AMC 8 2010
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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.
At Euclid Middle School, the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Miss Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest?
$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$
$\textbf{C}$
The total number of students taking the AMC 8 contest is $11+8+9=28$.
If $a @ b = \dfrac{a\times b}{a+b}$ for $a,b$ positive integers, then what is $5 @ 10$?
$\textbf{(A)}\ \dfrac{3}{10} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \dfrac{10}{3} \qquad\textbf{(E)}\ 50$
$\textbf{D}$
Substitute $a=5$ and $b=10$ into the expression for $a @ b$ to get $5 @ 10 = \dfrac{5\times 10}{5+10} = \dfrac{50}{15} = \dfrac{10}{3}$.
The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?
$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 89 \qquad \textbf{(E)}\ 100$
$\textbf{C}$
The highest price was $\$17$ in Month 1. The lowest price was $\$10$ in Month 3. 17 is $\dfrac{17}{10}-1=70\%$ more than 10.
What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$?
$\textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9$
$\textbf{C}$
By rearranging the numbers in increasing order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\dfrac{2+3}{2}=2.5.$ The average is $\dfrac{0+0+1+2+3+3+3+4}{8}=\dfrac{16}{8}=2.$ The sum of all three is $3+2.5+2=7.5$.
Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
$\textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40$
$\textbf{B}$
The light bulb is located $240-10=230$ centimeters above the floor. The height Alice can reach without a stool is $150+46=196$ centimeters above the floor. So the height of the stool is $230-196=34$ centimeters.
Which of the following figures has the greatest number of lines of symmetry?
$\textbf{(A)}\ \text{equilateral triangle}\newline$
$\textbf{(B)}\ \text{non-square rhombus}\newline$
$\textbf{(C)}\ \text{non-square rectangle}\newline$
$\textbf{(D)}\ \text{isosceles trapezoid}\newline$
$\textbf{(E)}\ \text{square}$
$\textbf{E}$
An equilateral triangle has $3$ lines of symmetry. A non-square rhombus has $2$ lines of symmetry. A non-square rectangle has $2$ lines of symmetry. An isosceles trapezoid has $1$ line of symmetry. A square has $4$ lines of symmetry. Therefore, the answer is the square.
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$
$\textbf{B}$
We need 4 pennies to pay for any amount of money less than 5 cents. For the amount of 5-9 cents, we need to add 1 nickle. For the amount of 10-24 cents, we need to add 2 dimes. For the amount of 25-100 cents, we need to add 3 quarters. So the total number of coins is $4+1+2+3=10$.
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$
$\textbf{D}$
Emily moves $12-8=4$ miles per hour faster than Emerson. From 1/2 mile behind to 1/2 mile ahead, it takes Emily $(\dfrac12+\dfrac12)/4=\dfrac14\ \text{hour}=15\ \text{minutes}$.
Ryan got $80\%$ of the problems correct on a $25$-problem test, $90\%$ on a $40$-problem test, and $70\%$ on a $10$-problem test. What percent of all the problems did Ryan answer correctly?
$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$
$\textbf{D}$
The total number of problems is $25+40+10=75$. And the total number of correct answers is $25\times80\%+40\times90\%+10\times70\%=63$. So Ryan answers $63/75=84\%$ problems correctly.
Six pepperoni circles will exactly fit across the diameter of a $12$-inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
$\textbf{(A)}\ \dfrac12 \qquad\textbf{(B)}\ \dfrac23 \qquad\textbf{(C)}\ \dfrac34 \qquad\textbf{(D)}\ \dfrac56 \qquad\textbf{(E)}\ \dfrac78$
$\textbf{B}$
The diameter of a pepperoni circle is $12/6=2$ inches. The area of a pepperoni circle is $\pi\left(\dfrac22\right)^2=\pi$ square inches. The area of the pizza is $\pi\left(\dfrac{12}{2}\right)^2=36\pi$ square inches. So the ratio is $\dfrac{24\pi}{36\pi}=\dfrac23$.
