Cagney can make $60 \div 20 = 3$ cupcakes per minute, and Lacey can make $60 \div 30 = 2.$

In $5$ minutes, they can make $$5(2 + 3) = 5 \cdot 5 = 25.$$

Thus, D is the correct answer.

Note that the one of the sides remains the same as the original square.

This means that one dimension is $8.$ The other dimension is the original side cut in half, which is $4.$

Thus, E is the correct answer.

It crawls a distance of $$|-2 - (-6)| = |-4| = 4$$ when it moves from $-2$ to $-6.$ It then travels a distance of $$|-6 - 5| = |-11| = 11$$ as it moves from $-6$ to $5.$

The total distance is then $$4 + 11 = 15.$$

Thus, E is the correct answer.

Note that both of the angles share the ray $AB.$ To minimize the desired degree, we want $D$ to be between $A$ and $C.$

This would make $$\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ}. \end{aligned}$$

Thus, C is the correct answer.

We have that there are $100 \div 2 = 50$ female cats. We then have that $50 \div 2 = 25$ cats that have kittens.

Since the average number of kittens per litter is $4,$ the total number of kittens is $$4 \cdot 25 = 100.$$

The total number of cats and kittens is then $$100 + 100 = 200.$$

Thus, B is the correct answer.

Let the two numbers be $x$ and $y$ such that $$xy = 9 \text{ and } \dfrac{1}{x} = \dfrac{4}{y}.$$

We get that $$y = \dfrac{9}{x}$$$$\dfrac{1}{x} = \dfrac{4x}{9}.$$$$x = \dfrac{3}{2},$$ since $x$ is positive.

Then $$y = 9 \div \dfrac{3}{2} = 6.$$

The desired sum is then $$6 + \dfrac{3}{2} = \dfrac{15}{2}.$$

Thus, D is the correct answer.

WLOG, let the number of marbles in the bag be $5.$ Since we only care about ratios, we can do this. Then there are $3$ blue marbles and $2$ red marbles.

Doubling the red marbles gives us $4$ of them. Then the fraction of red marbles is $$\dfrac{4}{4 + 3} = \dfrac{4}{7}.$$

Thus, C is the correct answer.

Let the three numbers be $a, b, c$ where $a \lt b \lt c.$ None of them are equal, since all three sums are different.

Then $$a + b = 12, a + c = 17,$$ and $$b + c = 19.$$

Adding all three equations together gives us $$2(a + b + c) = 48.$$

Then $$a + b + c = 24,$$ from which we can subtract $$a + c = 17,$$ to get $b = 7.$

Thus, D is the correct answer.

The pairs of numbers that sum to $7$ are $$(2, 5), (4, 3), \text{ and } (6, 1).$$

There is a $$\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}$$ chance that we get any of these pairs.

There are $3$ pairs, which means that the total probability that the rolls sum to $7$ is $$3 \cdot \dfrac{1}{9} = \dfrac{1}{3}.$$

Thus, D is the correct answer.

Let $a$ be the smallest possible sector angle and $d$ be the difference in the arithmetic sequence.

Then we have that $$\dfrac{12}{2}(a + a + 11d) = 12a + 66d$$ is the sum of the arithmetic sequence.

We have that this sums to $360,$ so $$12a + 66d = 360$$ $$2a + 11d = 60.$$

We want to minimize $a,$ so we maximize $d.$ If $d = 5,$ then $a$ is not an integer, so $d = 4$ and $a = 8.$

Thus, C is the correct answer.

Let $x$ be $BC.$ Note that $\triangle CEB$ and $\triangle CDA$ are similar due to angle-angle (tangent lines are perpendicular to radii).

Then $$\dfrac{x}{3} = \dfrac{8 + x}{5}.$$ Cross-multiplying gives us $$5x = 24 + 3x$$$$x = 12.$$

Thus, D is the correct answer.

On a typical year with $365$ days, moving one year into the future moves the day of the week forward one since $$365 = 52 \cdot 7 + 1.$$

On a leap year however, the day of the week gets moved forward twice since there is an extra day.

Fifty of the years between $2012$ and $1812$ are multiples of $4,$ but we have to discard $1900,$ which leaves us with $49$ leap years.

Therefore, moving back $200$ years means we have to go back $$200 + 49 = 249 = 35 \cdot 7 + 4$$ days, which means we move back $4$ days in the week.

This takes us back to Friday, which is the day that corresponds to February $7,$ $1812.$

Thus, A is the correct answer.

