Solution:

The total cost is \begin{align*} $1.5 + 5 \cdot $.25 &= $1.5 + $1.25\\ &= $2.75. \end{align*}

Thus, C is the correct answer.

Solution:

We just need to divide the total amount of sugar needed by the amount of sugar that the cup holds.

Therefore, She will need to refill the cup $$\dfrac{2\frac{1}{2}}{\frac{1}{4}} = \dfrac{5}{2} \cdot 5 = 10$$ times.

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 amount they paid was $105 + $125 + $175 = $405. This means that each should pay $405 \div 3 = $135. This means that Tom has to pay $30 more, and Dorothy has to pay $10 more.

Then $t = 30,$ $d = 10,$ and therefore, $$t - d = 20.$$

Thus, B is the correct answer.

Solution:

The two pairs of ages that add to $16$ are $(3, 13)$ and $(7, 9).$

Note, however, that we need two brothers younger than $10,$ but not the $5$-year old, to play baseball.

If the brothers that go to the moves are $7$ and $9,$ there are no choices for the brothers that play baseball.

This means that the $3$ and $13$ year olds go to the movies, and the $7$ and $9$ year olds play baseball.

Therefore, Joey is the $11$-year old.

Thus, D is the correct answer.

Solution:

There are $2$ cases. The first case is the student choose both math classes, the and the second case is they only take one.

Case 1: the student takes both math classes

There are $3$ classes left, which means that the student has $3$ choices for their final class.

Case 2: the student takes one math class

There are $2$ choices for which math class the student takes. Then, there are $3$ courses left, from which the students must choose $2.$

This is the same as choosing which course the student does not take, which can be done in $3$ ways.

Therefore, this case contributes $2 \cdot 3 = 6$ schedules that the student can take.

The total number of configurations in both cases is $3 + 6 = 9.$

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:

Let $A = (6, 5),$ $B = (8, -3),$ and $C = (9, 1).$ After reflecting, we get the points $E = (7, 1)$ and $D = (10, 5).$

Note that the area of the union is the area of $\triangle ABD$ minus the area of region $F.$

Sketching this diagram, we get: $$$$

The intersection of $\overline{DE}$ and $\overline{AC}$ occurs at $x = 8$ due to symmetry. We also get that $\overline{AC}$ can be represented as the line $$y=-\dfrac{4}{3}x + 13.$$

Plugging in $x = 8,$ we get the $y$-coordinate to be: \[\begin{align*} y&=-\dfrac{4}{3} \cdot 8 + 13 \\ &= -\dfrac{32}{3} + 13 \\ &=\dfrac{7}{3}. \end{align*}

The area of region $F$ is then $$\dfrac{1}{2} \cdot \left(5 - \dfrac{7}{3}\right) \cdot 4 = \dfrac{16}{3}.$$

Finally, the area of $\triangle ABD$ is $$\dfrac{1}{2} \cdot 4 \cdot 8 - \dfrac{16}{3} = \dfrac{32}{3}.$$

Thus, E is the correct answer.

Solution:

Note that the least common multiple of $3, 4,$ and $5$ is $60.$

We can split up the year into $6$ $60$-day periods, and count the number of times exactly two friends visit.

We have that Alice and Beatrix visit together $$\dfrac{60}{3 \cdot 4} - 1 = 5 - 1 = 4$$ times. We subtract $1$ since on the $60$ th day, all $3$ friends visit, which we don't want to count.

Similarly, Alice and Claire visit $$\dfrac{60}{3 \cdot 5} - 1 = 4 - 1 = 3$$ times and Beatrix and Claire visit $$\dfrac{60}{4 \cdot 5} - 1 = 3 - 1 = 2$$ times. This means that in any $60$-day period, exactly $2$ friends visit $$4 + 3 + 2 = 9$$ times. There are $6$ periods, which means that on $6 \cdot 9 = 54$ days, exactly $2$ friends visit.

Thus, B is the correct answer.

Solution:

We begin by finding the are of $ABCD.$ To do this, label the point $E= (\frac pq, \frac rs)$ and drop altitudes from $B$ and $C$ to the $x$-axis as follows:

Looking at the triangle $\triangle ABX,$ observe that the area is equal to: $$[\triangle ABX] = \dfrac 12 \cdot 1 \cdot 2 = 1$$ Similarly, looking at the trapezoid $XBCY,$ observe that the area is equal to: $$[XBCY] = \dfrac 12 \left(3-1\right)(2+3) = 5$$ And lastly, looking at the triangle $\triangle CYD,$ observe that the area is equal to: $$[\triangle CYD] = \dfrac 12 (1)(3) = \dfrac 32$$

Therefore the area of $ABCD$ is equal to: $$[ABCD] = 1+5+\dfrac 32 = \dfrac{15}{2}$$

The area of $\triangle ADE$ is then $\dfrac{15}{4}.$ This means that the height of this triangle is $$\dfrac{1}{2} \cdot 4h = \dfrac{15}{4}$$$$h = \dfrac{15}{8}.$$

We have that the slope of $\overline{CD}$ is $-3,$ which means that this line can be expressed with $$y = -3x + 12.$$ Plugging in the value $y=h,$ we get $$\dfrac{15}{8} = -3x + 12$$$$x = \dfrac{27}{8}.$$

We now have both the $x$ and $y$-coordinates of the intersection point. The desired sum is then $$15 + 27 + 8 + 8 = 58.$$

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:

The above figure illustrates the region that the square sweeps out. We can find the area of the shaded region by splitting it up.

We have four sectors and eight triangles that we can divide the shaded region into.

Note that two triangle regions can be combined to form a kite, which we can easily find the area of.

