Solution:

From $8\!:\!30$ am to $11\!:\!10$ am is $2$ hours $40$ minutes, or $160$ minutes, to complete one third of the job.

The whole job then takes $3 \cdot 160 = 480$ minutes, or $8$ hours. Adding $8$ hours to $8\!:\!30$ am gives $4\!:\!30$ pm.

Thus, D is the correct answer.

Solution:

Using a common denominator, $$\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3}{6} + \dfrac{4}{6} = \dfrac{7}{6}.$$

The reciprocal of $\dfrac{7}{6}$ is $\dfrac{6}{7}.$

Thus, A is the correct answer.

Solution:

Since $\tfrac{2}{3}$ of $10$ bananas is $\tfrac{20}{3}$ bananas, worth $8$ oranges, one banana is worth $$8 \div \dfrac{20}{3} = \dfrac{24}{20} = \dfrac{6}{5}$$ oranges.

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

Thus, C is the correct answer.

Solution:

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

Thus, B is the correct answer.

Solution:

Combining the fractions, $$\dfrac{2x}{3} - \dfrac{x}{6} = \dfrac{4x - x}{6} = \dfrac{x}{2}.$$

This is an integer exactly when $x$ is even. The example $x = 4$ is even but not a multiple of $3,$ which rules out every other statement.

Thus, B is the correct answer.

Solution:

Let $x$ be the sticker price in dollars. Store A charges $0.85x - 90$ dollars, and store B charges $0.75x$ dollars.

Since store A is $15$ dollars cheaper, $$0.85x - 90 = 0.75x - 15,$$ so $0.10x = 75$ and $x = 750.$

Thus, A is the correct answer.

Solution:

Rowing $1$ mile at $4$ miles per hour takes $\tfrac{1}{4}$ hour, or $15$ minutes. In that time $15 \cdot 10 = 150$ gallons of water enter the boat.

Since at most $30$ gallons may remain, LeRoy must bail $150 - 30 = 120$ gallons in $15$ minutes, a rate of $$\dfrac{120}{15} = 8$$ gallons per minute.

Thus, D is the correct answer.

Solution:

The cube with volume $1$ has side $1$ and surface area $6.$ The larger cube has surface area $12,$ so if its side is $s,$ then $6s^2 = 12,$ giving $s = \sqrt{2}.$

Its volume is $$(\sqrt{2})^3 = 2\sqrt{2}.$$

Thus, C is the correct answer.

Solution:

Since the sides and diagonal are in ratio $3:4:5,$ the height is $\tfrac{3}{5} \cdot 27 = 16.2$ inches and the width is $\tfrac{4}{5} \cdot 27 = 21.6$ inches.

The movie has aspect ratio $2:1,$ so its height is $\tfrac{21.6}{2} = 10.8$ inches.

Each darkened strip therefore has height $$\dfrac{16.2 - 10.8}{2} = 2.7$$ inches.

Thus, D is the correct answer.

Solution:

In one hour Doug paints $\tfrac{1}{5}$ of the room and Dave paints $\tfrac{1}{7},$ so together they paint $\tfrac{1}{5} + \tfrac{1}{7}$ of the room per hour.

Of the total time $t,$ one hour is spent at lunch, so they work for $t - 1$ hours. The fraction painted must equal $1,$ giving $$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.$$

Thus, D is the correct answer.

Solution:

The six faces of each cube sum to $1 + 2 + 4 + 8 + 16 + 32 = 63.$ From the pattern, the pairs of opposite faces are $1$ & $32,$ $2$ & $16,$ and $4$ & $8.$

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair $4 + 8 = 12$ is best. The top cube hides only its bottom face, so hide the $1.$

The greatest sum is $$3 \cdot 63 - 2 \cdot 12 - 1 = 189 - 24 - 1 = 164.$$

Thus, C is the correct answer.

Solution:

The value $f(x + 1)$ is defined when $0 \le x + 1 \le 2,$ that is, $-1 \le x \le 1,$ so the domain of $g$ is $[-1, 1].$

As $f(x + 1)$ ranges over $[0, 1],$ the value $1 - f(x + 1)$ ranges over $[0, 1]$ as well, so the range of $g$ is $[0, 1].$

Thus, B is the correct answer.

