AMC 8 2013
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.
Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
$\textbf{A}$
The least multiple of 6 greater than 23 is 24. So she needs to add $24-23=1$ more model car.
A sign at the fish market says, "$50\%$ off, today only: half-pound packages for just $\$3$ per package." What is the regular price for a full pound of fish, in dollars? (Assume that there are no deals for bulk)
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
$\textbf{D}$
The price of a full pound of fish today with a $50\%$ discount is $\$6$. So the regular price is $\$12$.
What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$?
$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$
$\textbf{E}$
We group the addends in pairs:
\begin{align*}
-1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500.
\end{align*}
Then the desired answer is $4 \times 500 =2000$.
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $\$2.50$ to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \$120\qquad\textbf{(B)}\ \$128\qquad\textbf{(C)}\ \$140\qquad\textbf{(D)}\ \$144\qquad\textbf{(E)}\ \$160$
$\textbf{C}$
Since Judi's 7 friends had to pay $\$2.50$ extra each to cover the total amount that Judi should have paid, $\$2.50\cdot7=\$17.50$ is the bill Judi would have paid if she had money. Hence, the total bill is $\$17.50\cdot8=\$140$.
Hammie is in the $6^\text{th}$ grade and weighs 106 pounds. Her quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
$\textbf{(A)}\ \text{median, by 60} \qquad\newline$
$\textbf{(B)}\ \text{median, by 20} \qquad\newline$
$\textbf{(C)}\ \text{average, by 5} \qquad\newline$
$\textbf{(D)}\ \text{average, by 15} \qquad\newline$
$\textbf{(E)}\ \text{average, by 20}$
$\textbf{E}$
Listing the elements from least to greatest, we have $\{5, 5, 6, 8, 106\}$. The median weight is 6 pounds. The average weight of the five kids is $(5+5+6+8+106)/5=26$. So the average is $26-6=20$ greater than the median.
The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$. What is the missing number in the top row?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
$\textbf{C}$
The missing number is the middle row is $600/30=20$. Furthermore, the missing number is the top row is $20/5=4$.
Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$
$\textbf{C}$
It takes $\dfrac{10}{6}=\dfrac53$ seconds for each car to pass. Since the total time is $2\ \text{minutes and}\ 45\ \text{seconds}=(2\times60+45)\ \text{seconds}=165\ \text{seconds}$, the total number of cars is $\dfrac{165}{5/3}=99\approx100$.
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
$\textbf{(A)}\ \dfrac18 \qquad \textbf{(B)}\ \dfrac14 \qquad \textbf{(C)}\ \dfrac38 \qquad \textbf{(D)}\ \dfrac12 \qquad \textbf{(E)}\ \dfrac34$
$\textbf{C}$
The probability of 3 consecutive heads, HHH, is $\left(\dfrac12\right)^3=\dfrac18$. The probability of HHT and THH are both $\left(\dfrac12\right)^3=\dfrac18$. So the probability of at least two consecutive heads is $\dfrac18\times3=\dfrac38$.
The Incredible Hulk can double the distance it jumps with each succeeding jump. If its first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will it first be able to jump more than 1 kilometer?
$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$
$\textbf{C}$
On the $n^\text{th}$ jump, the distance is $2^{n-1}$ meters. To jump more than $1\ \text{kilometer}=1000\ \text{meters}$, we have $2^{n-1}>1000$. We know that $2^{10}=1024$. So $n-1=10\rightarrow n=11$.
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$
$\textbf{C}$
To find either the LCM or the GCF of two numbers, always prime factorize first. The prime factorization of 180 is $$180 = 3^2 \times 5 \times 2^2$$ The prime factorization of 594 is $$594 = 3^3 \times 11 \times 2$$ To find the LCM, we have to find the greatest power of all the numbers there are. Hence, we get $$3^3\times5\times11\times2^2=5940$$For the GCF, use the least power of all of the numbers that are in both factorizations and multiply$$3^2 \times 2 = 18$$Thus the answer is $5940/18=330$.
