Solution:

Mike's $\$2$ tip is $10\%$ of his bill, so his bill is $2 \times 10 = 20$ dollars. Joe's $\$2$ tip is $20\%$ of his bill, so his bill is $2 \times 5 = 10$ dollars. The difference is $20 - 10 = 10$ dollars.

Thus, the correct answer is D.

Solution:

First $(1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3.$ Then $(-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0.$

Thus, the correct answer is C.

Solution:

From $2x + 7 = 3$ we get $x = -2.$ Substituting, $-2b - 10 = -2,$ so $-2b = 8$ and $b = -4.$

Thus, the correct answer is B.

Solution:

Let the width be $w,$ so the length is $2w.$ Then $x^2 = w^2 + (2w)^2 = 5w^2,$ giving $w^2 = \dfrac{x^2}{5}.$ The area is $w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.$

Thus, the correct answer is B.

Solution:

Alone, Dave pays for $6$ windows and receives one free to reach $7,$ costing $\$600;$ Doug pays for $7$ and receives one free to reach $8,$ costing $\$700.$ Separately they pay $\$1300.$ Together they need $15$ windows: buying $12$ yields $3$ free, for $\$1200.$ The savings are $1300 - 1200 = 100$ dollars.

Thus, the correct answer is A.

Solution:

The combined sum is $20 \cdot 30 + 30 \cdot 20 = 600 + 600 = 1200.$ The average of all $50$ numbers is $\dfrac{1200}{50} = 24.$

Thus, the correct answer is B.

Solution:

Let Mike ride $m$ miles. Josh rides $\frac{4}{5}$ the rate for $2$ times the time, so Josh's distance is $\frac{8}{5}m.$ Together they cover $13,$ so $m + \frac{8}{5}m = \frac{13}{5}m = 13,$ giving $m = 5.$

Thus, the correct answer is B.

Solution:

The triangles $ABH, BCE, CDF,$ and $DAG$ are congruent right triangles. In $\triangle BCE$ the hypotenuse is $BC = \sqrt{50}$ and $BE = 1,$ so $CE = \sqrt{50 - 1} = 7.$ Since $BH = CE = 7$ and $E$ lies on $BH$ with $BE = 1,$ the inner square's side is $EH = 7 - 1 = 6,$ giving area $6^2 = 36.$

Thus, the correct answer is C.

Solution:

The three X positions can be any of $\binom{5}{3} = 10$ equally likely choices, and exactly one of them produces XOXOX. So the probability is $\dfrac{1}{10}.$

Thus, the correct answer is B.

Solution:

Writing the equation as $4x^2 + (a+8)x + 9 = 0,$ there is one solution exactly when the discriminant $(a+8)^2 - 144 = 0.$ Then $a + 8 = \pm 12,$ so $a = 4$ or $a = -20,$ and their sum is $-16.$

Thus, the correct answer is A.

Solution:

The unit cubes have $6n^3$ faces total, of which the original surface accounts for $6n^2$ red faces. Then $\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4},$ so $n = 4.$

Thus, the correct answer is B.

Solution:

Since the base $2$ equals two radii, the radius is $1.$ The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four $60^\circ$ sectors of a circle of radius $1.$ Their total area is $4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.$

Thus, the correct answer is B.

Solution:

Taking $50$th roots, the condition becomes $130n \gt n^2 \gt 2^4 = 16.$ From $n^2 \gt 16$ we get $n \gt 4,$ and from $130n \gt n^2$ we get $n \lt 130.$ So $n$ ranges over the integers $5, 6, \ldots, 129,$ which is $125$ values.

Thus, the correct answer is E.

Solution:

The first and last digits must have the same parity so their average is a digit. Both odd gives $5 \cdot 5 = 25$ pairs. Both even, with a nonzero leading digit, gives $4 \cdot 5 = 20$ pairs. Each pair fixes the middle digit, for a total of $25 + 20 = 45$ numbers.

Thus, the correct answer is E.

Solution:

As a product of primes, $3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$ A cube divisor uses exponents that are multiples of $3:$ the exponent of $2$ can be $0, 3,$ or $6$ ($3$ choices), the exponent of $3$ can be $0$ or $3$ ($2$ choices), and the exponents of $5$ and $7$ must be $0.$ That gives $3 \cdot 2 \cdot 1 \cdot 1 = 6$ cubes.

