A cube has $12$ edges, $8$ corners, and $6$ faces. Adding these together yields $$12 + 8 + 6 = 26.$$

Thus, E is the correct answer.

Note that the smallest prime number is $2.$ This means that any even number would be our answer.

Thus, C is the correct answer.

We get that $120 - 30 = 90$ grams are not filler. The percentage is therefore $$100 \cdot \dfrac{90}{120} = 100 \cdot \dfrac{3}{4} = 75 \%.$$

Thus, D is the correct answer.

Let $b$ be the number of bicycles and $t$ be the number of tricycles. Then we can set up the following system of equations: $$\begin{gather*} b + t = 7, \\ 2b + 3t = 19. \end{gather*}$$ Multiplying the first equation by $2$ and subtracting from the second equation, we get $t = 5.$

Thus, C is the correct answer.

Since $20\%$ of the number is $12$, $10\%$ of the number is $6$.

Therefore $30\%$ of the same number is $3\cdot6=18$.

Thus, B is the correct answer.

We get that the side lengths of the squares are $$\sqrt{169} = 13$$$$\sqrt{144} = 12,$$ $$\sqrt{25} = 5.$$ respectively. Note that these lengths form a Pythagorean triple.

Therefore, the interior triangle is right. Its area is $$\dfrac{1}{2} \cdot 5 \cdot 12 = 30.$$

Thus, B is the correct answer.

The total point difference between the two is $$10 - 10 + 20 \cdot 2 = 40.$$ The average of this difference is $40 \div 4 = 10.$

Thus, A is the correct answer.

Since the cookies all have the same thickness and use the same amount of dough, the largest cookie shape produces the fewest cookies.

Art’s trapezoid has area $\frac12(3+5)\cdot3=12$. Roger’s rectangle has area $4\cdot2=8$. Paul’s parallelogram has area $3\cdot2=6$, and Trisha’s triangle has area $\frac12\cdot3\cdot4=6$.

Art has the largest cookie area, so Art gets the fewest cookies from one batch.

Thus, A is the correct answer.

Art makes $12$ cookies that sell for $60$ cents each, so one batch earns $12\cdot60=720$ cents.

The batch has $12\cdot12=144$ square inches of dough area, and each Roger cookie has area $4\cdot2=8$. Thus Roger makes $144\div8=18$ cookies.

To earn $720$ cents from $18$ cookies, each Roger cookie should cost $720\div18=40$ cents.

Thus, C is the correct answer.

Art’s cookie area is $12$ square inches, so the whole batch has area $12\cdot12=144$ square inches.

Trisha’s triangular cookie has area $\frac12\cdot3\cdot4=6$ square inches.

Therefore Trisha can make $144\div6=24$ cookies per batch.

Thus, E is the correct answer.

On Friday, the shoes are marked up by $10\%$, so the listed price becomes $40\cdot1.10=44$ dollars.

On Monday, the $10\%$ discount is taken from $\$44$, giving $44\cdot0.90=39.60$ dollars.

Thus, B is the correct answer.

If the face numbered $6$ is visible, then the visible product is divisible by $6$.

If the face numbered $6$ is on the bottom, then the visible faces include both $2$ and $3$, so their product is still divisible by $6$.

Every possible toss works, so the probability is $1$.

Thus, E is the correct answer.

A cube has exactly four painted faces exactly when it is attached to exactly two other cubes.

The $4$ top cubes touch only one other cube, so they have $5$ painted faces. The $4$ bottom corner cubes touch three other cubes, so they have $3$ painted faces.

The remaining $14-4-4=6$ cubes each touch exactly two other cubes, so they have exactly four painted faces.

Thus, B is the correct answer.

Since both $T$'s are $7,$ we get that $O$ is either $4$ or $5.$ Since $O$ is even, we get that $O = 4.$

Then, we get that $R = 4 + 4 = 8.$ We also know that $W + W$ doesn't carry over, since otherwise $O$ would be $5.$

Therefore, $W$ is less than $5$ and cannot be $4$ or $1.$ If $W = 0,$ then $U = 0,$ which gives two letters the same digit. If $W = 2,$ then $U = 4,$ which is also not allowed.

This makes $W = 3.$

Thus, D is the correct answer.

The front view requires an L-shape with three visible positions, so at least three cubes are needed.

