## AMC 8 2016

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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.

The longest professional tennis match lasted a total of 11 hours and 5 minutes. How many minutes is that?

$\textbf{(A) } 605 \qquad\textbf{(B) } 655\qquad\textbf{(C) } 665\qquad\textbf{(D) } 1005\qquad \textbf{(E) } 1105$

$\textbf{C}$

1 hour contains 60 minutes. Thus 11 hours and 5 minutes are equal to $60\times11+5=665$ minutes.

In rectangle $ABCD$, $AB=6$ and $AD=8$. Point $M$ is the midpoint of $\overline{AD}$. What is the area of $\triangle AMC$?

$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

$\textbf{A}$

The base of $\triangle AMC$ is $AM=4$ and the the height is $AB=6$. Hence, the area of $\triangle AMC$ is $\dfrac12\times4\times6=12$.

Four students take an exam. Three of their scores are $70, 80,$ and $90$. If the average of their four scores is $70$, then what is the remaining score?

$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$

$\textbf{A}$

Since the average score for 4 students is 70, the sum of their scores is $70\times4=280$. Since the scores of three of the them are given, the rest score is $280-70-80-90=40$.

When Cheenu was a boy he could run $15$ miles in $3$ hours and $30$ minutes. As an old man he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?

$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$

$\textbf{B}$

3 hours and 30 minutes are equal to $3\times60+30=210$ minutes. So the time for Cheenu to run 1 mile when he was a boy is $210/15=14$ minutes. As a old man, he walked 10 miles in 4 hours, which are $4\times60=240$ minutes. The time to walk for 1 mile is $240/10=24$ minutes. Therefore it takes him $24-14=10$ more minutes to travel a mile now compared to when he was a boy.

The number $N$ is a two-digit number.

$\bullet$ When $N$ is divided by $9$, the remainder is $1$.$\newline$

$\bullet$ When $N$ is divided by $10$, the remainder is $3$.

What is the remainder when $N$ is divided by $11$?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$

$\textbf{E}$

Given that the reminder is 3 when $N$ is divided by 10, we know the units digit of $N$ is 3. Try all the possible values from 13 to 93, the only value that gives a reminder of 1 when divided by 9 is 73. When 73 is divided by 11, the reminder is 7.

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$

$\textbf{B}$

The median of 19 numbers is the 10th number in order, which is 4.

Which of the following numbers is not a perfect square?

$\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}$

$\textbf{B}$

We notice that $1^{2016}=1^2$, $3^{2018}=\left(3^{1009}\right)^2$, $4^{2019}=\left(2^{2019}\right)^2$, and $5^{2020}=\left(5^{1010}\right)^2$. However, $2^{2017}$ can not be wrote in the form of a perfect square.

Find the value of the expression\[100-98+96-94+92-90+\cdots+8-6+4-2\]

$\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

$\textbf{C}$

We can group each subtracting pair together:\[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2)\]After subtracting, we have:\[2+2+2+\ldots+2+2=2\times(1+1+1+\ldots+1+1)\]There are $50$ even numbers, so the number of pairs is $\dfrac{50}{2}=25$. Therefore the sum is $2 \times 25=50$.

What is the sum of the distinct prime integer divisors of $2016$?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

$\textbf{B}$

The prime factorization of 2016 is $2016=2^5\times3^2\times7$. The sum of distinct prime integer divisors is $2+3+7=12$.

Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if\[2 * (5 * x)=1\]

$\textbf{(A) }\dfrac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\dfrac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14.$

$\textbf{D}$

$$2 * (5 * x)=2 * (3\times5-x)=2 * (15-x)=3\times2-(15-x)=x-9=1\Rightarrow x=10$$

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$

$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

$\textbf{B}$

Let the tenth digit of the number be $x$ and the units digits be $y$. The number is equal to $10x+y$, and the reversing number is $10y+x$. The sum of them is $$(10x+y)+(10y+x)=132\Rightarrow x+y=12$$ Hence, the $(x,y)$ pair could be $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$.

Jefferson Middle School has the same number of boys and girls. $\dfrac{3}{4}$ of the girls and $\dfrac{2}{3}$ of the boys went on a field trip. What fraction of the students on the field trip were girls?

$\textbf{(A) }\dfrac{1}{2}\qquad\textbf{(B) }\dfrac{9}{17}\qquad\textbf{(C) }\dfrac{7}{13}\qquad\textbf{(D) }\dfrac{2}{3}\qquad \textbf{(E) }\dfrac{14}{15}$

$\textbf{B}$

Let the number of boys in Jefferson Middle School is $x$. Therefore the number of girls in Jefferson Middle School is also $x$. When $\dfrac34x$ girls and $\dfrac23x$ boys went on a field trip, the fraction of girls is $\dfrac34x\div\left(\dfrac34x+\dfrac23x\right)=\dfrac{9}{17}$.

Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0$?

$\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$

$\textbf{D}$

There are $\dbinom{6}{2}=15$ ways to choose 2 numbers from the set. Since the product is 0, the number 0 must be chosen. The rest one can be chosen in 5 numbers, which gives us 5 ways. Hence, the probability is $\dfrac{5}{15}=\dfrac13$.

Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?

$\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735$

$\textbf{A}$

For the first 350 miles, Karl used $350/35=10$ gallons of gas, with $14-10=4$ gallons of gas remained. Adding 8 gallons of gas led to $4+8=12$ gallons of gas in the tank. When he arrived, his gas tank was half full. This means that $12-7=5$ gallons of gas were used in the second part of the trip, which is $5\times35=175$ miles. Hence, the total distance for the whole trip is $350+175=525$ miles.

