Solution:

Since $(-2)^{-2}=\frac{1}{(-2)^2}=\frac14$, the expression is $$2-\frac14=\frac74.$$

Thus, the correct answer is C.

Solution:

The time it takes to do $2$ tasks is $100$ minutes. Thus, it takes $50$ more minutes after $2:40,$ which is $3:30.$

Thus, the correct answer is B .

Solution:

Let the number written twice be $x$, and let the number written three times be $y$. Then $2x+3y=100$.

If $x=28$, then $3y=100-56=44$, impossible for an integer $y$. Therefore $y=28$, and $2x=100-84=16$, so $x=8$.

Thus, the correct answer is A.

Solution:

Since $\frac 13 > \frac 14 > \frac 15,$ we know Beth ate more than Cyril and Cyril ate more than Alex. Thus, those three are in order.

The amount Dan ate is $$1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}.$$ This is greater than $\frac 15$ and less than $\frac 14 ,$ so Dan is in between Cyril and Alex. This makes the order Beth, Cyril, Dan, Alex.

Thus, the correct answer is C .

Solution:

Marta finished 6th, so Jack finished 5th. Since Jack finished 2 places behind Todd, Todd finished 3rd. Since Todd finished 1 place behind Rand, Rand finished 2nd.

Hikmet finished 6 places behind Rand, so Hikmet finished 8th.

Thus, the correct answer is B.

Solution:

Marley plays basketball on Monday and golf on Wednesday. She cannot fit all three running days among Thursday, Friday, Saturday, and Sunday without having two consecutive running days, so Tuesday must be a running day.

From Thursday through Sunday, she must run twice, swim once, and play tennis once. Tennis cannot be the day after running or swimming, so tennis must be Thursday. Then the two remaining running days must be Friday and Sunday, leaving Saturday for swimming.

Thus, the correct answer is E.

Solution:

$$\begin{align*} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{align*}$$

Thus, the correct answer is A .

Solution:

The rotation puts the F under the $x$-axis with its lines going to the right.

Then, note that a half turn is the same as reflecting upon both axes in any order, so it undoes the reflection upon the $y$-axis and reflects it upon the $x$-axis. This means the last two turns just reflects it upon the $x$-axis.

The reflection puts the F above the $x$-axis with its lines going to the right.

Thus, the correct answer is E .

Solution:

The larger boundary is a quarter circle of radius $3$, so its area is $\frac14\pi\cdot3^2=\frac{9\pi}{4}$.

The inner boundary is the right half of a circle of radius $\frac32$, so its area is $\frac12\pi\left(\frac32\right)^2=\frac{9\pi}{8}$.

The shaded area is the difference, $\frac{9\pi}{4}-\frac{9\pi}{8}=\frac{9\pi}{8}$.

Thus, the correct answer is B.

Solution:

There are $1007$ odd numbers greater than $-2015.$

Our product is of an odd number of negative numbers, so the result is negative.

Also, we multiply by $-5$ in there, so the product is a multiple of $5,$ making it end in $5$ or $0.$ None of our factors are even, so the product can't be even.

Therefore, the product must end in $5.$

Thus, the correct answer is C .

Solution:

The available digits are $2,3,5,7$. There are $4$ one-digit numbers and $4^2=16$ two-digit numbers, for $20$ total choices.

All $4$ one-digit choices are prime. A two-digit prime cannot end in $2$ or $5$, so checking endings $3$ and $7$ gives the two-digit primes $23,37,53,73$.

Thus $8$ of the $20$ choices are prime, and the probability is $\frac{8}{20}=\frac25$.

Thus, the correct answer is B.

Solution:

The squared distance from $(x,-x)$ to $(5,5)$ is $$(x-5)^2+(-x-5)^2=2x^2+50.$$

Being inside or on the circle means $2x^2+50\le100$, so $x^2\le25$. Thus $-5\le x\le5$, giving $11$ integer values.

Thus, the correct answer is A.

Solution:

The triangle is a right triangle with legs of $12$ and $5.$ This makes the hypotenuse $13.$

Two of the altitudes are then $12$ and $5.$ Also, for any side, $A = \frac{bh}2$ where $b$ is the base and $h$ is the altitude.

The area is $\frac {12\cdot 5}2 = 30,$ so the other altitude $h$ can be found with $30 = \frac{13h}2.$ Thus, this altitude is $\frac{60}{13}.$

Therefore, the sum is $$12+5+\dfrac{60}{13} = \dfrac{281}{13} .$$

Thus, the correct answer is E .

Solution:

The equation is equal to $$(2x-(a+c))(x-b)$$$$= 2(x-b)\left(x- \dfrac{a+c}2\right).$$ This makes the roots equal to: $$b,\dfrac{a+c}2$$ and the sum is $$\dfrac{2b+a+c}2.$$

Therefore, we want to maximize $a,b,c,$ while making $b$ the highest.

As such, we can have $b=9,a=8,c=7$ and get a sum: $$9+7.5=16.5$$

Thus, the correct answer is D .

Solution:

If there are $h$ horses and $c$ cows, then there are $3h$ people, $9h$ ducks, and $4c$ sheep. The total is therefore $13h+5c$.

The listed values except $47$ can be written in that form: $$41=13\cdot2+5\cdot3,\quad 59=13\cdot3+5\cdot4,\quad 61=13\cdot2+5\cdot7,\quad 66=13\cdot2+5\cdot8.$$ For $47$, subtracting $0,13,26,39$ leaves $47,34,21,8$, none of which is divisible by $5$.

Thus, the correct answer is B.

Solution:

Let $(A,B,C)$ be the numbers assigned to Al, Bill, and Cal. We need $A$ to be a multiple of $B$, and $B$ to be a multiple of $C$, with all three numbers distinct.

