## AMC 8 2004

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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.

On a map, a $12$-centimeter length represents $72$ kilometers. How many kilometers does a $17$-centimeter length represent?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$

$\textbf{B}$

Since $12$ centimeters on the map represent $72$ kilometers, $1$ centimeter on the map represents $72/12=6$ kilometers. So $17$ centimeters on the map represent $6\times17=102$ kilometers.

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

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$

$\textbf{B}$

Since we have only 2 non-zero digits, we have 2 choices for the first digit. Then the rest non-zero digit has 3 places to go. The answer is $2\times3=6$.

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 $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 18$

$\textbf{A}$

Since 12 meals can serve 18 people, 1 meal is enough for 1.5 people. Hence, they should order $12/1.5=8$ meals.

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?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

$\textbf{B}$

We have $\dbinom43=4$ ways to choose 3 people from 4.

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?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

$\textbf{D}$

The remaining team will be the only undefeated one. The other $15$ teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.

After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%$. How many of the last $5$ shots did she make?

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

$\textbf{C}$

Sally made $20\times55\%=11$ shots in the beginning. Her final shots are $(20+5)\times56\%=14$. So she made 3 shots in the last 5.

An athlete's target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from $220$. To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?

$\textbf{(A)}\ 134\qquad\textbf{(B)}\ 155\qquad\textbf{(C)}\ 176\qquad\textbf{(D)}\ 194\qquad\textbf{(E)}\ 243$

$\textbf{B}$

The maximum heart rate of the 26-year-old athlete is $220-26=194$ beats per minute. Then the target heart rate is $194\times80\%\approx155$ beats per minutes.

Find the number of two-digit positive integers whose digits total $7$.

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 10$

$\textbf{B}$

The numbers are $16, 25, 34, 43, 52, 61$ and $70$, which gives us a total of $7$.

The average of the five numbers in a list is $54$. The average of the first two numbers is $48$. What is the average of the last three numbers?

$\textbf{(A)}\ 55\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 57\qquad\textbf{(D)}\ 58\qquad\textbf{(E)}\ 59$

$\textbf{D}$

The sum of the five numbers is $54\times5=270$. The sum of the first two numbers is $48\times2=96$. So the sum of the last three numbers is $270-96=174$, and the average is $174/3=58$.

Handy Aaron helped a neighbor $1\dfrac14$ hours on Monday, $50$ minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid $\$3$ per hour. How much did he earn for the week?

$\textbf{(A)}\ \$8 \qquad \textbf{(B)}\ \$9 \qquad \textbf{(C)}\ \$10 \qquad \textbf{(D)}\ \$12 \qquad \textbf{(E)}\ \$15$

$\textbf{E}$

Let's convert everything to minutes and add them together. On Monday he worked for $\dfrac54 \times 60 = 75$ minutes. On Tuesday he worked $50$ minutes. On Wednesday he worked for $2$ hours $25$ minutes, or $2\times60+25=145$ minutes. On Friday he worked $60/2=30$ minutes. This adds up to $75+50+145+30=300$ minutes, or $300/60=5$ hours. So he earned $5\times \$3 =\$15$.

The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:

$\text {The largest isn't first, but it is in one of the first three places.}\newline$

$\text {The smallest isn't last, but it is in one of the last three places.}\newline$

$\text {The median isn't first or last.}$

What is the average of the first and last numbers?

$\textbf{(A)}\ 3.5 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6.5 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$

$\textbf{C}$

From rule 1, the largest number, $12$, can be the second or third. From rule 2, the smallest number, $-2$, can be the third or fourth. From rule 3, the median, $6$ can be the second, third, or fourth. All three numbers mentioned above can not be the first or the last. Hence, the first and last numbers are $4$ and $9$, disregarding their order. Their average is $(4+9)/2 = 6.5$.

