2016 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

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

$605$

$655$

$665$

$1005$

$1105$

Answer: C

Concepts:unit conversion

Difficulty rating: 370

Small Hint:

Convert the hours to minutes first.

Big Hint:

Add the extra $5$ minutes after multiplying $11$ by $60$ .

Video solution:

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Written solution:

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 $\triangle AMC?$

$12$

$15$

$18$

$20$

$24$

Answer: A

Concepts:triangle area midpoint

Difficulty rating: 450

Small Hint:

Since $M$ is the midpoint of $\overline{AD}$ , find $AM$ .

Big Hint:

Use $AM$ as the base and the rectangle’s vertical side length as the height.

Video solution:

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Written solution:

Since $M$ is the midpoint of $\overline{AD},$ the base $AM$ has length $4.$ The altitude from $C$ to $\overline{AM}$ has length $AB=6.$ 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?

$40$

$50$

$55$

$60$

$70$

Answer: A

Concepts:mean

Difficulty rating: 450

Small Hint:

First find the total score needed for an average of $70$ .

Big Hint:

Subtract the three known scores from the needed total.

Video solution:

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Written solution:

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?

$6$

$10$

$15$

$18$

$30$

Answer: B

Concepts:rate unit conversion

Difficulty rating: 720

Small Hint:

Compare the two paces in minutes per mile.

Big Hint:

Convert $3$ hours $30$ minutes to minutes before dividing by $15$ .

Video solution:

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Written solution:

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 $\frac{210}{15} = 14$ 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 $\frac{240}{10} = 24$ 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:

When $N$ is divided by $9,$ the remainder is $1.$
When $N$ is divided by $10,$ the remainder is $3.$

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

$0$

$2$

$4$

$5$

$7$

Answer: E

Concepts:modular arithmetic systematic listing

Difficulty rating: 940

Small Hint:

The remainder when divided by $10$ tells you the units digit of $N$ .

Big Hint:

List the two-digit numbers ending in $3$ and test the condition modulo $9$ .

Video solution:

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Written solution:

The second condition says that $N$ ends in $3$ , so the possibilities are $13,$ $23,$ $33,$ $43,$ $53,$ $63,$ $73,$ $83,$ $93.$ Of these, only $73$ leaves remainder $1$ when divided by $9$ , because its digit sum is $10$ . Finally, $73=6\cdot11+7$ , so the requested remainder 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?

$3$

$4$

$5$

$6$

$7$

Answer: B

Concepts:median (data)data and graph interpretation

Difficulty rating: 770

Small Hint:

With $19$ names, the median is the $10$ th value after sorting.

Big Hint:

Count cumulative frequencies from the shortest names upward.

Video solution:

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Written solution:

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?

$1^{2016}$

$2^{2017}$

$3^{2018}$

$4^{2019}$

$5^{2020}$

Answer: B

Concepts:perfect square exponent

Difficulty rating: 870

Small Hint:

Even exponents automatically make perfect squares.

Big Hint:

Also remember that a perfect-square base raised to any positive integer power stays a square.

Video solution:

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Written solution:

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 $\begin{gathered} 100 - 98 + 96 - 94 + 92 - 90 \\ {}+ \cdots + 8 - 6 + 4 - 2. \end{gathered}$

$20$

$40$

$50$

$80$

$100$

Answer: C

Concepts:pairing and grouping

Difficulty rating: 900

Small Hint:

Group the expression into pairs of consecutive terms.

Big Hint:

Count how many pairs of the form $n-(n-2)$ appear.

Video solution:

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Written solution:

We can group the sum as follows: $\begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2). \end{gathered}$ 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 $2016?$

$9$

$12$

$16$

$49$

$63$

Answer: B

Concepts:prime factorization

Difficulty rating: 960

Small Hint:

Factor $2016$ into primes.

Big Hint:

The word “distinct” means to count each prime divisor only once.

Video solution:

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Written solution:

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

$\dfrac{1}{10}$

$2$

$\dfrac{10}{3}$

$10$

$14$

Answer: D

Concepts:custom operation linear equation

Difficulty rating: 1030

Small Hint:

Start by simplifying the inner operation $5\ast x$ .

