Solution:

Brianna is $42 \div 2 = 21$ years old. Caitlin is therefore $21 - 5 = 16$ years old.

Thus, B is the correct answer.

Solution:

$0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals.

The reciprocal of $2$ is $\frac{1}{2},$ but $2$ is not less than $\frac{1}{2}.$

Therefore, as $-2 < -\frac 12,$ we know that $-2$ is the only one of the answer choices that is less than its reciprocal.

Thus, A is the correct answer.

Solution:

The smallest whole number greater than $\frac{5}{3}$ is $2.$ The greatest whole number less than $2\pi$ is $6.$

The whole numbers within this range are $$2, 3, 4, 5, \text{ and } 6.$$

Thus, D is the correct answer.

Solution:

The only graph that shows all the data points is graph E .

Thus, E is the correct answer.

Solution:

To maximize the number of principals, assume that the first year of this period is the final year of some principal's term.

Then, there can be $2$ more principals for $6$ years, followed by another principal who works the final year.

This is $4$ principals.

Thus, C is the correct answer.

Solution:

We can subtract out the areas of the top unit square, the bottom right unit square, and the top right $4 \times 4$ square.

The area of $ABCD$ is therefore $$5^2 - 2 \cdot 1^2 - 4^2 = 7.$$

Thus, A is the correct answer.

Solution:

To get a negative product using three numbers, you can either multiply one negative number and two positives, or three negatives.

The only $2$ viable options are $$-8 \cdot -6 \cdot -4 = -192$$ and $$-8 \cdot 5 \cdot 7 = -280.$$ The latter is clearly the lesser value.

Thus, B is the correct answer.

Solution:

The sum of the numbers on one die is $$1 + 2 + 3 + 4 + 5 + 6 = 21.$$ Therefore, the sum of the numbers on all $3$ dice is $3 \cdot 21 = 63.$

The visible numbers add up to $$1 + 1 + 2 + 3 + 4 + 5 + 6 = 22.$$ This makes the sum of the unseen numbers $63 - 22 = 41.$

Thus, D is the correct answer.

Solution:

The only $3$-digit powers of $5$ are $125$ and $625.$ This means that the $2$ spot is filled with a $2.$

The only $3$-digit power of $2$ beginning with a $2$ is $256,$ so the outlined square is filled with a $6.$

Thus, D is the correct answer.

Solution:

Let $x$ be Ara and Shea's initial height. Then we get that $$\begin{align*} 1.2x &= 60 \\ x &= 50. \end{align*}$$

This means that Shea grew $10$ inches, which means that Ara grew $10 \div 2 = 5$ inches, making her $50 + 5 = 55$ inches tall.

Thus, E is the correct answer.

Solution:

Numbers ending in $1$, $2$, or $5$ all work in the ranges $11,21,31,41$, $12,22,32,42$, and $15,25,35,45$. This gives $12$ numbers.

The remaining working numbers are $24,33,36,44,48$. Numbers ending in $0$ do not work because division by $0$ is undefined.

Thus there are $12+5=17$ such numbers.

Thus, C is the correct answer.

Solution:

The total number of rows in the wall is $7,$ with each row being $1$ foot high.

To use the minimum number of bricks, rows $1, 3, 5,$ and $7$ will have the same pattern as the bottom row in the picture, which requires $50$ bricks to construct.

Rows $2, 4,$ and $6$ will have the same pattern as the upper row in the picture, which has $49$ $2$-foot bricks in the middle and $1$ $1$-foot bricks on each end, for a total of $51$ bricks.

When you add up $4$ rows of $50$ bricks and $3$ rows of $51$ bricks, you get a total of $$4 \cdot 50 + 3 \cdot 51 = 200 + 153 = 353$$ bricks.

Thus, D is the correct answer.

Solution:

We get $$\begin{align*} \angle CAT + \angle ATC + \angle ACT &= 180 \\ 36 + 2 \cdot \angle ATC &= 180 \\ \angle ATC &= 72^{\circ}. \end{align*}$$

Due to bisection, we also know that $$\angle RTC = 72 \div 2 = 36^{\circ}.$$

Finally, we see that $$\begin{align*} \angle RTC + \angle TCR + \angle CRT &= 180 \\ 36 + 72 + \angle CRT &= 180 \\ \angle CRT &= 72^{\circ}. \end{align*}$$

Thus, C is the correct answer.

Solution:

Note that the units digit of an exponent depends only upon the units digit of the base.

Experimenting, we get that $9$ to even power ends with a $1$ and to an odd power ends with a $9.$

Therefore, $19^{19}$ ends with a $9$ and $99^{99}$ also ends with a $9.$ Adding them together yields a number that ends in $8.$

Thus, D is the correct answer.

Solution:

The large equilateral triangle has side length $4,$ the middle one has side length $2,$ and the smaller one has side length $1.$

The perimeter is therefore $$\begin{gather*} CB + CD + DE + EF \\ FG + GA + AB \\ = 4 + 4 + 2 + 2 + 1 \\ + 1 + 1 = 15. \end{gather*}$$

Thus, C is the correct answer.

