### 2016 AMC 8 Exam Solutions

Problems used with permission of the Mathematical Association of America. Scroll down to view solutions, print PDF solutions, view answer key, or:

1.

The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?

$605$

$655$

$665$

$1005$

$1105$

###### Solution(s):

There are $60$ minutes in an hour, so the total time is $60 \cdot 11 + 5 = 665$ minutes.

Thus, **C** is the correct answer.

2.

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

$12$

$15$

$18$

$20$

$24$

###### Solution(s):

From the diagram, we can see that the base of $\triangle AMC$ is $4$ and the altitude is $4.$ The area is therefore $\dfrac{1}{2} \cdot 4 \cdot 6 = 12.$

Thus, **A** is the correct answer.

3.

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?

$40$

$50$

$55$

$60$

$70$

###### Solution(s):

From the average, we can calculate the sum of the scores to be $4 \cdot 70 = 280.$ This means that the remaining score is $280 - 70 - 80 - 90 = 40.$

Thus, **A** is the correct answer.

4.

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 walk a mile now compared to when he was a boy?

$6$

$10$

$15$

$18$

$30$

###### Solution(s):

To better compare the rates, we can change his speed into minutes per mile.

As a boy he ran $15$ miles in $3 \cdot 60 + 30 = 210$ minutes, which means that he ran at a pace of $210 / 15 = 14$ minutes per mile.

As an adult, he can walk $10$ miles in $4 \cdot 60 = 240$ minutes, which means he walks at a pace of $240 / 10 = 24$ minutes per mile.

Subtracting the two, we get that he takes $10$ more minutes to walk a mile as an adult.

Thus, **B** is the correct answer.

5.

The number $N$ is a two-digit number with the following properties:

$\quad$• When $N$ is divided by $9,$ the remainder is $1.$

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

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

$0$

$2$

$4$

$5$

$7$

###### Solution(s):

The two-digit numbers that leave a remainder of $1$ when divided by $9$ are: $\begin{align*}&10, 19, 28, 37, 46,\\& 55, 64, 73, 82, 91.\end{align*}$ The two-digit numbers that leave a remainder of $3$ when divided by $10$ are: $\begin{align*}&13, 23, 33, 43, 53,\\& 63, 73, 83, 93.\end{align*}$ Among these numbers, $73$ is the only common number. The remainder of $73$ when divided by $11$ is $7.$

Thus, **E** is the correct answer.

6.

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

$3$

$4$

$5$

$6$

$7$

###### Solution(s):

Since there are $19$ people, each with one corresponding name length, the middle length will be the tenth one. Counting from the left side, the tenth value that we arrive upon is $4.$

Thus, **B** is the correct answer.

7.

Which of the following numbers is not a perfect square?

$1^{2016}$

$2^{2017}$

$3^{2018}$

$4^{2019}$

$5^{2020}$

###### Solution(s):

Since any number with an even exponent is a perfect square, we can eliminate **A**, **C**, and **E**. Also, a square number to any power remains a square number, so that rules out **D**.

Thus, **B** is the correct answer.

8.

Find the value of the expression $\begin{align*} 100 - 98 + 96 - 94 + 92 - 90 \\ + \cdots 8 - 6 + 4 - 2. \end{align*}$

$20$

$40$

$50$

$80$

$100$

###### Solution(s):

We can group the sum as follows: $\begin{gather*} (100 - 98) + (96 - 94) \\ + \cdots + (4 - 2). \end{gather*}$ Note that each pair evaluates to $2$ and there are $25$ pairs. Therefore, the total sum is $2 \cdot 25 = 50.$

Thus, **C** is the correct answer.

9.

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

$9$

$12$

$16$

$49$

$63$

###### Solution(s):

We can prime factorize $2016$ as $2^5 \cdot 3^2 \cdot 7.$ This shows that the prime divisors of $2016$ are $2, 3,$ and $7.$ The sum of these is $12,$ so **B** is the correct answer.

10.

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

$\dfrac{1}{10}$

$2$

$\dfrac{10}{3}$

$10$

$14$

###### Solution(s):

We can simplify the equation as follows: $\begin{align*} 1 &= 2 * (5 * x) \\ &= 2 * (3 \cdot 5 - x) \\ &= 2 * (15 - x) \\ &= 3 \cdot 2 - (15 - x) \\ &= x - 9. \end{align*}$ Solving yields $x = 10.$

Thus, **D** is the correct answer.