Thus, the correct answer is E.

Solution:

If the number is $10a + b,$ then $(10a + b) - (a + b) = 9a.$ The units digit of $9a$ is $6$ only when $a = 4,$ since $9 \cdot 4 = 36.$ The digit $b$ can then be anything from $0$ to $9,$ giving the ten numbers $40$ through $49.$

Thus, the correct answer is D.

Solution:

Every number is an endpoint of two segments, so the five segment sums total $2(3 + 5 + 6 + 7 + 9) = 60.$ The middle term of a five-term arithmetic sequence equals its mean, which is $\dfrac{60}{5} = 12.$

Thus, the correct answer is D.

Solution:

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game $2$ and A ends up with the series (three wins) leaves the equally likely sequences

$$\text{BBAAA}, \quad \text{ABBAA}, \quad \text{ABABA}, \quad \text{ABAAB}, \quad \text{ABAAA}.$$

Only in BBAAA does team B win the first game, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is A.

Solution:

When lowered, the rotated square's two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ Working out the geometry, the square's bottom vertex settles at height $\frac{1}{2}.$ The point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt{2}$ higher, so its height is $\frac{1}{2} + \sqrt{2}.$

Thus, the correct answer is D.

Solution:

Extend the four sides of length $1$ to form a square. Each short side $\dfrac{\sqrt{2}}{2}$ is the hypotenuse of an isosceles right triangle with legs $\dfrac{1}{2},$ and cutting these four corners from a square of side $1 + 2 \cdot \frac{1}{2} = 2$ gives the octagon. Its area is $2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}.$

Thus, the correct answer is A.

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ the quotient is $\dfrac{6n}{n(n+1)/2} = \dfrac{12}{n+1},$ which is an integer exactly when $n + 1$ divides $12.$ The divisors of $12$ that are at least $2$ are $2, 3, 4, 6, 12,$ giving $n = 1, 2, 3, 5, 11$ — five values.

Thus, the correct answer is B.

Solution:

The elements common to $S$ and $T$ are the multiples of $\operatorname{lcm}(4,6) = 12.$ Now $S$ contains multiples of $4$ up to $4 \cdot 2005 = 8020,$ while $T$ reaches up to $6 \cdot 2005 = 12{,}030,$ so the common elements are the multiples of $12$ not exceeding $8020.$ There are $\left\lfloor \dfrac{8020}{12} \right\rfloor = 668$ of them.

Thus, the correct answer is D.

Solution:

Let $O$ be the center. From $2 \cdot AC = BC$ and $AC + BC = AB,$ we get $AC = \dfrac{AB}{3},$ so $CO = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}.$ Triangles $DCO$ and $DAB$ share the apex $D$ with bases $CO$ and $AB$ on the same line, so $[\triangle DCO] = \dfrac{CO}{AB}[\triangle DAB] = \dfrac{1}{6}[\triangle DAB].$ Because $O$ is the midpoint of $DE,$ $[\triangle DCE] = 2\,[\triangle DCO] = \dfrac{1}{3}[\triangle DAB].$

Thus, the correct answer is C.

Solution:

The condition $P(n) = \sqrt{n}$ means $n$ is the square of a prime $q,$ and likewise $n + 48 = p^2$ for a prime $p.$ Then $48 = p^2 - q^2 = (p - q)(p + q).$ Checking the same-parity factorizations of $48,$ only $p - q = 2,\ p + q = 24$ yields primes, giving $(p, q) = (13, 11)$ and $n = 121.$ So there is exactly one such $n.$

Thus, the correct answer is B.

Solution:

Triangles $ADE$ and $ABC$ share angle $A,$ so $\dfrac{[ADE]}{[ABC]} = \dfrac{AD \cdot AE}{AB \cdot AC} = \dfrac{19 \cdot 14}{25 \cdot 42} = \dfrac{266}{1050} = \dfrac{19}{75}.$ Since $[BCED] = [ABC] - [ADE],$ we get $\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}.$

Thus, the correct answer is D.