Answer: E

Solution(s):

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

Thus, E is the correct answer.

Answer: C

Solution(s):

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

Thus, C is the correct answer.

Answer: D

Solution(s):

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.

Answer: C

Solution(s):

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.

Answer: B

Solution(s):

Let $x$ the be the number. Then $$.2x = 12$$$$x = 60.$$ From this we get that $$.3 \cdot 60 = 18.$$

Thus, B is the correct answer.

Answer: B

Solution(s):

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.

Answer: A

Solution(s):

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.

Answer: A

Solution(s):

Note that the person who has the largest cookie would make the fewest cookies.

Art's cookie has an area of $$\dfrac{1}{2} (3 + 5) \cdot 3 = 12 \text{ in}^2$$

Roger's is $2 \cdot 4 = 8 \text{ in}^2,$ Paul's is $2 \cdot 3 = 6 \text{ in}^2,$ and Trisha's is $$\dfrac{1}{2} \cdot 3 \cdot 4 = 6 \text{ in}^2$$

Thus, A is the correct answer.

Answer: C

Solution(s):

From Problem $8,$ we know that Art's cookie has an area of $12 \text{ in}^2$ Since there are $12$ cookies in a batch, each batch has $12 \cdot 12 = 144 \text{ in}^2$ of dough.

Since Roger's cookie has an area of $8 \text{ in}^2,$ Roger can make $144 \div 8 = 18$ cookies.

Art would make $12 \cdot 60 = 720 ¢$ from his cookies, so Roger would need to charge $720 \div 18 = 40 ¢$ per cookie.

Thus, C is the correct answer.

Answer: E

Solution(s):

From Problem $9,$ we know that a batch has $144 \text{ in}^2$ of cookie dough. We also know that Trisha's cookies have an area of $6 \text{ in}^2$

Therefore, Trisha can make $144 \div 6 = 24$ cookies per batch.

Thus, E is the correct answer.

Answer: B

Solution(s):

On Friday, the shoes would cost 40 \cdot 1.1 = 44 $. On Monday, they would cost 44 \cdot 0.9 = 39.6 $.

Thus, B is the correct answer.

Answer: E

Solution(s):

Note that if $6$ is not the bottom case, then the product is automatically divisible by $6$ since $6$ is a factor.

If $6$ is on the bottom, then $2$ and $3$ are seen, also ensuring that the product is divisible by $6.$

Therefore, every possible scenario allows for the product to be divisible by $6.$

Thus, E is the correct answer.

Answer: B

Solution(s):

The $4$ cubes on top have $5$ exposed faces, so they don't work. The $4$ corners in the bottom row have $3$ exposed sides, so they don't work either.

Every other cube has $4$ exposed sides, so those work. There are $14 - 4 - 4 = 6$ cubes that work.

Thus, B is the correct answer.

Answer: D

Solution(s):

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 $W = 2,$ then $U = 4,$ which is also not allowed.

This makes $W = 3.$

Thus, D is the correct answer.

Answer: B

Solution(s):

We need at least $3$ cubes to form the front view. Then we also need another cube to ensure that the side views are correct.

Thus, B is the correct answer.

Answer: D

Solution(s):

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 passenger 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.

Answer: E

Solution(s):

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.

Answer: D

Solution(s):

Note that Sarah invites only the dots that are most $2$ line segments away from Sarah.

There are $4$ people who are completely disconnected from Sarah, and there are $2$ people who are $3$ lines away from Sarah.

Sarah will not invite any of these $4 + 2 = 6$ people.

Thus, D is the correct answer.

Answer: C

Solution(s):

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.

Answer: D

Solution(s):

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.

Answer: B

Solution(s):

Drop perpendiculars from $B$ and $C$ to $\overline{AD}$ and let them hit at $E$ and $F.$

Then, using the Pythagorean theorem, we get that $$AE = 6 \text{ and } FD = 15.$$

Then the area of $ABCD$ can be expressed as $$[ABCD] =$$$$[ABE] + [CDF] + [BCEF].$$ Note that:$$[ABE] = \dfrac{1}{2} \cdot 6 \cdot 8 = 24$$$$[CDF] = \dfrac{1}{2} \cdot 15 \cdot 8 = 60$$ $$[BCEF] = 8\cdot BC$$ Substituting, we get that $$\begin{align*} 164 &= 24+60+8BC\\ 80 &= 8BC \\ BC &= 10. \end{align*}$$

Thus, B is the correct answer.

Answer: C

Solution(s):

The shaded area of $A$ is $$2^2 - 1^2 \pi = 4 - \pi \approx .86 \text{ cm}^2$$

The shaded area of $B$ is $$2^2 - 4 \cdot \dfrac{1}{2}^2 \pi = 4 - \pi \approx .86 \text{ cm}^2$$

Note that the diagonal of the square in $C$ is equal to the diameter of the circle.

Therefore, the side length is $$2 \div \sqrt{2} = \sqrt{2}.$$ This makes the area of the shaded region $$1^2 \pi - \sqrt{2}^2 = \pi - 2 \approx 1.14 \text{ cm}^2$$

Thus, C is the correct answer.

Answer: A

Solution(s):

We can find the positions of the mouse and cat individually. Note that the mouse's position repeats every $8$ moves and the cat's every $4$ moves.

$247$ has a remainder of $3$ when divided by $4,$ which means the cat is in the same position as after the $3$rd move, which is the bottom right square.

$247$ has a remainder of $7$ when divided by $8,$ which means the cat is in the same position as after the $7$th move, which is the bottom left segment.

Thus, A is the correct answer.

Answer: B

Solution(s):

Note that any point along a circle is the same distance from its center. This means that travelling from $A$ to $B,$ the distance from $X$ remains constant.

We also know that as we move from $B$ to the midpoint of $\overline{BC},$ we get closer to $X.$

Similarly, as we move from the midpoint towards $C,$ we get farther from $X.$ This can be represented by a downwards facing semicircle.

Thus, B is the correct answer.

Answer: C

Solution(s):

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.