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 $24$ hours. If she is using it constantly, the battery will last for only $3$ hours. Since the last recharge, her phone has been on $9$ hours, and during that time she has used it for $60$ minutes. If she doesn't talk any more but leaves the phone on, how many more hours will the battery last?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

$\textbf{B}$

When not being used, the cell phone uses up $\dfrac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\dfrac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8\times\dfrac{1}{24} + 1\times\dfrac{1}{3}= \dfrac{2}{3}$ of its battery has been used up. To drain the remaining $\dfrac{1}{3}$, the phone can last for $\dfrac{1}{3}\div\dfrac{1}{24}=8$ more hours without being used.

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

I. Bill is the oldest.$\newline$

II. Amy is not the oldest.$\newline$

III. Celine is not the youngest.

Rank the friends from the oldest to the youngest.

$\textbf{(A)}\ \text{Bill, Amy, Celine}\qquad\newline$ $\textbf{(B)}\ \text{Amy, Bill, Celine}\qquad\newline$

$\textbf{(C)}\ \text{Celine, Amy, Bill}\newline$

$\textbf{(D)}\ \text{Celine, Bill, Amy} \qquad \newline$

$\textbf{(E)}\ \text{Amy, Celine, Bill}$

$\textbf{E}$

If Bill is the oldest, then Amy is not the oldest, and both statements I and II are true, so statement I is not the true one.$\newline$

If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements II and III are true, so statement II is not the true one.$\newline$

Therefore, statement III is the true statement, and both I and II are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is $\text{Amy, Celine, Bill}$.

What is the area enclosed by the geoboard quadrilateral below?

$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 18\dfrac{1}{2} \qquad \textbf{(C)}\ 22\dfrac{1}{2} \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 41$

$\textbf{C}$

The area of the quadrilateral is the area of the whole graph subtracts the area of the blank. The area of the whole graph is $10\cdot10=100$. The area of the bottom left blank triangle is $\dfrac12\cdot4\cdot5=10$. The area of the right blank triangle is $\dfrac12\cdot6\cdot10=30$. The top blank quadrilateral can be divided into two parts: the left trapezoid of area $\dfrac12(5+6)\cdot3=16.5$ and the right triangle of area $\dfrac12\cdot6\cdot7=21$. Hence, the area of the quadrilateral we want is $100-10-30-16.5-21=22.5$.

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

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

$\textbf{C}$

The side length of the new created "hexagon" is 4 white tiles. Considering the overlapping of the 6 corners, the number of white tiles we need to add for the outer ring is $4\times6-6=18$. Hence, the difference between white and black tiles is $18+6-13=11$.

Two $600$ mL pitchers contain orange juice. One pitcher is $1/3$ full and the other pitcher is $2/5$ 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?

$\textbf{(A)}\ \dfrac18 \qquad \textbf{(B)}\ \dfrac{3}{16} \qquad \textbf{(C)}\ \dfrac{11}{30} \qquad \textbf{(D)}\ \dfrac{11}{19}\qquad \textbf{(E)}\ \dfrac{11}{15}$

$\textbf{C}$

The first pitcher contains $600 \times \dfrac13 = 200$ mL of orange juice. The second pitcher contains $600 \times \dfrac25 = 240$ mL of orange juice. In the large pitcher, there is a total of $200+240=440$ mL of orange juice in the $600+600=1200$ mL of fluids, giving a fraction of $\dfrac{440}{1200} =\dfrac{11}{30}$.

Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

$\textbf{D}$

Think about 6 square in a line $\square\square\square\square\square\square$. We need to insert 2 boards to separate them into 3 groups like $\square\square|\square\square\square|\square$. The number of squares in each group represents the number of pencils each person get. Since there are 5 slot between 6 squares, we have $\dbinom52=10$ ways.

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 $1$ through $10$. Each throw hits the target in a region with a different value. The scores are: Alice $16$ points, Ben $4$ points, Cindy $7$ points, Dave $11$ points, and Ellen $17$ points. Who hits the region worth $6$ points?

