2008 AMC 10A — Problems & Online Timed Exam

Timed

1:15:00

1.

A bakery owner turns on his doughnut machine at $8{:}30$ am. At $11{:}10$ am the machine has completed one third of the day’s job. At what time will the doughnut machine complete the job?

$1{:}50$ pm

$3{:}00$ pm

$3{:}30$ pm

$4{:}30$ pm

$5{:}50$ pm

Answer: D

Concepts:date and time ratio and proportion

Difficulty rating: 770

Small Hint:

Find how long one third of the job takes, in minutes

Big Hint:

The whole job takes three times as long, measured from the $8{:}30$ am start

Solution:

From $8{:}30$ am to $11{:}10$ am is $2$ hours and $40$ minutes, or $160$ minutes, to finish one third of the job.

The entire job therefore takes $3 \cdot 160 = 480$ minutes, or $8$ hours.

Eight hours after $8{:}30$ am is $4{:}30$ pm.

Thus, the correct answer is D.

2.

A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $2:1.$ The ratio of the rectangle’s length to its width is $2:1.$ What percent of the rectangle’s area is inside the square?

$12.5$

$25$

$50$

$75$

$87.5$

Answer: A

Concepts:area ratio percentage

Difficulty rating: 840

Small Hint:

Let the square have side $s$ and express the rectangle’s dimensions in terms of $s$

Big Hint:

The rectangle is $2s$ wide and $4s$ long; compare its area to $s^2$

Solution:

Let the side of the square be $s,$ so its area is $s^2.$

The width of the rectangle is $2s,$ and its length is $2 \cdot 2s = 4s,$ giving an area of $8s^2.$

The fraction inside the square is $\dfrac{s^2}{8s^2} = \dfrac{1}{8} = 12.5\%.$

Thus, the correct answer is A.

3.

For the positive integer $n,$ let $\langle n \rangle$ denote the sum of all the positive divisors of $n$ with the exception of $n$ itself. For example, $\langle 4 \rangle = 1 + 2 = 3$ and $\langle 12 \rangle = 1 + 2 + 3 + 4 + 6 = 16.$ What is $\langle\langle\langle 6 \rangle\rangle\rangle?$

$6$

$12$

$24$

$32$

$36$

Answer: A

Concepts:sum of factors function

Difficulty rating: 940

Small Hint:

$\langle 6 \rangle$ is the sum of the divisors of $6$ other than $6$

Big Hint:

Compute $\langle 6 \rangle$ first; the result feeds back into the same operation

Solution:

The positive divisors of $6$ other than $6$ are $1, 2,$ and $3,$ so $\langle 6 \rangle = 1 + 2 + 3 = 6.$

Since applying the operation to $6$ again returns $6,$ we get $\langle\langle\langle 6 \rangle\rangle\rangle = 6.$

(A number equal to the sum of its proper divisors is called a perfect number, and $6$ is the smallest.)

Thus, the correct answer is A.

4.

Suppose that $\dfrac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $\dfrac{1}{2}$ of $5$ bananas?

$2$

$\dfrac{5}{2}$

$3$

$\dfrac{7}{2}$

$4$

Answer: C

Concepts:ratio and proportion rate

Difficulty rating: 1040

Small Hint:

Find the worth of a single banana in oranges

Big Hint:

$\dfrac{2}{3}$ of $10$ is $\dfrac{20}{3}$ bananas, and these equal $8$ oranges

Solution:

Since $\dfrac{2}{3}$ of $10$ bananas is $\dfrac{20}{3}$ bananas worth $8$ oranges, one banana is worth $8 \div \dfrac{20}{3} = \dfrac{6}{5}$ oranges.

Now $\dfrac{1}{2}$ of $5$ bananas is $\dfrac{5}{2}$ bananas, worth $\dfrac{5}{2} \cdot \dfrac{6}{5} = 3$ oranges.

Thus, the correct answer is C.

5.

