Solution:

The sum inside the parentheses is $\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}.$

Its reciprocal is $\dfrac54,$ so the expression equals $10\cdot\dfrac54=\dfrac{25}{2}.$

Thus, the correct answer is C.

Solution:

Since a child ticket is half an adult ticket, $5$ adult and $4$ child tickets equal $5+\tfrac12\cdot4=7$ adult tickets, so one adult ticket costs $\dfrac{24.50}{7}=\$3.50.$

The second purchase equals $8+\tfrac12\cdot6=11$ adult tickets, costing $11\cdot 3.50=\$38.50.$

Thus, the correct answer is B.

Solution:

If orange comes first, then blue and yellow cannot be adjacent, forcing the order orange, blue, red, yellow.

If blue comes first, yellow can be in the third or fourth position (never second, to avoid adjacency), giving blue, orange, yellow, red and blue, orange, red, yellow.

These are the only $3$ valid orderings.

Thus, the correct answer is B.

Solution:

The rate is $\dfrac{b}{ac}$ gallons per cow per day.

So $d$ cows over $e$ days produce $\dfrac{b}{ac}\cdot d\cdot e=\dfrac{bde}{ac}$ gallons.

Thus, the correct answer is A.

Solution:

The remaining $25\%$ scored $100.$ Since $45\%$ scored at most $80$ and $75\%$ scored at most $90,$ the median is $90.$

The mean is $0.10(70)+0.35(80)+0.30(90)+0.25(100)=7+28+27+25=87.$

The difference is $90-87=3.$

Thus, the correct answer is C.

Solution:

Let the larger number be $10a+b.$ Then $$(10a+b)-(10b+a)=9(a-b)=5(a+b),$$ which simplifies to $2a=7b.$

The only nonzero digits satisfying this are $a=7$ and $b=2,$ so the number is $72$ and its reverse is $27.$

Their sum is $72+27=99.$

Thus, the correct answer is D.

Solution:

Writing the terms as powers of $3,$ they are $3^{1/2},$ $3^{1/3},$ $3^{1/6}.$ The common ratio is $\dfrac{3^{1/3}}{3^{1/2}}=3^{-1/6}.$

The fourth term is $3^{1/6}\cdot3^{-1/6}=3^{0}=1.$

Thus, the correct answer is A.

Solution:

For a price $P\gt100,$ the reductions are $\dfrac{P}{10},$ $20,$ and $\dfrac{18}{100}(P-100).$

Coupon 1 beats coupon 2 when $\dfrac{P}{10}\gt20,$ that is $P\gt200.$ Coupon 1 beats coupon 3 when $\dfrac{P}{10}\gt\dfrac{18}{100}(P-100),$ that is $P\lt225.$

The only listed price in $(200,225)$ is $\$219.95.$

Thus, the correct answer is C.

Solution:

The integers $a,a+1,a+2,a+3,a+4$ have average $a+2,$ so $b=a+2.$

The integers starting at $b$ have average $b+2=(a+2)+2=a+4.$

Thus, the correct answer is B.

Solution:

The equilateral triangle has area $\dfrac{\sqrt3}{4}.$ Each isosceles triangle has base $1$ and height $h,$ so $3\cdot\dfrac12 h=\dfrac{\sqrt3}{4},$ giving $h=\dfrac{\sqrt3}{6}.$

A congruent side is the hypotenuse from the apex to a base endpoint: $$\sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}=\sqrt{\dfrac14+\dfrac{1}{12}}=\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}.$$

Thus, the correct answer is B.

Solution:

Let $d$ be the remaining distance after one hour and $t$ the remaining time until the flight. At $35$ mph he would be an hour late, so $d=35(t+1).$ At $50$ mph he is half an hour early, so $d=50\left(t-\tfrac12\right).$

Setting these equal gives $35t+35=50t-25,$ so $t=4$ and $d=175.$

The total distance is $175+35=210$ miles.

Thus, the correct answer is C.

Solution:

Let the circles have radii $R$ (with the $30^\circ$ arc) and $r$ (with the $60^\circ$ arc). The common chord has length $2R\sin15^\circ=2r\sin30^\circ,$ so $\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.$

The smaller central angle gives the larger radius, so $R\gt r.$ The area ratio is $$\left(\dfrac{R}{r}\right)^2=\dfrac{1}{4\sin^2 15^\circ}=\dfrac{1}{2(1-\cos30^\circ)}=\dfrac{1}{2-\sqrt3}=2+\sqrt3.$$

Thus, the correct answer is D.

