Small Hint:

Convert each frosting rate to cupcakes per minute

Big Hint:

Add the two rates, then multiply by $5$ minutes

Solution:

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.

Small Hint:

One side of each rectangle stays $8$

Big Hint:

The other side is half of the square side length

Solution:

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.

Small Hint:

Add the distance from $-2$ to $-6$ and from $-6$ to $5$

Big Hint:

Distance on a number line uses absolute differences

Solution:

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.

Small Hint:

Both angles share ray $BA$

Big Hint:

The smallest angle occurs when the two given angles overlap as much as possible

Solution:

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.

Small Hint:

Half the adult cats are female, and half of those have litters

Big Hint:

Multiply the number of litters by the average litter size

Solution:

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.

Small Hint:

Let the two numbers be $x$ and $y$

Big Hint:

Use $xy=9$ and the reciprocal condition to relate $x$ and $y$

Solution:

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.

Small Hint:

Start with a $3:2$ ratio of blue to red marbles

Big Hint:

Doubling the red marbles changes the ratio to $3:4$

Solution:

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.

Small Hint:

Add the three pair sums

Big Hint:

The total of the three pair sums is twice the sum of the three numbers

Solution:

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.

Small Hint:

List the even-odd pairs that sum to $7$

Big Hint:

Each listed value appears on two faces of its die

Solution:

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.

Small Hint:

Let the arithmetic sequence be $a,a+d,\ldots,a+11d$

Big Hint:

Use $12a+66d=360$ and maximize $d$ while keeping $a$ positive

Solution:

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.

Small Hint:

Draw radii to the two tangent points

Big Hint:

The two right triangles with vertex $C$ are similar

Solution:

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.

Small Hint:

Count day shifts modulo $7$

Big Hint:

In $200$ years, include leap days but exclude $1900$

Solution:

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.

Small Hint:

Write the final average as a weighted sum

Big Hint:

Later entries receive larger weights

Solution:

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.

Small Hint:

Rows alternate between $16$ and $15$ black squares

Big Hint:

There are $16$ rows of one type and $15$ of the other

Solution:

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.

Small Hint:

Place $A$ and $B$ on coordinate axes

Big Hint:

Find the intersection of the two drawn lines

Solution:

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.

Small Hint:

Use relative speeds on the circular track

Big Hint:

Find when the middle and fast runners lap the slow runner by whole laps

Solution:

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.

Small Hint:

Factor $a^3-b^3$

Big Hint:

The equation becomes a quadratic in $\frac{a}{b}$

Solution:

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.

Small Hint:

Each arc is a $120^\circ$ arc of a unit circle

Big Hint:

Rearrange the sector pieces around the regular hexagon

Solution:

Each arc has radius $1$ and length $\frac{2\pi}{3},$ so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle.

The regular hexagon is made of six equilateral triangles of side $2,$ so its area is $$6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.$$ Adding the unit circle gives a total enclosed area of $\pi+6\sqrt3.$

Thus, E is the correct answer.

Small Hint:

Let the lunch break be $m$ minutes

Big Hint:

Use Monday for $p+h$, Tuesday for $h$, and Wednesday for $p$

Solution:

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.

Small Hint:

Corner squares form a $4$-cycle under rotation

Big Hint:

For each $4$-cycle, count colorings with no white square receiving a white predecessor

Solution:

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.