Which of the following is equal to the product

$\begin{aligned} &\dfrac{8}{4} \cdot \dfrac{12}{8} \cdot \dfrac{16}{12} \cdots \dfrac{4n+4}{4n} \\ &\quad \cdots \dfrac{2008}{2004}? \end{aligned}$

$251$

$502$

$1004$

$2008$

$4016$

Answer: B

Concepts:telescoping fraction

Difficulty rating: 1050

Small Hint:

Write out the first few factors and look for cancellation

Big Hint:

Each numerator cancels the next denominator, leaving $\dfrac{2008}{4}$

Solution:

Every denominator except the first cancels with the numerator of the previous fraction, so the whole product telescopes to $\dfrac{2008}{4} = 502.$

Thus, the correct answer is B.

6.

A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of $3$ kilometers per hour, bikes at a rate of $20$ kilometers per hour, and runs at a rate of $10$ kilometers per hour. Which of the following is closest to the triathlete’s average speed, in kilometers per hour, for the entire race?

$3$

$4$

$5$

$6$

$7$

Answer: D

Concepts:distance rate and time harmonic mean

Difficulty rating: 1100

Small Hint:

Let each segment have length $x$ and add the three travel times

Big Hint:

Average speed is the total distance $3x$ divided by the total time

Solution:

Let each segment have length $x.$ The total time is $\dfrac{x}{3} + \dfrac{x}{20} + \dfrac{x}{10} = \dfrac{29}{60}x$ hours for the distance $3x.$

The average speed is $\dfrac{3x}{\frac{29}{60}x} = \dfrac{180}{29} \approx 6.2,$ which is closest to $6.$

Thus, the correct answer is D.

7.

The fraction

$\dfrac{\left(3^{2008}\right)^2 - \left(3^{2006}\right)^2}{\left(3^{2007}\right)^2 - \left(3^{2005}\right)^2}$

simplifies to which of the following?

$1$

$\dfrac{9}{4}$

$3$

$\dfrac{9}{2}$

$9$

Answer: E

Concepts:exponent factoring

Difficulty rating: 1210

Small Hint:

$\left(3^{2008}\right)^2 = 9^{2008}$

Big Hint:

Factor $9^{2005}$ out of both the numerator and the denominator

Solution:

Since $\left(3^{k}\right)^2 = 9^{k},$ the fraction is $\dfrac{9^{2008} - 9^{2006}}{9^{2007} - 9^{2005}}.$

Factoring $9^{2005}$ from each part gives $\dfrac{9^{2005}\left(9^3 - 9\right)}{9^{2005}\left(9^2 - 1\right)} = \dfrac{9\left(9^2 - 1\right)}{9^2 - 1} = 9.$

Thus, the correct answer is E.

8.

Heather compares the price of a new computer at two different stores. Store A offers $15\%$ off the sticker price followed by a $\$90$ rebate, and store B offers $25\%$ off the same sticker price with no rebate. Heather saves $\$15$ by buying the computer at store A instead of store B. What is the sticker price of the computer, in dollars?

$750$

$900$

$1000$

$1050$

$1500$

Answer: A

Concepts:percentage linear equation

Difficulty rating: 1190

Small Hint:

Let $x$ be the sticker price and write each store’s final price

Big Hint:

Store A costs $0.85x - 90$ and store B costs $0.75x,$ with A cheaper by $\$15$

Solution:

Let $x$ be the sticker price. Heather pays $0.85x - 90$ at store A and $0.75x$ at store B.

Since store A is $\$15$ cheaper, $0.85x - 90 = 0.75x - 15,$ which gives $0.1x = 75,$ so $x = 750.$

Thus, the correct answer is A.

9.

Suppose that

$\dfrac{2x}{3} - \dfrac{x}{6}$

is an integer. Which of the following statements must be true about $x?$

It is negative.

It is even, but not necessarily a multiple of $3.$

It is a multiple of $3,$ but not necessarily even.

It is a multiple of $6,$ but not necessarily a multiple of $12.$

It is a multiple of $12.$

Answer: B

Concepts:fraction parity divisibility

Difficulty rating: 1170

Small Hint:

Combine $\dfrac{2x}{3} - \dfrac{x}{6}$ over a common denominator

Big Hint:

The expression simplifies to $\dfrac{x}{2};$ decide what that being an integer forces about $x$

Solution:

Combining over a common denominator, $\dfrac{2x}{3} - \dfrac{x}{6} = \dfrac{4x - x}{6} = \dfrac{x}{2}.$

For $\dfrac{x}{2}$ to be an integer, $x$ must be even.