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
$\textbf{D}$
It takes him $\dfrac25\ \text{hour}=24\ \text{minutes}$ when the speed is 5 miles per hour, $\dfrac23\ \text{hour}=40\ \text{minutes}$ when the speed is 3 miles per hour, and $\dfrac24\ \text{hour}=30\ \text{minutes}$ when the speed is 4 miles per hour. So the total time is $24+40+30=94\ \text{minutes}$. If he had always walked at 4 miles per hour, the total time is $3\times30=90\ \text{minutes}$, which is 4 minutes less than before.
At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50$, you get a second pair at a $40\%$ discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the 150 dollar regular price did he save?
$\textbf{(A)}\ 25\% \qquad \textbf{(B)}\ 30\% \qquad \textbf{(C)}\ 33\% \qquad \textbf{(D)}\ 40\% \qquad \textbf{(E)}\ 45\%$
$\textbf{B}$
He spent $50+50\times60\%+50\times50\%=105$ for three pairs of sandals. Comparing to the regular price of $\$150$, $\$45$ is saved. The percentage saved is $\dfrac{45}{150}=30\%$.
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$
$\textbf{A}$
Let the correct two-digit number be $\overline{ab}$. We can rewrite the correct score as $10a+b$. Clara misinterpreted it as $10b+a$. The difference between the two scores is $(10a+b)-(10b+a)=9(a-b)$, which is a multiple of 9. The only answer choice that is a multiple of 9 is $45$.
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \dfrac14 \qquad \textbf{(B)}\ \dfrac13 \qquad \textbf{(C)}\ \dfrac38 \qquad \textbf{(D)}\ \dfrac12 \qquad \textbf{(E)}\ \dfrac23$
$\textbf{C}$
The probability that the matched color is green is $\dfrac12\times\dfrac14=\dfrac18$. The probability that the matched color is red is $\dfrac12\times\dfrac24=\dfrac14$. So the total probability that the colors match is $\dfrac18+\dfrac14=\dfrac38$.
If $3^p + 3^4 = 90$, $2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$, $r$, and $s$?
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$
$\textbf{B}$
By solving the equations, we get
\begin{align*}
3^p=90-3^4=9&\rightarrow p=2\\
2^r=76-44=32&\rightarrow r=5 \\
6^s=1421-5^3=1296&\rightarrow s=4
\end{align*}
So the product of $p$, $r$, and $s$ is $2\times5\times4=40$.
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$-graders to $6^\text{th}$-graders is $5:3$, and the ratio of $8^\text{th}$-graders to $7^\text{th}$-graders is $8:5$. What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$
$\textbf{E}$
Let the number of $8^\text{th}$-graders be $x$. Thus the number of $6^\text{th}$-graders is $\dfrac35x$, and the number of $7^\text{th}$-graders is $\dfrac58x$. Hence, $x$ must be a multiple of both 5 and 8. The least common multiple of 5 and 8 is 40. So the smallest value of $x$ is 40. Then we get the total number of students $x+\dfrac35x+\dfrac58x=40+24+25=89$.
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$
$\textbf{B}$
The average of the six consecutive positive integers is $2013/6=335.5$, which is also the average of the 3rd and 4th integer. Hence, the 4th integer is 336, and the 6th integer is 338.
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$
$\textbf{B}$
The empty cuboid inside of the fort is 10 feet long, 8 feet wide and 4 feet high. So its volume is $10\times8\times4=320\ \text{ft}^3$. When the empty part is full of blocks, the volume of the fort is $12\times10\times5=600\ \text{ft}^3$. Hence, the volume of the walls and floor is $600-320=280\ \text{ft}^3$. Since the volume of each one-foot cubical block is $1\ \text{ft}^3$, we get the number of blocks $280/1=280$.
Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from the highest score to the lowest score?
