AMC 8 2003

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Instructions

This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
You will receive 1 point for each correct answer. There is no penalty for wrong answers.
No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
Figures are not necessarily drawn to scale.
You will have 40 minutes working time to complete the test.

Question 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?

$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26$

Solution

$\textbf{E}$

There are $12$ edges, $8$ corners, and $6$ faces on a cube. Adding them up gets $12+8+6= 26$.

Question 2

Which of the following numbers has the smallest prime factor?

$\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61$

Solution

$\textbf{C}$

The smallest prime factor is $2$, and since $58$ is the only even number, the answer is $58$.

Question 3

A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?

$\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%$

Solution

$\textbf{D}$

The percentage of filler is $30/120=25\%$. So the percentage of non-filler is $1-25\%=75\%$.

Question 4

A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$

Solution

$\textbf{C}$

Let the number of bicycles and tricycle be $a$ and $b$. Hence, we have
\begin{align*}
a+b&=7\\
2a+3b&=19
\end{align*}

By solving the equations we get $a=2$ and $b=5$. So the number of tricycles is 5.

Question 5

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

$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30$

Solution

$\textbf{B}$

If $20\%$ of a number is 12, $10\%$ of the number is 6. So $30\%$ of the number is $6\times3=18$.

Question 6

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

$\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800$

Solution

$\textbf{B}$

The sides of the triangle are 5, 12, and 13, which means it is a right triangle. So the area of the right triangle is $\dfrac12\times5\times12=30$.

Question 7

Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 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?

$\mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$

Solution

$\textbf{A}$

The total score of Jenny is $10-10+20+20=40$ points higher than Blake. So her average is $40/4=10$ points higher than Blake.

Question 8

$\textbf{Bake Sale}$

(Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures)

Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.

$\circ$ Art's cookies are trapezoids.

$\circ$ Roger's cookies are rectangles.

$\circ$ Paul's cookies are parallelograms.

$\circ$ Trisha's cookies are triangles.

Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough?

$\textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.}$

Solution

$\textbf{A}$

The area of each cookie is 12, 8, 6, and 6. Since each friend uses the same amount of dough, Art makes the fewest cookies.

Question 9

Each friend uses the same amount of dough, and Art makes exactly $12$ cookies. Art's cookies sell for $60$ cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90$

Solution

$\textbf{C}$

The total area of cookies made by Art is $12\times12=144$. So Roger can made $144/8=18$ cookies. Art earns $12\times60=720$ cents. To earn the same amount of money, the price of Roger's cookie is $720/18=40$ cents.

Question 10

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

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$

Solution

$\textbf{E}$

The total number of cookies Trisha made is $144/6=24$.

Question 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 $10\%$. 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 dollars on Thursday?

$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44$

Solution

$\textbf{B}$

The price of the shoes is $40\times(1+10\%)$ dollars on Friday, then $40\times(1+10\%)(1-10\%)=40\cdot0.99=39.60$ dollars on Monday.

Question 12

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

$\textbf{(A)}\ \dfrac{1}{3}\qquad\textbf{(B)}\ \dfrac{1}{2}\qquad\textbf{(C)}\ \dfrac{2}{3}\qquad\textbf{(D)}\ \dfrac{5}{6}\qquad\textbf{(E)}\ 1$

Solution

$\textbf{E}$

If the face with number 6 is not on the bottom, the product is apparently divisible by 6. If the face with number 6 is on the bottom, then the faces with number 2 and 3 can be seen, which means the product is also divisible by 6. Therefore, the probability that the product is divisible by 6 is $100\%$.

Question 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 red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution

$\textbf{B}$

Those cubes who have 4 red faces must be adjacent to 2 cubes in the figure. The top 4 cubes are connected to only 1 cube each. The 4 bottom corner cubes are adjacent to 3 cubes each. Hence, only 6 bottom middle cubes have 4 red faces each.

Question 14

In this addition problem, each letter stands for a different digit.

If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution

$\textbf{D}$

If $T=7$, $FO$ must be 14 or 15. Since $O$ is an even number, we know $F=1$, $O=4$. Then the units digit $R=8$. The $(W,U)$ pair could be $(0,0), (1,2), (2,4), (3,6)$ or $(4,8)$. Since each letter stands for different digit, the only possible $(W,U)$ pair is $(3,6)$. So $W=3$.

Question 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?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution

$\textbf{B}$

We need at least 4 cubes. The figure above is one of the example.

Question 16

Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24$

Solution

$\textbf{D}$

We have 2 choices for the driver's seat, then 3 choices for the front passenger seat, 2 choices for the rear left passenger seat, and 1 choice for the rear right seat. So the total number of arrangement is $2\times3\times2\times1=12$.

Question 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?

