Small Hint:

Simplify the part without the question mark first

Big Hint:

Find the operation that makes $6\ ?\ 3=2$

Solution:

We have that $$4 - (2 - 1) = 4 - 1 = 3.$$

Then $$(6 \ ? \ 3) + 3 = 5$$$$6 \ ? \ 3 = 2.$$

The only choice that works is $\div.$

Thus, A is the correct answer.

Small Hint:

Each hour mark is $30^\circ$ apart

Big Hint:

At $10$ o’clock, the hands are two hour marks apart

Solution:

At $10$ o’clock, we have that the hour hand is at $10$ and the minute hand is at $12.$

This means that the hands are a $\frac{1}{6}$ of the entire circle apart. This is equal to $$360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}.$$

Thus, C is the correct answer.

Small Hint:

Compute the easiest-looking sums first

Big Hint:

Only one listed triplet fails to total $1$

Solution:

We have that A is the same as $$\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.$$

B reduces to $$2 - 2 + 1 = 1.$$

C adds up to $$0.1 + 0.3 + 0.6 = 1.$$

D however adds to $$1.1 - 2.1 + 1.0 = 0.$$

To make sure, we check that E has a sum of $$-\dfrac{8}{2} + 5 = 5 - 4 = 1.$$

Thus, D is the correct answer.

Small Hint:

Read both distances at $4$ hours

Big Hint:

Subtract Bjorn’s distance from Alberto’s

Solution:

Looking at the graph, we have that Alberto has biked $60$ miles and Bjorn $45$ miles after $4$ hours.

The difference between the two distances is $60 - 45 = 15$ miles.

Thus, A is the correct answer.

Small Hint:

Keep the perimeter the same

Big Hint:

The square side length is one-fourth of the old perimeter

Solution:

The current length of the fence is $$2(50 + 10) = 2 \cdot 60 = 120$$ feet. If all the sides become the same, then each side length is $120 \div 4 = 30$ feet. As such, the area of the square is then $30^2 = 900$ square feet. Note that the original area of the garden is $50 \cdot 10 = 500$ square feet.

Therefore, the difference is $900 - 500 = 400$ square feet.

Thus, D is the correct answer.

Small Hint:

Eliminate anyone known to have more than someone else

Big Hint:

Use each comparison to show who is above Moe

Solution:

Let $B, C, F, J,$ and $M$ represent the amount of money that Bo, Coe, Flo, Jo, and Moe have respectively.

Then $$J \lt F,\quad B \lt F.$$

We also have $$B \gt M,\quad C \gt M.$$

Finally, we are given $$M \lt J \lt B.$$

From the first and second inequalities, we can get that $F \gt M.$

This shows that everyone has more money than Moe.

Thus, E is the correct answer.

Small Hint:

First find the distance between mileposts $40$ and $160$

Big Hint:

Move $\frac{3}{4}$ of that distance past milepost $40$

Solution:

There are $160 - 40 = 120$ miles between the third and tenth exits.

This means $\frac{3}{4}$ of the way is at the $$40 + \dfrac{3}{4} \cdot 120 = 130$$ milepost.

Thus, E is the correct answer.

Small Hint:

Fold around the square marked $Y$

Big Hint:

Track which face lands opposite $W$

Solution:

Consider the cube with the yellow face facing upwards.

Then we can fold the white, green, and orange faces down.

Since the blue face is attached to the green face, it will end up folding backwards.

This means that the blue face will end up facing opposite the white face.

Thus, A is the correct answer.

Small Hint:

Add the three bed counts first

Big Hint:

Subtract the overlaps that were counted twice

Solution:

Note there are $50 + 100 = 150$ plants that are in two beds and there are no plants in all three beds.

The total number of plants is then $$500 + 450 + 350 - 150 = 1150.$$ We subtract to get rid of the plants that we counted twice.

Thus, C is the correct answer.

Small Hint:

Not green means red or yellow

Big Hint:

Use time not green over total cycle time

Solution:

During a given cycle, the light is not green for $30 + 5 = 35$ seconds.

Then the probability that it is not green is $$\dfrac{35}{60} = \dfrac{7}{12}.$$

Thus, E is the correct answer.

Small Hint:

The center square is counted in both sums

Big Hint:

Maximize the center number, then split the doubled total equally

Solution:

Let $x$ be the number in the center. The horizontal sum plus the vertical sum counts every number once, except the center number is counted twice.

So twice the common sum is $1+4+7+10+13+x$ $=35+x$. This is largest when $x=13$.

The largest common sum is therefore at most $\dfrac{35+13}{2}=24$, and it is attainable: put $13$ in the center, $1$ and $10$ at the ends of one row, and $4$ and $7$ at the ends of the other. Both sums are $24$.

Thus, D is the correct answer.

Small Hint:

Use $11$ wins and $4$ losses as one batch

Big Hint:

The lost fraction is losses over total games

Solution:

The team can be viewed as winning $11$ games and losing $4$ games in each batch of $15$ games.

The percent lost is $\dfrac4{15}\cdot100\%\approx 27\%$.

Thus, B is the correct answer.

Small Hint:

Convert each average into a total age

Big Hint:

Subtract the girls’ and boys’ totals from the camp total

Solution:

The sum of the ages of everybody at the camp is $40 \cdot 17 = 680.$

The sum of the ages of the girls is $20 \cdot 15 = 300$ and of the boys is $15 \cdot 16 = 240.$

As such, we know that the ages of the adults must be $680 - 300 - 240 = 140.$

And therefore, the average age of the adults is then $$140 \div 5 = 28.$$

Thus, C is the correct answer.