Solution:

We can see that the length is $1000 \div 25 = 40$ m, and the perimeter is $1000 \div 10 = 100$ m.

Note that the perimeter is $2$ times the sum of the length and width.

This means that the width is $$100 \div 2 - 40 = 10 \text{ m},$$ and the area is $$40 \cdot 10 = 400 \text{ m}^2.$$

Thus, C is the correct answer.

Solution:

We can calculate it as follows. $$\begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(1 / 2)^2}{3} - \dfrac{1^2}{(4 / 3)} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3}. \end{gather*}$$

Thus, A is the correct answer.

Solution:

Assume that the pegs on this grid are separated by $1$ unit.

Note that region $I$ is a parallelogram with base $1$ and height $1,$ makings its area $1 \cdot 1 = 1.$

We can split region $II$ into $2$ triangles. Both with base $1$ and height $1.$ This makes the sum of the areas $$2 \cdot \dfrac{1}{2} \cdot 1 \cdot 1 = 1.$$ This shows that both regions have the same area.

Note that each region has $2$ sides that are of length $\sqrt{2}.$ Region $I$ has $2$ unit sides, whereas region $II$ only has $1.$

The other side of region $II$ is clearly greater than $1,$ which shows that region $II$ has the greater perimeter.

Thus, E is the correct answer.

Solution:

Create a rectangle that covers the bottom half of the figure as shown below.

Then, we get that $$[ABCD] = [ABD] + [BCD].$$

We also know that $$[ABD] = [BDEF] - [ABE]$$$$- [ADF].$$

$ABE$ and $ADF$ are both quartercircles that form a semicircle with the same area as $CBD.$

This means that $$[ABD] = [BDEF] - [BCD]$$ and $$[ABCD] = [BCD] + [BDEF]$$$$- [BCD]$$ $$= [BDEF] = 10 \cdot 5 = 50.$$

Thus, C is the correct answer.

Solution:

The number of pennies must have the same remainder as $102$ modulo $5$, so there are either $2$ or $7$ pennies. Seven pennies would leave only two coins for nickels, dimes, and quarters, impossible because at least one of each type is needed.

So there are $2$ pennies. The remaining $7$ coins are worth $100$ cents. If $n,d,q$ are the numbers of nickels, dimes, and quarters, then $n+d+q=7$ and $5n+10d+25q=100$.

Dividing the value equation by $5$ and subtracting the coin-count equation gives $d+4q=13$. The only positive solution is $q=3$, $d=1$, and $n=3$.

Thus there must be $1$ dime.

Thus, A is the correct answer.

Solution:

There are $8$ equally likely outcomes if we record Keiko's coin and Ephraim's two coins.

If Keiko gets heads, Ephraim must get exactly one head; this happens in $2$ outcomes. If Keiko gets tails, Ephraim must get no heads; this happens in $1$ outcome.

So $3$ of the $8$ equally likely outcomes work.

Thus, B is the correct answer.

Solution:

The original surface area is just $$6 \cdot 2^2 = 6 \cdot 4 = 24.$$

Note that the top face of the unit cube plus the visible area of the top face of the larger cube is the same as the area of one face of the larger cube.

This means that the unit square on top only adds $4$ unit squares to the total surface area, making the increase $4.$

The percent increase is therefore $$100 \cdot \dfrac{4}{24} = 100 \cdot \dfrac{1}{6} \approx 16.7\%.$$

Thus, C is the correct answer.

Solution:

The sum of the first four numbers is $4 \cdot 5 = 20.$ The sum of the last four numbers is $4 \cdot 8 = 32.$

The sum of all seven numbers is $7 \cdot 6\frac{4}{7} = 46.$ We know that the number common to both sets is included in both of first two sums.

This means that the sum of the first two sums includes every number once, except for the common number which is included twice.

The third sum, however, only includes every number once. This means that the sum of the first two sums minus the third sum yields our desired number.

Therefore, the common number is $$20 + 32 - 46 = 52 - 46 = 6.$$

Thus, B is the correct answer.

Solution:

In $\triangle AFG$, the two base angles are equal and $\angle A=20^\circ$, so

$$\angle AFG=\angle AGF=\frac{180^\circ-20^\circ}{2}=80^\circ.$$

Since $\angle AFG$ and $\angle BFD$ form a straight angle, $\angle BFD=100^\circ$.

In $\triangle BFD$, $\angle B+\angle D=180^\circ-100^\circ=80^\circ$.

Thus, D is the correct answer.

Solution:

Let the rectangle have side lengths $2a$ and $2b$, so its area is $4ab=72$ and $ab=18$.

The three right triangles outside the desired triangle have areas $\frac12(2a)(b)=ab$, $\frac12(2b)(a)=ab$, and $\frac12(a)(b)=\frac12ab$.

Their total area is $\frac52ab=45$. Therefore the desired triangle has area $72-45=27$.

Thus, B is the correct answer.