The valid triples are $$(4,2,1),(6,2,1),(8,2,1),(10,2,1),(6,3,1),(9,3,1),(8,4,1),(10,5,1),(8,4,2).$$ There are $9$ favorable assignments.

The total number of assignments is $10\cdot9\cdot8=720$, so the probability is $\frac9{720}=\frac1{80}$.

Thus, the correct answer is C.

Solution:

The octahedron can be viewed as two congruent pyramids whose shared base is the rhombus through the centers of the four side faces. This rhombus has diagonals $4$ and $5$, so its area is $\frac12\cdot4\cdot5=10$.

Each pyramid has height $\frac32$, half the prism's height. Thus the total volume is $$2\cdot\frac13\cdot10\cdot\frac32=10.$$

Thus, the correct answer is B.

Solution:

A coin ends as tails if and only if it has $3$ flips that are tails, which happens with probability $\frac 18.$ Thus, the probability of any coin being heads is $\frac 78 .$

As the probability that a given coin flip is $\frac 78,$ and there are $64$ coin flips in total, the expected number of coins that are now heads is: $$64 \cdot \dfrac 78 = 56.$$

Thus, the correct answer is D .

Solution:

The center of the circle through $X,Y,Z,W$ lies on the perpendicular bisectors of $XY$ and $ZW$. These are also the perpendicular bisectors of $AB$ and $AC$, so the same point is the circumcenter of right triangle $ABC$.

Therefore the center is the midpoint $O$ of hypotenuse $AB$, so $OA=OB=OC=6$. Let $a=\frac12BC$ and $b=\frac12CA$. Then $a^2+b^2=6^2$.

From the square on $AB$, $OX^2=6^2+12^2=180$. From the square on $AC$, the corresponding radius also gives $OW^2=b^2+(a+2b)^2$. Hence $$b^2+(a+2b)^2=180.$$ Together with $a^2+b^2=36$, this gives $a=b=3\sqrt{2}$.

Thus $AC=BC=6\sqrt{2}$, and the perimeter is $12+12\sqrt{2}$.

Thus, the correct answer is C.

Solution:

The first two edges can be chosen in $3\cdot2=6$ ways. These two edges determine an initial face of the cube. After those moves, there is one unvisited vertex on that initial face.

That remaining vertex must be visited next; otherwise Erin would later reach it after all of its neighbors had already been visited, and the path could not continue. The last four vertices are then on the opposite face and can be visited in two cyclic orders.

Of those two orders, exactly one ends at a vertex not adjacent to the starting point. Hence there are $6$ valid paths.

Thus, the correct answer is A.

Solution:

Suppose Dash takes $d+1$ jumps. Then the number of steps $t$ is one of $$5d+1,5d+2,5d+3,5d+4,5d+5.$$ Cozy takes $19$ more jumps, so Cozy takes $d+20$ jumps, which means $t$ is either $2d+39$ or $2d+40$.

Matching these possibilities, the integer solutions are $$5d+3=2d+39,\quad 5d+1=2d+40,\quad 5d+4=2d+40.$$ They give $t=63,66,64$, respectively.

Thus $s=63+66+64=193$, and the sum of its digits is $13$.

Thus, the correct answer is D.

Solution:

By symmetry, $AG=HC=HJ=1$, and triangles $AFG$ and $BGH$ are congruent, so $FG=GH$. Let $FG=b$, and let $CD=d$.

The similar triangles in the pentagon give $$\frac1b=1+b$$ and $$\frac{1+b}{1}=d.$$ The first equation is $b^2+b-1=0$, so $b=\frac{-1+\sqrt{5}}{2}$. Then $d=1+b=\frac{1+\sqrt{5}}{2}$.

Therefore $$FG+JH+CD=b+1+d=1+\sqrt{5}.$$

Thus, the correct answer is D.

Solution:

The number of trailing zeros is the number of factors of $5$. For $5\le n\le9$, $n!$ has $k=1$ zero. We need $(2n)!$ to have $3$ zeros, which happens when $15\le2n\le19$. Thus $n=8,9$.

For $10\le n\le14$, $n!$ has $k=2$ zeros. We need $(2n)!$ to have $6$ zeros, which happens when $25\le2n\le29$. Thus $n=13,14$.

These are the four least possible values, so $s=8+9+13+14=44$. The sum of the digits of $s$ is $8$.

Thus, the correct answer is B.

Solution:

When Aaron reaches $(k,-k)$, he has just completed the square spiral containing all grid points with coordinates between $-k$ and $k$. Therefore $$p_{(2k+1)^2-1}=(k,-k).$$

With $k=22$, this gives $p_{2024}=(22,-22)$. Since $2024-2015=9$, stepping backward along the bottom edge subtracts $9$ from the $x$-coordinate, so $$p_{2015}=(13,-22).$$

Thus, the correct answer is D.

Solution:

The condition is $$abc=2(ab+ac+bc).$$ Since $abc\le6bc$, we have $a\le6$. Also $a=1$ and $a=2$ give no positive solutions, so test $a=3,4,5,6$.

For $a=3$, $(b-6)(c-6)=36$, giving $(b,c)=(7,42),(8,24),(9,18),(10,15),(12,12)$. For $a=4$, $(b-4)(c-4)=16$, giving $(5,20),(6,12),(8,8)$.

For $a=5$, $(3b-10)(3c-10)=100$, and the only solution with $a\le b\le c$ is $(b,c)=(5,10)$. For $a=6$, $(b-3)(c-3)=9$, and the only solution with $a\le b\le c$ is $(6,6)$.

The total number of triples is $5+3+1+1=10$.

Thus, the correct answer is B.