Solution:

Let $r$ and $R$ be the radii of the smaller and larger circles, and let $E$ be the center of the smaller circle. By symmetry $E$ lies on the bisector of $\angle AOB,$ so $OE$ makes a $30^\circ$ angle with $OA.$

Dropping the radius $ED$ perpendicular to $OA$ gives a $30$-$60$-$90$ triangle with $OE = 2 \cdot ED = 2r.$ Since the circles are internally tangent, $OE = R - r.$

Then $R - r = 2r,$ so $R = 3r$ and $\tfrac{r}{R} = \tfrac{1}{3}.$ The ratio of areas is $$\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}.$$

Thus, B is the correct answer.

Solution:

The region is a rhombus centered at $\left(6, -\tfrac{7}{2}\right).$ Setting $2y + 7 = 0$ gives $|3x - 18| \le 3,$ so $x \in [5, 7],$ a horizontal diagonal of length $2.$

Setting $3x - 18 = 0$ gives $|2y + 7| \le 3,$ so $y \in [-5, -2],$ a vertical diagonal of length $3.$

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

Thus, A is the correct answer.

Solution:

The units digit of $2008^2$ is $4.$ Since $2008$ is a multiple of $4,$ the units digit of $2^{2008}$ is $6.$ Thus $k$ has units digit $0,$ and so does $k^2.$

Both $2008^2$ and $2^{2008}$ are multiples of $4,$ so $k$ is a multiple of $4.$ Therefore the units digit of $2^k$ is $6.$

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

Thus, D is the correct answer.

Solution:

The three terms are $3\log a + 7\log b,$ $5\log a + 12\log b,$ and $8\log a + 15\log b.$ Setting the two consecutive differences equal, $$2\log a + 5\log b = 3\log a + 3\log b,$$ so $\log a = 2\log b.$

The first term is then $(3 \cdot 2 + 7)\log b = 13\log b,$ and the common difference is $(2 \cdot 2 + 5)\log b = 9\log b.$

The $12$th term is $$(13 + 11 \cdot 9)\log b = 112\log b = \log(b^{112}),$$ so $n = 112.$

Thus, D is the correct answer.

Solution:

If $a_1$ is even, then $a_2 = a_1/2 \lt a_1,$ so the condition fails.

If $a_1 \equiv 1 \pmod 4,$ then $a_2 = 3a_1 + 1$ is a multiple of $4,$ so $a_3 = (3a_1 + 1)/2$ and $a_4 = (3a_1 + 1)/4 \le a_1,$ and again the condition fails.

If $a_1 \equiv 3 \pmod 4,$ then $a_2$ is even but not a multiple of $4,$ so $a_3 = (3a_1 + 1)/2 \gt a_1,$ and $a_3$ is odd, giving $a_4 = 3a_3 + 1 \gt a_3 \gt a_1.$ The condition holds.

Exactly $\tfrac{2008}{4} = 502$ values of $a_1 \le 2008$ satisfy $a_1 \equiv 3 \pmod 4.$

Thus, D is the correct answer.

Solution:

Let $A = (a, 0, 0),$ $B = (0, b, 0),$ $C = (0, 0, c).$ Assigning the sides, $$a^2 + b^2 = 25, \quad b^2 + c^2 = 36, \quad a^2 + c^2 = 49.$$

Adding gives $a^2 + b^2 + c^2 = 55,$ so $a^2 = 19,$ $b^2 = 6,$ and $c^2 = 30.$

The volume is $$\dfrac{1}{6}abc = \dfrac{1}{6}\sqrt{19 \cdot 6 \cdot 30} = \dfrac{1}{6}\sqrt{3420} = \sqrt{95}.$$

Thus, C is the correct answer.

Solution:

Each term is $x^{a + b + c}$ with $0 \le a \le 27$ and $0 \le b, c \le 14.$ To get $x^{28}$ we need $a = 28 - b - c.$

There are $(14 + 1)^2 = 225$ choices for $(b, c).$ For every choice except $(b, c) = (0, 0),$ the required $a = 28 - b - c$ lies in $[0, 27],$ giving a valid term.

The coefficient of $x^{28}$ is therefore $225 - 1 = 224.$

Thus, C is the correct answer.