Big Hint:

Apply the definition $a\ast b=3a-b$ twice, from the inside out.

Video solution:

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Written solution:

First, $5\ast x=15-x$ . Applying the operation again gives $\begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9. \end{aligned}$ Thus $x-9=1$ , so $x=10$ .

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 $132.$

$5$

$7$

$9$

$11$

$12$

Answer: B

Concepts:digits place value

Difficulty rating: 1100

Small Hint:

Write the two-digit number as $10a+b$ .

Big Hint:

Adding the reversed number gives $11(a+b)$ .

Video solution:

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Written solution:

Let $a$ be the tens digit and $b$ the units digit. Adding the number and its reversal gives $\begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132, \end{aligned}$ so $a+b=12$ . The valid digit pairs are $(3,9),$ $(4,8),$ $(5,7),$ $(6,6),$ $(7,5),$ $(8,4),$ $(9,3),$ giving $7$ numbers.

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?

$\dfrac{1}{2}$

$\dfrac{9}{17}$

$\dfrac{7}{13}$

$\dfrac{2}{3}$

$\dfrac{14}{15}$

Answer: B

Concepts:fraction ratio and proportion

Difficulty rating: 1170

Small Hint:

Choose a convenient equal number of boys and girls.

Big Hint:

Using $12$ boys and $12$ girls makes both fractions whole numbers.

Video solution:

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Written solution:

Suppose there are $12$ girls and $12$ boys. Then $9$ girls and $8$ boys go on the field trip, so the fraction of field-trip students who are girls is $\dfrac{9}{9+8}=\dfrac9{17}.$

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 $0?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: D

Concepts:basic probability counting pairs

Difficulty rating: 1020

Small Hint:

A product is $0$ exactly when one of the selected numbers is $0$ .

Big Hint:

Count selections containing $0$ out of all ways to choose two different numbers.

Video solution:

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Written solution:

There are $\binom62=15$ ways to select two different numbers. A product is $0$ exactly when one selected number is $0$ , and there are $5$ choices for the other number. Thus the probability is $\dfrac5{15}=\dfrac13.$

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?

$525$

$560$

$595$

$665$

$735$

Answer: A

Concepts:rate

Difficulty rating: 1170

Small Hint:

Track gallons of gas rather than miles at first.

Big Hint:

After the $350$ -mile drive and the refill, compare the new amount of gas to half a tank.

Video solution:

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Written solution:

If Karl drove $350$ miles, then he used $\frac{350}{35}$ 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 traveled a total distance of $350+175=525$ miles.

Thus, A is the correct answer.

15.

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

$8$

$16$

$32$

$64$

$128$

Answer: C

Concepts:difference of squares power of 2

Difficulty rating: 1310

Small Hint:

Factor $13^4-11^4$ as a difference of squares.

Big Hint:

After factoring, count only the powers of $2$ in the product.

Video solution:

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Written solution:

We can factor this expression using difference of squares.

$\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435. \end{aligned}$

Because $435$ is odd, $32$ 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?

$1\dfrac{1}{4}$

$3\dfrac{1}{3}$

$4$

$5$

$25$

Answer: D

Concepts:relative speed distance rate and time ratio and proportion

Difficulty rating: 1240

Small Hint:

A $25\%$ faster pace means Annie runs $\frac{5}{4}$ as far as Bonnie in the same time.

Big Hint:

Annie first passes Bonnie when she has gained one full lap.

Video solution:

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Written solution:

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 finishes 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?

$30$

$7290$

$9000$

$9990$

$9999$

Answer: D

Concepts:multiplication principle complementary counting

Difficulty rating: 1020

Small Hint:

First count all four-digit passwords when leading zeroes are allowed.

Big Hint:

Then subtract the passwords whose first three digits are forced to be $9,$ $1,$ $1$ .

Video solution:

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Written solution:

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?

$36$

$42$

$43$

$60$

$72$

Answer: C

Concepts:basic counting

Difficulty rating: 1170

Small Hint:

Think about how many sprinters are eliminated in each race.

Big Hint:

To leave one champion from $216$ sprinters, all but one sprinter must be eliminated.

Video solution:

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Written solution:

Note that each race eliminates $5$ people. For there to be a winner, $215$ must be eliminated. Therefore, $\frac{215}{5} = 43$ 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 even integers?