The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$. In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$
$\textbf{B}$
Let the height of the taller tree be $h$. Thus the height of the shorter tree is $h-16$. Since the ratio of the shorter tree to the taller tree is $\dfrac{3}{4}$, we have $\dfrac{h-16}{h}=\dfrac{3}{4}$. Solving for $h$ gives us $h=64$.
Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed from the bag so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$
$\textbf{D}$
There are $500\times(1-80\%)=100$ blue balls in the bag. When $75\%$ of balls are red, the number of red balls is 3 times as mush as blue balls. So we have 300 red balls at last. Hence, we need to remove 100 red balls from the bag.
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
$\textbf{E}$
Let the length of the shortest side be $x$. So the lengths of the rest two sides are $x+1$ and $x+2$. The perimeter is $x+(x+1)+(x+2)=3x+3$. Since $\dfrac{x}{3x+3}=0.3\rightarrow x=9$. So the length of the longest side is $9+2=11$.
What is the sum of the prime factors of $2010$?
$\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$
$\textbf{C}$
The prime factorization of $2010$ is $2010=2\times 3 \times 5 \times 67$. The sum of the prime factors is $2+3+5+67=77$.
A jar contains five different colors of gumdrops: $30\%$ are blue, $20\%$ are brown, $15\%$ red, $10\%$ yellow, and the other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
$\textbf{(A)}\ 35 \qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$
$\textbf{C}$
The percentage of green gumdrops is $1-30\%-20\%-15\%-10\%=25\%$. So the total number of gumdrops is $30\div25\%=120$. When half of the blue gumdrops are replaced with brown gumdrops, the percentage of brown gumdrops is $30\%\times0.5+20\%=35\%$. So the number of brown gumdrops is $120\times35\%=42$.
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?
$\textbf{(A)}\ \dfrac{\sqrt{\pi}}{2} \qquad\textbf{(B)}\ \sqrt{\pi} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ 2\pi \qquad\textbf{(E)}\ \pi^{2}$
$\textbf{B}$
Let the side length of the square be $s$ and the radius of the circle be $r$. Thus we have $s^2=\pi r^2$. Dividing each side by $r^2$, we get $\dfrac{s^2}{r^2}=\pi$. Hence, we have the ratio $\dfrac{s}{r}=\sqrt{\pi}$.
The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5$. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}$?
$\textbf{(A)}\ \dfrac25 \qquad \textbf{(B)}\ \dfrac12 \qquad \textbf{(C)}\ \dfrac35 \qquad \textbf{(D)}\ \dfrac23 \qquad \textbf{(E)}\ \dfrac34$
$\textbf{D}$
The area of the octagon is 10. Since $\overline{PQ}$ bisects the area of the octagon, the area below $\overline{PQ}$ is 5. So the area of the triangle is 4. This means the height of the triangle is $\dfrac85$. Now we have $\overline{QY}=\dfrac35$ and $\overline{XQ}=\dfrac25$. The ratio is $\dfrac{XQ}{QY}=\dfrac23$.
A decorative window is made up of a rectangle with semicircles on either end. The ratio of $AD$ to $AB$ is $3:2$, and $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles?
$\textbf{(A)}\ 2:3 \qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi \qquad\textbf{(D)}\ 9: \pi \qquad\textbf{(E)}\ 30 : \pi$
$\textbf{C}$
The length of $AD=45$ inches. The area of the rectangle is $45\times30=1350$ square inches. The sum of the areas of two semicircles is the area of a circle with radius 15, which is $\pi\times15^2=225\pi$ square inches. So the ratio is $1350:225\pi=6:\pi$.
The two circles pictured have the same center $C$. Chord $\overline{AD}$ is tangent to the inner circle at $B$, $AC$ is $10$, and chord $\overline{AD}$ has length $16$. What is the area between the two circles?
