Solution:

The mileage charge is $\$0.25\cdot 5=\$1.25$.

Adding the fixed charge gives $\$1.50+\$1.25=\$2.75$.

Thus, C is the correct answer.

Solution:

Alice needs $2\frac12=\frac52$ cups of sugar.

Each full measuring cup gives $\frac14$ cup, so the number of fillings is $\frac52\div\frac14=\frac52\cdot4=10$.

Thus, B is the correct answer.

Solution:

We have by the formula for the area of a triangle that $$40 = \dfrac{1}{2} \cdot 10 \cdot BE.$$ This gives us $$40 = 5BE$$$$BE = 8.$$ Thus, E is the correct answer.

Solution:

Note that if they scored twice as many runs as their opponents, then they scored an even number of runs.

This means in the games where they scored $2, 4, 6, 8,$ and $10$ runs, their opponents scored $1, 2, 3, 4,$ and $5$ runs respectively.

This sums to $$1 + 2 + 3 + 4 + 5 = 15.$$

In the other games, their opponents scored $2, 4, 6, 8,$ and $10$ runs.

This sums to $$2 + 4 + 6 + 8 + 10 = 30.$$

The total number of runs is then $$15 + 30 = 45.$$ Thus, C is the correct answer.

Solution:

The total cost was $\$105+\$125+\$175=\$405$, so each person's share was $\$405\div3=\$135$.

Tom paid $\$30$ less than his share, and Dorothy paid $\$10$ less than her share, so $t=30$ and $d=10$.

Therefore $t-d=30-10=20$.

Thus, B is the correct answer.

Solution:

The age pairs that sum to $16$ are $(3,13)$, $(5,11)$, and $(7,9)$.

The $5$-year-old stayed home, so the movie pair cannot be $(5,11)$. If $7$ and $9$ went to the movies, then the only remaining brother younger than $10$ would be the $3$-year-old, not enough for baseball.

Thus the movie pair was $(3,13)$, the baseball pair was $(7,9)$, and Joey is the remaining brother, age $11$.

Thus, D is the correct answer.

Solution:

English is required, so choose the other $3$ courses from the $5$ courses Algebra, Geometry, History, Art, and Latin.

There are $\binom53=10$ such choices, but one of them, History-Art-Latin, contains no mathematics course.

Therefore the number of valid programs is $10-1=9$.

Thus, C is the correct answer.

Solution:

Factoring out a $2^{2012},$ we get: $$\dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{5}{3}.$$

Thus, C is the correct answer.

Solution:

Let $x$ be the number of two-point shots and $y$ be the number of three-point shots.

Then, Shenille scores $.3x$ two-points shots and $.2y$ three-point shots, for a total score of $$2 \cdot .3x + 3 \cdot .2y = .6x + .6y.$$

We know that $$x + y = 30$$$$.6(x + y) = 18.$$

Thus, B is the correct answer.

Solution:

Let the total number of flowers be $x.$ There are $.6x$ pink flowers and $.4x$ red flowers.

Then there are $$\dfrac{1}{3} \cdot .6x = .2x$$ pink roses, which means there are $.4x$ pink carnations.

There are also $$\dfrac{3}{4} \cdot .4x = .3x$$ red carnations. This means there are $.3x + .4x = .7x$ carnations. This is $70\%$ of the total flowers.

Thus, E is the correct answer.

Solution:

Let $x$ be the number of students. Then the number of ways to pick a two-person committee is $$\binom{x}{2} = \dfrac{x(x - 1)}{2}.$$ We know that this equals $10,$ so $$x^2 - x = 20$$$$x^2 - x - 20 = 0.$$ Factoring yields $$(x - 5)(x + 4) = 0$$$$x = 5,$$ since there cannot be a negative number of students.

Then, the number of ways to pick a $3$-person committee is $$\binom{5}{3} = \binom{5}{2} = 10.$$

Thus, A is the correct answer.

Solution:

Note that $\triangle DBE \sim \triangle ABC$ and $\triangle FEC \sim \triangle ABC$ due to the parallel lines.

