AMC 8 2006
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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.
Mindy made three purchases for $\$1.98$, $\$5.04$, and $\$9.89$. What was her total, to the nearest dollar?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$
$\textbf{D}$
The three prices round to $\$2$, $\$5$, and $\$10$, which have a sum of $\$17$.
On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$
$\textbf{C}$
The AMC 8 contest only rewards 1 point for each correct answer with no penalty for wrong answers or no answer. So his score is 13.
Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?
$\textbf{(A)}\ \dfrac{1}{2}\qquad\textbf{(B)}\ \dfrac{3}{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
$\textbf{A}$
When Elisa started, she finished a lap in $\dfrac{25}{10}=2.5$ minutes. Now, she finishes a lap in $\dfrac{24}{12}= 2$ minutes. The difference is $2.5-2=\dfrac{1}{2}$.
Initially, a spinner points west. Chenille moves it clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions. In what direction does the spinner point after the two moves?
$\textbf{(A)}\ \text{north} \qquad \textbf{(B)}\ \text{east} \qquad \textbf{(C)}\ \text{south} \qquad \textbf{(D)}\ \text{west} \qquad \textbf{(E)}\ \text{northwest}$
$\textbf{B}$
Note that full revolutions do not matter, so this is equivalent to going clockwise $\dfrac{1}{4}$ revolutions and then counterclockwise $\dfrac{3}{4}$ revolutions, making it ultimately go counterclockwise $\dfrac{1}{2}$, having the spinner point $\text{east}$.
Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$
$\textbf{D}$
Drawing segments $AC$ and $BD$, the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. So the area of $ABCD$ is half the area of the larger square. The answer is $\dfrac{60}{2}=30$.
The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
$\textbf{C}$
If the two rectangles were separate, the perimeter would be $2\times2\times(2+4)=24$ inches. When placed next to each other, their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = 20$ inches.
Circle $X$ has a radius of $\pi$. Circle $Y$ has a circumference of $8 \pi$. Circle $Z$ has an area of $9 \pi$. List the circles in order from smallest to largest radius.
$\textbf{(A)}\ X, Y, Z\qquad\textbf{(B)}\ Z, X, Y\qquad\textbf{(C)}\ Y, X, Z\qquad\textbf{(D)}\ Z, Y, X\qquad\textbf{(E)}\ X, Z, Y$
$\textbf{B}$
Using the formulas of circles, the circumference $C=2 \pi r$ and the area $A= \pi r^2$, we find that circle $Y$ has a radius of $4$ and circle $Z$ has a radius of $3$. Thus, the order from smallest to largest radius is $Z, X, Y$.
The table shows some of the results of a survey by radiostation KAMC. What percentage of the males surveyed listen to the station?
$\textbf{(A)}\ 39\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 55\qquad\textbf{(E)}\ 75$
$\textbf{E}$
While $200-96=104$ males take the survey, $136-58=78$ males listen to the station. So the percentage of the males surveyed listen to the station is $\dfrac{78}{104}=75\%$.
What is the product of $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}$ ?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$
$\textbf{C}$
We notice that the numerator in each fraction cancels out with the denominator of the one following it. This leaves us with only two numbers that didn't cancel: $\dfrac{2006}{2}=1003$.
Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area 12. What should his graph look like?
$\textbf{A}$
The relationship between $w$ and $l$ is $l=\dfrac{12}{w}$, which is an inverse proportional function. The only graph that could represent a inverted relationship is A.
How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$
$\textbf{C}$
There is $1$ integer whose digits sum to $1$: $10$.$\newline$
There are $4$ integers whose digits sum to $4$: $13, 22, 31, \text{and } 40$.$\newline$
There are $9$ integers whose digits sum to $9$: $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$.$\newline$
There are $3$ integers whose digits sum to $16$: $79, 88, \text{and } 97$.$\newline$
Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$.$\newline$
Thus, the answer is $1 + 4 + 9 + 3 = 17$.
Antonette gets $70\%$ on a 10-problem test, $80\%$ on a 20-problem test and $90\%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$
$\textbf{D}$
The total number of correct answers she got on the 60-problem test is $10\times70\%+20\times80\%+30\times90\%=50$. So the percentage of correct answer is $50/60\approx83\%$.
Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?
$\textbf{(A)}\ 10: 00\qquad\textbf{(B)}\ 10: 15\qquad\textbf{(C)}\ 10: 30\qquad\textbf{(D)}\ 11: 00\qquad\textbf{(E)}\ 11: 30$
$\textbf{D}$
Since Cassie leaves half an hour earlier then Brian, when Brian starts, the distance between them will be $62-\dfrac{12}{2}=56$ miles. Every hour, they will get $12+16=28$ miles closer. $\dfrac{56}{28}=2$, so 2 hours from $9:00$ AM is when they meet, which is $11: 00$ AM.
