her cell phone uses up 124\frac{1}{24} of its battery per hour. When it is in use, it uses up 13\frac{1}{3} of its battery per hour.

Niki's phone has been on for 99 hours, with 88 of those hours being idle and 11 hour being used to talk on the phone.

This means that the phone has used up 23\frac{2}{3} of its battery. In order to drain the remaining 13\frac{1}{3} of the battery, the phone can last for 88 more hours without being used.

Thus, B is the correct answer.

13.

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.

I. Bill is the oldest.

II. Amy is not the oldest.

III. Celine is not the youngest.

Rank the friends from the oldest to youngest.

Bill, Amy, Celine\text{Bill, Amy, Celine}

Amy, Bill, Celine\text{Amy, Bill, Celine}

Celine, Amy, Bill\text{Celine, Amy, Bill}

Celine, Bill, Amy\text{Celine, Bill, Amy}

Amy, Celine, Bill\text{Amy, Celine, Bill}

Answer: E
Solution(s):

If Bill is the oldest, we get that statement I and II are true, which is not allowed. This tell us that statement I is false.

If Amy is not the oldest, then Celine is the oldest since Bill isn't. This makes statement III true, which is also not allowed.

Thus, statement III must the true one and I and II are false. Using this, we get the order to be Amy, Celine, Bill.

Thus, E is the correct answer.

14.

What is the area enclosed by the geoboard quadrilateral below?

1515

181218\frac{1}{2}

221222\frac{1}{2}

2727

4141

Answer: C
Solution(s):

We can use Pick's Theorem which tells us that the area of a polygon whose vertices are lattice points is given by A=I+12B1, A = I + \dfrac{1}{2}B - 1, where II is the number of interior lattice points and BB is the number of boundary lattice points.

Alternatively, one could also decompose the quadrilateral into triangles, but it is also possible to prove Pick's Theorem by recursively decomposing arbitrary polygons into triangles with lattice point vertices, though that is left an exercise for the reader.

Counting, we get that B=5B = 5 and I=21.I = 21. Plugging these in, we get A=52+211=2212. A = \dfrac{5}{2} + 21 - 1 = 22 \dfrac{1}{2}.

Thus, C is the correct answer.

15.

Thirteen dark and six bright hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of bright tiles with the same size and shape as the others, what will be the difference between the total number of bright tiles and the total number of dark tiles in the new figure?

55

77

1111

1212

1818

Answer: C
Solution(s):

Note that the first ring around the middle tile has 66 tiles. The second ring has 1212 tiles.

Following this pattern, we get that the third ring has 1818 tiles. The total number of dark tiles is 1+12=131 + 12 = 13 and bright tiles is 6+18=24.6 + 18 = 24.

The difference is therefore 2413=11.24 - 13 = 11.

Thus, C is the correct answer.

16.

Two 600600 mL pitchers contain orange juice. One pitcher is 13\frac 13 full and the other pitcher is 25\frac 25 full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?

18\dfrac{1}{8}

316\dfrac{3}{16}

1130\dfrac{11}{30}

1119\dfrac{11}{19}

1115\dfrac{11}{15}

Answer: C
Solution(s):

The first pitcher contains 60013=200600 \cdot \dfrac{1}{3} = 200 mL of orange juice. The second one has 60025=240600 \cdot \dfrac{2}{5} = 240 mL.

The large container then has 200+240=440200 + 240 = 440 mL of orange juice. The total amount of mixture is 2600=12002 \cdot 600 = 1200 mL.

Then the fraction of orange juice is 4401200=1130.\dfrac{440}{1200} = \dfrac{11}{30}.

Thus, C is the correct answer.

17.

Three friends have a total of 66 identical pencils, and each one has at least one pencil. In how many ways can this happen?

11

33

66

1010

1212

Answer: D
Solution(s):

Recall that stars and bars gives us the number of ways to put nn objects into kk bins, where each bin must have at least one object, by (n1k1). \binom{n - 1}{k - 1}.

In our scenario, we must split 66 pencils amongst 33 people, where each person must have at least one pencil. Therefore, the number of ways is (6131)=(52)=10. \binom{6 - 1}{3 - 1} = \binom{5}{2} = 10.

Thus, D is the correct answer.

18.

Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers 11 through 10.10. Each throw hits the target in a region with a different value. The scores are: Alice 1616 points, Ben 44 points, Cindy 77 points, Dave 1111 points, and Ellen 1717 points. Who hits the region worth 66 points?

Alice\text{Alice}

Ben\text{Ben}

Cindy\text{Cindy}

Dave\text{Dave}

Ellen\text{Ellen}

Answer: A
Solution(s):

The only way to get Ben's score is with a 11 and 33 since he can't hit 22 twice.