$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$
$\textbf{C}$
Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$. Thus $AB=BD=8$. From the Pythagorean Theorem, $CB=6$. Thus the area between the two circles is $\pi\times10^2-\pi\times6^2=64\pi$.
In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$
$\textbf{A}$
Given that $2/5$ of the people are wearing gloves and $3/4$ of the people are wearing hats, the total number of people in the room must be a multiple of 20. At least $\dfrac25+\dfrac34-1=\dfrac3{20}$ of the people are wearing both gloves and hats. Since we want the minimum number, we can assume that the total number of people is 20. So there are $20\times\dfrac3{20}=3$ people wearing both gloves and hats.
Hui is an avid reader. She bought a copy of the best seller $\textit{Math is Beautiful}$. On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$
$\textbf{C}$
Let $x$ be the number of pages in the book. After the first day, Hui had $\dfrac{4x}{5}-12$ pages left to read. After the second day, she had $\dfrac{3}{4}\left(\dfrac{4x}{5}-12\right)-15 = \dfrac{3x}{5}-24$ left. After the third day, she had $\dfrac{2}{3}\left(\dfrac{3x}{5}-24\right)-18=\dfrac{2x}{5}-34$ left. This is equivalent to $62.$ So we have $\dfrac{2x}{5}-34=62\rightarrow x=240$.
The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$
$\textbf{E}$
Let the hundreds, tens, and units digits of the original three-digit number be $a$, $b$, and $c$, respectively. We are given that $a=c+2$. The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$. The hundreds, tens, and units digits of the reversed three-digit number are $c$, $b$, and $a$, respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$. Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$. Thus, the units digit in the final result is 8.
Semicircles $POQ$ and $ROS$ pass through the center of circle $O$. What is the ratio of the combined areas of the two semicircles to the area of circle $O$?
$\textbf{(A)}\ \dfrac{\sqrt 2}{4}\qquad\textbf{(B)}\ \dfrac{1}{2}\qquad\textbf{(C)}\ \dfrac{2}{\pi}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{\sqrt 2}{2}$
$\textbf{B}$
The radius of the semicircle is 1. So the combined areas of the two semicircles is $\pi\times1^2=\pi$. Using Pythagorean Theorem, the radius of the large circle is $\sqrt{1^2+1^2}=\sqrt2$. So the area of the large circle is $\pi\times\left(\sqrt2\right)^2=2\pi$. The ratio of the combined areas of the two semicircles to the area of the large circle is $\dfrac{\pi}{2\pi}=\dfrac12$.
What is the correct ordering of the three numbers, $10^8$, $5^{12}$, and $2^{24}$?
$\textbf{(A)}\ 2^{24}<10^8<5^{12}\newline$
$\textbf{(B)}\ 2^{24}<5^{12}<10^8\newline$
$\textbf{(C)}\ 5^{12}<2^{24}<10^8\newline$
$\textbf{(D)}\ 10^8<5^{12}<2^{24}\newline$
$\textbf{(E)}\ 10^8<2^{24}<5^{12}$
$\textbf{A}$
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. So we can just compare $10^2=100$, $5^3=125$, and $2^6=64$. Since $64<100<125$, we have $2^{24}<10^8<5^{12}$.
Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
$\textbf{E}$
Considering the six steps, Jo has to land on the last step. For the rest 5 steps, them can be landed or not. So there are $2^5=32$ ways to land on the other five steps if we ignore the restriction that Jo can only climb 1, 2, or 3 steps at a time. Now we need to subtract the number of ways to climb the steps while taking a leap of $4$, $5$, or $6$. The eight possible ways for this are ($4$, $1$, $1$), ($4$, $2$), ($1$, $4$, $1$), ($1$, $1$, $4$), ($2$, $4$), ($1$, $5$), ($5$, $1$), and ($6$). So there are $32-8=24$ ways for Jo to climb the stairs.