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?
$\textbf{(A)}\ 7,600\qquad\textbf{(B)}\ 11,400\qquad\textbf{(C)}\ 12,500\qquad\textbf{(D)}\ 15,200\qquad\textbf{(E)}\ 22,800$
$\textbf{B}$
Bob spent 15 more seconds than Chandra to read a page. So he need $15\cdot760=11,400$ more seconds to finish the book.
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$
$\textbf{C}$
Bob reads a page in 45 seconds while Chandra reads a page in 30 seconds. This means Bob reads $2$ pages for every $3$ pages that Chandra reads. Therefore, Chandra should read $\dfrac{3}{2+3}=\dfrac{3}{5}$ of the book, which is $\dfrac{3}{5}\times760=456$ pages.
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$
$\textbf{E}$
The ratio of time for Alice, Bob, and Chandra to read a page is $20:45:30=4:9:6$. So the amount of pages Alice, Bob, and Chandra will read is in the ratio $\dfrac14:\dfrac19:\dfrac16=9:4:6$. Therefore, Alice, Bob, and Chandra read $760\times\dfrac{9}{19}=360$, $760\times\dfrac{4}{19}=160$, and $760\times\dfrac{6}{19}=240$ pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply $160 \times 45 =7200$.
Jeff rotates spinners $P$, $Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number?
$\textbf{(A)}\ \dfrac{1}{4}\qquad\textbf{(B)}\ \dfrac{1}{3}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{3}{4}$
$\textbf{B}$
Noticing that spinner $Q$ always gives an even number. Similarly, spinner $R$ always gives an odd number. In order to get an odd number as the sum, spinner $P$ must give an even number. The probability that spinner $P$ gives an even number is $\dfrac13$.
A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?
$\textbf{(A)}\ \dfrac{1}{9}\qquad\textbf{(B)}\ \dfrac{1}{4}\qquad\textbf{(C)}\ \dfrac{4}{9}\qquad\textbf{(D)}\ \dfrac{5}{9}\qquad\textbf{(E)}\ \dfrac{19}{27}$
$\textbf{D}$
The surface area of the large cube is $6\times3\cdot3=54$ square inches, which means 54 $1\ \text{inch}\times1\ \text{inces}$ square faces. Each of the eight black cubes has 3 faces on the outside, making $3\times8=24$ black faces. So the cube has $54-24=30$ white faces. The ratio of white faces is $\dfrac{30}{54}= \dfrac{5}{9}$.
Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$. Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$, and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$. What is the length of $\overline{BD}$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$
$\textbf{D}$
Since triangle $ABD$ is congruent to triangle $ECD$, we have $\overline{AB}=\overline{CE}=11$. Since triangle $ABC$ is isosceles, $\overline{BC}=\overline{AB}=11$. Because point $D$ is the midpoint of $\overline{BC}$, $\overline{BD}=5.5$.
A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica (the sixth player) win?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
$\textbf{C}$
There are $\dbinom{6}{2}=15$ games in total, and each game will have only 1 winner. So the number of games Monica won is $15-4-3-2-2-2=2$.
An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a depth of $37$ cm. A rock with volume $1000 \text{ cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
$\textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5$
$\textbf{A}$
The area of the water surface is $100\times40=4000\ \text{cm}^2$. So the water level will increase by $1000/4000=0.25\ \text{cm}$.
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35$
$\textbf{D}$
Let the numbers of the bottom row from the left to the right be $a$, $b$, and $c$. Then the second row will be $a+b$ and $b+c$, and the top cell will be $a+2b+c$. To obtain the smallest sum, place $1$ in the center cell of the bottom row, and $2$ and $3$ in the outer ones. The top number will be $7$. For the largest sum, place $9$ in the center cell and $7$ and $8$ in the outer ones. This top number will be $33$. The difference is $33 - 7 =26$.
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$
$\textbf{A}$
If there were two more coins in the box, the number of coins would be divisible by both $6$ and $5$. The smallest number that is divisible by $6$ and $5$ is $30$, so the smallest possible number of coins in the box is $28$, and the remainder when divided by $7$ is $0$.
In the multiplication problem below $A$, $B$, $C$, $D$ are different digits. What is $A+B$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9$
$\textbf{A}$
$CDCD = CD \times 101$, so $ABA = 101$. Therefore, $A = 1$, $B = 0$, $A+B=1$.
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$
$\textbf{B}$
Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime, 2, the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, and the constant sum is $59+2=61$. Thus, the first card's hidden number is $61-44=17$, and the last card's hidden number is $61-38=23$. The average of the hidden primes is $\dfrac{2+17+23}{3}=14$.