The example $x = 4$ shows that $x$ need not be a multiple of $3$ and rules out the other statements.

Thus, the correct answer is B.

10.

Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3.$ What is the area of $S_3?$

$\dfrac{1}{2}$

$1$

$2$

$3$

$4$

Answer: E

Concepts:square (geometry)Pythagorean Theorem area ratio

Difficulty rating: 1240

Small Hint:

The side of $S_1$ is $4;$ use the midpoints to find the side of $S_2$

Big Hint:

Connecting midpoints of a square gives a new square with half the area

Solution:

The side of $S_1$ is $4.$ By the Pythagorean theorem, the side of $S_2$ is $\sqrt{2^2 + 2^2} = 2\sqrt{2},$ so its area is $8.$

By the same reasoning, $S_3$ has half the area of $S_2,$ namely $4.$

Thus, the correct answer is E.

11.

While Steve and LeRoy are fishing $1$ mile from shore, their boat springs a leak, and water comes in at a constant rate of $10$ gallons per minute. The boat will sink if it takes in more than $30$ gallons of water. Steve starts rowing toward the shore at a constant rate of $4$ miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?

$2$

$4$

$6$

$8$

$10$

Answer: D

Concepts:rate unit conversion

Difficulty rating: 1280

Small Hint:

Find how many minutes Steve needs to row one mile

Big Hint:

In that time $150$ gallons enter, but the boat can hold only $30$

Solution:

At $4$ miles per hour, Steve rows $1$ mile in $15$ minutes. During that time $15 \cdot 10 = 150$ gallons enter.

To stay under $30$ gallons, LeRoy must bail $150 - 30 = 120$ gallons in $15$ minutes, or $\dfrac{120}{15} = 8$ gallons per minute.

Thus, the correct answer is D.

12.

In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?

$2.85r$

$3r$

$3.4r$

$3.85r$

$4.25r$

Answer: C

Concepts:percentage algebraic manipulation

Difficulty rating: 1260

Small Hint:

Write the blue and green counts in terms of $r$

Big Hint:

From $r = 1.25b$ we get $b = 0.8r,$ and $g = 1.6r$

Solution:

Since $r = 1.25b,$ the number of blue marbles is $b = \dfrac{r}{1.25} = 0.8r.$

The number of green marbles is $g = 1.6r.$

The total is $r + 0.8r + 1.6r = 3.4r.$

Thus, the correct answer is C.

13.

Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $t?$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t + 1) = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t + 1 = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1$

$(5 + 7)t = 1$

Answer: D

Concepts:rate linear equation

Difficulty rating: 1280

Small Hint:

Together they paint $\dfrac{1}{5} + \dfrac{1}{7}$ of the room each working hour

Big Hint:

They actually work for $t - 1$ hours because of the one-hour lunch break

Solution:

Working together, Doug and Dave paint $\dfrac{1}{5} + \dfrac{1}{7}$ of the room per hour.

Because they break for one hour, they work for only $t - 1$ hours, and this must complete the whole room:

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.$

Thus, the correct answer is D.

14.

Older television screens have an aspect ratio of $4:3.$ That is, the ratio of the width to the height is $4:3.$ The aspect ratio of many movies is not $4:3,$ so they are sometimes shown on a television screen by “letterboxing” — darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $2:1$ and is shown on an older television screen with a $27$ -inch diagonal. What is the height, in inches, of each darkened strip?

$2$

$2.25$

$2.5$

$2.7$

$3$

Answer: D

Concepts:ratio and proportion Pythagorean Theorem rectangle

Difficulty rating: 1410

Small Hint:

The screen’s sides satisfy $h : w : 27 = 3 : 4 : 5$

Big Hint:

The lit region has aspect ratio $2:1,$ so its height is half the screen width

Solution:

Since the screen is $4:3$ with a $27$ -inch diagonal, $h : w : 27 = 3 : 4 : 5,$ giving height $h = \dfrac{3}{5}\cdot 27 = 16.2$ and width $w = \dfrac{4}{5}\cdot 27 = 21.6.$

The lit $2:1$ region has the full width $21.6$ and height $\dfrac{21.6}{2} = 10.8.$

The two strips share the remaining height, so each has height $\dfrac{16.2 - 10.8}{2} = 2.7.$

Thus, the correct answer is D.

