AMC 8 2014
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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.
Harry and Terry are each told to calculate $8-(2+5)$. Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$. If Harry's answer is $H$ and Terry’s answer is $T$, what is $H - T$?
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$
$\textbf{A}$
Henry's answer is $H=8-(2+5)=1$. Terry's answer is $T=8-2+5=11$. So $H-T=1-11=-10$.
Paul owes Paula 35 cents and has a pocket full of 5-cent coins, 10-cent coins, and 25-cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
$\textbf{E}$
The way to pay with the largest number is 7 5-cent coins. And the way to pay with the smallest number of coins is 1 25-cent coin and 1 10-cent coin. So the difference is $7-2=5$.
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?
$\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280$
$\textbf{B}$
The number of pages is $36\times3+44\times3+10=250$.
The sum of two prime numbers is 85. What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$
$\textbf{E}$
Since the sum of two prime numbers is an odd number, one of the prime number must be even, and the other one odd. The only even prime number is 2. Therefore, the other prime number must be $85-2=83$. The product of these two prime numbers is $2\times83=166$.
Margie's car can go 32 miles on a gallon of gas, and gas currently costs $\$4$ per gallon. How many miles can Margie drive on $\$20$?
$\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad\textbf{(E) }640$
$\textbf{C}$
Margie can buy at most $20/4=5$ gallons of gas. So she can drive $5\times32=160$ miles.
Six rectangles each with a common base width of 2 have lengths of 1, 4, 9, 16, 25, and 36. What is the sum of the areas of the six rectangles?
$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad\textbf{(E) }202$
$\textbf{D}$
The sum of the area is $$2\times1+2\times4+2\times9+2\times16+2\times25+2\times36=2\times(1+4+9+16+25+36)=2\times91=182$$
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?
$\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad\textbf{(E) }2 : 1$
$\textbf{B}$
If we subtract 4 girls from the class, the number of boys and girls should be equal. Hence, the number of boys is $(28-4)/2=12$. The ratio of girls to boys is $(12+4):12=4:3$.
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $\$\underline{1}\underline{A}\underline{2}$. What is the missing digit A of this 3-digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
$\textbf{D}$
The three-digit number should be a multiple of 11. We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$. Thus, we get $1+2-A$ is a multiple of 11. The only multiple that works here is $0$, as $11 \times 0 = 0$. Thus, $A = 3$.
In $\triangle ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$?
$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$
$\textbf{D}$
Since $BD=DC$, $\triangle BCD$ is isosceles. So $\angle CBD=\angle BCD=70^\circ$, $\angle ADB=\angle BCD+\angle CBD=140^\circ$.
The first AMC 8 was given in 1985 and it has been given annually since that time. Samantha turned 12 years old the year that she took the seventh AMC 8. In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$
$\textbf{A}$
The seventh AMC 8 was given in 1991 when Samantha was 12 years old. So she was born in $1991-12=1979$.
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }10$
$\textbf{A}$
If Jack ignore the restriction of the dangerous area, he has $\dbinom{5}{3}\dbinom{2}{2}=10$ ways to choose the path. If Jack MUST pass the dangerous area, he has $\dbinom{2}{1}\dbinom{1}{1}=2$ ways to reach the dangerous area, and then $\dbinom{3}{2}\dbinom{1}{1}=3$ ways to reach Jill's house. So the total ways passing through the dangerous area is $2\times3=6$. Hence, he has $10-6=4$ ways to reach Jill's house without passing through the dangerous area.
A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly as a fraction?
$\textbf{(A) }\dfrac{1}{9}\qquad\textbf{(B) }\dfrac{1}{6}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad\textbf{(E) }\dfrac{1}{2}$
$\textbf{B}$
There is a $\dfrac{1}{3}$ chance that the reader will choose the correct baby picture for the first person. Next, the second person gives a $\dfrac{1}{2}$ chance, and the last person leaves only 1 choice. Thus, the probability is $\dfrac{1}{3\times 2}=\dfrac{1}{6}.$
If $n$ and $m$ are integers and $n^2+m^2$ is even, which of the following is impossible?
