Answer: E

Small Hint:

A cube has a fixed number of edges, corners, and faces.

Big Hint:

Add the counts for edges, corners, and faces.

Solution:

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

Small Hint:

Look for a number with prime factor $2.$

Big Hint:

Among the choices, only one is even.

Solution:

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

Small Hint:

First find how many grams are not filler.

Big Hint:

Compare the non-filler grams to $120.$

Solution:

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

Small Hint:

If all $7$ children had bicycles, count the wheels.

Big Hint:

Each tricycle adds one wheel compared with a bicycle.

Solution:

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

Small Hint:

$20\%$ is one fifth.

Big Hint:

Use $10\%$ as a stepping stone from $20\%$ to $30\%.$

Solution:

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.

Answer: B

Small Hint:

The square areas give the side lengths of the three squares.

Big Hint:

The side lengths form a $5-12-13$ right triangle.

Solution:

The side lengths of the squares are $$\sqrt{169} = 13,$$$$\sqrt{144} = 12,$$ and $$\sqrt{25} = 5.$$ 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

Small Hint:

Add Jenny’s four score differences from Blake’s.

Big Hint:

Divide the total difference by $4$ to compare averages.

Solution:

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

Small Hint:

Same dough and thickness means larger cookie area gives fewer cookies.

Big Hint:

Compare the areas of the four cookie shapes.

Solution:

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.

Answer: C

Small Hint:

First find the total money Art earns from a batch.

Big Hint:

Use the number of Roger’s cookies from the shared dough amount.

Solution:

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.

Answer: E

Small Hint:

Trisha’s cookie area is half of Art’s cookie area.

Big Hint:

Same dough amount means half-size cookies make twice as many.

Solution:

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.

Answer: B

Small Hint:

The Monday discount is taken from Friday’s increased price.

Big Hint:

Multiply the original price by $1.10,$ then by $0.90.$

Solution:

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.

Answer: E

Small Hint:

Check the only case where the face numbered $6$ is hidden.

Big Hint:

If $6$ is hidden, the visible faces still include $2$ and $3.$

Solution:

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.

Answer: B

Small Hint:

A cube has four painted faces exactly when it touches two other cubes.

Big Hint:

Separate the top cubes, bottom corner cubes, and remaining cubes.

Solution:

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.

Answer: D

Small Hint:

Use the hundreds column after $T = 7.$

Big Hint:

The ones column and the fact that $O$ is even force $O.$

Solution:

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.

Answer: B

Small Hint:

Try to satisfy both views with only three cubes.

Big Hint:

A four-cube construction is shown; you only need to rule out three.

Solution:

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

Suppose there were exactly three cubes. The two cubes that appear one above the other must share a face, so they have the same depth. The third cube appears beside the lower one. To share a face with either of the first two cubes, it must share the lower cube’s depth as well; otherwise it would be disconnected from both. Thus all three cubes would occupy one depth column, giving a one-column side view instead of the required L-shape.

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

Thus, B is the correct answer.

Answer: D

Small Hint:

Choose the driver first.

Big Hint:

After the driver is chosen, arrange the other three people in the remaining seats.

Solution:

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.

Answer: E

Small Hint:

Jim’s siblings must each share eye color or hair color with Jim.

Big Hint:

The two siblings must also share at least one characteristic with each other.

Solution:

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

Small Hint:

Sarah invites vertices at distance $1$ or $2$ from her.

Big Hint:

Count the dots not within two line segments of Sarah.

Solution:

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.

Answer: C

Small Hint:

Use the least common multiple of $15,$ $20,$ and $25.$

Big Hint:

Count the multiples of that LCM between $1000$ and $2000.$

Solution:

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

Small Hint:

At $4{:}20,$ the minute hand points at $4.$

Big Hint:

The hour hand has moved one third of the way from $4$ to $5.$

Solution:

At $4{:}20,$ the hour hand will be $\frac{1}{3}$ 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 $4.$

At $20$ minutes, the minute hand points to $4,$ so the acute angle between the hands is $10^{\circ}.$

Thus, D is the correct answer.

Answer: B

Small Hint:

Drop perpendiculars from $B$ and $C$ to the base.

Big Hint:

Use the $6-8-10$ and $8-15-17$ right triangles.

Solution:

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.

Answer: C

Small Hint:

Figures $A$ and $B$ leave the same shaded area.

Big Hint:

For $C,$ compare $\pi-2$ with $4-\pi.$

Solution:

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 copies of figure $A$ with half the side length, so each copy has one-fourth the area and the 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.

Answer: A

Small Hint:

The cat repeats every $4$ moves.

Big Hint:

The mouse repeats every $8$ moves; use the remainders of $247.$

Solution:

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.

Answer: B

Small Hint:

Along the semicircle, the distance from $X$ is constant.

Big Hint:

On the straight segment, the ship first gets closer to $X,$ then farther away.

Solution:

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.