Solution:

Susan pays $4 \cdot 0.75 \cdot 20 = 60$ dollars.

Pam pays $5 \cdot 0.70 \cdot 20 = 70$ dollars.

The difference is $70 - 60 = 10$ dollars.

Thus, the correct answer is C.

Solution:

The numerator is $6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8.$

The denominator is $6\#2 = 6 + 2 - 6 \cdot 2^2 = 8 - 24 = -16.$

The quotient is $$\dfrac{8}{-16} = -\dfrac{1}{2}.$$

Thus, the correct answer is A.

Solution:

The brick has volume $40 \cdot 20 \cdot 10 = 8000$ cubic centimeters.

If the water rises by $h$ centimeters, the added volume is $100 \cdot 40 \cdot h = 4000h$ cubic centimeters.

Setting these equal gives $4000h = 8000,$ so $h = 2.$

Thus, the correct answer is D.

Solution:

Let the smaller integer be $x.$ Then the larger is $x + 2,$ and $$x + 2 = 3x,$$ so $x = 1.$

The two integers are $1$ and $3,$ and their sum is $4.$

Thus, the correct answer is A.

Solution:

Let $p$ and $n$ be the prices in cents of a pencil and a notebook. Then $$7p + 8n = 415$$$$5p + 3n = 177.$$

Solving this system gives $p = 9$ and $n = 44.$

So $16$ pencils and $10$ notebooks cost $16(9) + 10(44) = 584$ cents, or $\$5.84.$

Thus, the correct answer is B.

Solution:

From $2002$ to $2003,$ the increase is $$\dfrac{6}{60} = \dfrac{1}{10} = 10\%.$$

The other increases are $\dfrac{4}{66},$ $\dfrac{6}{70},$ $\dfrac{2}{76},$ and $\dfrac{7}{78},$ each less than $\dfrac{1}{10}.$

So the largest percentage increase was between $2002$ and $2003.$

Thus, the correct answer is A.

Solution:

After federal taxes, Mr. Public keeps $80\%$ of his inheritance.

State taxes take $10\%$ of that, which is $8\%$ of the inheritance.

The total tax is $20\% + 8\% = 28\%$ of the inheritance, so the inheritance is $$\dfrac{10{,}500}{0.28} = 37{,}500.$$

Thus, the correct answer is D.

Solution:

Since $\triangle ABC$ is isosceles, $\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ.$

Since $\triangle ADC$ is isosceles, $\angle DAC = \tfrac12(180^\circ - 140^\circ) = 20^\circ.$

Therefore $\angle BAD = \angle BAC - \angle DAC = 70^\circ - 20^\circ = 50^\circ.$

Thus, the correct answer is D.

Solution:

The equations become $3^a = 3^{4(b+2)}$ and $5^{3b} = 5^{a-3}.$

So $a = 4(b + 2)$ and $3b = a - 3.$

Solving gives $a = -12$ and $b = -5,$ so $ab = 60.$

Thus, the correct answer is E.

Solution:

Let $N$ be the number of children and $T$ the total age of the family.

Then $20 = \dfrac{T}{N + 2}$ and $16 = \dfrac{T - 48}{N + 1}.$

These give $20N + 40 = T$ and $16N + 64 = T,$ so $20N + 40 = 16N + 64.$

Hence $4N = 24$ and $N = 6.$

Thus, the correct answer is E.

Solution:

Each vertex belongs to exactly three faces, so summing the numbers over all six faces gives $$3(1 + 2 + \cdots + 8) = 3 \cdot 36 = 108.$$

There are six faces, so the common sum is $108 \div 6 = 18.$

Thus, the correct answer is C.

Solution:

Each tourist independently chooses one of the two guides, giving $2^6 = 64$ arrangements.

Exactly two of these leave a guide with no tourists, so the answer is $64 - 2 = 62.$

Thus, the correct answer is D.

Solution:

Let $x$ and $y$ be Yan's distances from home and from the stadium, and let $w$ be his walking speed.

Walking to the stadium takes $\dfrac{y}{w}.$ Walking home and biking takes $\dfrac{x}{w} + \dfrac{x + y}{7w}.$

Setting these equal gives $7y = 8x + y,$ so $8x = 6y$ and $\dfrac{x}{y} = \dfrac{3}{4}.$

Thus, the correct answer is B.

Solution:

Let the sides be $3x, 4x, 5x.$ The triangle is right-angled, so its hypotenuse is a diameter.

Thus $5x = 2 \cdot 3 = 6,$ giving $x = \tfrac65.$

The area is $$\tfrac12 (3x)(4x) = 6x^2 = 6 \cdot \tfrac{36}{25} = 8.64.$$

Thus, the correct answer is A.

