Solution:

The troop buys $1000 \div 5 = 200$ groups of five bars, costing $200 \cdot 2 = 400$ dollars.

They sell $1000 \div 2 = 500$ pairs of bars, earning $500 \cdot 1 = 500$ dollars.

The profit is $\$500 - \$400 = \$100.$

Thus, A is the correct answer.

Solution:

The statement translates to $$\dfrac{x}{100}\cdot x = 4,$$ so $x^2 = 400.$

Since $x$ is positive, $x = 20.$

Thus, D is the correct answer.

Solution:

After the first day, $1 - \dfrac13 = \dfrac23$ of the paint remains.

On the second day, $\dfrac13 \cdot \dfrac23 = \dfrac29$ of the original amount is used.

The fraction available on the third day is $$\dfrac23 - \dfrac29 = \dfrac49.$$

Thus, D is the correct answer.

Solution:

Each inner expression evaluates to $$5 \diamond 12 = \sqrt{5^2 + 12^2} = 13$$ and similarly $(-12) \diamond (-5) = \sqrt{144 + 25} = 13.$

Then $$13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}.$$

Thus, D is the correct answer.

Solution:

Buying all the CDs costs three times as much as buying one third of them, namely $3 \cdot \dfrac15 = \dfrac35$ of her money.

The fraction left over is $1 - \dfrac35 = \dfrac25.$

Thus, C is the correct answer.

Solution:

Lisa needs an A on at least $0.8 \cdot 50 = 40$ quizzes.

She already has $22,$ so she needs $40 - 22 = 18$ A's among the remaining $20$ quizzes.

That leaves at most $20 - 18 = 2$ quizzes with a grade lower than an A.

Thus, B is the correct answer.

Solution:

Let the smaller circle have radius $r,$ so its area is $\pi r^2.$

The smaller square, which circumscribes this circle, has side $2r,$ and its diagonal $2\sqrt2\,r$ is the diameter of the larger circle. So the larger circle has radius $\sqrt2\,r.$

The larger square circumscribes the larger circle, so it has side $2\sqrt2\,r$ and area $8r^2.$

The desired ratio is $$\dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}.$$

Thus, B is the correct answer.

Solution:

The four quarter circles on a tile together make one full circle of radius $\dfrac12,$ with area $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}.$

So each tile has shaded area $1 - \dfrac{\pi}{4}$ square feet.

There are $8 \cdot 10 = 80$ tiles, so the total shaded area is $$80\left(1 - \dfrac{\pi}{4}\right) = 80 - 20\pi.$$

Thus, A is the correct answer.

Solution:

The first die is odd (a $1$ or $3$) with probability $\dfrac23$ and even with probability $\dfrac13.$ The second die is odd (a $5$) with probability $\dfrac13$ and even with probability $\dfrac23.$

The sum is odd when the two parities differ: $$\dfrac13 \cdot \dfrac13 + \dfrac23 \cdot \dfrac23 = \dfrac19 + \dfrac49 = \dfrac59.$$

Thus, D is the correct answer.

Solution:

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $AC = BC,$ $H$ is the midpoint of $AB,$ so $BH = 1$ and $CH^2 = 7^2 - 1^2 = 48.$

Applying the Pythagorean theorem in $\triangle CHD,$ where $HD = BH + BD = 1 + BD,$ gives $$8^2 = 48 + (1 + BD)^2,$$ so $(1 + BD)^2 = 16.$

Then $1 + BD = 4,$ so $BD = 3.$

Thus, A is the correct answer.

Solution:

The sequence begins $2005, 133, 55, 250, 133, \ldots,$ so after the first term it repeats the cycle $133, 55, 250$ of length $3.$

The terms from position $2$ onward follow this cycle. Since $2005 = 2 + 3 \cdot 667 + 2,$ the $2005$th term matches the third entry of the cycle, $250.$

Thus, E is the correct answer.

Solution:

The product is prime exactly when one die shows a prime ($2, 3,$ or $5$) and the other eleven all show $1.$

The probability that any single die is the prime one is $\dfrac36 = \dfrac12,$ and each of the other eleven shows $1$ with probability $\dfrac16.$ Accounting for which of the twelve dice is prime, the probability is $$12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} = 6 \cdot \left(\dfrac16\right)^{11} = \left(\dfrac16\right)^{10}.$$

Thus, E is the correct answer.

Solution:

Between $1$ and $2005$ there are $668$ multiples of $3,$ $501$ multiples of $4,$ and $167$ multiples of $12.$

Every multiple of $12$ is both a multiple of $3$ and of $4,$ so removing them from each group gives $$(668 - 167) + (501 - 167) = 835$$ numbers that are multiples of $3$ or $4$ but not $12.$

Thus, C is the correct answer.

Solution:

Take $\overline{CD}$ as the base. Since $C$ is the midpoint of $\overline{BD}$ and $BC = 2,$ we have $CD = 2.$

The height of $\triangle CDM$ is the distance from $M$ to line $BD.$ Because $M$ is the midpoint of $\overline{AC},$ this distance is half the height of $\triangle ABC,$ which is $\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}.$

The area is $$\dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}.$$

Thus, C is the correct answer.

Solution:

There are $\binom82 = 28$ equally likely pairs.