Solution:

By the Angle Bisector Theorem, $AD:DB = CA:CB = 3:4,$ so $AD = \tfrac{15}{7}$ and $BD = \tfrac{20}{7}.$ The areas of $\triangle ADC$ and $\triangle BCD$ share base $CD,$ so they are in ratio $3:4,$ namely $\tfrac{18}{7}$ and $\tfrac{24}{7}.$

Splitting $\triangle ABC$ along $CD,$ which meets each leg at $45^\circ,$ gives $$\dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6,$$ so $CD = \tfrac{12\sqrt{2}}{7}.$

Using $r = \text{area}/s$ with $s$ the semiperimeter, $$\dfrac{r_a}{r_b} = \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} = \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}},$$ where the messy $CD$ terms cancel after factoring.

Rationalizing, $\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7},$ so $$\dfrac{r_a}{r_b} = \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} = \dfrac{3}{28}(10 - \sqrt{2}).$$

Thus, E is the correct answer.

Solution:

Call a permutation balanced if $a_1 + a_2 = a_4 + a_5.$ Reversing the entries swaps the two strict cases, so heavy-tailed and heavy-headed permutations are equally numerous.

The total $1 + 2 + 3 + 4 + 5 = 15$ is odd, so in a balanced permutation $a_3$ must be odd, one of $1, 3, 5.$ For each choice, the remaining four numbers split uniquely into two equal-sum pairs.

Any of the four can be $a_1$ (fixing $a_2$), and either remaining number can be $a_4$ (fixing $a_5$), giving $3 \cdot 4 \cdot 2 = 24$ balanced permutations.

The other $120 - 24 = 96$ permutations split evenly, so there are $\tfrac{96}{2} = 48$ heavy-tailed permutations.

Thus, D is the correct answer.

Solution:

Take one mat with outer corners $P$ and $Q,$ and let $R$ be the point of the table's edge diametrically opposite $P.$ Then $PR = 8$ is a diameter, so $\triangle PQR$ has a right angle at $Q,$ with $PQ = x.$

Along $QR,$ the inner corners of neighboring mats meet in an isosceles triangle with two sides of length $x$ and vertex angle $120^\circ,$ whose base is $\sqrt{3}\,x.$ Hence $QR = \sqrt{3}\,x + 2.$

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

Taking the positive root, $$x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$$

Thus, C is the correct answer.

Solution:

Adding $1 + i$ to both sides, the left side becomes $$z^4 + 4z^3 i - 6z^2 - 4zi + 1 = (z + i)^4,$$ so $(z + i)^4 = 1 + i.$

The four solutions for $w = z + i$ are equally spaced on a circle of radius $|1 + i|^{1/4} = (2^{1/2})^{1/4} = 2^{1/8},$ and they form a square. Subtracting $i$ merely translates it.

A square inscribed in a circle of radius $2^{1/8}$ has diagonal $2 \cdot 2^{1/8} = 2^{9/8},$ so its side is $\tfrac{2^{9/8}}{\sqrt{2}} = 2^{5/8}.$

The area is $$\left(2^{5/8}\right)^2 = 2^{5/4}.$$

Thus, D is the correct answer.

Solution:

Place $C = (0, 0),$ $B = (2, 2\sqrt{3})$ so that $\angle C = 60^\circ$ and $BC = 4,$ and let $A = (x, 0)$ with $x \gt 0.$ Then $D = (1, \sqrt{3})$ is the midpoint of $BC.$

The lines $AD$ and $AB$ have slopes $\tfrac{\sqrt{3}}{1 - x}$ and $\tfrac{2\sqrt{3}}{2 - x}.$ Using the tangent-difference formula and simplifying, $$\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$$

Setting the derivative to zero gives $x^2 = 8,$ so $x = 2\sqrt{2}.$ Substituting, $$\tan(\angle BAD) = \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} = \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} = \dfrac{\sqrt{3}}{4\sqrt{2} - 3}.$$

Thus, D is the correct answer.

Solution:

Let $z_n = a_n + b_n i.$ Then $$z_{n+1} = (\sqrt{3}\,a_n - b_n) + (\sqrt{3}\,b_n + a_n)i = (a_n + b_n i)(\sqrt{3} + i),$$ so $z_{n+1} = z_n(\sqrt{3} + i)$ and $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ De Moivre's theorem gives $(\sqrt{3} + i)^{99} = 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ As $2970^\circ$ is coterminal with $90^\circ,$ this equals $2^{99} i.$

Thus $2 + 4i = z_1 \cdot 2^{99} i,$ so $$z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$$

Then $a_1 = \tfrac{4}{2^{99}}$ and $b_1 = -\tfrac{2}{2^{99}},$ so $$a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$$

Thus, D is the correct answer.