2003 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?

1212

1616

2020

2222

2626

Answer: E
Solution(s):

A cube has 1212 edges, 88 corners, and 66 faces. Adding these together yields 12+8+6=26. 12 + 8 + 6 = 26.

Thus, E is the correct answer.

2.

Which of the following numbers has the smallest prime factor?

5555

5757

5858

5959

6161

Answer: C
Solution(s):

Note that the smallest prime number is 2.2. This means that any even number would be our answer.

Thus, C is the correct answer.

3.

A burger at Ricky C's weighs 120120 grams, of which 3030 grams are filler. What percent of the burger is not filler?

60%60 \%

65%65 \%

70%70 \%

75%75 \%

90%90 \%

Answer: D
Solution(s):

We get that 12030=90120 - 30 = 90 grams are not filler. The percentage is therefore 10090120=10034=75%. 100 \cdot \dfrac{90}{120} = 100 \cdot \dfrac{3}{4} = 75 \%.

Thus, D is the correct answer.

4.

A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 77 children and 1919 wheels. How many tricycles were there?

22

44

55

66

77

Answer: C
Solution(s):

Let bb be the number of bicycles and tt be the number of tricycles. Then we can set up the following system of equations: b+t=7,2b+3t=19. \begin{gather*} b + t = 7, \\ 2b + 3t = 19. \end{gather*} Multiplying the first equation by 22 and subtracting from the second equation, we get t=5.t = 5.

Thus, C is the correct answer.

5.

If 20%20 \% of a number is 12,12, what is 30%30\% of the same number?

1515

1818

2020

2424

3030

Answer: B
Solution(s):

Let xx the be the number. Then .2x=12 .2x = 12 x=60. x = 60. From this we get that .360=18. .3 \cdot 60 = 18.

Thus, B is the correct answer.

6.

Given the areas of the three squares in the figure, what is the area of the interior triangle?

1313

3030

6060

300300

18001800

Answer: B
Solution(s):

We get that the side lengths of the squares are 169=13 \sqrt{169} = 13144=12, \sqrt{144} = 12, 25=5. \sqrt{25} = 5. respectively. Note that these lengths form a Pythagorean triple.

Therefore, the interior triangle is right. Its area is 12512=30. \dfrac{1}{2} \cdot 5 \cdot 12 = 30.

Thus, B is the correct answer.

7.

Blake and Jenny each took four 100100-point tests. Blake averaged 7878 on the four tests. Jenny scored 1010 points higher than Blake on the first test, 1010 points lower than him on the second test, and 2020 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?

1010

1515

2020

2525

4040

Answer: A
Solution(s):

The total point difference between the two is 1010+202=40. 10 - 10 + 20 \cdot 2 = 40. The average of this difference is 40÷4=10.40 \div 4 = 10.

Thus, A is the correct answer.

8.

Who gets the fewest cookies from one batch of cookie dough?

Art\text{Art}

Roger\text{Roger}

Paul\text{Paul}

Trisha\text{Trisha}

There is a tie for fewest.\text{There is a tie for fewest.}

Answer: A
Solution(s):

Note that the person who has the largest cookie would make the fewest cookies.

Art's cookie has an area of 12(3+5)3=12 in2 \dfrac{1}{2} (3 + 5) \cdot 3 = 12 \text{ in}^2

Roger's is 24=8 in2,2 \cdot 4 = 8 \text{ in}^2, Paul's is 23=6 in2,2 \cdot 3 = 6 \text{ in}^2, and Trisha's is 1234=6 in2 \dfrac{1}{2} \cdot 3 \cdot 4 = 6 \text{ in}^2

Thus, A is the correct answer.

9.

Art's cookies sell for 60¢60 ¢ each. To earn the same amount from a single batch, how much should one of Roger's cookies cost?

18¢18 ¢

25¢25 ¢

40¢40 ¢

75¢75 ¢

90¢90 ¢

Answer: C
Solution(s):

From Problem 8,8, we know that Art's cookie has an area of 12 in212 \text{ in}^2 Since there are 1212 cookies in a batch, each batch has 1212=144 in212 \cdot 12 = 144 \text{ in}^2 of dough.

Since Roger's cookie has an area of 8 in2,8 \text{ in}^2, Roger can make 144÷8=18144 \div 8 = 18 cookies.

Art would make 1260=720¢12 \cdot 60 = 720 ¢ from his cookies, so Roger would need to charge 720÷18=40¢720 \div 18 = 40 ¢ per cookie.

Thus, C is the correct answer.

10.

How many cookies will be in one batch of Trisha's cookies?

1010

1212

1616

1818

2424

Answer: E
Solution(s):

From Problem 9,9, we know that a batch has 144 in2144 \text{ in}^2 of cookie dough. We also know that Trisha's cookies have an area of 6 in26 \text{ in}^2

Therefore, Trisha can make 144÷6=24144 \div 6 = 24 cookies per batch.