$\textbf{(A) }n$ and $m$ are even $\newline$
$\textbf{(B) }n$ and $m$ are odd $\newline$
$\textbf{(C) }n+m$ is even $\newline$
$\textbf{(D) }n+m$ is odd $\newline$
$\textbf{(E) }$none of these are impossible
$\textbf{D}$
Since $n^2+m^2$ is even, either both $n^2$ and $m^2$ are even, or they are both odd. Therefore, $n$ and $m$ are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, $n+m$ must be even. D is impossible.
Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$?
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$
$\textbf{B}$
Since the rectangle $ABCD$ and triangle $DCE$ share the same side $CD$ and have the same area, we have $CE=2BC=12$. According to the Pythagorean Theorem, $DE=\sqrt{CD^2+CE^2}=\sqrt{5^2+12^2}=13$.
The circumference of the circle with center $O$ is divided into 12 equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$?
$\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$
$\textbf{C}$
$\angle IOG$ corresponds to arc $IG$, which is 1/6 of the circumference. So $\angle IOG=360/6=60^\circ$, which makes triangle $OIG$ equilateral. Hence, $y=60$. Similarly, $\angle AOE$ corresponds to arc $AE$, which is 1/3 of the circumference. So $\angle AOE=360/3=120^\circ$, $x=(180-120)/2=30$. Finally, we get $x+y=30+60=90$.
The "Middle School Eight" basketball conference has 8 teams. Every season, each team plays every other conference team twice (home and away), and each team also plays 4 games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad\textbf{(E) }160$
$\textbf{B}$
Each team plays with 7 teams within the conference twice. So there are 14 plays each team played within the conference. However, the total number of games within the conference is not $8\times14=112$ due to double counting. Hence, the real number of games within the conference is $112/2=56$. Each team also plays 4 games outside the conference, which make a total of $4\times8=32$ games without double counting. So the total number of games in a season is $56+32=88$.
George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\dfrac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\dfrac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad\textbf{(E) }12$
$\textbf{B}$
On a normal day, it takes him 1/3 hour to get to school. However, today it took $\dfrac{1/2 \text{ mile}}{2 \text{ mph}}=\dfrac14$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\dfrac{1/2 \text{ mile}}{1/12 \text { hour}}=6 \text{ mph}$.
Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?
$\textbf{(A) }$ all 4 are boys$\newline$
$\textbf{(B) }$ all 4 are girls$\newline$
$\textbf{(C) }$ 2 are girls and 2 are boys$\newline$
$\textbf{(D) }$ 3 are of one gender and 1 is of the other gender$\newline$
$\textbf{(E) }$ all of these outcomes are equally likely
$\textbf{D}$
The probability that all 4 are boys is $\left(\dfrac{1}{2}\right)^2=\dfrac1{16}$. By symmetry, the probability that all 4 are girls is also $\dfrac1{16}$. For C, we have $\dbinom{4}{2}=6$ ways to choose which 2 are girls, and the rest 2 are boys. So the probability is $\dfrac{6}{2^4}=\dfrac38$. D is the only situation if we exclude the situation of A, B, and C. So the probability that 3 are of one gender and 1 is of the other gender is $1-\dfrac1{16}-\dfrac1{16}-\dfrac38=\dfrac12$. Apparently, D has te largest probability.
A cube with 3-inch edges is to be constructed from 27 smaller cubes with 1-inch edges. Twenty-one of the cubes are colored red and 6 are colored white. If the 3-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
$\textbf{(A) }\dfrac{5}{54}\qquad\textbf{(B) }\dfrac{1}{9}\qquad\textbf{(C) }\dfrac{5}{27}\qquad\textbf{(D) }\dfrac{2}{9}\qquad\textbf{(E) }\dfrac{1}{3}$
$\textbf{A}$
The surface area of the 3-inch cube is $3\times3\times6=54$ square inches. For the least possible surface area that is white, we should put 1 white cube in the center, and the other 5 to the center place of the surface with only 1 face exposed. This gives 5 square inches of white surface area. So the fraction of white surface is $\dfrac5{54}$.
Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle with a radius of $1$ is centered at $A$, a circle with a radius of $2$ is centered at $B$, and a circle with a radius of $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
$\textbf{(A) }3.5\qquad\textbf{(B) }4.0\qquad\textbf{(C) }4.5\qquad\textbf{(D) }5.0\qquad\textbf{(E) }5.5$
$\textbf{B}$
The area of the region inside the rectangle but outside three circles is the area of the rectangle subtracts the area of three sectors. The area of the rectangle is $3\times5 =15$. The area of three sectors is $\dfrac14\times\pi\times1^2=\dfrac14\pi$, $\dfrac14\times\pi\times2^2=\pi$, and $\dfrac14\times\pi\times3^2=\dfrac94\pi$. So the area in the rectangle but outside the circles is $$15-\dfrac14\pi-\pi-\dfrac94\pi=15-\dfrac72\pi\approx4$$
The 7-digit numbers $\underline{74A52B1}$ and $\underline{326AB4C}$ are each multiples of 3. Which of the following could be the value of $C$?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad\textbf{(E) }8$
$\textbf{A}$
The sum of all digits of a multiple of 3 is also a multiple of 3. So $$7+4+A+5+2+B+1=A+B+19$$ is a multiple of 3, which means $A+B+1$ is a multiple of 3. Similarly, $$3+2+6+A+B+4+C=A+B+C+15$$ is a multiple of 3, or we can say $A+B+C$ is a multiple of 3. To eliminate the unknowns $A$ and $B$, by subtracting, we found that $$(A+B+C)-(A+B+1)=C-1$$ is a multiple of 3. So $C$ could be 1, 4, or 7.
A 2-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad\textbf{(E) }9$
$\textbf{E}$
Let the 2-digit number be $\overline{ab}$. We have $$ab+a+b=10a+b\rightarrow ab=9a$$ Since $a$ can not be 0, we get the units digit $b=9$.
Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all 2-digit primes.$\newline$
Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.$\newline$
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.$\newline$
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?
$\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad\textbf{(E) }23$
$\textbf{A}$
Since the number of date can not be greater than 31, we list all 2-digit primes less than 31: $\{11, 13, 17, 19, 23, 29\}$. 23 and 29 can be excluded because the sum of 23 (or 29) and another prime is greater than 31. 19 can be excluded because if 19 is chosen, at least one of the rest two primes must fall into the set of $\{13,17\}$, then $13+19$ or $17+19$ is greater than 31. Here we have 11, 13 and 17 left.
Let $a$, $b$, $c$ be the uniform number of Ashley, Bethany, and Caitlin. According to the order of date, we have $$a+c<b+c<a+b$$ Hence, we get $c<a<b$. The uniform number of Caitlin is 11.
One day the Beverage Barn sold 252 cans of soda to 100 customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
$\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad\textbf{(E) }4.5$
$\textbf{C}$
In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\dfrac{203}{51}$ gives us $3$ and a remainder of $50$. Seeing this, the largest number of cans the $50$th person could have is $3$, which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$. Thus our answer is C.
A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?
Note: 1 mile = 5280 feet
$\textbf{(A) }\dfrac{\pi}{11}\qquad\textbf{(B) }\dfrac{\pi}{10}\qquad\textbf{(C) }\dfrac{\pi}{5}\qquad\textbf{(D) }\dfrac{2\pi}{5}\qquad\textbf{(E) }\dfrac{2\pi}{3}$
$\textbf{B}$
His path contains $5280/40=132$ semicircles. The radius of each semicircle is $40/2=20$ feet. Hence, the circumference of each semicircle is $20\pi$ feet. The total length of the path is $132\times20\pi=2640\pi$ feet, or $2640\pi/5280=0.5\pi$ miles. Since the speed is 5 miles per hour, the time it takes to cover the stretch is $0.5\pi/5=\pi/10$ hour.