15.

Yesterday Han drove $1$ hour longer than Ian at an average speed $5$ miles per hour faster than Ian. Jan drove $2$ hours longer than Ian at an average speed $10$ miles per hour faster than Ian. Han drove $70$ miles more than Ian. How many more miles did Jan drive than Ian?

$120$

$130$

$140$

$150$

$160$

Answer: D

Concepts:distance rate and time system of equations algebraic manipulation

Difficulty rating: 1440

Small Hint:

Let Ian drive $t$ hours at $r$ miles per hour

Big Hint:

Expanding $(r + 5)(t + 1) - rt = 70$ gives $5t + r = 65$

Solution:

Let Ian drive $t$ hours at rate $r,$ covering $rt$ miles.

Han drove $(r + 5)(t + 1) - rt$ $= 5t + r + 5 = 70,$ so $5t + r = 65.$

Jan drove $(r + 10)(t + 2) - rt$ $= 10t + 2r + 20$ $= 2(5t + r) + 20$ $= 2 \cdot 65 + 20$ $= 150$ miles more than Ian.

Thus, the correct answer is D.

16.

Points $A$ and $B$ lie on a circle centered at $O,$ and $\angle AOB = 60^\circ.$ A second circle is internally tangent to the first and tangent to both $OA$ and $OB.$ What is the ratio of the area of the smaller circle to that of the larger circle?

$\dfrac{1}{16}$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Answer: B

Concepts:tangent circles special right triangle area ratio

Difficulty rating: 1580

Small Hint:

The small circle’s center lies on the bisector of $\angle AOB$

Big Hint:

The center is $2r$ from $O$ by a $30$ - $60$ - $90$ triangle, and also $R - r$ from $O$

Solution:

Let the radii be $r$ and $R.$ The small circle’s center $E$ lies on the bisector of $\angle AOB,$ so $OE$ makes a $30^\circ$ angle with $OA.$

The perpendicular from $E$ to $OA$ has length $r,$ and in the resulting $30$ - $60$ - $90$ triangle $OE = 2r.$

Since $OE = R - r,$ we get $2r = R - r,$ so $R = 3r$ and the area ratio is $\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}.$

Thus, the correct answer is B.

17.

An equilateral triangle has side length $6.$ What is the area of the region containing all points that are outside the triangle and not more than $3$ units from a point of the triangle?

$36 + 24\sqrt{3}$

$54 + 9\pi$

$54 + 18\sqrt{3} + 6\pi$

$\left(2\sqrt{3} + 3\right)^2 \pi$

$9\left(\sqrt{3} + 1\right)^2 \pi$

Answer: B

Concepts:area decomposition sector equilateral triangle

Difficulty rating: 1680

Small Hint:

The region is three rectangles along the sides plus three sectors at the corners

Big Hint:

The three corner sectors are each $120^\circ,$ and together they make one full circle of radius $3$

Solution:

Along each of the three sides is a $6 \times 3$ rectangle, contributing $3 \cdot 6 \cdot 3 = 54.$

At each vertex is a $120^\circ$ sector of radius $3;$ the three together form a full circle of area $\pi \cdot 3^2 = 9\pi.$

The total area is $54 + 9\pi.$

Thus, the correct answer is B.

18.

A right triangle has perimeter $32$ and area $20.$ What is the length of its hypotenuse?

