## 2016 AMC 8 Solutions

Typeset by: LIVE, by Po-Shen Loh

https://live.poshenloh.com/past-contests/amc8/2016/solutions

Problems © Mathematical Association of America. Reproduced with permission.

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.

Problems: https://live.poshenloh.com/past-contests/amc8/2016