Solution:

The two meetings lasted $45+90=135$ minutes, and the work day is $9\cdot60=540$ minutes.

The fraction of the day spent in meetings is $$\frac{135}{540}=\frac14=25\%.$$

Thus, the correct answer is C.

Solution:

The region splits into an $8\times2$ vertical rectangle and a $2\times3$ horizontal foot, whose width is $5-2=3.$

The total area is $$8\cdot2+3\cdot2=16+6=22.$$

Thus, the correct answer is A.

Solution:

The price $x$ must divide both totals, so $x$ is a common divisor of $48$ and $64.$

Since $\gcd(48,64)=16,$ the common divisors are $1, 2, 4, 8,$ and $16.$ There are $5$ possible values.

Thus, the correct answer is E.

Solution:

Since $31=4\cdot7+3,$ the first three days of the month each occur five times, and the other four days occur four times.

Mondays and Wednesdays are equal in number exactly when both fall in the five-time group or both fall in the four-time group.

If the first day is Monday, the five-time days are Mon, Tue, Wed (both appear five times). If the first day is Thursday or Friday, the five-time days miss both Monday and Wednesday (both appear four times). Every other starting day includes exactly one of Monday or Wednesday.

So the first day can be Monday, Thursday, or Friday, giving $3$ possibilities.

Thus, the correct answer is B.

Solution:

The correct value is $a-(b-(c-(d+e)))=a-b+c-d-e.$ With $a, b, c, d=1, 2, 3, 4,$ this equals $1-2+3-4-e=-2-e.$

Larry dropped the parentheses and computed $1-2-3-4+e=-8+e.$

Setting $-2-e=-8+e$ gives $2e=6,$ so $e=3.$

Thus, the correct answer is D.

Solution:

Assume $100$ students. The number of "Yes" answers rises from $50$ to $70,$ so at least $70-50=20$ students switched from "No" to "Yes"; thus $x\ge20.$

Since only $30$ students answer "No" at the end, at least $50-30=20$ of the original "Yes" students still answer "Yes," so at most $80$ students switched; thus $x\le80.$

Both extremes are achievable, so the difference is $80-20=60.$

Thus, the correct answer is D.

Solution:

Let $t$ be the number of minutes driven in the rain. She covers $20\cdot\frac{t}{60}$ miles in the rain and $30\cdot\frac{40-t}{60}$ miles in the sun.

Setting the total to $16$ gives $$\frac{20t+30(40-t)}{60}=16,$$ so $1200-10t=960$ and $t=24.$

Thus, the correct answer is C.

Solution:

With $n$ schools there are $3n$ students. Carla placed $64$th, so $3n\ge64$ and $n\ge22.$

The scores are distinct and Andrea is the median, so $3n$ is odd, forcing $n$ odd and $n\ge23.$

Andrea's position is $\dfrac{3n+1}{2},$ and she beat Beth ($37$th), so $\dfrac{3n+1}{2}\lt37,$ giving $3n\lt73$ and $n\le24.$ The only odd value is $n=23.$

Thus, the correct answer is B.

Solution:

To be smallest, $n$ uses only the primes of $20,$ so $n=2^a\cdot5^b$ with $a\ge2$ and $b\ge1.$

Since $n^2=2^{2a}5^{2b}$ is a perfect cube, $3\mid a$ and $3\mid b.$ Since $n^3=2^{3a}5^{3b}$ is a perfect square, $2\mid a$ and $2\mid b.$ Hence $6\mid a$ and $6\mid b.$

The smallest choice is $a=b=6,$ so $n=2^6\cdot5^6=10^6=1{,}000{,}000,$ which has $7$ digits.

Thus, the correct answer is E.

Solution:

The numbers $1$ through $99$ sum to $\dfrac{99\cdot100}{2}=4950.$

The average condition is $$\frac{4950+x}{100}=100x,$$ so $4950+x=10000x$ and $9999x=4950.$

Thus $x=\dfrac{4950}{9999}=\dfrac{50}{101}.$

Thus, the correct answer is B.

Solution:

A four-digit palindrome has the form $\overline{abba}=1001a+110b$ with $1\le a\le9$ and $0\le b\le9.$

Since $1001=7\cdot11\cdot13$ is divisible by $7$ and $110$ is not, the number is divisible by $7$ exactly when $7\mid b,$ that is $b=0$ or $b=7.$

For each $a,$ that is $2$ of the $10$ choices of $b,$ a probability of $\dfrac{2}{10}=\dfrac15.$

Thus, the correct answer is E.

Solution:

Let $L=\log_2 x.$ Converting each term to base $2:$

$\log_{\sqrt2}\sqrt{x}=\dfrac{L/2}{1/2}=L,$ $\log_2 x=L,$ $\log_4 x^2=\dfrac{2L}{2}=L,$ $\log_8 x^3=\dfrac{3L}{3}=L,$ and $\log_{16}x^4=\dfrac{4L}{4}=L.$

The equation becomes $5L=40,$ so $L=8$ and $x=2^8=256.$

Thus, the correct answer is D.