A sum of at least $\$20$ comes from both twenties ($1$ way), a twenty paired with any of the six smaller bills ($2 \cdot 6 = 12$ ways), or both tens ($1$ way).

The probability is $$\dfrac{1 + 12 + 1}{28} = \dfrac{14}{28} = \dfrac12.$$

Thus, D is the correct answer.

Solution:

Let $r_1$ and $r_2$ be the roots of $x^2 + px + m = 0,$ so $m = r_1 r_2$ and $p = -(r_1 + r_2).$

The roots of $x^2 + mx + n = 0$ are $2r_1$ and $2r_2,$ so $n = 4r_1 r_2$ and $-m = 2(r_1 + r_2).$

Then $n = 4m$ and $m = -2(r_1 + r_2) = 2p,$ so $p = \dfrac{m}{2}.$ Therefore $$\dfrac{n}{p} = \dfrac{4m}{m/2} = 8.$$

Thus, D is the correct answer.

Solution:

Chaining the equations, $$4^{abcd} = \left(\left(\left(4^a\right)^b\right)^c\right)^d = \left(\left(5^b\right)^c\right)^d = \left(6^c\right)^d = 7^d = 8.$$

Since $8 = 4^{3/2},$ we conclude $a \cdot b \cdot c \cdot d = \dfrac32.$

Thus, B is the correct answer.

Solution:

The seven digits are chosen from $\{2, 3, 4, 5, 6, 7, 8, 9\},$ and once chosen they must be written in increasing order, so only the choice of digits matters.

Choosing seven of these eight digits is the same as choosing the one digit to leave out, which can be done in $8$ ways.

Thus, D is the correct answer.

Solution:

The percentage scoring $95$ is $100 - 10 - 25 - 20 - 15 = 30.$

The mean is $$0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 86.$$

Since $35\%$ scored below $85$ and $35\%$ scored above $85,$ the middle student scored $85,$ so the median is $85.$

The difference is $86 - 85 = 1.$

Thus, B is the correct answer.

Solution:

By symmetry, each of the five digits appears equally often in each place, so the average digit in every place is $$\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8.$$

The average number is therefore $$4.8(1 + 10 + 100 + 1000 + 10000) = 4.8 \cdot 11111 = 53332.8.$$

Thus, C is the correct answer.

Solution:

Both events draw from $\binom{40}{4}$ equally likely selections, so $\dfrac{q}{p}$ is the ratio of their favorable counts.

Exactly $10$ draws give four slips of the same number, one for each value.

For two $a$'s and two $b$'s, choose the two values in $\binom{10}{2}$ ways, then two of the four $a$-slips and two of the four $b$-slips: $$\binom{10}{2}\binom{4}{2}\binom{4}{2} = 45 \cdot 6 \cdot 6 = 1620.$$

Therefore $\dfrac{q}{p} = \dfrac{1620}{10} = 162.$

Thus, A is the correct answer.

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ divisibility is equivalent to $$\dfrac{n!}{n(n+1)/2} = \dfrac{2(n-1)!}{n+1}$$ being an integer.

If $n + 1$ is not prime (and $n \ge 1$), its factors appear among $1, 2, \ldots, n-1$ or in the factor $2,$ so the fraction is an integer. If $n + 1$ is an odd prime, it divides neither $(n-1)!$ nor $2,$ so the fraction is not an integer.

The odd primes at most $25$ are $3, 5, 7, 11, 13, 17, 19, 23,$ giving $8$ failing values of $n.$ Hence $24 - 8 = 16$ values work.

Thus, C is the correct answer.

Solution:

Let $AB = a$ and $DC = c.$ The midsegment $\overline{FE}$ has length $\dfrac{a + c}{2},$ and $ABEF$ and $FECD$ have the same height.

Their areas are proportional to the averages of their parallel sides, so $$\dfrac{a + \frac{a+c}{2}}{\frac{a+c}{2} + c} = \dfrac{3a + c}{a + 3c} = 2.$$

Then $3a + c = 2a + 6c,$ so $a = 5c$ and $\dfrac{AB}{DC} = 5.$

Thus, C is the correct answer.

Solution:

Write $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $$m^2 = x^2 - y^2 = 99(a^2 - b^2) = 99(a+b)(a-b).$$

Since $99 = 9 \cdot 11,$ for $m^2$ to be a perfect square we need $11 \mid (a+b)(a-b).$ As $a + b \le 17,$ this forces $a + b = 11,$ and then $a - b$ must itself be a perfect square.

With $a - b \le 8,$ the only workable case is $a - b = 1,$ giving $(a, b) = (6, 5).$ Then $x = 65,$ $y = 56,$ and $m^2 = 99 \cdot 11 = 33^2,$ so $m = 33.$

Therefore $x + y + m = 65 + 56 + 33 = 154.$

Thus, E is the correct answer.

Solution:

The pairs summing to $125$ are $(25, 100), (26, 99), \ldots, (62, 63),$ which is $62 - 25 + 1 = 38$ pairs. From each pair, $B$ may contain at most one element.

The numbers $1$ through $24$ cannot pair with anything in range to sum to $125,$ so all $24$ of them may be included.

Thus $B$ has at most $38 + 24 = 62$ elements, and the set $\{1, 2, \ldots, 62\}$ achieves this.

Thus, C is the correct answer.