Solution:

Susan pays $(4)(0.75)(20)=60$ dollars.

Pam pays $(5)(0.70)(20)=70$ dollars.

So Pam pays $70-60=10$ more dollars than Susan.

Thus, the correct answer is C.

Solution:

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

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

Setting this equal to the brick's volume gives $8000=4000h,$ so $h=2.$

Thus, the correct answer is D.

Solution:

Let the smaller integer be $x.$ Then the larger is $x+2.$

So $x+2=3x,$ which gives $x=1.$

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

Thus, the correct answer is A.

Solution:

Kate rode for $\tfrac12$ hour at $16$ mph, covering $8$ miles.

She walked for $\tfrac32$ hours at $4$ mph, covering $6$ miles.

She covered $14$ miles in $2$ hours, so her average speed was $7$ mph.

Thus, the correct answer is A.

Solution:

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

He pays $10\%$ of that in state taxes, which is $8\%$ of the inheritance.

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

Thus, the correct answer is D.

Solution:

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

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

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

Thus, the correct answer is D.

Solution:

Let $D$ be the common difference. Then $a=c-2D,$ $b=c-D,$ $d=c+D,$ and $e=c+2D,$ so $$a+b+c+d+e=5c.$$

Thus $5c=30,$ giving $c=6.$

The other terms cannot be determined: the sequences $4,5,6,7,8$ and $10,8,6,4,2$ both satisfy the conditions but differ in every term except the middle one.

Thus, the correct answer is C.

Solution:

Consider the two chords meeting at the number $5.$ They run to $12$ and to $10,$ so the arc they subtend extends from $10$ to $12.$

That arc spans two of the twelve hour-marks, so its measure is $\tfrac{2}{12}\cdot 360^\circ=60^\circ.$

By the Inscribed Angle Theorem, the vertex angle is half the arc, or $\tfrac12\cdot 60^\circ=30^\circ.$ By symmetry every vertex angle equals $30^\circ.$

Thus, the correct answer is C.

Solution:

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

Walking to the stadium takes $\dfrac{y}{w}.$ Walking home then biking takes $\dfrac{x}{w}+\dfrac{x+y}{7w}=\dfrac{8x+y}{7w}.$

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 $3x,$ $4x,$ and $5x.$ The triangle is right, so its hypotenuse is a diameter.

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

The area is $\tfrac12\cdot 3x\cdot 4x=6x^2=6\cdot\tfrac{36}{25}=\tfrac{216}{25}=8.64.$

Thus, the correct answer is A.

Solution:

Because of the cycling property, each digit that appears is used the same number of times in the hundreds, tens, and units places.

Let $k$ be the sum of the units digits over all terms. Then $S=100k+10k+k=111k=3\cdot 37\cdot k.$

So $S$ is always divisible by $37.$ It need not be divisible by anything larger: the sequence $123,231,312$ gives $S=666=2\cdot 3^2\cdot 37.$

Thus, the correct answer is D.

Solution:

Exactly half of the integers from $0$ to $2007$ are odd.

A product $ad$ is odd only when both factors are odd, with probability $\tfrac12\cdot\tfrac12=\tfrac14,$ and even with probability $\tfrac34.$ The same holds for $bc.$

Then $ad-bc$ is even when both products are odd or both are even: $$\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58.$$

Thus, the correct answer is E.

Solution:

The mouse is closest to the cheese at the foot of the perpendicular from $(12,10)$ to the line.

This perpendicular has slope $\tfrac15,$ so its equation is $y=10+\tfrac15(x-12)=\tfrac15 x+\tfrac{38}{5}.$

Setting $\tfrac15 x+\tfrac{38}{5}=-5x+18$ gives $x=2$ and $y=8.$ Thus $(a,b)=(2,8)$ and $a+b=10.$

Thus, the correct answer is B.

Solution:

The five factors are distinct integers multiplying to $45.$ If any factor had absolute value more than $5,$ the remaining four (distinct) would have product at least $|(-3)(-1)(1)(3)|=9,$ forcing the total above $45.$

So the factors come from $\pm 1,\pm 3,\pm 5.$ The product of all six of these is $-225=(-5)(45),$ so the five factors are $-3,-1,1,3,5.$

Then $a,b,c,d,e$ are $9,7,5,3,1$ in some order, and their sum is $25.$

Thus, the correct answer is C.

Solution:

The mean is $\dfrac{28+n}{5}.$

If $n\lt 6,$ the median is $6,$ so $28+n=30$ and $n=2.$

If $6\lt n\lt 9,$ the median is $n,$ so $28+n=5n$ and $n=7.$

If $n\gt 9,$ the median is $9,$ so $28+n=45$ and $n=17.$

The sum of all possible values is $2+7+17=26.$

Thus, the correct answer is E.

Solution:

The three distinct digits form an increasing arithmetic progression. Counting by common difference: $8$ with difference $1,$ $6$ with difference $2,$ $4$ with difference $3,$ and $2$ with difference $4,$ for $20$ sets.

Of these, $4$ sets contain $0$ (namely $\{0,1,2\},\{0,2,4\},\{0,3,6\},\{0,4,8\}$); each yields $2\cdot 2!=4$ valid numbers since $0$ cannot lead.