Solution:

A cosine plus a sine equals $2$ only when each equals $1.$ So $\cos(2A-B)=1$ and $\sin(A+B)=1,$ giving $2A-B=0^\circ$ and $A+B=90^\circ.$

Solving, $A=30^\circ$ and $B=60^\circ,$ so $\triangle ABC$ is a $30\text{-}60\text{-}90$ right triangle with the right angle at $C.$

With hypotenuse $AB=4,$ the side $BC$ opposite the $30^\circ$ angle is half the hypotenuse, so $BC=2.$

Thus, the correct answer is C.

Solution:

Each of $a+b,$ $d+e,$ and $c$ is at most $M$ (note $c\le c+d\le M$). Adding, $2010=(a+b)+c+(d+e)\le3M,$ so $M\ge670.$

If $M=670,$ then $c=670,$ but then $b+c\ge671\gt M,$ a contradiction. Hence $M\ge671.$

The value $671$ is reached by $(a,b,c,d,e)=(669,1,670,1,669),$ whose consecutive-pair sums are $670,671,671,670.$

Thus, the correct answer is B.

Solution:

We need exactly two of $i^x,$ $(1+i)^y,$ $z$ equal, with the third different. The three cases are the three possible equal pairs.

Case $i^x=(1+i)^y:$ since $|i^x|=1$ but $|(1+i)^y|=2^{y/2}\gt1$ for $y\ge1,$ we need $y=0,$ so $(1+i)^0=1$ and $i^x=1,$ i.e. $x\in\{0,4,8,12,16\}.$ Then $z$ is any of the $19$ values other than $1.$ This gives $5\cdot19=95$ triples.

Case $i^x=z:$ the only nonnegative-integer value of $i^x$ is $1$ (with $x$ a multiple of $4$), so $z=1$ and $(1+i)^y\ne1,$ meaning $y\ge1.$ This gives $5\cdot19=95$ triples.

Case $(1+i)^y=z:$ since $(1+i)^2=2i,$ the power $(1+i)^y$ is a nonnegative integer below $20$ only for $y=0$ (value $1$) or $y=8$ (value $16$). If $y=0,\ z=1,$ we need $i^x\ne1,$ so $x$ is not a multiple of $4$ ($15$ values). If $y=8,\ z=16,$ then $i^x$ is never $16,$ so $x$ is free ($20$ values). This gives $15+20=35$ triples.

Altogether $95+95+35=225.$

Thus, the correct answer is D.

Solution:

Factor $abc+ab+a=a(bc+b+1).$ Since $2010$ is a multiple of $3,$ each of $a, b, c$ is uniform modulo $3.$

If $3\mid a$ (probability $\tfrac13$), the product is divisible by $3.$

If $3\nmid a$ (probability $\tfrac23$), we need $3\mid bc+b+1.$ Checking residues, this holds exactly when $(b,c)\equiv(1,1)$ or $(2,0)\pmod3,$ a probability of $\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29.$

The total probability is $$\frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}.$$

Thus, the correct answer is E.

Solution:

Write $a_{ij}$ for the entry in row $i,$ column $j.$ The conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}\in\{4,5,6\}.$

If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}$ and $\{5,6,7,8\}$ split as complementary pairs filling the rest of the last row and column: $\binom42=6$ splits times $2$ orders for $\{2,3\}$ gives $12$ arrays. By symmetry $a_{22}=6$ also gives $12.$

If $a_{22}=5,$ then $\{a_{12},a_{13},a_{23}\}$ and $\{a_{21},a_{31},a_{32}\}$ are complementary subsets of $\{2,3,4,6,7,8\}$ subject to the ordering constraints, forcing the first set to be $\{2,3,4\}$ or $\{6,7,8\};$ this gives $\binom63-2=18$ arrays.

Altogether $12+12+18=42.$

Thus, the correct answer is D.

Solution:

This is a continuous (geometric) probability. Anchor the second jump from $P=(0,0)$ to $Q=(1,0),$ and let $\alpha,\beta$ be the directions of the first and third jumps, so the start is $A=(\cos\alpha,\sin\alpha)$ and the end is $B=(1+\cos\beta,\sin\beta).$

Taking $0\le\alpha\le\pi$ and $0\le\beta\le2\pi,$ the requirement $AB\le1$ holds exactly when $\alpha\le\beta\le\pi.$

In the $\alpha\beta$-rectangle of area $2\pi^2,$ the favorable region is a triangle of area $\tfrac{\pi^2}{2},$ so the probability is $\dfrac{\pi^2/2}{2\pi^2}=\dfrac14.$

Thus, the correct answer is C.

Solution:

Let the Raiders score $a, ar, ar^2, ar^3$ (increasing geometric, $r\gt1$) and the Wildcats $a, a+d, a+2d, a+3d$ (increasing arithmetic), tied in the first quarter at $a.$

Every quarter score is a positive integer and each total is under $100,$ so the ratio and first term are small. Testing $r=2$ gives Raiders $5, 10, 20, 40$ with total $75.$

The Wildcats then total $74=4a+6d=20+6d,$ so $d=9,$ giving $5, 14, 23, 32.$ The Raiders won $75$ to $74.$

The first-half total is $(5+10)+(5+14)=34.$

Thus, the correct answer is E.

