2002 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.

A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?

22

33

44

55

66

Answer: D
Solution:

A line and circle can only intersect each other at at most two points. Two lines can only intersect at one point.

Therefore, the maximum number of intersections is 22+1=5. 2 \cdot 2 + 1 = 5.

Thus, D is the correct answer.

2.

How many different combinations of $5 bills and $2 bills can be used to make a total of $17? Order does not matter in this problem.

22

33

44

55

66

Answer: A
Solution:

We cannot use 44 or more dollar $5 bills since that would go over the total.

We can use three $5 bills and one $2 bill. We cannot use two $5 bills since we cannot make an odd amount of money with $2 bills.

Finally, we can use one $5 bill and six $2 bills. As above, we cannot use zero $5 bills. Therefore, there are 22 ways to get the total.

Thus, A is the correct answer.

3.

What is the smallest possible average of four distinct positive even integers?

33

44

55

66

77

Answer: C
Solution:

To get the smallest possible average, we want to use the smallest 44 positive even integers.

This can be achieved as follows: 2+4+6+84=204 \dfrac{2 + 4 + 6 + 8}{4} = \dfrac{20}{4}=5. = 5.

Thus, C is the correct answer.

4.

The year 20022002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 20022002 that is a palindrome?

00

44

99

1616

2525

Answer: B
Solution:

We don't want to increase the thousands digit, so we can keep that as 2.2.

This means that we have to increase the tens and hundreds digits to 1,1, to yield the next palindrome of 2112.2112. The product of its digits is 4.4.

Thus, B is the correct answer.

5.

Carlos Montado was born on Saturday, November 9,2002.9, 2002. On what day of the week will Carlos be 706706 days old?

Monday

Wednesday

Friday

Saturday

Sunday

Answer: C
Solution:

The days of the week cycle every 77 days. After 700700 days, the day of the week will still be Saturday.

After 66 more days, the day of the week will be Friday.

Thus, C is the correct answer.

6.

A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 2020 milliliters per minute and drains at the rate of 1818 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?

Answer: A
Solution:

The birdbath gains 2018=220 - 18 = 2 milliliters of water per minute.

The birdbath will continue to gain water until it is fill, when the volume will remain constant.

Thus, A is the correct answer.

7.

The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?E?

55

1212

1515

1616

2020

Answer: E
Solution:

There are a total of 6+8+4+2+5=25 6 + 8 + 4 + 2 + 5 = 25 students in the class. The percent that chose EE is 100525=1005=20%. 100 \cdot \dfrac{5}{25} = \dfrac{100}{5} = 20 \%.

Thus, E is the correct answer.

8.

Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, 6¢ each, Peru 4¢ each, and Spain 5¢ each. (Brazil and Peru are South American countries and France and Spain are in Europe.)

Number of Stamps by Decade

How many of his European stamps were issued in the ‘8080s?

99

1515

1818

2424

4242

Answer: D
Solution:

Note that France and Spain are the European countries. The number of ‘8080s stamps from these countries respectively is 1515 and 99 for a total of 15+9=24 15 + 9 = 24 stamps.

Thus, D is the correct answer.

9.

Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, 6¢ each, Peru 4¢ each, and Spain 5¢ each. (Brazil and Peru are South American countries and France and Spain are in Europe.)

Number of Stamps by Decade

His South American stamps issued before the ‘7070s cost him

$0.40

$1.06

$1.80

$2.38

$2.64

Answer: B
Solution:

Note that Brazil and Peru are the South American countries.

Brazil's ‘5050 s and ‘6060 s total 1111 stamps with a cost of 116=66¢.11 \cdot 6 = 66¢.

Peru's ‘5050 s and ‘6060 s total 1010 stamps with a cost of 104=40¢.10 \cdot 4 = 40¢.

Therefore, the total cost is 66¢+40¢=106¢ 66¢ + 40¢ = 106¢ = $1.06.

Thus, B is the correct answer.

10.

Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, 6¢ each, Peru 4¢ each, and Spain 5¢ each. (Brazil and Peru are South American countries and France and Spain are in Europe.)

Number of Stamps by Decade

The average price of his ‘7070 s stamps is closes to

3.5¢3.5¢

4¢

4.5¢4.5¢

5¢

5.5¢5.5¢

Answer: E
Solution:

The total price of all the ‘7070 s is 126+126+64+135 12 \cdot 6 + 12 \cdot 6 + 6 \cdot 4 + 13 \cdot 5 =72+72+24+65 = 72 + 72 + 24 + 65 =233¢.= 233¢. The total number of stamps is 12+12+6+13=43. 12 + 12 + 6 +13 = 43. Therefore, the average is 233÷435.4¢. 233 \div 43 \approx 5.4¢.