This tells us that $DB = DE$ and $FE = FC.$ We have that the perimeter of $ADEF$ is $$AD + DE + EF + AF$$$$= AD + DB + FC + AF$$ $$= AB + AC$$$$= 56.$$

Thus, C is the correct answer.

Solution:

Note that for the number to not be divisible by $5,$ the units digits cannot be either $0$ or $5.$

Let $x$ be the hundreds and units digit and $y$ be the tens digit. Then we want $$2x + y \lt 20.$$ Casing on the $9$ options of $x,$ we get:

If $x$ is $1, 2, 3,$ or $4,$ then $y$ can be anything since $y \lt 10.$

If $x = 6,$ then $y \lt 8,$ which gives us $8$ solutions.

If $x = 7,$ then $y \lt 6,$ which gives us $6$ solutions.

If $x = 8,$ then $y \lt 4,$ which gives us $4$ solutions.

If $x = 9,$ then $y \lt 2,$ which gives us $2$ solutions.

This gives us a total of $$4 \cdot 10 + 8 + 6 + 4 + 2 = 60$$ solutions.

Thus, B is the correct solution.

Solution:

Removing the cubes does not remove any edges from the original cube. It only adds edges.

After removing each cube, we can see that $9$ extra edges are added to the solid.

$8$ cubes are removed, which means $8 \cdot 9 = 72$ edges are added to the original $12$ edges, for a total of $72 + 12 = 84$ edges.

Thus, D is the correct answer.

Solution:

Let $h_1$ be the length of the altitude to the side of length $10$ and similarly define $h_2$ for the other given side.

We have that $$10h_1 = 15h_2$$$$h_1 = \dfrac{3}{2}h_2.$$

The third altitude is the average of the other two, which makes its length $$\dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2.$$

Let the third side have length $x.$ Then $$\dfrac{5}{4}h_2x = 15h_2$$$$x = 12.$$

Thus, D is the correct answer.

Solution:

The reflected triangle has vertices $(7,1)$, $(8,-3)$, and $(10,5)$.

The line through $(6,5)$ and $(9,1)$ is $y=-\frac43x+13$, so it meets $x=8$ at $y=\frac73$. By symmetry, the union is two congruent triangles with vertical base from $(8,-3)$ to $(8,\frac73)$.

That base has length $\frac73+3=\frac{16}{3}$, and each triangle has horizontal height $2$. Hence the union area is $2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}$.

Thus, E is the correct answer.

Solution:

Pairwise visits occur every $\operatorname{lcm}(3,4)=12$, $\operatorname{lcm}(3,5)=15$, and $\operatorname{lcm}(4,5)=20$ days.

In the next $365$ days, these pair counts are $\left\lfloor\frac{365}{12}\right\rfloor=30$, $\left\lfloor\frac{365}{15}\right\rfloor=24$, and $\left\lfloor\frac{365}{20}\right\rfloor=18$.

All three visit every $60$ days, which happens $\left\lfloor\frac{365}{60}\right\rfloor=6$ times. Subtracting those days from each pair count gives $(30-6)+(24-6)+(18-6)=54$.

Thus, B is the correct answer.

Solution:

Let the cutting line meet $\overline{CD}$ at $G$. Drop perpendiculars from $B$, $C$, and $G$ to the $x$-axis as in the diagram.

The areas of $\triangle ABF$, trapezoid $BCEF$, and $\triangle CDE$ are $1$, $5$, and $\frac32$, respectively, so $[ABCD]=\frac{15}{2}$.

Thus $\triangle ADG$ has area $\frac{15}{4}$. Since $AD=4$, the height of $G$ is $\frac{15}{8}$.

The line $CD$ has equation $y=-3x+12$, so $\frac{15}{8}=-3x+12$, giving $x=\frac{27}{8}$. Therefore $p+q+r+s=27+8+15+8=58$.

Thus, B is the correct answer.