Solution:

The common ratio is $r=\dfrac{a_2}{a_1}=\cot x.$ Then $a_4=a_3\cdot r=\tan x\cot x=1.$

From $a_3=a_1r^2,$ we get $\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},$ so $\sin^2 x=\cos^3 x,$ i.e. $(\cos^2 x)(1+\cos x)=1.$

Hence $1+\cos x=\dfrac{1}{\cos^2 x}.$ Also $r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},$ so $r^4=\dfrac{1}{\cos^2 x}=1+\cos x.$

Therefore $1+\cos x=a_4\cdot r^4=a_8,$ so $n=8.$

Thus, the correct answer is E.

Solution:

Since $1, 3, 5, 7$ are roots of $P(x)-a,$ write $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$ with $Q$ having integer coefficients.

Evaluating at $x=2, 4, 6, 8$ (where $P=-a$) gives $$-2a=-15\,Q(2)=9\,Q(4)=-15\,Q(6)=105\,Q(8).$$

So $15, 9,$ and $105$ all divide $2a,$ hence $\operatorname{lcm}(15,9,105)=315$ divides $2a.$ Since $315$ is odd, $315\mid a,$ so $a\ge315.$

The value $a=315$ is attainable, so the smallest is $315.$

Thus, the correct answer is B.

Solution:

Let $a=AB, b=BC, c=CD, d=DA$ and $k=bc=ad.$ Writing each triangle's area in terms of the circumradius and using $[ABC]+[CDA]=[BCD]+[ABD]$ gives $(ab+cd)\cdot AC=2k\cdot BD.$

Ptolemy's theorem gives $AC\cdot BD=ac+bd.$ Eliminating $AC,$ $$BD^2=\frac{(ac+bd)(ab+cd)}{2k}=\frac12\left(a^2+b^2+c^2+d^2\right).$$

The sides are distinct integers below $15$ with $bc=ad,$ so neither $11$ nor $13$ can appear (each is prime and would need a matching factor on the other side).

To maximize, take the largest side $14.$ Writing the others as $s_1\gt s_2\gt s_3$ with $14s_3=s_1s_2,$ the best case is $s_2=7,$ giving $s_1=2s_3$ and $(a,b,c,d)=(14,12,7,6).$ Then $$2BD^2=14^2+12^2+7^2+6^2=425,$$ so $BD=\sqrt{\dfrac{425}{2}}.$

Thus, the correct answer is D.

Solution:

Write $P(x)=(x-h_1)^2-k_1^2$ and $Q(x)=(x-h_2)^2-k_2^2,$ with minimum values $-k_1^2$ and $-k_2^2.$

The zeros of $P(Q(x))$ occur where $Q(x)=h_1\pm k_1;$ their four solutions are symmetric about $h_2,$ so $h_2$ is the average $\tfrac{-23-21-17-15}{4}=-19.$ Then $Q(-15)-Q(-17)=(16-k_2^2)-(4-k_2^2)=12,$ and this difference equals $2k_1,$ so $k_1=6.$

Symmetrically, $h_1=\tfrac{-59-57-51-49}{4}=-54,$ and $P(-49)-P(-51)=(25-k_1^2)-(9-k_1^2)=16=2k_2,$ so $k_2=8.$

The sum of the minimum values is $-k_1^2-k_2^2=-36-64=-100.$

Thus, the correct answer is A.

Solution:

Let $f(x)$ be the left-hand side. On each interval between consecutive asymptotes $2009, 2010, 2011,$ the function $f$ is decreasing, and $f\lt1$ for all $x\lt2009.$

On each of $(2009,2010),$ $(2010,2011),$ and $(2011,\infty),$ the solution is the part from the left asymptote up to a value $x_i$ where $f(x_i)=1.$ So the solution set consists of three intervals with left endpoints $2009, 2010, 2011$ and right endpoints $x_1, x_2, x_3.$

The total length is $(x_1-2009)+(x_2-2010)+(x_3-2011)=x_1+x_2+x_3-6030.$

Clearing denominators in $f(x)=1$ gives $$x^3-(2009+2010+2011+3)x^2+\cdots=0,$$ whose roots are $x_1, x_2, x_3.$ By Vieta, $x_1+x_2+x_3=6033,$ so the sum of lengths is $6033-6030=3.$

Thus, the correct answer is C.

Solution:

Since $2010=2\cdot3\cdot5\cdot67,$ write the product as $2^A3^B5^C67^D$ times a factor coprime to all four primes; then $m=\min(A,B,C,D).$

Prime $2:$ $\operatorname{pow}(n)$ is a power of $2$ only when $n=2^k.$ Since $2^{12}=4096\lt5300\lt2^{13},$ the values $k=1,\ldots,12$ contribute $A=1+2+\cdots+12=78.$