Thus, E is the correct answer.

11.

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

1111

1212

1313

1414

1515

Answer: C
Solution:

Note that the number of tiles the nn th square requires is n2n^2 since each side has nn tiles.

Therefore, the 66 th square will need 62=366^2 = 36 tiles and 77 th will need 72=49.7^2 = 49. The difference is 4936=13.49 - 36 = 13.

Thus, C is the correct answer.

12.

A board game spinner is divided into three regions labeled A,A, BB and C.C. The probability of the arrow stopping on region AA is 13\frac{1}{3} and on region BB is 12.\frac{1}{2}. The probability of the arrow stopping on region CC is

112\dfrac{1}{12}

16\dfrac{1}{6}

15\dfrac{1}{5}

13\dfrac{1}{3}

25\dfrac{2}{5}

Answer: B
Solution:

The total probability is 1.1. We need to subtract the probability of the spinner landing on BB and AA to get C,C, which is 11312=16. 1 - \dfrac{1}{3} - \dfrac{1}{2} = \dfrac{1}{6}.

Thus, B is the correct answer.

13.

For his birthday, Bert gets a box that holds 125125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?

250250

500500

625625

750750

10001000

Answer: E
Solution:

The larger box will have approximately 125222=1000 125 \cdot 2 \cdot 2 \cdot 2 = 1000 jellybeans.

Thus, E is the correct answer.

14.

A merchant offers a large group of items at 30%30\% off. Later, the merchant takes 20%20\% off these sale prices. The total discount is

35%35\%

44%44\%

50%50\%

56%56\%

60%60\%

Answer: B
Solution:

let xx be the original price of the items. 30%30\% is .7x..7x. Another 20%20\% off is .8.7x=.56x..8 \cdot .7x = .56x.

This means that the price is 44%44\% less than the original price.

Thus, B is the correct answer.

15.

Which of the following polygons has the largest area?

A\text{A}

B\text{B}

C\text{C}

D\text{D}

E\text{E}

Answer: E
Solution:

The number of boxes enclosed by each polygon can be obtained by dividing the polygon into unit squares and right triangles with sidelength 11 and adding up their values.

The unit squares count as 11 and the triangles count as .5..5.

AA has a total area of 5,5, BB has 5,5, CC has 5,5, DD has 4.5,4.5, and EE has 5.5.5.5.

Thus, E is the correct answer.

16.

Right isosceles triangles are constructed on the sides of a 3453-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?

X+Z=W+YX + Z = W + Y

W+X=ZW + X = Z

3X+4Y=5Z3X + 4Y = 5Z

X+W=12(Y+Z)X + W = \dfrac{1}{2}(Y + Z)

X+Y=ZX + Y = Z

Answer: E
Solution:

We can find the area of all the triangles since we know both legs.

W=342=6X=332=4.5Y=442=8Z=552=12.5 \begin{gather*} W = \dfrac{3 \cdot 4}{2} = 6 \\ X = \dfrac{3 \cdot 3}{2} = 4.5 \\ Y = \dfrac{4 \cdot 4}{2} = 8 \\ Z = \dfrac{5 \cdot 5}{2} = 12.5 \end{gather*}

Plugging these values into the answer choices, we see that X+Y=ZX+Y=Z is the only that is true.

Thus, E is the correct answer.

17.

In a mathematics contest with ten problems, a student gains 55 points for a correct answer and loses 22 points for an incorrect answer. If Olivia answered every problem and her score was 29,29, how many correct answers did she have?

55

66

77

88

99

Answer: C
Solution:

Let xx be the number of correct answers. Then she answered 10x10 - x questions incorrectly.

This gives her a total score of 5x2(10x)=7x20. 5x - 2(10 - x) = 7x - 20.

We know that this equals 29,29, and solving yields 7x20=29 7x - 20 = 29 x=7. x = 7.

Thus, C is the correct answer.

18.

Gage skated 11 hr 1515 min each day for 55 days and 11 hr 3030 min each day for 33 days. How long would he have to skate the ninth day in order to average 8585 minutes of skating each day for the entire time?

11 hr

11 hr 1010 min

11 hr 2020 min

11 hr 4040 min

22 hr

Answer: E
Solution:

Gage has skated a total of 575+390=645 min. 5 \cdot 75 + 3 \cdot 90 = 645 \text{ min.}

For an average of 8585 minutes over 99 days, Gage must have skated a total of 859=765 min. 85 \cdot 9 = 765 \text{ min.}

This means that Gage must skate 765645=120 min. 765 - 645 = 120 \text{ min.} on the last day. Note that 120120 minutes is the same as 22 hours.

Thus, E is the correct answer.

19.