$\textbf{(A)}\ \text{Nadeen and Austin}\qquad\newline$
$\textbf{(B)}\ \text{Benjamin and Sue}\qquad\newline$
$\textbf{(C)}\ \text{Benjamin and Austin}\qquad\newline$
$\textbf{(D)}\ \text{Nadeen and Tevyn}\newline$
$\textbf{(E)}\ \text{Austin and Sue}$

Solution

$\textbf{E}$

Jim has brown eyes and blonde hair. If his family characteristic is brown eyes, we can find only one sibling of Jim, Nadeen, who has brown eyes. So Jim's family characteristic can not be brown eyes. Hence, we know his family characteristic is blonde hair. So his siblings are Austin and Sue.

Question 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?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution

$\textbf{D}$

Those classmates who will not be invited are labeled with red circles.

Question 19

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

$\textbf{C}$

Find the least common multiple of $15, 20, 25$ by turning the numbers into their prime factorization.
\begin{align*}
15 &= 3 \cdot 5\\
20 &= 2^2 \cdot 5\\
25 &= 5^2
\end{align*}
Hence, we get the least common multiple $2^2\cdot3\cdot5^2=300$. The multiples of $300$ bewteen 1000 and 2000 are 1200, 1500, and 1800.

Question 20

What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution

$\textbf{D}$

Imagine the clock as a circle. The minute hand will point to number 4 at 4:20 PM, while the hour hand moves $\dfrac{20}{60}=\dfrac13$ away from number 4 to number 5. The central angle between number 4 and 5 is 30 degrees. So the angle between the minute hand and the hour hand is $30\times\dfrac13=10$ degrees.

Question 21

The area of trapezoid $ABCD$ is $164\text{ cm}^2$. The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$, in centimeters?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution

$\textbf{B}$

Using the formula for the area of a trapezoid, we have $$164=\dfrac12\cdot8(BC+AD)$$ Thus $$BC+AD=41$$ Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$. From the Pythagorean Theorem, we get $AE=6$ and $DF=15$. So $$AD=BC+21$$ Substituting back into our original equation we have $$BC+BC+21=41\rightarrow BC=10$$

Question 22

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

$\textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}$

Solution

$\textbf{C}$

First we have to find the area of the shaded region in each of the figures. In figure A the area of the shaded region is the area of the circle subtracted from the area of the square. That is $$2^2-\pi\times1^2 =4-\pi\approx4-3.14=0.86$$ In figure B the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is $$2^2-4 \times\left(\pi\times \left( \dfrac{1}{2} \right)^2 \right)=4-4\times\dfrac{\pi}{4} =4-\pi\approx4-3.14=0.86$$ In figure C the area of the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. So the side length of the square is $\sqrt2$. The area of the shaded region is $$\pi\times1^2 -\left(\sqrt2\right)^2=\pi-2\approx3.14-2=1.14$$ Clearly the largest area that we found among the three shaded regions is $\pi-2$. The answer is C.

Question 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 247th move?

Solution

$\textbf{A}$

The period of the cat's movement is 4. And the period of the mouse is 8. Since 248 is a multiple of both 4 and 8, we know that the positions of the cat and the mouse after 247th move are the same as -1th move, which means the cat moves counterclockwise and the mouse moves clockwise for 2 steps from figure 1. Hence, the result is $\textbf{A}$.

Question 24

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

Solution

$\textbf{B}$

When the ship is on the travel from point $A$ to $B$, the distance from the ship to the island $X$ is a constant, which means a horizontal line on the graph. From point $B$ to the middle point of $BC$ the distance decreases, and then from the middle point of $BC$ to point $C$ the distance increases. So the answer is $\textbf{B}$.

Question 25

In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC$, $AB = AC$, and when $\triangle ABC$ is folded over side $\overline{BC}$, point $A$ coincides with $O$, the center of square $WXYZ$. What is the area of $\triangle ABC$, in square centimeters?

$\textbf{(A)}\ \dfrac{15}4\qquad\textbf{(B)}\ \dfrac{21}4\qquad\textbf{(C)}\ \dfrac{27}4\qquad\textbf{(D)}\ \dfrac{21}2\qquad\textbf{(E)}\ \dfrac{27}2$

Solution

$\textbf{C}$

The side lengths of square $WXYZ$ must be 5 cm, since the area is $25\ \text{cm}^2$. First, we should determine the height of $\triangle{ABC}$. The distance from $A$ to $BC$ is the same as the distance from $O$ to $BC$. The distance from $O$ to line $WZ$ must be 2.5 cm, since line $WX=5$ cm, and the distance from $O$ to $WZ$ is half of that. The distance from line $WZ$ to line $BC$ must be 2 cm, since the side lengths of the small squares are 1 cm, and there are two squares from line $WZ$ to line $BC$. So, the height of $\triangle{ABC}$ must be 4.5, which is 2.5 + 2. The length of $BC$ can be determined by subtracting 2 from 5, since the length of $WZ$ is 5, and the two squares in the corners give us 2 together. This gives us the base for $\triangle{ABC}$, which is 3 cm. Then, we multiply 4.5 by 3 and divide by 2, to get the area of the $\triangle ABC$ $\dfrac12\times4.5\times3=\dfrac{27}{4}\ \text{cm}^2$.