Small Hint:

Convert the total painting time to seconds first

Big Hint:

There are $60$ seconds in one minute

Solution:

It will take Casey $2\cdot300=600$ seconds to do the job. Since $60$ seconds make one minute, this is $600\div60=10$ minutes.

Thus, D is the correct answer.

Small Hint:

List the factor pairs of $24$

Big Hint:

Find the pair whose sum is $11$

Solution:

The factor pairs of $24$ are $(1,24),$ $(2,12),$ $(3,8),$ and $(4,6)$.

The only pair with sum $11$ is $(3,8)$, so the larger number is $8$.

Thus, D is the correct answer.

Small Hint:

Find one-third of Granny Smith’s money

Big Hint:

Elberta has $\$2$ more than Anjou

Solution:

Anjou has $\$63\div3=\$21$. Elberta has $\$2$ more than Anjou, so Elberta has $\$21+\$2=\$23$.

Thus, E is the correct answer.

Small Hint:

The number must end in $2$ or $4$

Big Hint:

Make the earlier digits as small as possible

Solution:

The units digit must be $2$ or $4$. If it is $2$, arranging the remaining digits as small as possible gives $13492$. If it is $4$, the smallest arrangement is $12394$, which is smaller. Therefore the smallest even number is $12394$, whose tens digit is $9$.

Thus, E is the correct answer.

Small Hint:

Use distance equals rate times time

Big Hint:

Compare the distance with $2$ miles

Solution:

In $10$ seconds, the thunder was able to travel $10 \cdot 1088 = 10880$ feet.

Note that $2$ miles is about $2 \cdot 5280 = 10560$ feet, which is close to what we found above.

Thus, C is the correct answer.

Small Hint:

The first to fourth trees contain three equal gaps

Big Hint:

The first to last trees contain five equal gaps

Solution:

There are $3$ gaps between the first and fourth trees, which means that one gap is $60 \div 3 = 20$ feet long.

There are $5$ gaps between the first and last trees. This means that they are $5 \cdot 20 = 100$ feet apart.

Thus, B is the correct answer.

Small Hint:

Use the kite diagonals on the grid

Big Hint:

The area is half the product of the diagonals

Solution:

Recall that the area of a kite is the product of its diagonals divided by $2.$

Therefore, the area of the small kite is $$\dfrac{6 \cdot 7}{2} = \dfrac{42}{2} = 21.$$

Thus, A is the correct answer.

Small Hint:

The bracing is the sum of the two diagonals

Big Hint:

Tripling the grid triples both diagonal lengths

Solution:

The long diagonal is $7$ units, and the short one is $6$ units.

In the large kite, one unit is $3$ inches, so the total amount of bracing material needed is $$3(7 + 6) = 3 \cdot 13 = 39.$$

Thus, E is the correct answer.

Small Hint:

Tripling both dimensions multiplies area by $9$

Big Hint:

The waste area equals the large kite area

Solution:

The area of the entire grid would be $$3 \cdot 7 \cdot 3 \cdot 6 = 378.$$ The area of the kite is one-half this area from the formula for the area of the kite, so the wasted material is $378 \div 2 = 189.$

Thus, D is the correct answer.

Small Hint:

Four quarters have face value $\$1$

Big Hint:

$2000\%$ means $20$ times face value

Solution:

Four state quarters have face value $\$1$.

Since $2000\%=20$, the collector pays $20$ times face value, or $\$20$.

Thus, A is the correct answer.

Small Hint:

View $ABCD$ as a trapezoid

Big Hint:

The parallel vertical sides have lengths $4$ and $2$

Solution:

We can see that $ABCD$ is a trapezoid since $\overline{DC} \parallel \overline{AB}.$

Recall that the formula for the area of a trapezoid is $$A = \dfrac{1}{2} (b_1 + b_2) h.$$

Plugging in $$b_1 = DC = 2,$$$$b_2 = AB = 4,$$ and $$h = CB = 6,$$ we get that $$A = \dfrac{1}{2} (2 + 4) \cdot 6$$$$= \dfrac{1}{2} \cdot 6 \cdot 6 = 18.$$

Thus, C is the correct answer.

Small Hint:

Evaluate $6\otimes4$ first

Big Hint:

Then use the result as the first input with $3$

Solution:

We can evaluate it as follows $$\begin{align*} (6 \otimes 4) \otimes 3 &= \dfrac{6 + 4}{6 - 4} \otimes 3 \\ &= 5 \otimes 3 \\&= \dfrac{5 + 3}{5 - 3} \\&= 4. \end{align*}$$

Thus, A is the correct answer.

Small Hint:

Find how many students remain after the first three pies

Big Hint:

Cherry is half of the remaining group

Solution:

The number of students that prefer cherry or lemon pie is $$36 - 12 - 8 - 6 = 10.$$ Half of these like cherry pie, which is $10 \div 2 = 5.$

The number of degrees for cherry pie would then be $$360^{\circ} \cdot \dfrac{5}{36} = 5 \cdot 10^{\circ} = 50^{\circ}.$$

Thus, D is the correct answer.

Small Hint:

Choose the two vegetables as an unordered pair

Big Hint:

Multiply meat choices, vegetable pairs, and dessert choices

Solution:

He has $3$ choices for the meat and $4$ choices for dessert.

He must choose $2$ of the $4$ vegetables. Since order does not matter, there are $\dfrac{4\cdot3}{2}=6$ vegetable choices.