With only three cubes, all cubes would have to occupy those three front-view positions. Then the side view would have only one depth column, not the required L-shape.

The shown four-cube construction has both required views, so the minimum is $4$.

Thus, B is the correct answer.

There are $2$ options for who sits in the driver's seat. There are $3$ options for the other front seat, and $2$ options for the first back seat.

The last person has to sit in the last seat, for a total of $$2 \cdot 3 \cdot 2 = 12$$ possible seating arrangements.

Thus, D is the correct answer.

Note that Nadeen, Austin, and Sue are the only individuals who share a characteristic with Jim. We need to find which of the $3$ are completely different from the others.

Austin and Sue both have blue eyes, which makes Nadeen the odd one out. Therefore, Austin and Sue are Jim's siblings.

Thus, E is the correct answer.

Sarah invites her friends and the classmates who are friends with at least one of her friends. In graph terms, she invites dots that are $1$ or $2$ line segments away from Sarah.

From the graph, $4$ dots are disconnected from Sarah’s component, and $2$ more dots in Sarah’s component are $3$ segments away.

Those $4+2=6$ classmates will not be invited.

Thus, D is the correct answer.

If a number $x$ has these three numbers as factors, then their least common multiple must also divide $x.$

These numbers have the following prime factorizations: $$15 = 3 \cdot 5,$$$$20 = 2^2 \cdot 5,$$$$25 = 5^2.$$

From these values, we get that the least common multiple is $$2^2 \cdot 3 \cdot 5^2 = 300.$$

Therefore, the multiples of $300$ between $1000$ and $2000$ are $1200, 1500,$ and $1800.$

Thus, C is the correct answer.

At $4 : 20,$ the hour hand will be a $\frac{1}{3}$ of the way between $4$ and $5.$

Each hour represents $360 \div 12 = 30^{\circ}.$ This means the hour hand will be $30 \div 3 = 10^{\circ}$ past $4.$

Note that at $20$ minutes, the minute hand will be at $4.$ This means the degree formed is $10^{\circ}.$

Thus, D is the correct answer.

Drop perpendiculars from $B$ and $C$ to $\overline{AD}$, meeting it at $E$ and $F$.

In the left and right right triangles, $AE=\sqrt{10^2-8^2}=6$ and $FD=\sqrt{17^2-8^2}=15$.

The two side triangles have areas $\frac12\cdot6\cdot8=24$ and $\frac12\cdot15\cdot8=60$. The middle rectangle has area $8\cdot BC$.

Thus $164=24+60+8BC$, so $8BC=80$ and $BC=10$.

Thus, B is the correct answer.

In figure $A$, the shaded area is the area of a $2$ by $2$ square minus a circle of radius $1$, so it is $4-\pi$.

Figure $B$ is made of four half-size copies of figure $A$, so its total shaded area is also $4-\pi$.

In figure $C$, the circle has radius $1$, and the inscribed square has diagonal $2$, so its area is $2$. The shaded area is $\pi-2$.

Since $\pi>3$, we have $\pi-2>4-\pi$, so figure $C$ has the largest shaded area.

Thus, C is the correct answer.

The cat’s position repeats every $4$ moves, and the mouse’s position repeats every $8$ moves.

Since $247\equiv3\pmod 4$, the cat is in the same position as after the $3$rd move: the lower right square.

Since $247\equiv7\pmod 8$, the mouse is in the same position as after the $7$th move: the left side of the lower left square.

Thus, A is the correct answer.

Every point on the semicircular path from $A$ to $B$ is the same distance from the center $X$, so the graph starts horizontally.

On the straight path from $B$ to $C$, the ship first gets closer to $X$ and then farther away from $X$.

Only graph $B$ starts flat and then decreases before increasing.

Thus, B is the correct answer.

We get that the side lengths of $WXYZ$ are $5$ cm, since $\sqrt{25} = 5.$

We also know that the distance from $\overline{WZ}$ to $\overline{BC}$ is $2$ since it is the sum of the side lengths of $2$ unit squares.

Finally, the distance from $A$ to $\overline{BC}$ is the same as the distance from $\overline{BC}$ to $O,$ which is $$2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}$$

Now, we can find $BC,$ which is $$WZ - 2 = 5 - 2 = 3 \text{ cm}$$

Therefore, the area of $\triangle ABC$ is $$\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2$$

Thus, C is the correct answer.