Solution:

Note that the units digit of represents the remainder when the number is divided by the base.

The question then boils down to finding all numbers, $b,$ such that $2013$ leaves a remainder of $3$ when divided by $b.$

This means that $b$ must divide $2010.$ Also note that $b \geq 4,$ since otherwise the remainder cannot be $3.$

The prime factorization of $2010$ is $$2010 = 2 \cdot 3 \cdot 5 \cdot 67.$$ Then, $2010$ has $$(1 + 1)^4 = 2^4 = 16$$ factors. It has $3$ factors less than $4,$ namely $1, 2,$ and $3.$ This means there are $16 - 3 = 13$ valid values for $b.$

Thus, C is the correct answer.

Solution:

Consider one quarter of the swept region and multiply its area by $4$.

The sector has angle $45^\circ$ and radius $\frac{\sqrt2}{2}$, so its area is $\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}$.

The two right-triangle pieces in that quarter have areas $\frac12\cdot\frac12\cdot\frac12=\frac18$ and $\frac12\left(\frac{\sqrt2-1}{2}\right)^2$.

Multiplying the quarter-area sum by $4$ gives $2-\sqrt2+\frac{\pi}{4}$.

Thus, C is the correct answer.

Solution:

Work backward. If $n$ coins remain for the $12$th pirate, then before pirate $k$ took a share, the chest had $\frac{12}{12-k}$ times as many coins as it had afterward.

Therefore the initial number of coins is $n\cdot\frac{12^{11}}{11!}$.

Since $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$, the smallest $n$ that makes the initial number an integer is $5^2\cdot7\cdot11=1925$.

Thus, the $12$th pirate receives $1925$ coins, and D is the correct answer.

Solution:

The centers of the six radius-$1$ spheres form a regular hexagon of side length $2$, so each is $2$ units from the large sphere's center. Hence the large sphere has radius $3$.

Let the eighth sphere have radius $r$, and let its center be distance $x$ from the large sphere's center. Internal tangency gives $x+r=3$, so $x=3-r$.

Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, $(r+1)^2=2^2+(3-r)^2$. Thus $2r+1=13-6r$, so $r=\frac32$.

Thus, B is the correct answer.

Solution:

By power of a point from $C$, $CB\cdot CX=AC^2-AB^2=97^2-86^2$.

This equals $(97-86)(97+86)=11\cdot183=2013=3\cdot11\cdot61$.

Both $CX$ and $BX$ are integers, so $BC=BX+CX$ is an integer factor paired with $CX$. Also $CX<BC<86+97=183$, so the only possible pair is $CX=33$, $BC=61$.

Thus, D is the correct answer.

Solution:

Label Central's players $A,B,C$ and Northern's players $X,Y,Z$.

Player $A$'s six-round opponent string contains two each of $X,Y,Z$, so it can be chosen in $\frac{6!}{2!2!2!}=90$ ways.

For a fixed $A$-string, say $XXYYZZ$, player $B$'s string must also contain two each of $X,Y,Z$ and cannot match $A$'s opponent in any position.

If $B$'s first two entries are $Y,Z$ in either order, then the middle two entries must be $X,Z$ in either order and the last two must be $X,Y$ in either order, giving $2^3=8$ strings. The two remaining possibilities are $YYZZXX$ and $ZZXXYY$, for $10$ total $B$-strings.

Once $A$ and $B$ are scheduled, $C$'s schedule is forced. Hence there are $90\cdot10=900$ schedules.

Thus, E is the correct answer.

Solution:

If no three diagonals were concurrent, each choice of $4$ vertices would determine one interior intersection, giving $\binom84=70$.

The $4$ long diagonals through opposite vertices all meet at the center, so the center was counted $\binom42=6$ times and should be counted once. Subtract $5$.

There are also $8$ symmetric points where $3$ diagonals meet. Each was counted $\binom32=3$ times and should be counted once, so subtract $8(3-1)=16$.

The number of distinct interior intersection points is $70-5-16=49$.

Thus, A is the correct answer.