This gives $3\cdot6\cdot4=72$ possible meals.

Thus, C is the correct answer.

Small Hint:

Find how many potatoes remain when Christen starts

Big Hint:

After that, their combined rate is $8$ per minute

Solution:

Homer had peeled $3 \cdot 4 = 12$ potatoes by the time Christen joined him, leaving $44 - 12 = 32$ potatoes.

Together, Homer and Christen peel $5 + 3 = 8$ potatoes per minute, taking them $32 \div 8 = 4$ minutes to peel the rest.

In $4$ minutes, Christen peeled $4 \cdot 5 = 20$ potatoes.

Thus, A is the correct answer.

Small Hint:

After folding, the paper is a $4\times2$ rectangle

Big Hint:

The small rectangles are $4\times1$

Solution:

The square is folded in half to create a $4\times2$ rectangle.

Cutting the folded paper in half parallel to the fold gives two small $4\times1$ rectangles and one large $4\times2$ rectangle.

The ratio of the perimeter of a small rectangle to the perimeter of the large rectangle is $$\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56.$$

Thus, E is the correct answer.

Small Hint:

Most listed increases are doublings

Big Hint:

Compare the few increases that are not exactly $100\%$

Solution:

Most of the listed increases are doublings, which are $100\%$ increases.

The exceptions among the answer choices are from $2$ to $3$, from $3$ to $4$, and from $11$ to $12$.

These percent increases are $\dfrac{300-200}{200}=50\%$, $\dfrac{500-300}{300}=66\dfrac23\%$, and $\dfrac{125000-64000}{64000}$, which is about $95\%$.

The smallest is from question $2$ to question $3$.

Thus, B is the correct answer.

Small Hint:

The product is a multiple of $5$ exactly when a die shows $5$

Big Hint:

It is easier to count no die showing $5$

Solution:

The only way for the product to be a multiple of $5$ is if at least one of the rolls is $5.$

We can do complementary counting. The probability that neither rolls is a $5$ is $$\dfrac{5}{6} \cdot \dfrac{5}{6} = \dfrac{25}{36}.$$

Therefore, the probability that at least one roll is a $5$ is $$1 - \dfrac{25}{36} = \dfrac{11}{36}.$$

Thus, D is the correct answer.

Small Hint:

Twice the speed means half the time for the same distance

Big Hint:

The solid line should be higher and shorter

Solution:

Car $N$ travels at twice the speed of car $M,$ so it is represented by a point that is twice as high on the speed axis as car $M.$

Since car $N$ travels at the same distance as car $M$ but at a faster speed, it takes half the time to complete the trip.

Therefore, the line representing car $N$’s speed is half the length of the line representing car $M$’s speed.

Both lines are horizontal because the speeds are constant. Examining the given graphs, the only one that meets these conditions is graph D .

Thus, D is the correct answer.

Small Hint:

Quay only sees Kaleana’s score

Big Hint:

Use each hidden-score statement relative to Kaleana

Solution:

Since Quay only knows Kaleana’s score, we get that Quay and Kaleana have the same score, so $\text{K} = \text{Q}.$

Marty knows that his score is higher than Kaleana’s, so $\text{M} \gt \text{K}.$ Shana knows that her score is lower than Kaleana’s, so $\text{S} \lt \text{K}.$

Substituting $\text{Q}$ for $\text{K},$ we get the following inequality: $\text{S} \lt \text{Q} \lt \text{M}.$

Thus, A is the correct answer.

Small Hint:

The five numbers sum to $75$

Big Hint:

Make the other four numbers as small as possible

Solution:

The median of the set of numbers is the third largest number, which is $18.$ There are two numbers less than $18$ and two numbers greater than it.

The mean of the set is $15,$ so the sum of all the numbers is $5 \cdot 15 = 75.$ In order to maximize the largest number with this sum, the other numbers must be as small as possible.

The two numbers less than $18$ must be positive and distinct, so they must be $1$ and $2.$

The number immediately after $18$ must also be as small as possible, so it must be $19.$

Therefore, the remaining number, the maximum possible value in the set, is $$75 - 1 - 2 - 18 - 19 = 35.$$

Thus, D is the correct answer.

Small Hint:

Check scores near $100$

Big Hint:

With $19$ correct answers, only two scores are possible

Solution:

With $20$ correct answers, the score is $100$.

With $19$ correct answers, the score is either $95$ or $96$, depending on whether the remaining question is incorrect or unanswered.

With $18$ correct answers, the remaining two questions can contribute $0$, $1$, or $2$ points, giving $90$, $91$, or $92$.

Thus the listed scores $90,91,92,95$ are possible, while $97$ is not.

Thus, E is the correct answer.

Small Hint:

Classify triangles by side lengths

Big Hint:

Use symmetry to avoid recounting congruent cases

Solution:

Using the side length of the small equilateral triangles as one unit, the possible triangle side-length patterns are:

equilateral with sides $2,2,2$; equilateral with sides $1,1,1$; small isosceles with sides $1,1,\sqrt3$; and larger isosceles with sides $1,\sqrt3,2$.

These four patterns all occur in the figure, and symmetry shows every triangle formed by three of the six points matches one of them.

Thus there are $4$ noncongruent triangles.

Thus, D is the correct answer.

Small Hint:

Account for all red triangles first

Big Hint:

Then track how the unmatched blue triangles must pair

Solution:

Each half has $3$ red, $5$ blue, and $8$ white triangles.