Small Hint:

Drop an altitude from $B$

Big Hint:

The two side right triangles have horizontal leg $4$

Solution:

Let $H$ be where the altitude from $B$ to $\overline{AD}$ intersects $\overline{AD}.$

Then we have that $$AH = \dfrac{16 - 8}{2} = 4,$$ since $AB = CD.$

We then have that $$AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.$$

Then the perimeter is $$8 + 16 + 2 \cdot 5 = 34.$$

Thus, D is the correct answer.

Small Hint:

The old number of plates is $5\cdot3\cdot4$

Big Hint:

To maximize a product, make the three factors as balanced as possible

Solution:

There are currently $$5 \cdot 3 \cdot 4 = 60$$ license plates that can be made.

If both letters are added to the first set, then there are $$7 \cdot 3 \cdot 4 = 84$$ possible plates.

If they are both added to the second, there are $$5 \cdot 5 \cdot 4 = 100$$ plates.

If they are added to the third, there are $$5 \cdot 3 \cdot 6 = 90$$ choices.

If one is added to the first set and the other to the second set, there are $$6 \cdot 4 \cdot 4 = 96$$ plates.

If the other is added to the third set, we get $$6 \cdot 3 \cdot 5 = 90$$ possible plates.

Finally, if the letters are added to the second and third sets, there are $$5 \cdot 4 \cdot 5 = 100$$ plates.

We see that $100$ is the greatest number of plates that we can achieve. This is an additional $100 - 60 = 40$ plates.

Thus, D is the correct answer.

Small Hint:

Compute Tori’s current correct answers by category

Big Hint:

Compare that total with $60\%$ of $75$

Solution:

She answered $10 \cdot 0.7 = 7$ arithmetic questions correctly. She answered $30 \cdot 0.4 = 12$ algebra ones correctly. She also got $35 \cdot 0.6 = 21$ geometry questions correctly.

This means she got a total of $7 + 12 + 21 = 40$ questions correct.

As such, in order to get a $60\%,$ Tori must have answered $$75 \cdot 0.6 = 45$$ questions correctly. This means that she would have needed to answer an additional $45 - 40 = 5$ questions correctly.

Thus, B is the correct answer.

Small Hint:

Find how many full recipes are needed for $216$ cookies

Big Hint:

Each full recipe uses $2$ eggs

Solution:

Since the students eat an average of $2$ cookies each, they will eat a total of $108 \cdot 2 = 216$ cookies.

Each recipe makes $15$ cookies, which means we need $$\left\lceil\dfrac{216}{15}\right\rceil = 15$$ full recipes to make enough cookies.

Each pan requires $2$ eggs, which means we need $15 \cdot 2 = 30$ eggs. There are $6$ eggs in a half-dozen, so we need $30 \div 6 = 5$ half-dozens.

Thus, C is the correct answer.

Small Hint:

A $25\%$ drop leaves $75\%$ attendance

Big Hint:

Round the needed number of recipes up

Solution:

There will now only be $$108 \cdot \left(1 - \dfrac{1}{4}\right) = 108 \cdot \dfrac{3}{4} = 81$$ people at the party. This means we need $$\left\lceil \dfrac{81 \cdot 2}{15} \right\rceil = \left\lceil \dfrac{162}{15} \right\rceil = 11$$ recipes.

Thus, E is the correct answer.

Small Hint:

First round the number of pans up

Big Hint:

Convert tablespoons of butter to sticks and round up

Solution:

To make $216$ cookies, they have to make $$\left\lceil \dfrac{216}{15} \right\rceil = 15$$ pans. Since each pan requires $3$ tablespoons of butter, all the pans will need $15 \cdot 3 = 45$ tablespoons.

They will then need $$\left\lceil \dfrac{45}{8} \right\rceil = 6$$ sticks of butter.

Thus, B is the correct answer.

Small Hint:

A front view keeps only the taller stack in each column

Big Hint:

For each column, take the larger front/back height

Solution:

Note that the height of the column is the maximum height of the front and back columns.

In the left column, we have that the back column is taller with height $2.$

In the middle column, we have that the front column is taller with height $3.$

Finally, the right column has height $4.$

Thus, B is the correct answer.

Small Hint:

Use supplementary angles at the two intersections

Big Hint:

Then use a triangle angle sum near the $40^\circ$ angle

Solution:

Label the vertices as below.

We then have that $$\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}$$ by supplementary angles. Then we have $$\angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ again by supplementary angles. Using the sum of the interior angles of a triangle is $180^{\circ},$ we get $$\begin{aligned} \angle ECD &= 180^{\circ} - 70^{\circ} - 40^{\circ} \\ &= 70^{\circ}. \end{aligned}$$ Then, using vertical angles, we have $$\angle ACB = \angle ECD = 70^{\circ}.$$ Finally, using the sum of the interior angles of a triangle, we get $$\angle A = 180^{\circ} - 80^{\circ} - 70^{\circ} = 30^{\circ}.$$

Thus, B is the correct answer.

Small Hint:

Turn each trade into a value equation

Big Hint:

Convert fish to bread, then bread to rice

Solution:

Three fish are worth two loaves of bread, so one fish is worth $\dfrac23$ of a loaf.

One loaf is worth $4$ bags of rice, so one fish is worth $\dfrac23\cdot4=\dfrac83=2\dfrac23$ bags of rice.

Thus, D is the correct answer.

Small Hint:

Each of the three regions has area one-third of the square

Big Hint:

Use the area of $\triangle BMC$ to find $BM$

Solution:

The area of the square is $3^2 = 9,$ which means that the area of one region is $9 \div 3 = 3.$

This means the area of $\triangle BMC$ is $3,$ which means that $$\dfrac{1}{2} \cdot 3 \cdot BM = 3$$$$BM = 2.$$

Since $\triangle BMC$ is right, we have $$CM = \sqrt{2^2 + 3^2} = \sqrt{13}.$$

Thus, C is the correct answer.