The area of the four sectors is half the area of the circle. The radius of the circle can be calculating by realizing that the radius is one half the diameter of the square.

Therefore, $$r = \dfrac{\sqrt{2}}{2}$$$$\dfrac{1}{2} \pi r^2 = \dfrac{\pi}{4}.$$

One of the diagonals of the kite is $r.$ Let the other be $x.$

Then we get the equation $$x + x\sqrt{2} = 1$$$$x = \sqrt{2} - 1.$$

This equation comes from splitting up a side of the square into $x,$ and the legs of two isosceles right triangles whose hypotenuse is $x.$

Finally, the area of all $4$ kites is $$4 \cdot \dfrac{1}{2} \cdot (\sqrt{2} - 1) \cdot \dfrac{\sqrt{2}}{2} = 2 - \sqrt{2}.$$

The total area is then $$2 - \sqrt{2} + \dfrac{\pi}{4}.$$

Thus, C is the correct answer.

Solution:

Let $x$ be the number of coins initially in the treasure chest. Note that after the $k^{\text{th}}$ pirate, $\dfrac{12 - k}{k}$ of the coins are left.

This means that $$x \cdot \dfrac{11}{12} \cdot \dfrac{10}{12} \cdot \dfrac{9}{12} \cdot \dfrac{8}{12} \cdot \dfrac{7}{12} \cdot \dfrac{6}{12}$$ $$\cdot \dfrac{5}{12} \cdot \dfrac{4}{12} \cdot \dfrac{3}{12} \cdot \dfrac{2}{12} \cdot \dfrac{1}{12}$$ is an integer.

We want to minimize $x$ while keeping the above expression an integer. Cancelling out the common factors of the numerator and denominator will tell us what $x$ needs to be.

After doing this, we get $$11 \cdot 5 \cdot 7 \cdot 5 = 1925$$ is left in the numerator. If we set $x$ equal to the resulting denominator, we have that the $12$th pirate gets $1925$ coins.

Thus, D is the correct answer.

Solution:

We can consider the cross-section of the largest sphere that contains all the six smaller spheres.

From this we can see that the radius of the largest sphere is $2 + 1 = 3.$

Now to find the radius, $r,$ of the eight sphere, we can construct a right triangle connecting the centers of a small sphere, the seventh sphere, and the eight sphere.

The distance between the small sphere and the seventh sphere is $2.$ The other leg is $3 - r,$ the hypotenuse is $1 + r.$

We can apply the Pythagorean Theorem to get $$2^2 + (3 - r)^2 = (1 + r)^2.$$ Simplifying yields $$4 + 9 - 6r = 1 + 2r$$ $$12 = 8r$$$$r = \dfrac{3}{2}.$$

Thus, B is the correct answer.

Solution:

Let $Z$ and $Y$ be the intersections of $\overline{AC}$ and the circle as shown in the diagram.

We have that $$CY = 97 - 86 = 11$$ and $$CZ = 86 + 97 = 183.$$

We can then apply Power of a Point to get $$CX \cdot CB = CY \cdot CZ$$ $$$$$$CX \cdot CB = 11 \cdot 183$$ $$= 11 \cdot 3 \cdot 61.$$

Since $BX$ and $CX$ are integers, we also have that $CB$ is an integer. Also, by the Triangle Inequality, $$BC \lt 86 + 97 = 183.$$

Using these two facts combined with $CX \lt CB,$ we have that the only pair of values that work is $CX = 31$ and $BC = 61.$

Thus, D is the correct answer.

Solution:

Label the players from Central High School $A, B,$ and $C,$ and the players from Northern High School $X, Y,$ and $Z.$

There are $$\dfrac{6!}{2!2!2!} = 90$$ ways to figure out the order in which $A$ plays his matches.

We just need to figure out the order in which $B$ plays their matches. Then $C$'s matches are fixed.

WLOG, let the order of matches $A$ plays be $$XXYYZZ.$$

To figure out the possible arrangements for $B,$ we can case on the positions of $X$'s and $Y$'s.

If $X$ goes in the middle two spots, then $Y$ has to go in the last two spots, Otherwise, the $Z$'s will overlap.

Similarly, if the $X$'s go in the last two spots, then there is only spot to put the $Y$'s.

If one $X$ goes in the middle $2,$ and the other $X$ goes in the last two, then there are $2$ options for the $Y$'s go.

The $Y$'s have $2$ options in the first $2$ spots and the other is forced to be in the remaining spot in the last $2.$

There are $2 \cdot 2$ ways to place the $X$'s in the above configuration. There are then a total of $$1 + 1 + 2 \cdot 2 \cdot 2 = 10$$ ways to determine the schedule for $B$'s matches.

We then have to multiply by $90$ for the number of configurations for $A$'s matches.

The total number of ways to order the games is then $$90 \cdot 10 = 900.$$

Thus, E is the correct answer.

Solution:

We can over count the number of intersections and then subtract out the ones we double counted.

Assuming that every set of $4$ points contribute one intersection, we have $$\binom{8}{4} = 70$$ intersections.

Four diagonals intersect in the center, which means that we need to subtract out $$\binom{4}{2} - 1 = 5$$ because of this. We subtract one since we need to still keep one of them.

As shown in the diagram, multiple lines can also intersect in places such as $\overline{AD}, \overline{CG},$ and $\overline{BE}.$

There are $8$ of these intersections, which means that we have to subtract out $$8 \left(\binom{3}{2} - 1\right) = 16$$ more intersections.

This means that the total number of intersections is $$70 - 5 - 16 = 49.$$

Thus, A is the correct answer.