Cindy can achieve her score with 1+6,2+5, or 3+4. 1 + 6, 2 + 5, \text{ or } 3 + 4. Ben already hit 11 and 3,3, so Cindy must have hit 22 and 5.5.

Similarly, Dave must have hit 44 and 7.7. Finally, since 77 is already used, Alice is forced to have hit 66 and 1010 with Ellen hitting 88 and 9.9.

Thus, A is the correct answer.

19.

A whole number larger than 22 leaves a remainder of 22 when divided by each of the numbers 3,4,5,3, 4, 5, and 6.6. The smallest such number lies between which two numbers?

40 and 4940\ \text{and}\ 49

60 and 7960 \text{ and}\ 79

100 and 129100\ \text{and}\ 129

210 and 249210\ \text{and}\ 249

320 and 369320\ \text{and}\ 369

Answer: B
Solution(s):

Let xx be the number. Then x2x - 2 is divisible by 3,4,5,3, 4, 5, and 6.6.

The least common multiple of these numbers is 60,60, which makes xx 62.62.

Thus, B is the correct answer.

20.

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are 66 empty chairs, how many people are in the room?

1212

1818

2424

2727

3636

Answer: D
Solution(s):

Note that 14\dfrac{1}{4} of the chairs are empty. Since this is equal to 6,6, there are 34=183 \cdot 4 = 18 taken chairs.

If 1818 is 23\dfrac{2}{3} the number of people, then the total number of people is 18÷23=27.18 \div \dfrac{2}{3} = 27.

Thus, D is the correct answer.

21.

Spinners AA and BB are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

Answer: D
Solution(s):

For the product to be even, then at least one of the spinners must land on an even number.

We can use complementary counting and calculate the probability of both spinners landing on odds.

This happens with a probability of 1223=13. \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{3}.

Then the probability of landing on at least one even is 113=23.1 - \dfrac{1}{3} = \dfrac{2}{3}.

Thus, D is the correct answer.

22.

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is 25.\dfrac{2}{5}. What fraction of the people in the room are married men?

13\dfrac{1}{3}

38\dfrac{3}{8}

25\dfrac{2}{5}

512\dfrac{5}{12}

35\dfrac{3}{5}

Answer: B
Solution(s):

WLOG, let there be 55 women in the room. Then there are 525=25 \cdot \dfrac{2}{5} = 2 single women.

This means that there are 52=35 - 2 = 3 married women, which is also the number of married men.

There are a total of 5+3=85 + 3 = 8 people in the room. The fraction of married men is 38.\dfrac{3}{8}.

Thus, B is the correct answer.

23.

Tess runs counterclockwise around rectangular block JKLM.JKLM. She lives at corner J.J. Which graph could represent her straight-line distance from home?

Answer: D
Solution(s):

Tess cannot be represented by a constant horizontal line because this would imply that she is running in a circle centered at her starting point.

This means that B and E are incorrect as they involve straight lines.

A is incorrect because the maximum distance is at the end of Tess's journey, and C is wrong because it has 22 maximums.

Thus, D is the correct answer.

24.

In the figure, ABCDABCD is a rectangle and EFGHEFGH is a parallelogram. Using the measurements given in the figure, what is the length dd of the segment that is perpendicular to HE\overline{HE} and FG?\overline{FG}?

6.86.8

7.17.1

7.67.6

7.87.8

8.18.1

Answer: C
Solution(s):

First, we can calculate the area of the parallelogram. This can be done be subtracting out the areas of the 44 triangles to get 108212(34+65) 10 \cdot 8 - 2 \cdot \dfrac{1}{2}(3 \cdot 4 + 6 \cdot 5) =80(12+30) =80 - (12 + 30) =38.= 38.

We can also calculate the area of the parallelogram by multiplying the base by the altitude, d.d. Therefore 38=5d 38 = 5dd=385=7.6 d = \dfrac{38}{5} = 7.6 (we get the 55 from the Pythagorean theorem).

Thus, C is the correct answer.

25.

Two 4×44 \times 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

164π16-4\pi

162π16-2\pi

284π28-4\pi

282π28-2\pi

322π32-2\pi

Answer: D
Solution(s):

Note that the union of the two squares is a square with side length 2.2. This makes the area of the overlapping squares 42+4222=28. 4^2 + 4^2 - 2^2 = 28.

We need to remove the area of the circle, which has radius 2\sqrt{2} (we get this since the diameter is the diagonal of a square with side length 22).

Therefore, the area of the shaded region is 28π(2)2=282π. 28 - \pi \left(\sqrt{2}\right)^2 = 28 - 2 \pi.

Thus, D is the correct answer.