11.

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 $132.$

$5$

$7$

$9$

$11$

$12$

###### Solution(s):

Let $ab$ be the two-digit number in question. Then, it follows that the number obtained by reversing its digits is $ba.$ Therefore, in order for $ab$ to satisfy the property in the question: $\begin{align*}10(a + b) + a + b &=132\\11(a + b)&=132\\a+b&=12.\end{align*}$ The only possible solutions $(a,b$ to this equation, where $a,b$ are both one digit, are: $(3,9), (4,8), (5,7), (6,6),$$(7,5), (8,4),(9,3).$ As such, there are $7$ solutions.

Thus, **B** is the correct answer.

12.

Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students on the field trip were girls?

$\dfrac{1}{2}$

$\dfrac{9}{17}$

$\dfrac{7}{13}$

$\dfrac{2}{3}$

$\dfrac{14}{15}$

###### Solution(s):

To more easily compare, we can convert the fractions to have the same denominator:$\dfrac{3}{4} = \dfrac{9}{12}$ $\dfrac{2}{3} = \dfrac{8}{12}$ This shows that the ratio of girls to boys is $9 : 8,$ which means that the fraction of girls on the field trip is $\dfrac{9}{17}.$

Thus, **B** is the correct answer.

13.

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 $0?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

###### Solution(s):

The only way for the product to be $0$ is if one of the number chosen is $0.$ If the first number chosen is $0,$ then there are $5$ options for the second number.

Similarly, there are $5$ combinations if $0$ was chosen second.

Therefore, there are $10$ total pairs where the product is $0.$ The total number of pairs is $6 \cdot 5 = 30,$ so the probability is $\dfrac{10}{30} = \dfrac{1}{3}.$

Thus, **D** is the correct answer.

14.

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?

$525$

$560$

$595$

$665$

$735$

###### Solution(s):

If Karl drove $350$ miles, then he used $350 / 35$ gallons of gas.

When he bought more gas, he added $8$ gallons to $14 - 10 = 4$ gallons, attaining a total of $12$ gallons.

If his tank was half full when he arrived, he used $12 - 7 = 5$ gallons, which equates to $5 \cdot 35 = 175$ miles.

Therefore, he travelled a total distance of $350 + 175 = 525\text{ miles.}$

Thus, **A** is the correct answer.

15.

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

$8$

$16$

$32$

$64$

$128$

###### Solution(s):

We can factor this expression using difference of squares.

$\begin{align*} (13^2 + 11^2)&(13^2 - 11^2) \\ &= 290 \cdot 48 \\ &= 32 \cdot 145 \cdot 3 \end{align*}$

This shows that $32$ is the largest power of $2$ that divides the expression.

Thus, **C** is the correct answer.

16.

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?

$1\dfrac{1}{4}$

$3\dfrac{1}{3}$

$4$

$5$

$25$

###### Solution(s):

Since Annie is $25\%$ faster than Bonnie, for every lap Bonnie finishes, Annie completes $1 \dfrac{1}{4}$ laps. Therefore, Annie gains a quarter lap every time Bonnie finished a lap.

With this in mind, for Annie to completely lap Bonnie, Bonnie must finish $4$ laps, which means that Annie finished $5$ laps.

Thus, **D** is the correct answer.

17.

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?

$30$

$7290$

$9000$

$9990$

$9999$

###### Solution(s):

The total number of passwords with no conditions is $10^4.$ The condition removes $10$ possible passwords since the first $3$ are determined, and the last one can be anything. Therefore, the number of acceptable passwords is $10,000 - 10 = 9990.$

Thus, **D** is the correct answer.

18.

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?

$36$

$42$

$43$

$60$

$72$

###### Solution(s):

Note that each race eliminates $5$ people. For there to be a winner, $215$ must be eliminated. Therefore, $215 / 15 = 43$ races are required to eliminate this number of people.

Thus, **C** is the correct answer.

19.