Thus, E is the correct answer.

11.

Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%.10 \%. Over the weekend, Lou advertises the sale: "Ten percent off the listed price. Sale starts Monday." How much does a pair of shoes cost on Monday that cost $40 on Thursday?

$36

$39.60

$40

$40.40

$44

Answer: B
Solution(s):

On Friday, the shoes would cost 40 \cdot 1.1 = 44 $. On Monday, they would cost 44 \cdot 0.9 = 39.6 $.

Thus, B is the correct answer.

12.

When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by 6?6?

13\frac 13

12\frac 12

23\frac 23

56\frac 56

11

Answer: E
Solution(s):

Note that if 66 is not the bottom case, then the product is automatically divisible by 66 since 66 is a factor.

If 66 is on the bottom, then 22 and 33 are seen, also ensuring that the product is divisible by 6.6.

Therefore, every possible scenario allows for the product to be divisible by 6.6.

Thus, E is the correct answer.

13.

Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted purple. The figure is then separated into individual cubes. How many of the individual cubes have exactly four purple faces?

44

66

88

1010

1212

Answer: B
Solution(s):

The 44 cubes on top have 55 exposed faces, so they don't work. The 44 corners in the bottom row have 33 exposed sides, so they don't work either.

Every other cube has 44 exposed sides, so those work. There are 1444=614 - 4 - 4 = 6 cubes that work.

Thus, B is the correct answer.

14.

In this addition problem, each letter stands for a different digit. TWO+TWOFOUR\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array} If T=7T = 7 and the letter OO represents an even number, what is the only possible value for W?W?

00

11

22

33

44

Answer: D
Solution(s):

Since both TT's are 7,7, we get that OO is either 44 or 5.5. Since OO is even, we get that O=4.O = 4.

Then, we get that R=4+4=8.R = 4 + 4 = 8. We also know that W+WW + W doesn't carry over, since otherwise OO would be 5.5.

Therefore, WW is less than 55 and cannot be 44 or 1.1. If W=2,W = 2, then U=4,U = 4, which is also not allowed.

This makes W=3.W = 3.

Thus, D is the correct answer.

15.

A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?

33

44

55

66

77

Answer: B
Solution(s):

We need at least 33 cubes to form the front view. Then we also need another cube to ensure that the side views are correct.

Thus, B is the correct answer.

16.

Ali, Bonnie, Carlo and Dianna are going to drive together to a nearby theme park. The car they are using has four seats: one driver's seat, one front passenger seat and two back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?

22

44

66

1212

2424

Answer: D
Solution(s):

There are 22 options for who sits in the driver's seat. There are 33 options for the other front seat, and 22 options for the first passenger seat.

The last person has to sit in the last seat, for a total of 232=12 2 \cdot 3 \cdot 2 = 12 possible seating arrangements.

Thus, D is the correct answer.

17.

The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?

Nadeen and Austin

Benjamin and Sue

Benjamin and Austin

Nadeen and Tevyn

Austin and Sue

Answer: E
Solution(s):

Note that Nadeen, Austin, and Sue are the only individuals who share a characteristic with Jim. We need to find which of the 33 are completely different from the others.

Austin and Sue both have blue eyes, which makes Nadeen the odd one out. Therefore, Austin and Sue are Jim's siblings.

Thus, E is the correct answer.

18.

Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?

11

44

55

66

77

Answer: D
Solution(s):

Note that Sarah invites only the dots that are most 22 line segments away from Sarah.

There are 44 people who are completely disconnected from Sarah, and there are 22 people who are 33 lines away from Sarah.

Sarah will not invite any of these 4+2=64 + 2 = 6 people.

Thus, D is the correct answer.

19.

How many integers between 10001000 and 20002000 have all three of the numbers 15,20,15, 20, and 2525 as factors?

11

22

33

44

55

Answer: C
Solution(s):

If a number xx has these three numbers as factors, then their least common multiple must also divide x.x.

These numbers have the following prime factorizations: 15=35, 15 = 3 \cdot 5,20=225, 20 = 2^2 \cdot 5,25=52. 25 = 5^2.

From these values, we get that the least common multiple is 22352=300. 2^2 \cdot 3 \cdot 5^2 = 300.

Therefore, the multiples of 300300 between 10001000 and 20002000 are 1200,1500,1200, 1500, and 1800.1800.

Thus, C is the correct answer.

20.

What is the measure of the acute angle formed by the hands of the clock at 4:204 : 20 a.m.?

00^{\circ}

55^{\circ}

88^{\circ}

1010^{\circ}

1212^{\circ}

Answer: D
Solution(s):

At 4:20,4 : 20, the hour hand will be a 13\frac{1}{3} of the way between 44 and 5.5.