Solution:

All singles: assign $5$ friends to $5$ rooms in $5!=120$ ways.

One pair: choose the pair in $\binom52=10$ ways, then place the $4$ groups into rooms in $5\cdot4\cdot3\cdot2=120$ ways, giving $10\cdot120=1200.$

Two pairs: choose the solo friend in $5$ ways and split the rest into two pairs in $3$ ways ($15$ groupings), then place the $3$ groups into rooms in $5\cdot4\cdot3=60$ ways, giving $15\cdot60=900.$

The total is $120+1200+900=2220.$

Thus, the correct answer is B.

Solution:

Let $d=b-a\gt0,$ so $b=a+d$ and $c=a+2d.$ Since $a,c,b$ is geometric, $$\dfrac{c}{a}=\dfrac{b}{c}\implies(a+2d)^2=a(a+d),$$ which simplifies to $3ad+4d^2=0,$ so $3a+4d=0.$

Then $a=-4k$ and $d=3k$ for a positive integer $k,$ giving $c=a+2d=2k.$ The smallest value is $c=2$ (with $a=-4,$ $b=-1,$ $c=2$).

Thus, the correct answer is C.

Solution:

Write $\overline{abcba}=10001a+1010b+100c.$ Summing over all palindromes, each value of $a\in\{1,\dots,9\}$ occurs with $10\cdot10=100$ choices of $b,c,$ and each value of $b$ or $c$ occurs with $9\cdot10=90$ choices of the other two digits.

Using $\sum a=\sum b=\sum c=45,$ $$S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)=45\cdot1{,}100{,}000=49{,}500{,}000.$$

The sum of the digits of $S$ is $4+9+5=18.$

Thus, the correct answer is B.

Solution:

By carrying out the multiplication, $8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04,$ which has $k-2$ ones.

The digit sum is $7+(k-2)+0+4=k+9.$ Setting $k+9=1000$ gives $k=991.$

Thus, the correct answer is D.

Solution:

Place the box with a corner at the origin. Each small sphere sits in a corner with center $1$ unit from three faces. The four top small-sphere centers form a square of side $2,$ whose center lies on the box axis; a corner of that square is $\sqrt2$ from the center.

The big sphere's center is on the axis, at distance $2+1=3$ from each top small center. The vertical gap between them is $\sqrt{3^2-\sqrt2^2}=\sqrt7.$

The big center is at height $\dfrac h2$ and the top small centers at height $h-1,$ so $(h-1)-\dfrac h2=\sqrt7,$ giving $\dfrac h2=1+\sqrt7$ and $h=2+2\sqrt7.$

Thus, the correct answer is A.

Solution:

Working from the outside, $f$ is defined exactly when $\log_4\!\left(\log_{1/4}\!\left(\log_{16}\!\left(\log_{1/16}x\right)\right)\right)\gt0,$ which is equivalent to $\log_{1/4}\!\left(\log_{16}\!\left(\log_{1/16}x\right)\right)\gt1.$

Since the base $\tfrac14\lt1,$ this means $0\lt\log_{16}\!\left(\log_{1/16}x\right)\lt\tfrac14,$ hence $1\lt\log_{1/16}x\lt16^{1/4}=2.$

As $\tfrac{1}{16}\lt1,$ this reverses to $\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1,$ i.e. $\tfrac{1}{256}\lt x\lt\tfrac{1}{16}.$ The length is $\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256},$ so $m+n=15+256=271.$

Thus, the correct answer is C.

Solution:

If an integer $x$ is a root, then $k=-\left(5x+\dfrac{12}{x}\right),$ so $x\ne0.$ For $x\ge2,$ $|k|=5|x|+\dfrac{12}{|x|}$ increases, and $|x|=39$ gives $|k|\approx195.3\lt200,$ while $|x|=40$ gives $|k|\gt200.$

Thus $x$ ranges over $\pm1,\pm2,\dots,\pm39,$ which is $78$ values. If two integers $a\ne b$ gave the same $k,$ then $5a+\tfrac{12}{a}=5b+\tfrac{12}{b}$ forces $5ab=12,$ which has no integer solutions, so all $78$ values of $k$ are distinct.

Thus, the correct answer is E.

