Solution(s):

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.

Solution(s):

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.

Solution(s):

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 $$.1 + .3 + .6 = 1.$$

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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

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

Then $$J \lt F \text{ and } B \lt F.$$

We also have $$B \gt M \text{ and } C \gt M.$$

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.

Solution(s):

There are $160 - 40 = 120$ mileposts 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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

Let $x$ be the number in the middle. Then the sum of the horizontal and vertical sums is $$1 + 4 + 7 + 10 + 13 + x,$$ since $x$ is counted twice. To maximize this, we want to maximize $x,$ so we let $x = 13.$

Then, from the above, expression, we have that the largest possible value of the sum of the horizontal and vertical sums is $35 + 13 = 48.$

The sum of one direction is then $48 \div 2 = 24.$

Thus, D is the correct answer.

Solution(s):

The total number of games is $11 + 4 = 15,$ so the percent of games lost is $$\dfrac{4}{15} \cdot 100 \% = \dfrac{4 \cdot 20}{3} \% \approx 27 \%.$$

Thus, B is the correct answer

Solution(s):

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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

She answered $10 \cdot .7 = 7$ arithmetic questions correctly. She answered $30 \cdot .4 = 12$ algebra ones correctly. She also got $35 \cdot .6 = 21$ 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 $$75 \cdot .6 = 45$$ questions correctly. This means that she would have needed to answer an additional $45 - 40 = 5$ questions correctly.

Thus, B is the correct answer.

Solution(s):

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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

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.

Solution(s):

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 $$\angle ECD = 180^{\circ} - 70^{\circ} - 40^{\circ}$$$$= 70^{\circ}.$$ 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.

Solution(s):

Let $f$ be the worth of a fish, $l$ for a loaf of bread, and $r$ for rice.

Then we are given $$3f = 2l \text{ and } l = 4r.$$

Substituting, we get $$3f = 2 \cdot 4r = 8r$$$$f = \dfrac{8}{3}r.$$

Thus, D is the correct answer.

Solution(s):

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.

Solution(s):

Note that to find the remainder when divided by $5,$ we only care about the units digit.

This means we only have to observe how the powers of the units digit work, namely the powers of $9.$

Then, looking at powers of $9,$ we see that the units digit alternates between $9$ and $1:$ $$9, 81, 729, \cdots$$ This means that $1999^{2000}$ ends in a $1$ since the power is even.

The remainder when divided by $5$ is then $1,$ since it is $1$ more than a multiple of $10.$

Thus, D is the correct answer.

Solution(s):

Note that all the triangles are isosceles right triangles. We can find the following side lengths: $$CD = \dfrac{CG}{2} = 3 = DG,$$ $$DE = \dfrac{DG}{2} = \dfrac{3}{2} = EG,$$ and $$EF = \dfrac{EG}{2} = \dfrac{3}{4}.$$

We then have that the areas of the $3$ triangles are $$\dfrac{1}{2} \cdot 3^2 = \dfrac{9}{2},$$ $$\dfrac{1}{2} \cdot \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{8},$$ and $$\dfrac{1}{2} \cdot \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{32}.$$

The sum of these areas is $$\dfrac{9}{2} + \dfrac{9}{8} + \dfrac{9}{32} = \dfrac{189}{32} \approx 5.9.$$

This can act as underestimate for the areas of all $100$ triangles, since it does not include some areas.

Now, we have that the other $97$ triangles are all contained with $\triangle FGH.$

This means that we can use the area of $\triangle FGH$ is an overestimate for all the other triangles.

The area of $\triangle FGH$ is $$\dfrac{1}{2} \cdot \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{32}.$$ Then $$\dfrac{189}{32} + \dfrac{9}{32} = \dfrac{198}{32} \approx 6.2.$$

Then, the true value is between $5.9$ and $6.2,$ which means that $6$ is nearest it.

Thus, A is the correct answer.