$\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \newline$
$\textbf{(B)}\ \text{Hannah, Bridget, Cassie} \newline$
$\textbf{(C)}\ \text{Cassie, Bridget, Hannah} \newline$
$\textbf{(D)}\ \text{Cassie, Hannah, Bridget} \newline$
$\textbf{(E)}\ \text{Bridget, Cassie, Hannah}$
$\textbf{D}$
Everyone know Hannah's score. When Cassie says 'I didn't get the lowest score in our class', we know that Hannah's score is lower than Cassie's. When Bridget says 'I didn't get the highest score', we know that Bridget's score is lower than Hannah. Hence, we get the order of scores $\text{Cassie}>\text{Hannah}>\text{Bridget}$.
A $1\times 2$ rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle?
$\textbf{(A)}\ \dfrac\pi2 \qquad \textbf{(B)}\ \dfrac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \dfrac{4\pi}3 \qquad \textbf{(E)}\ \dfrac{5\pi}3$
$\textbf{C}$
By symmetry, the center of the semicircle is located at the midpoint of the long side of the rectangle. The radius of the semicircle, by the Pythagorean Theorem, is $\sqrt{1^2+1^2}=\sqrt{2}$. So the area of the semicircle is $\dfrac12\cdot\pi\left(\sqrt2\right)^2=\pi$.
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$
$\textbf{E}$
She has $\dbinom{3}{1}=3$ ways to the southwest corner of the City Park, 1 way to pass the diagonal, and $\dbinom{4}{2}=6$ ways to the school. So the number of different routes is $3\times1\times6=18$.
Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?
$\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932$
$\textbf{E}$
We have 33 rows of horizontal toothpicks with 60 per row, and 61 columns of vertical toothpick with 32 per column. So the total number of toothpicks is $33\times60+61\times32=3932$.
Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$
$\textbf{B}$
The area of semicircle on $\overline{AB}$ is $\dfrac12\times\pi\times\left(\dfrac12\overline{AB}\right)^2=8\pi$. So $\overline{AB}=8$. The arc of the semicircle on $\overline{AC}$ is $\dfrac12\times\pi\times\overline{AC}=8.5\pi$. So $\overline{AC}=17$. Since $\triangle ABC$ is a right triangle, $\overline{BC}=\sqrt{17^2-8^2}=15$. Hence, the radius of the semicircle on $\overline{BC}$ is $15/2=7.5$.
Squares $ABCD$, $EFGH$, and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE$, respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?
$\textbf{(A)}\ \dfrac{1}{4}\qquad\textbf{(B)}\ \dfrac{7}{24}\qquad\textbf{(C)}\ \dfrac{1}{3}\qquad\textbf{(D)}\ \dfrac{3}{8}\qquad\textbf{(E)}\ \dfrac{5}{12}$
$\textbf{C}$
The area of the shaded pentagon is the area of $\triangle AJX$ subtracts the area of rectangle $BCIX$. Let the side length of squares $ABCD$, $EFGH$, and $GHIJ$ be 1. The area of $\triangle AJX$ is $\dfrac12(1.5)(2)=1.5$. The area of rectangle $BCIX$ is $(1)(0.5)=0.5$. So the area of the shaded pentagon is $1.5-0.5=1$. The sum of the area of the three squares is 3. Hence, the ratio is $\dfrac13$.
A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?
$\textbf{(A)}\ 238\pi\qquad\textbf{(B)}\ 240\pi\qquad\textbf{(C)}\ 260\pi\qquad\textbf{(D)}\ 280\pi\qquad\textbf{(E)}\ 500\pi$
$\textbf{A}$
The radius of the ball is $r=2\ \text{inches}$. The track of the center of the ball in the first semicircular is also a semicircular, but its radius is $R_1-r=98\ \text{inches}$. Similarly, the radius in the third semicircular is $R_3-r=78\ \text{inches}$. However, the radius on the second semicircular is $R_2+r=62\ \text{inches}$. Hence, distance the center of the ball traveled is $\pi\times(98+62+78)=238\pi$ inches.