$360$

$388$

$412$

$416$

$424$

Answer: E

Concepts:arithmetic sequence

Difficulty rating: 1100

Small Hint:

For an odd number of consecutive terms, the average is the middle term.

Big Hint:

There are $12$ even integers larger than the middle term.

Video solution:

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Written solution:

The average of these numbers is $\frac{10{,}000}{25} = 400.$ 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 $c?$

$20$

$30$

$60$

$120$

$180$

Answer: A

Concepts:least common multiple

Difficulty rating: 1390

Small Hint:

Since $b$ appears in both least common multiples, start with possible values of $b$ .

Big Hint:

Try to make $b$ absorb the common factor $3$ from $12$ and $15$ .

Video solution:

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Written solution:

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 we may take $a = 4$ and $c = 5.$ The least common multiple in this scenario is $20.$

Thus, A is the correct answer.

21.

A box 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?

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

Answer: B

Concepts:basic probability symmetry

Difficulty rating: 1490

Small Hint:

The stopping result is determined by which color appears last in the full ordering of the chips.

Big Hint:

Count equally likely positions for the two green chips among the five total chip positions.

Video solution:

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Written solution:

The $3$ reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are $10$ equally likely ways to choose the two positions of the green chips, and $4$ of them have a green chip in the last position.

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

$2$

$2 \dfrac{1}{2}$

$3$

$3 \dfrac{1}{2}$

$4$

Answer: C

Concepts:similarity triangle area

Difficulty rating: 1640

Small Hint:

Use the intersection point of the two diagonals that form the bat wings.

Big Hint:

Similar triangles give the height of the small triangle cut off at the top.

Video solution:

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Written solution:

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 $\triangle BCE$ is $\dfrac12\cdot1\cdot4=2.$ 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, $[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.$ It follows that $[\triangle ECH]=2-\dfrac12=\dfrac32.$ Since the figure is symmetric, the total area of the bat wings is $2\cdot\dfrac32=3.$

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 $\angle CED?$

$90$

$105$

$120$

$135$

$150$

Answer: C

Concepts:equilateral triangle inscribed angle angle chasing

Difficulty rating: 1510

Small Hint:

The centers and one intersection point form an equilateral triangle.

Big Hint:

Use the fact that angles subtending diameters are right angles.

Video solution:

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Written solution:

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

The segments $\overline{DB}$ and $\overline{AC}$ are diameters, so $\angle DEB=\angle AEC=90^\circ.$ Therefore, $\begin{aligned} \angle CED &= \angle DEB+\angle AEC\\ &\qquad-\angle AEB\\ &=90^\circ+90^\circ-60^\circ\\ &=120^\circ. \end{aligned}$

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 $P?$

$1$

$2$

$3$

$4$

$5$

Answer: A

Concepts:divisibility casework

Difficulty rating: 1580

Small Hint:

The divisibility by $5$ condition determines $S$ .

Big Hint:

Once $S$ is known, list the possible two-digit endings $QR$ that make $PQR$ divisible by $4$ .

Video solution:

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Written solution:

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

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

If $QR=12,$ then $R=2,$ so $RST=25T$ cannot be divisible by $3$ using the remaining digits. If $QR=32,$ then $R=2$ again, giving the same obstacle. Thus $QR=24,$ and then $RST=45T.$ Among the remaining digits, only $T=3$ makes $453$ divisible by $3$ .

Therefore, $PQRST = 12453.$

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?

$4 \sqrt{3}$

$\dfrac{120}{17}$

$10$

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

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

Answer: B

Concepts:triangle area Pythagorean Theorem

Difficulty rating: 1610

Small Hint:

Draw the radius to the point where the semicircle touches a side of the triangle.

Big Hint:

Compare the area of the right half of the triangle using two different bases.

Video solution:

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Written solution:

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.

The right half of the isosceles triangle has area $[\triangle BOC]=\dfrac12\cdot8\cdot15=60.$ The radius $OE$ is perpendicular to the tangent side $BC$ , so the same area is $\dfrac12\cdot OE\cdot17.$ Hence $OE=\dfrac{120}{17}.$

Thus, B is the correct answer.