What is the largest power of $2$ that is a divisor of $13^4 - 11^4$?

$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$

$\textbf{C}$

Using difference of squares

\begin{align*}

13^4-11^4&=\left(13^2\right)^2-\left(11^2\right)^2\\&=\left(13^2+11^2\right)\left(13^2-11^2\right)\\&=\left(13^2+11^2\right)(13+11)(13-11)\\&=290\times24\times2\\&=2^5\times3\times5\times29

\end{align*}

The largest power of 2 that is a divisor of $13^4 - 11^4$ is $2^5=32$.

Annie and Bonnie are running laps around a $400$-meter oval track. They started together, but Annie has pulled ahead because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$

$\textbf{D}$

We notice that when Bonnie runs 4 laps, Annie runs $4\times125\%=5$ laps, which means that Annie is ahead by 1 lap and passes Bonnie.

An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$

$\textbf{D}$

Without any restriction, we have $10^4=10000$ possible passwords. If the passwords start with 911, the last digit has 10 choice. Hence, the answer is $10000-10=9990$.

In an All-Area track meet, $216$ sprinters enter a $100-$meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?

$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$

$\textbf{C}$

We need $216/6=36$ races in round one, which give us 36 winners. Then we need $36/6=6$ races in round two to give us 6 winners. In the final round we get the champion. There are $36+6+1=43$ races in total.

The sum of $25$ consecutive even integers is $10,000$. What is the largest of these $25$ consecutive integers?

$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$

$\textbf{E}$

The average number of the 25 consecutive even integers, which is $10000/25=400$, is the 13th integer in the middle of the sequence. So the last integer is $400+2\times(25-13)=424$.

The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$

$\textbf{A}$

To find the least possible value of the least common multiple of $a$ and $c$, we want $b$ to be as large as possible. The greatest common factor of $12$ and $15$ is 3. Therefore $b=3$, $a=4$, $c=5$. The least common multiple of 4 and 5 is 20.

A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

$\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}$

$\textbf{B}$

The probability that 3 reds are drawn is equal to the probability that only 3 reds are drawn (case 1) plus the probability that 3 reds and 1 green are drawn (case 2).

In case 1, the probability is $\dfrac35\times\dfrac24\times\dfrac13=\dfrac1{10}$.

In case 2, the green one can not be the last of the 4 drawn chips. If the green chip is the first drawn, the probability is $\dfrac25\times\dfrac34\times\dfrac23\times\dfrac12=\dfrac1{10}$. Similarly, if the green chip is the second drawn or the third draw, by the same calculation we find the probabilities are also $\dfrac1{10}$. Hence, the probability that 3 reds and 1 greed are drawn is $3\times\dfrac{1}{10}=\dfrac3{10}$.

Finally, the probability that 3 reds are drawn is $\dfrac1{10}+\dfrac3{10}=\dfrac25$.

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$. The area of the "bat wings" (shaded area) is

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

$\textbf{C}$

Let the point of intersection at the center of the bat wings be $O$. we notice that $\triangle OBC\sim\triangle OEF$, and $BC=\dfrac13EF$. If the area of $\triangle OBC$ is $x$, then the area of $\triangle OEF$ is $9x$. Let the area of the bat wings be $y$. The area of $\triangle BCE$ is $$x+\dfrac12y=\dfrac12BC\cdot DE=\dfrac12\times1\times4=2$$ The area of $\triangle CEF$ is $$9x+\dfrac12y=\dfrac12EF\cdot DE=\dfrac12\times3\times4=6$$ Hence, we get

\begin{align*}

x+\dfrac12y&=2\\9x+\dfrac12y&=6

\end{align*}

The area of the bat wings is $y=3$.

Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$?

$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

$\textbf{C}$

We notice that $\angle CED=\angle CEB+\angle AED-\angle AEB$. Since both $\triangle CEB$ and $\triangle AED$ are right triangles, $\angle CEB=\angle AED=90^\circ$. $\triangle AEB$ is equilateral, so $\angle AEB=60^\circ$. Hence, $\angle CED=\angle CEB+\angle AED-\angle AEB=90^\circ+90^\circ-60^\circ=120^\circ$.

The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$?

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

$\textbf{A}$

Since $QRS$ is divisible by $5$, $S$ must be 5. We notice that $PQR$ is divisible by $4$, so $R$ could be either $2$ or $4$. Given that $RST$ is divisible by $3$, we know that $R+S+T$ is a multiple of $3$. If $R=2$, $R+S+T=2+5+T=7+T$ could not be a multiple of $3$ when $T=1,3,4$. Hence, $R$ must be $4$. $R+S+T=4+5+T=9+T$ is a multiple of $3$ only when $T=3$. And $PQR$ is divisible by $4$ only when $Q=2$. Finally, we get $P=1$.

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$

$\textbf{B}$

The area of $\triangle ABC$ is $\dfrac12AB\cdot CD=\dfrac12\times16\times15=120$. It is twice the area of $\triangle ADC$, which is $\dfrac12AC\cdot DE$. The base $AC=\sqrt{AD^2+CD^2}=\sqrt{8^2+15^2}=17$. The height $DE$ is equal to the radius $r$ of the semicircle. Hence, we have $$\dfrac12AC\cdot DE=\dfrac12\times17r=60\Rightarrow r=\dfrac{120}{17}$$