$\dfrac{57}{4}$

$\dfrac{59}{4}$

$\dfrac{61}{4}$

$\dfrac{63}{4}$

$\dfrac{65}{4}$

Answer: B

Concepts:Pythagorean Theorem system of equations algebraic manipulation

Difficulty rating: 1580

Small Hint:

Let the legs be $y, z$ and the hypotenuse $x,$ and record perimeter, area, and Pythagorean relations

Big Hint:

Square $y + z = 32 - x$ and use $y^2 + z^2 = x^2$ with $yz = 40$

Solution:

Let the legs be $y, z$ and the hypotenuse $x.$ Then $y^2 + z^2 = x^2,$ $y + z = 32 - x,$ and $yz = 40.$

Squaring the second equation, $\begin{aligned} &(32 - x)^2 \\ &\quad = y^2 + z^2 + 2yz = x^2 + 80. \end{aligned}$

This gives $1024 - 64x = 80,$ so $x = \dfrac{59}{4}.$

Thus, the correct answer is B.

19.

Rectangle $PQRS$ lies in a plane with $PQ = RS = 2$ and $QR = SP = 6.$ The rectangle is rotated $90^\circ$ clockwise about $R,$ then rotated $90^\circ$ clockwise about the point that $S$ moved to after the first rotation. What is the length of the path traveled by point $P?$

$\left(2\sqrt{3} + \sqrt{5}\right)\pi$

$6\pi$

$\left(3 + \sqrt{10}\right)\pi$

$\left(\sqrt{3} + 2\sqrt{5}\right)\pi$

$2\sqrt{10}\pi$

Answer: C

Concepts:transformation arc Pythagorean Theorem

Difficulty rating: 1840

Small Hint:

Point $P$ traces one quarter-circle arc for each rotation

Big Hint:

The first radius is $PR = \sqrt{40};$ the second radius is the side of length $6$

Solution:

In the first rotation, $P$ moves on a quarter circle about $R$ with radius $PR = \sqrt{2^2 + 6^2} = 2\sqrt{10}.$ The arc length is $\dfrac{1}{4}\left(2\pi \cdot 2\sqrt{10}\right) = \sqrt{10}\,\pi.$

In the second rotation, $P$ moves on a quarter circle about the new position of $S$ with radius $6.$ The arc length is $\dfrac{1}{4}(2\pi \cdot 6) = 3\pi.$

The total path length is $\left(3 + \sqrt{10}\right)\pi.$

Thus, the correct answer is C.

20.

Trapezoid $ABCD$ has bases $AB$ and $CD$ and diagonals intersecting at $K.$ Suppose that $AB = 9,$ $DC = 12,$ and the area of $\triangle AKD$ is $24.$ What is the area of trapezoid $ABCD?$

$92$

$94$

$96$

$98$

$100$

Answer: D

Concepts:trapezoid similarity area ratio

Difficulty rating: 1710

Small Hint:

Triangles $AKB$ and $CKD$ are similar with ratio $9 : 12$

Big Hint:

$\triangle AKD$ and $\triangle BKC$ have equal areas; scale the others from $[AKD] = 24$

Solution:

Triangles $AKB$ and $CKD$ are similar with ratio $\dfrac{9}{12} = \dfrac{3}{4}.$

Since $\triangle AKD$ and $\triangle KCD$ have bases $AK$ and $KC$ on the same line and share the same altitude from $D,$ $\dfrac{[KCD]}{[AKD]} = \dfrac{KC}{AK} = \dfrac{4}{3},$ so $[KCD] = 32.$ Similarly $[AKB] = 18.$

Also $[BKC] = [AKD] = 24.$ The total is $24 + 32 + 18 + 24 = 98.$

Thus, the correct answer is D.

21.

A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C,$ as shown. What is the area of quadrilateral $ABCD?$

$\dfrac{\sqrt{6}}{2}$

$\dfrac{5}{4}$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{3}$

Answer: A

Concepts:cube geometry rhombus diagonal

Difficulty rating: 1770

Small Hint:

All four sides of $ABCD$ are equal, so it is a rhombus

Big Hint:

Its diagonals are a space diagonal $AC = \sqrt{3}$ and a face diagonal $BD = \sqrt{2}$

Solution:

Each side of $ABCD$ joins a vertex of the cube to the midpoint of an edge, so all four sides are equal and $ABCD$ is a rhombus.

Its diagonals are the space diagonal $AC = \sqrt{3}$ and the face diagonal $BD = \sqrt{2}.$

The area of a rhombus is half the product of its diagonals: $\dfrac{1}{2}\cdot\sqrt{3}\cdot\sqrt{2} = \dfrac{\sqrt{6}}{2}.$

Thus, the correct answer is A.