The other $16$ sets each yield $3!=6$ numbers. The total is $4\cdot 4+16\cdot 6=112.$

Thus, the correct answer is C.

Solution:

Squaring both equations gives $\sin^2 a+2\sin a\sin b+\sin^2 b=\tfrac53$ and $\cos^2 a+2\cos a\cos b+\cos^2 b=1.$

Adding and using $\sin^2\theta+\cos^2\theta=1$ twice, $$2+2(\sin a\sin b+\cos a\cos b)=\tfrac83.$$

So $\cos(a-b)=\sin a\sin b+\cos a\cos b=\tfrac13.$

Thus, the correct answer is B.

Solution:

Since $f$ has real coefficients, the conjugates $-2i$ and $2-i$ are also roots. Thus $$f(x)=(x^2+4)(x^2-4x+5)=x^4-4x^3+9x^2-16x+20.$$

Then $a+b+c+d=-4+9-16+20=9.$ Equivalently, $a+b+c+d=f(1)-1=(1+4)(1+1)-1=9.$

Thus, the correct answer is D.

Solution:

The altitude $h$ from $A$ in $\triangle ABC$ satisfies $2007=\tfrac12\cdot 223\cdot h,$ so $h=18.$ Thus $A$ lies on $y=18$ or $y=-18.$

Line $DE$ has equation $x-y-300=0.$ The condition on $\triangle ADE$ similarly places $A$ on one of two lines parallel to $DE.$

The four possible positions of $A$ are the vertices of a parallelogram whose center is the intersection of $y=0$ with line $DE,$ namely $(300,0).$ Hence the sum of the four $x$-coordinates is $4\cdot 300=1200.$

Thus, the correct answer is E.

Solution:

Slicing removes two equal segments of length $x$ from each edge. Each octagon then has side length $x\sqrt2,$ and the edge satisfies $1=2x+x\sqrt2,$ so $$x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}.$$

Each removed corner is a tetrahedron with three mutually perpendicular legs of length $x,$ so its volume is $\tfrac16 x^3.$ There are $8$ corners, giving total volume $$8\cdot\tfrac16 x^3=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3=\frac{10-7\sqrt2}{3}.$$

Thus, the correct answer is B.

Solution:

The product of the zeros is $\tfrac ca$ and the sum of the zeros is $-\tfrac ba.$ Equating them gives $c=-b.$

Then the sum of the coefficients is $a+b+c=a,$ which is the coefficient of $x^2.$

The other choices fail in general: for $f(x)=-2x^2-4x+4$ the common value is $-2,$ but the coefficient of $x$ is $-4,$ the $y$-intercept is $4,$ the $x$-intercepts are $-1\pm\sqrt3,$ and their mean is $-1.$

Thus, the correct answer is A.

Solution:

For $n\le 2007,$ $S(n)\le S(1999)=28,$ and then $S(S(n))\le S(28)=10.$ So any solution has $n\ge 2007-28-10=1969.$

Also $n,$ $S(n),$ and $S(S(n))$ are congruent modulo $9,$ and $2007$ is a multiple of $9,$ so all three must be multiples of $3.$

Checking the multiples of $3$ between $1969$ and $2007$ (many are eliminated because $n+S(n)$ already exceeds $2007$) leaves $1977,1980,1983,$ and $2001.$ That is $4$ values.

Thus, the correct answer is D.

Solution:

Let $A=(p,\log_a p)$ and $B=(q,2\log_a q).$ Since $AB$ is horizontal, $\log_a p=2\log_a q=\log_a q^2,$ so $p=q^2.$

The side length is $6=|p-q|=|q^2-q|,$ whose only positive solution is $q=3.$

Since $C=(q,3\log_a q),$ the vertical side gives $BC=6=\log_a q=\log_a 3.$ Thus $a^6=3,$ so $a=\sqrt[6]{3}.$

Thus, the correct answer is A.

Solution:

On each interval where $\sin nx\ge 0,$ the graphs of $\sin x$ and $\sin nx$ meet twice, unless they share the value $1$ there, in which case they meet once. Counting the humps and the endpoint at $(\pi,0)$ gives

$F(n)=n+1$ when $n$ is even or $n\equiv 3\pmod 4,$ and $F(n)=n$ when $n\equiv 1\pmod 4.$

Thus $$\sum_{n=2}^{2007}F(n)=\sum_{n=2}^{2007}(n+1)-\#\{n\equiv 1\!\!\pmod 4\}.$$ The first sum is $2{,}017{,}033,$ and there are $501$ values $n\equiv 1\pmod 4$ in the range, giving $2{,}017{,}033-501=2{,}016{,}532.$

Thus, the correct answer is D.

Solution:

Let $c_n$ be the number of spacy subsets of $\{1,\ldots,n\}.$ A spacy subset either omits $n$ (there are $c_{n-1}$ of these) or contains $n,$ in which case it omits $n-1$ and $n-2$ (there are $c_{n-3}$ of these).

Hence $c_n=c_{n-1}+c_{n-3},$ with $c_1=2,$ $c_2=3,$ $c_3=4.$

The sequence continues $6,9,13,19,28,41,60,88,129,$ so $c_{12}=129.$

Thus, the correct answer is E.