$\textbf{(A)}\ \text{Alice}\qquad \textbf{(B)}\ \text{Ben}\qquad \textbf{(C)}\ \text{Cindy}\qquad \textbf{(D)}\ \text{Dave} \qquad \textbf{(E)}\ \text{Ellen}$

$\textbf{A}$

The only way to get Ben's score is with $1+3=4$. Cindy's score can be made of $1+6$, $2+5$ or $3+4$, but since Ben already hit $1$ and $3$, Cindy hit $2+5=7$. Similarly, Dave's darts were in the region $4+7=11$. Lastly, because there is no $7$ left, $\text{Alice}$ must have hit the regions $6+10=16$ and Ellen $8+9=17$.

A whole number larger than $2$ leaves a remainder of $2$ when divided by each of the numbers $3, 4, 5,$ and $6$. The smallest such number lies between which two numbers?

$\textbf{(A)}\ 40\ \text{and}\ 49 \qquad \textbf{(B)}\ 60 \text{ and } 79 \qquad \textbf{(C)}\ 100\ \text{and}\ 129 \qquad \textbf{(D)}\ 210\ \text{and}\ 249\qquad \textbf{(E)}\ 320\ \text{and}\ 369$

$\textbf{B}$

If the target whole number subtracts 2, it will be divisible by 3, 4, 5, and 6. The least common multiple of 3, 4, 5, and 6 is 60. Hence, the target whole number is 62, which lies between 60 and 79.

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 $6$ empty chairs, how many people are in the room?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

$\textbf{D}$

The number of chairs seated is 3 times as much as the number of empty chairs. Since there are 6 empty chairs, the number of chairs seated is $6\times3=18$. 18 is also 2/3 of the people in the room. So the total number of people in the room is 27.

Spinners $A$ and $B$ 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?

$\textbf{(A)}\ \dfrac14\qquad \textbf{(B)}\ \dfrac13\qquad \textbf{(C)}\ \dfrac12\qquad \textbf{(D)}\ \dfrac23\qquad \textbf{(E)}\ \dfrac34$

$\textbf{D}$

If the product is odd, both the two spinners' numbers are odd. The probability that the product is odd is $\dfrac12\times\dfrac23=\dfrac13$. Hence, the probability that the product is even is $1-\dfrac13=\dfrac23$.

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\dfrac25$. What fraction of the people in the room are married men?

$\textbf{(A)}\ \dfrac13\qquad \textbf{(B)}\ \dfrac38\qquad \textbf{(C)}\ \dfrac25\qquad \textbf{(D)}\ \dfrac{5}{12}\qquad \textbf{(E)}\ \dfrac35$

$\textbf{B}$

Assuming there are $5$ women in the room, of which $5 \times \dfrac25 = 2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\dfrac38$.

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

$\textbf{D}$

From $J$ to $K$ she is running away from home. From $K$ to $L$ she is running away from home with a lower slope in the graph. From $L$ to $M$ she is getting closer to home. And finally the distance turns to 0 when she comes back to point $J$. The answer is D.

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

$\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1$

$\textbf{C}$

The area of the parallelogram $EFGH$ can be found in two ways. The first is by taking rectangle $ABCD$ and subtracting four triangles at the corners to create parallelogram $EFGH$. This is\[(4+6)(5+3) - 2 \times \dfrac12 \times 6 \times 5 - 2 \times \dfrac12 \times 3 \times 4 = 80 - 30 - 12 = 38\]The second way is by multiplying the base of the parallelogram such as $\overline{FG}$ with its altitude $d$, which is perpendicular to both bases. $\triangle FGC$ is a $3-4-5$ right triangle so $\overline{FG} = 5$. Set these two representations of the area together.\[5d = 38 \rightarrow d = 7.6\]

Two $4 \times 4$ 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?

$\textbf{(A)}\ 16-4\pi\qquad \textbf{(B)}\ 16-2\pi \qquad \textbf{(C)}\ 28-4\pi \qquad \textbf{(D)}\ 28-2\pi \qquad \textbf{(E)}\ 32-2\pi$

$\textbf{D}$

The area of the overlapping square is $2\times2=4$. The diameter of the circle is $2\sqrt2$. Hence, the area of the circle is $\pi\times\left(\sqrt2\right)^2=2\pi$. The area of the shaded region is $2\times4\times4-4-2\pi=28-2\pi$.