Solution:

It will take case $2 \cdot 300 = 600$ seconds to do the job. This is the same as $600 \div 60 = 10$ minutes.

Thus, D is the correct answer.

Solution:

Let the numbers be $x$ and $y.$ Then we know that $xy = 24$ and $x + y = 11.$

From this, we get that $x = 11 - y,$ and substituting yields $$\begin{align*} (11 - y)y &= 24 \\ y^2 - 11y + 24 &= 0 \\ (y - 3)(y - 8) &= 0. \end{align*}$$ This means that $y$ is either $3$ or $8,$ and $x$ is either $8$ or $3$ respectively. The larger number is $8.$

Thus, D is the correct answer.

Solution:

Anjous has $62 \div 3 = $21. This means that Elberta has $21 + $2 = $23.

Thus, E is the correct answer.

Solution:

To get a smaller number, we want the larger digits to appear towards the right of the number.

This means that the units digit is $4$ (it has to be either $2$ or $4$ since the number is even).

The next largest number is $9,$ which then has to go in the tens place.

Thus, E is the correct answer.

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.

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.

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.

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 $$3(7 + 6) = 3 \cdot 13 = 39 \text{ in.}$$

Thus, E is the correct answer.

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.

Solution:

Four state quarters is the same as on $1.

$2000\%$ of this is $1 \cdot \dfrac{2000}{100} = $1 \cdot 20 = $20.

Thus, A is the correct answer.

Solution:

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

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.

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.

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.

Solution:

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

He has $4$ options for the first vegetable and $3$ for the second. Since order doesn't matter teh number of options for the vegetables is $$\dfrac{4 \cdot 3}{2} = \dfrac{12}{2} = 6.$$

This gives him a total of $$3 \cdot 6 \cdot 4 = 72$$ combinations of meals.

Thus, C is the correct answer.

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.

Solution:

The square is folded in half to create a $4 \times 2$ rectangle.

Cutting the folded paper in half yields $2$ smaller $4 \times 1$ rectangles and one $4 \times 2$ rectangle.

The ratio of the perimeter of the small to rectangle to that of the large rectangle is therefore $$\dfrac{2(4 + 1)}{2(4 + 2)} = \dfrac{5}{6}.$$

Thus, E is the correct answer.

Solution:

Note that must of the increases are doubling. The only exceptions are $2$ to $3,$ $3$ to $4,$ and $11$ to $12.$

$11$ to $12$ is close to doubling, so we can ignore that. $2$ to $3$ is a $$\dfrac{300 - 200}{200} = \dfrac{1}{2} = 50 \%$$ increase. $3$ to $4$ is a $$\dfrac{500 - 300}{300} = \dfrac{2}{3} \approx 67 \%$$ increase.

Thus, B is the correct answer.

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.

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.

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.

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.

Solution:

The highest possible score occurs if you get everything question right for a total of $20 \cdot 5 = 100$ points.

The next highest score is if there are $19$ right answers and one unanswered question for a total of $$19 \cdot 5 + 1 \cdot 1 = 96.$$

This means that any score between these $2$ numbers is impossible.

Thus, E is the correct answer.

Solution:

Split this triangle into half (draw a line from $S$ to $X$).

Note that any noncongruent triangle except for $\triangle RST$ itself can be formed from these $4$ points. Any other triangle can be reflected or rotated to get one of these triangles.

Counting the number of triangles in this half, we get $3,$ and adding on $\triangle RST,$ we get a total of $4$ triangles.

Thus, D is the correct answer.

Solution:

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

$2$ pairs of red triangles are used, leaving $1$ red triangle on each half. $3$ pairs of blue triangles are used, leaving $2$ blue triangles on each half.

$2$ pairs of red-white triangles are used, using $1$ red triangle and $1$ white triangle on each half (one side cannot use both red triangles since they each only have $1$).

This leaves $4$ pairs of blue-white triangles since any more blue-blue pairs are not allowed (otherwise there would be more than $3$ pairs).

This leaves $5$ white-white pairs.

Thus, B is the correct answer.

Solution:

Let $x$ be the number that we are looking for.

Note that $x$ cannot be $4$ times another number, since the smallest value of $x$ would then be $$2457 \cdot 4 = 9828,$$ which is past the range of the set.

Now, let's see if $x$ is $3$ times another number. The smallest possible value would then be $$2457 \cdot 3 = 7371,$$ and the largest would be $$2475 \cdot 3 = 7425,$$ since multiplying any other number in the set $3$ would go past the range of the set.

We see that $7425$ is indeed a number in the set.

Thus, D is the correct answer.