How many whole numbers between 9999 and 999999 contain exactly one 0?0?

7272

9090

144144

162162

180180

Answer: D
Solution:

Note that the 00 digit can either be the tens or the units digit. This gives us 22 options for this.

There are 99 options for each of the other digits for a total of 299=162 2 \cdot 9 \cdot 9 = 162 numbers.

Thus, D is the correct answer.

20.

The area of triangle XYZXYZ is 88 square inches. Points AA and BB are midpoints of congruent segments XY\overline{XY} and XZ.\overline{XZ}. Altitude XC\overline{XC} bisects YZ.\overline{YZ}. The area (in square inches) of the shaded region is

1121 \frac{1}{2}

22

2122 \frac{1}{2}

33

3123 \frac{1}{2}

Answer: D
Solution:

Note that the area of XYC\triangle XYC is 8÷2=48 \div 2 = 4 since XC\overline{XC} splits XYZ\triangle XYZ into two congruent triangles.

We also know that the unshaded region is 14\frac{1}{4} the area of the whole triangle, since all the sides are 12\frac{1}{2} the length of the larger triangle.

This means that the shaded region is 34\frac{3}{4} the area of the whole triangle, which is 434=3.4 \cdot \frac{3}{4} = 3.

Thus, D is the correct answer.

21.

Harold tosses a coin four times. The probability that he gets at least as many heads as tails is

516\dfrac{5}{16}

38\dfrac{3}{8}

12\dfrac{1}{2}

58\dfrac{5}{8}

1116\dfrac{11}{16}

Answer: E
Solution:

The probability that there are at least as many head as tails is the same as the probability that there are at least as many tails as heads.

The only overlap between these two scenarios is when the number of heads and the number of tails is equal.

Let pp be the desired probability. Then from our analysis we get that p+pq=1, p + p - q = 1, where qq is the probability of getting the same number of heads and tails.

To find q,q, there are (42)=6\binom{4}{2} = 6 ways to choose which coins are heads, and there are a total of 24=162^4 = 16 possibilities.

Therefore, q=616=38, q = \dfrac{6}{16} = \dfrac{3}{8}, and 2p38=1 2p - \dfrac{3}{8} = 1 p=1116. p = \dfrac{11}{16}.

Thus, E is the correct answer.

22.

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.

1818

2424

2626

3030

3636

Answer: C
Solution:

We can count the number of unexposed sides to find how many sides contribute to the surface area.

Three cubes have 11 side unexposed, two cubes have 22 sides unexposed, and one cube has 33 sides unexposed.

This gives us a total of 31+22+13=10 3 \cdot 1 + 2 \cdot 2 + 1 \cdot 3 = 10 unexposed sides, which gives us 6610=3610 6 \cdot 6 - 10 = 36 - 10 =26= 26 exposed sides.

Each exposed side contributes 12=11^2 = 1 to the surface area, for a total surface area of 126=26.1 \cdot 26 = 26.

Thus, C is the correct answer.

23.

A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?

13\dfrac{1}3

49\dfrac{4}9

12\dfrac{1}2

59\dfrac{5}9

58\dfrac{5}8

Answer: B
Solution:

Notice that there are repeating 3×33 \times 3 regions with the same pattern (they might be rotated differently).

In this region, there are three unit squares, and two triangles that combine to form another unit square.

This makes the area of the darker region 44 and the whole region 9.9. The desired fraction is then 49.\dfrac{4}{9}.

Thus, B is the correct answer.

24.

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 88 ounces of pear juice from 33 pears and 88 ounces of orange juice from 22 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

3030

4040

5050

6060

7070

Answer: B
Solution:

We can set up a proportion to find the amount of juice Miki can extract from the fruits: p12=83 \dfrac{p}{12} = \dfrac{8}{3} for the pears and o12=82 \dfrac{o}{12} = \dfrac{8}{2} for the oranges. Solving yields p=32 and o=48. p = 32 \text{ and } o = 48.

The percent of the whole that is pear juice is 1003232+48=10025 100 \cdot \dfrac{32}{32 + 48} = 100 \cdot \dfrac{2}{5} =40%.= 40 \%.

Thus, B is the correct answer.

25.

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?

110\dfrac{1}{10}

14\dfrac{1}{4}

13\dfrac{1}{3}

25\dfrac{2}{5}

12\dfrac{1}{2}

Answer: B
Solution:

WLOG, assume that everyone gave Ott $1. This means that Moe had $5, Loki had $4, and Nick had $3 originally.

Ott now has $3, and the total amount of money is $5 + $4 + $3 = $12.

This means that Ott has 312=14 \dfrac{3}{12} = \dfrac{1}{4} of the group's money.

Thus, B is the correct answer.