Solution:

Reflect $B$ across line $AC$ to get $B',$ and reflect $C$ across line $AB$ to get $C'.$ Then $BE=B'E$ and $CD=C'D,$ so $BE+DE+CD=B'E+ED+DC',$ a broken path from $B'$ to $C'.$

This is minimized when the path is the straight segment $B'C'.$ We have $AB'=AB=10,$ $AC'=AC=6,$ and $\angle B'AC'=3\cdot40^\circ=120^\circ.$

By the Law of Cosines, $$B'C'^2=10^2+6^2-2\cdot10\cdot6\cos120^\circ=136+60=196,$$ so $B'C'=14.$

Thus, the correct answer is D.

Solution:

Write $x=n+r$ with integer $n$ ($1\le n\le2013$) and $0\le r\lt1.$ Then $f(x)=n\left(2014^{\,r}-1\right),$ and $f(x)\le1$ becomes $2014^{\,r}\le1+\dfrac1n,$ i.e. $0\le r\le\log_{2014}\dfrac{n+1}{n}.$

Each $n$ contributes an interval of length $\log_{2014}\dfrac{n+1}{n},$ so the total is $$\sum_{n=1}^{2013}\log_{2014}\dfrac{n+1}{n}=\log_{2014}\!\left(\dfrac21\cdot\dfrac32\cdots\dfrac{2014}{2013}\right)=\log_{2014}2014=1.$$

Thus, the correct answer is A.

Solution:

Because $2^2\lt5\lt2^3,$ each interval $(5^n,5^{n+1})$ contains either two or three powers of $2.$ The chain $5^n\lt2^m\lt2^{m+2}\lt5^{n+1}$ holds exactly when the interval contains three consecutive powers of $2,$ and then there is a unique such $m.$

Let $d$ and $t$ be the numbers of intervals $(5^n,5^{n+1})$ for $0\le n\le866$ containing two and three powers of $2,$ respectively. Since $2^{2013}\lt5^{867}\lt2^{2014}$ there are $2013$ powers of $2$ in total, giving $d+t=867$ and $2d+3t=2013.$

Solving, $t=2013-2\cdot867=279.$

Thus, the correct answer is B.

Solution:

Reading the block in pairs of digits (base $100$), $\dfrac{1}{9801}=\dfrac{1}{99^2}$ expands as $00,01,02,\ldots,$ since $\dfrac{1}{(100-1)^2}=\sum_{k\ge1}k\cdot100^{-k}.$ Carrying works out so that every two-digit block $00,01,\ldots,97$ appears, the block $98$ is skipped, and $99$ appears, before the period repeats.

If the blocks $00$ through $99$ all appeared, the digit sum would be $(0+1+\cdots+9)\cdot20=900.$ Removing the missing $98$ subtracts $9+8,$ giving $900-9-8=883.$

Thus, the correct answer is B.

Solution:

If $f_{n-1}(x)=\pm k,$ then $f_n(x)=k-1.$ So if $f_0(x)=\pm k$ for a nonnegative integer $k,$ then $f_k(x)=0,$ after which the sequence alternates $0,-1,0,\ldots$ Thus $f_{100}(x)=0$ exactly when $f_0(x)=2k$ for some integer $-50\le k\le50.$

Now $f_0(x)=x+|x-100|-|x+100|$ equals $x+200$ for $x\lt-100,$ $-x$ for $-100\le x\lt100,$ and $x-200$ for $x\ge100.$ Its graph is piecewise linear with turning points $(-100,100)$ and $(100,-100).$

A line $y=2k$ meets this graph three times for $-49\le k\le49$ and twice for $k=\pm50.$ The total is $99\cdot3+2\cdot2=301.$

Thus, the correct answer is C.

Solution:

Since $(0,0)$ is the midpoint of $A=(4,3)$ and $B=(-4,-3),$ the segment $AB$ is the latus rectum, so the directrix is parallel to $AB$ at distance $5$ on the far side, namely $4y-3x+25=0.$

Equating distances to focus and directrix gives $(4x+3y)^2=25\big(25+2(4y-3x)\big).$ Writing $4x+3y=5s$ forces $5\mid s,$ and $s=5t$ forces $t$ odd; with $t=2u+1$ the integer points are $$x=-6u^2+2u+4,\qquad y=8u^2+14u+3.$$

Then $|4x+3y|=|50u+25|\le1000$ iff $|2u+1|\le39,$ i.e. $-20\le u\le19.$ That gives $40$ lattice points.

Thus, the correct answer is B.