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

$360$

$388$

$412$

$416$

$424$

###### Solution(s):

The average of these numbers is $10,000 / 25 = 400.$ The largest number is $12$ even numbers away, which means that it equals $400 + 12 \cdot 2 = 424.$

Thus, **E** is the correct answer.

20.

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 $c?$

$20$

$30$

$60$

$120$

$180$

###### Solution(s):

We know that $b$ has to divide both $12$ and $15,$ so it must equal either $1$ or $3.$

If $b = 1,$ then $a = 12$ and $c = 15,$ making their least common multiple $60.$ If $b = 3,$ then the smallest value of $a$ is $12$ and $c$ is $5.$ The least common multiple in this scenario is $20.$

Thus, **A** is the correct answer.

21.

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?

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

###### Solution(s):

The only way for the $3$ reds to be drawn first is if there is only one green drawn in the first $4$ draws. The green can be in any of the first $4$ spots, yielding $4$ possibilities.

We can find the total number of possibilities to be $10$ by listing them out. Therefore, the desired probability is $\dfrac{4}{10} = \dfrac{2}{5}.$

Thus, **B** is the correct answer.

22.

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

$2$

$2 \dfrac{1}{2}$

$3$

$3 \dfrac{1}{2}$

$5$

###### Solution(s):

Define $I$ to be the midpoint of $\overline{AD}$ and $G$ to be the midpoint of $\overline{EF}.$ Also define $H$ to be the intersection of $\overline{CF}$ and $\overline{BE}.$

The area of $\triangle BCE$ equals $\dfrac{1}{2} \cdot 1 \cdot 4 = 2.$ By symmetry, we can see that $\triangle BCH$ and $\triangle EFH$ are similar. Since their bases are in a $1 : 3$ ratio, so are their altitudes. This means that $3IH = HG,$ which implies that $IH = 1.$

Therefore, the area of $\begin{align*}\triangle BCH &= \dfrac{1}{2} \cdot 1 \cdot 1\\ &= \dfrac{1}{2}.\end{align*}$ This implies that the area of $\begin{align*}\triangle ECH &= 2 - \dfrac{1}{2} \\&= \dfrac{3}{2}.\end{align*}$ Since the figure is symmetric, the total area of the bat wings is $2 \cdot \dfrac{3}{2} = 3.$

Thus, **C** is the correct answer.

23.

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 $\angle CED?$

$90$

$105$

$120$

$135$

$150$

###### Solution(s):

We know that $AE = EB = AB$ since they are all radii of congruent circles, so they form an equilateral triangle, which means that $\angle AEB = 60^{\circ}.$

Also, since $\overline{DB}$ and $\overline{AC}$ are diameters, $\angle DEB = \angle AEC = 90^{\circ}.$ Therefore, $\begin{align*}\angle CED = \angle DEB &+ \angle AEC\\ &- \angle AEB,\end{align*}$ which equals $120^{\circ}.$

Thus, **C** is the correct answer.

24.

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 $P?$

$1$

$2$

$3$

$4$

$5$

###### Solution(s):

Since $QRS$ is divisible by $5,$ we know that $S = 5.$

Since $PQR$ is divisible by $4,$ $QR$ equals either $12, 24,$ or $32.$

This means that $RST$ will equal either $25T$ or $45T.$ Note that $RST = 25T$ would require $T$ to be $2$ or $5,$ each of which has solutions. Therefore the remaining digits when considering $RST = 45T,$ $453$ is the only number divisible by \(3.\

Therefore, $PQRST = 12435.$

Thus, **A** is the correct answer.

25.

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?

$4 \sqrt{3}$

$\dfrac{120}{17}$

$10$

$\dfrac{17\sqrt{2}}{2}$

$\dfrac{17\sqrt{3}}{2}$

###### Solution(s):

Let $O$ be the center of the circle, which is the midpoint of $\overline{AB}.$

We then get that $BC = 17$ via the Pythagorean theorem.

In addition, we can also calculate the area of $\triangle BOC$ as: $\begin{align*}\triangle BOC &= \dfrac{1}{2} \cdot 8 \cdot 15 \\&= 60.\end{align*}$ As the area of $\triangle BOC=60=\dfrac{1}{2} OE \cdot CB$ we can see that $OE = \dfrac{120}{17}.$

Thus, **B** is the correct answer.