Solution:

Consider the isosceles right triangle joining the center of the radius-$2$ circle to the centers of two adjacent small circles. Its legs have length $2 + 1 = 3,$ so its hypotenuse is $3\sqrt2.$

The side of the square exceeds this hypotenuse by $2$ (one radius on each end), so $s = 2 + 3\sqrt2.$

The area is $$(2 + 3\sqrt2)^2 = 4 + 12\sqrt2 + 18 = 22 + 12\sqrt2.$$

Thus, the correct answer is B.

Solution:

Half the integers from $0$ to $2007$ are odd, so each of $ad$ and $bc$ is odd with probability $\tfrac12 \cdot \tfrac12 = \tfrac14$ and even with probability $\tfrac34.$

The difference $ad - bc$ is even when both products have the same parity: $$\tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58.$$

Thus, the correct answer is E.

Solution:

Since $n^3 = 75m = 3 \cdot 5^2 \cdot m,$ every prime factor must occur a multiple of three times.

The smallest such $m$ is $3^2 \cdot 5 = 45,$ giving $n^3 = 3^3 \cdot 5^3$ and $n = 15.$

Then $m + n = 45 + 15 = 60.$

Thus, the correct answer is D.

Solution:

Put the figure on coordinates with $A = (-2, 6),$ $B = (2, 6),$ $C = (2, 2),$ $G = (2, -6),$ and $H = (-2, -6).$

Line $AG$ is $y = -3x,$ and line $CH$ is $y = 2x - 2.$

Their intersection is $M = \left(\tfrac25, -\tfrac65\right).$

Applying the shoelace formula to $A, B, C, M$ gives area $$\dfrac{|{-35.2}|}{2} = \dfrac{88}{5}.$$

Thus, the correct answer is C.

Solution:

Let $s$ be the side, $w$ the brush width, and $x$ the leg of one unpainted isosceles right triangle. Each triangle has area $\tfrac18 s^2,$ so $\tfrac12 x^2 = \tfrac18 s^2$ and $x = \tfrac{s}{2}.$

The leg plus the brush width is half the diagonal: $x + w = \tfrac{\sqrt2}{2} s.$ Thus $w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.$

Therefore $$\dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.$$

Thus, the correct answer is C.

Solution:

Squaring $a + a^{-1} = 4$ gives $a^2 + 2 + a^{-2} = 16,$ so $a^2 + a^{-2} = 14.$

Squaring again gives $a^4 + 2 + a^{-4} = 196,$ so $a^4 + a^{-4} = 194.$

Thus, the correct answer is D.

Solution:

Each face of the outer cube has area $24 \div 6 = 4,$ so its side is $2,$ and the sphere has diameter $2.$

This diameter is the space diagonal of the inner cube, so $l\sqrt3 = 2,$ giving $l^2 = \tfrac43.$

The inner cube's surface area is $6 l^2 = 6 \cdot \tfrac43 = 8.$

Thus, the correct answer is C.

Solution:

Each digit appears as a hundreds digit, a tens digit, and a units digit the same number of times across the sequence.

If $k$ is the sum of the units digits of all terms, then $S = 111k = 3 \cdot 37 \cdot k,$ so $S$ is always divisible by $37.$

The sequence $123, 231, 312$ gives $S = 666 = 2 \cdot 3^2 \cdot 37,$ which has no larger prime factor forced, so $37$ is the answer.

Thus, the correct answer is D.

Solution:

Since $(m + n)(m - n) = 96$ and $96$ is even, both factors must be even.

The even factor pairs are $(48, 2),$ $(24, 4),$ $(16, 6),$ and $(12, 8),$ giving $(m, n) = (25, 23),$ $(14, 10),$ $(11, 5),$ and $(10, 2).$

So there are $4$ ordered pairs.

Thus, the correct answer is B.

Solution:

Rectangle $ABFE$ has area $AE \cdot AB = 2 \cdot 4\sqrt2 = 8\sqrt2.$

Right triangles $ACO$ and $BDO$ each have hypotenuse $2\sqrt2$ and a leg of $2,$ so each is isosceles right with area $2.$

Angles $CAE$ and $DBF$ are each $45^\circ,$ so sectors $CAE$ and $DBF$ each have area $\tfrac18 \pi \cdot 2^2 = \tfrac{\pi}{2}.$

The shaded area is $$8\sqrt2 - 2 \cdot 2 - 2 \cdot \tfrac{\pi}{2} = 8\sqrt2 - 4 - \pi.$$

Thus, the correct answer is B.

Solution:

If $n \le 2007,$ then $S(n) \le 28$ and $S(S(n)) \le 10,$ so $n \ge 2007 - 28 - 10 = 1969.$

Since $n,$ $S(n),$ and $S(S(n))$ all leave the same remainder modulo $9$ and $2007$ is a multiple of $9,$ each must be a multiple of $3.$

Checking the multiples of $3$ between $1969$ and $2007,$ the condition holds for $1977, 1980, 1983,$ and $2001.$

So there are $4$ values of $n.$

Thus, the correct answer is D.