Each hour represents 360÷12=30.360 \div 12 = 30^{\circ}. This means the hour hand will be 30÷3=1030 \div 3 = 10^{\circ} past 4.4.

Note that at 2020 minutes, the minute hand will be at 4.4. This means the degree formed is 10.10^{\circ}.

Thus, D is the correct answer.

21.

The area of trapezoid ABCDABCD is 164 cm2.164\text{ cm}^2. The altitude is 88 cm, ABAB is 1010 cm, and CDCD is 1717 cm. What is BC,BC, in centimeters?

99

1010

1212

1515

2020

Answer: B
Solution(s):

Drop perpendiculars from BB and CC to AD\overline{AD} and let them hit at EE and F.F.

Then, using the Pythagorean theorem, we get that AE=6 and FD=15. AE = 6 \text{ and } FD = 15.

Then the area of ABCDABCD can be expressed as [ABCD]= [ABCD] =[ABE]+[CDF]+[BCEF]. [ABE] + [CDF] + [BCEF]. Note that:[ABE]=1268=24 [ABE] = \dfrac{1}{2} \cdot 6 \cdot 8 = 24 [CDF]=12158=60 [CDF] = \dfrac{1}{2} \cdot 15 \cdot 8 = 60 [BCEF]=8BC[BCEF] = 8\cdot BC Substituting, we get that 164=24+60+8BC80=8BCBC=10.\begin{align*} 164 &= 24+60+8BC\\ 80 &= 8BC \\ BC &= 10. \end{align*}

Thus, B is the correct answer.

22.

The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

AA only

BB only

CC only

both AA and BB

all are equal

Answer: C
Solution(s):

The shaded area of AA is 2212π=4π.86 cm22^2 - 1^2 \pi = 4 - \pi \approx .86 \text{ cm}^2

The shaded area of BB is 224122π=4π.86 cm2 2^2 - 4 \cdot \dfrac{1}{2}^2 \pi = 4 - \pi \approx .86 \text{ cm}^2

Note that the diagonal of the square in CC is equal to the diameter of the circle.

Therefore, the side length is 2÷2=2. 2 \div \sqrt{2} = \sqrt{2}. This makes the area of the shaded region 12π22=π21.14 cm2 1^2 \pi - \sqrt{2}^2 = \pi - 2 \approx 1.14 \text{ cm}^2

Thus, C is the correct answer.

23.

In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.

If the pattern is continued, where would the cat and mouse be after the 247247th move?

Answer: A
Solution(s):

We can find the positions of the mouse and cat individually. Note that the mouse's position repeats every 88 moves and the cat's every 44 moves.

247247 has a remainder of 33 when divided by 4,4, which means the cat is in the same position as after the 33rd move, which is the bottom right square.

247247 has a remainder of 77 when divided by 8,8, which means the cat is in the same position as after the 77th move, which is the bottom left segment.

Thus, A is the correct answer.

24.

A ship travels from point AA to point BB along a semicircular path, centered at Island X.X. Then it travels along a straight path from BB to C.C. Which of these graphs best shows the ship’s distance from Island XX as it moves along its course?

Answer: B
Solution(s):

Note that any point along a circle is the same distance from its center. This means that travelling from AA to B,B, the distance from XX remains constant.

We also know that as we move from BB to the midpoint of BC,\overline{BC}, we get closer to X.X.

Similarly, as we move from the midpoint towards C,C, we get farther from X.X. This can be represented by a downwards facing semicircle.

Thus, B is the correct answer.

25.

In the figure, the area of square WXYZWXYZ is 25 cm2.25 \text{ cm}^2. The four smaller squares have sides 11 cm long, either parallel to or coinciding with the sides of the large square. In ABC,\triangle ABC, AB=AC,AB = AC, and when ABC\triangle ABC is folded over side BC,\overline{BC}, point AA coincides with O,O, the center of square WXYZ.WXYZ. What is the area of ABC,\triangle ABC, in square centimeters?

154\dfrac{15}{4}

214\dfrac{21}{4}

274\dfrac{27}{4}

212\dfrac{21}{2}

272\dfrac{27}{2}

Answer: C
Solution(s):

We get that the side lengths of WXYZWXYZ are 55 cm, since 25=5.\sqrt{25} = 5.

We also know that the distance from WZ\overline{WZ} to BC\overline{BC} is 22 since it is the sum of the side lengths of 22 unit squares.

Finally, the distance from AA to BC\overline{BC} is the same as the distance from BC\overline{BC} to O,O, which is 2+52=92 cm 2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}

Now, we can find BC,BC, which is WZ2=52=3 cm WZ - 2 = 5 - 2 = 3 \text{ cm}

Therefore, the area of ABC\triangle ABC is 12392=274 cm2 \dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2

Thus, C is the correct answer.