22.

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be $6.$ To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts $1.$ If it comes up tails, he takes half of the previous term and subtracts $1.$ What is the probability that the fourth term in Jacob’s sequence is an integer?

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

Answer: D

Concepts:basic probability tree diagram parity

Difficulty rating: 1880

Small Hint:

Build a tree of the possible sequences; each flip splits into two branches

Big Hint:

Track which fourth terms are integers, remembering that halving an odd number breaks integrality

Solution:

Starting from $6,$ the second terms are $11$ (heads) and $2$ (tails).

Continuing the tree, the eight equally likely fourth terms are $41, 9.5, 8, 1.25, 5, 0.5, -1, -1.$

Of these, $41, 8, 5, -1, -1$ are integers, so the probability is $\dfrac{5}{8}.$

Thus, the correct answer is D.

23.

Two subsets of the set $S = \{a, b, c, d, e\}$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?

$20$

$40$

$60$

$160$

$320$

Answer: B

Concepts:subsets combinations multiplication principle

Difficulty rating: 1770

Small Hint:

First choose the two elements that belong to both subsets

Big Hint:

Each of the remaining elements goes to exactly one subset; then correct for the unordered pair

Solution:

Choose the two common elements in $\binom{5}{2} = 10$ ways.

Each of the remaining $3$ elements must lie in exactly one subset, giving $2^3 = 8$ assignments, for $80$ ordered pairs.

Since the order of the two subsets does not matter, divide by $2$ to get $\dfrac{80}{2} = 40.$

Thus, the correct answer is B.

24.

Let $k = 2008^2 + 2^{2008}.$ What is the units digit of $k^2 + 2^k?$

$0$

$2$

$4$

$6$

$8$

Answer: D

Concepts:units digit modular arithmetic pattern recognition

Difficulty rating: 1910

Small Hint:

Find the units digit of $k$ first

Big Hint:

$2^{2008}$ ends in $6$ and $2008^2$ ends in $4;$ then determine $k \bmod 4$ to handle $2^k$

Solution:

The units digit of $2^n$ cycles $2, 4, 8, 6,$ so $2^{2008}$ ends in $6.$ Also $2008^2$ ends in $4.$

Thus $k$ ends in $0,$ so $k^2$ ends in $0.$

Both $2008^2$ and $2^{2008}$ are multiples of $4,$ so $k \equiv 0 \pmod 4,$ which makes $2^k$ end in $6.$

The units digit of $k^2 + 2^k$ is $0 + 6 = 6.$

Thus, the correct answer is D.

25.

A round table has radius $4.$ Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x.$ Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x?$

$2\sqrt{5} - \sqrt{3}$

$3$

$\dfrac{3\sqrt{7} - \sqrt{3}}{2}$

$2\sqrt{3}$

$\dfrac{5 + 2\sqrt{3}}{2}$

Answer: C

Concepts:circle inscribed angle Pythagorean Theorem quadratic

Difficulty rating: 2150

Small Hint:

Take one mat with outer corners $P, Q,$ and let $R$ be the point of the circle opposite $P,$ so $\triangle PQR$ is right-angled with hypotenuse $8$

Big Hint:

The inner corners meet in $120^\circ$ isosceles triangles, giving $QR = \sqrt{3}\,x + 2$

Solution:

Pick a mat with outer corners $P$ and $Q,$ and let $R$ be the point on the circle diametrically opposite $P.$ Then $\triangle PQR$ is right-angled at $Q$ with hypotenuse $PR = 8.$

The inner corners of adjacent mats meet in isosceles triangles with vertex angle $120^\circ$ and sides $x,$ whose base is $\sqrt{3}\,x.$ Together with the two mat widths, $QR = \sqrt{3}\,x + 2.$

By the Pythagorean theorem, $\left(\sqrt{3}\,x + 2\right)^2 + x^2 = 64,$ which simplifies to $x^2 + \sqrt{3}\,x - 15 = 0.$

Taking the positive root, $x = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

Thus, the correct answer is C.