Let the order of the numbers be $$a, b, c, d, e.$$

Then the iterative average is $$\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$$$$= \dfrac{a + b + 2c + 4d + 8e}{16}.$$

To minimize this, we make the order $$5, 4, 3, 2, 1,$$ which gives us a sum of $$\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$$

To maximize it, we have to reverse this order to get an average of $$\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$$

The difference between these is $$\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$$

Thus, C is the correct answer.

Note that there are $15$ rows with $15$ black tiles and $16$ rows with $16$ black tiles.

This can be seen by observing that the first row has $16$ black tiles, and all the other rows alternate with $15$ and $16$ tiles.

Then, due to the alternating pattern, there will be a total of $16$ rows with $16$ tiles, and the other rows have $15$ tiles.

The total number of black squares is then $$15^2 + 16^2 = 481.$$

Thus, B is the correct answer.

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let $A$ be the origin and $B = (1, 0).$ Then the slope of the line through $A$ is $-\dfrac{1}{2},$ which makes the equation of the line $$y = -\dfrac{1}{2}x.$$ The slope of the line through $B$ is $2.$ The $y$-intercept is $-2.$ This makes the equation of this line $$y = 2x - 2.$$

Equating the equations, we get $$2x - 2 = -\dfrac{1}{2}x$$$$x = \dfrac{4}{5}.$$

This makes the $y$-coordinate of $C$ $$-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.$$

The area of triangle $ABC$ is then $$\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.$$

Thus, B is the correct answer.

Let us find the amount of time that it takes for the runner running at $4.8$ meters per second to lap the slowest person.

We must have that $$4.8x - 4.4x = 500$$$$x = 1250,$$ where $x$ is the amount of time it takes for the faster runner to lap the other.

Note that $4.4 \cdot 1250 = 5500,$ which means that these two runners always intersect at the starting line.

We now have to find the least time, $t,$ such that $t$ is a multiple of $1250$ and the fastest runner ends up at the starting line.

Every $1250$ seconds, the fastest runner runs $1250 \cdot 5 = 6250$ meters. Then in $2500$ seconds, the fastest runner runs $12500$ meters, which is a whole number of laps.

Thus, C is the correct answer.

Recall that we can factor $$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$ Canceling out this factor gives us that $$\dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}.$$

Cross-multiplying and rearranging gives us $$70a^2 - 149ab + 70b^2 = 0.$$ Since $b \neq 0,$ we can divide through by $b^2$ to get $$70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0.$$

Applying the quadratic formula and noting that $a \gt b$ gives us that $\dfrac{a}{b} = \dfrac{10}{7}.$

Since $a$ and $b$ are relatively prime, we have that $a = 10$ and $b = 7.$ Their difference is $10 - 7 = 3.$

Thus, C is the correct answer.

Note that the region enclosed by the curve but outside the hexagon consists of $3$ sectors with angle $240^{\circ}.$

This means that together they form $2$ whole circles with radius $1.$ Now to find the area of the region inside both the hexagon and the curve.

This area can be found by finding the area of the hexagon and subtracting out the areas of the sectors outside the curve.

There are three $\dfrac{1}{3}$ circles that together form a whole circle. The area of the hexagon can be given by $$\dfrac{3}{2} \cdot 2^2 \cdot \sqrt{3} = 6\sqrt{3}.$$

The desired area is then $$2\pi + (6\sqrt{3} - \pi) = \pi + 6\sqrt{3}.$$

Thus, E is the correct answer.

Let the lunch break be $m$ minutes, Paula's rate be $p$ percent per minute, and the helpers' combined rate be $h$ percent per minute.

Monday gives $(p+h)(480-m)=50$. Tuesday gives $h(372-m)=24$. Since the remaining work on Wednesday was $26$ percent, Wednesday gives $p(672-m)=26$.

Adding the Tuesday and Wednesday equations and subtracting the Monday equation gives $108h-192p=0$, so $h=\dfrac{16}{9}p$.

Substituting into Monday gives $\dfrac{25}{9}p(480-m)=50$, while Wednesday gives $p(672-m)=26$. Solving these two equations gives $m=48$.

Thus, D is the correct answer.

The center square must initially be black, contributing probability $\dfrac12$. The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle.

For one $4$-cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are $BBBB$, the four rotations of $BBBW$, and the two rotations of $BWBW$, for $7$ colorings out of $16$.

The same count applies to the edge-middle cycle, independently. Therefore the probability is $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$.

Thus, A is the correct answer.

Note that $EF = \dfrac{1}{2}AD$ since it is a midsegment of $\triangle ABD.$ Similarly, $HG = \dfrac{1}{2}AD$ and $FG = \dfrac{1}{2}BC.$