Small Hint:

Find Brianna’s age first

Big Hint:

Caitlin is $5$ years younger than Brianna

Solution:

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

Thus, B is the correct answer.

Small Hint:

Check whether $0$ has a reciprocal

Big Hint:

Compare $-2$ with $-\frac{1}{2}$

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.

Small Hint:

Locate $\frac{5}{3}$ and $2\pi$ between whole numbers

Big Hint:

Count the whole numbers strictly inside the interval

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 $$2, 3, 4, 5, 6.$$

Thus, D is the correct answer.

Small Hint:

Match the four percentages to the four years

Big Hint:

The plotted values should rise from about $5$ to $30$

Solution:

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

Thus, E is the correct answer.

Small Hint:

Let the $8$-year window start at the end of a term

Big Hint:

A principal can appear for only part of the $8$-year period

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.

Small Hint:

Break the shaded L into rectangles

Big Hint:

A $4\times4$ square minus a $3\times3$ square also works

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

Thus, A is the correct answer.

Small Hint:

A negative product needs an odd number of negative factors

Big Hint:

Compare using three negatives versus one negative and two positives

Solution:

A negative product comes either from three negative factors or from one negative factor and two positive factors.

With three negative factors, the minimum is $$-8 \cdot -6 \cdot -4 = -192.$$ With one negative factor, use the most negative number and the two largest positive numbers to get $$-8 \cdot 5 \cdot 7 = -280.$$ Since $-280<-192,$ the minimum possible product is $-280.$

Thus, B is the correct answer.

Small Hint:

Each die has $21$ total dots

Big Hint:

Subtract the visible dots from the total on all three dice

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.

Small Hint:

List the three-digit powers of $5$

Big Hint:

The across entry is a three-digit power of $2$ starting with $2$

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.

Small Hint:

Recover the original common height from Shea’s new height

Big Hint:

Ara grew half as many inches as Shea

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.

Small Hint:

Group numbers by their units digit

Big Hint:

Remember numbers ending in $0$ do not work

Solution:

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

The remaining working numbers are $24,$ $33,$ $36,$ $44,$ and $48.$ 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.

Small Hint:

Start with all $2$-foot blocks

Big Hint:

Only every other row needs an extra block to stagger joins

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 one $1$-foot brick 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.

Small Hint:

First find the two equal base angles of $\triangle CAT$

Big Hint:

Use the angle bisector at $T$

Solution:

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

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

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

Thus, C is the correct answer.

Small Hint:

Only the units digit matters

Big Hint:

Odd powers of a number ending in $9$ end in $9$

Solution:

Note that the units digit of a power depends only upon the units digit of the base.

Experimenting, we get that $9$ to an 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.

Small Hint:

Use the midpoint information to find smaller equilateral side lengths

Big Hint:

Add the outside boundary lengths only

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.

Small Hint:

The length comes from walking it $25$ times

Big Hint:

The perimeter comes from walking it $10$ times

Solution:

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

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

This means that the width is $$100 \div 2 - 40 = 10$$ meters, and the area is $$40 \cdot 10 = 400$$ square meters.

Thus, C is the correct answer.

Small Hint:

Evaluate each $\otimes$ expression from the inside out

Big Hint:

Keep the parentheses; the operation is not associative

Solution:

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

Thus, A is the correct answer.

Small Hint:

Decompose both quadrilaterals into unit right triangles

Big Hint:

Compute all four side lengths of each quadrilateral

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 $1,$ making 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.

Small Hint:

Move the curved pieces to make a rectangle

Big Hint:

The rearranged rectangle has dimensions $5$ and $10$

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 quarter-circles that form a semicircle with the same area as $BCD.$

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

Thus, C is the correct answer.

Small Hint:

Use cents modulo $5$ to find the number of pennies

Big Hint:

After the pennies, solve for nickels, dimes, and quarters

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 $n,$ $d,$ and $q$ 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.

Small Hint:

List Keiko’s result and Ephraim’s two-coin result

Big Hint:

Count matching head totals among the equally likely outcomes

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, and the probability is $\dfrac{3}{8}.$

Thus, B is the correct answer.

Small Hint:

The original cube has surface area $6\cdot2^2$

Big Hint:

The small cube hides one unit square but adds five

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.

Small Hint:

Convert each average into a sum

Big Hint:

The shared number is counted twice in the two four-number sums

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 the 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.

Small Hint:

Use the isosceles triangle $AFG$ first

Big Hint:

Then use the angle sum of $\triangle BFD$

Solution:

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

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

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.

Small Hint:

Let the rectangle sides be $2a$ and $2b$

Big Hint:

Subtract the three outside right triangles from the rectangle

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.