Solution:

Each classroom has $18-2=16$ more students than rabbits.

Across $4$ classrooms, the difference is $4\cdot16=64$.

Thus, C is the correct answer.

Solution:

The smaller side of the rectangle is equal to the diameter, which is $2r=2\cdot 5=10.$

Due to the $2:1$ ratio, the long side is $10\cdot 2=20.$

Therefore, the area is $20\cdot 10=200.$

Thus, the correct answer is E .

Solution:

Notice that the line in question is perfectly horizontal. This means that if we were to construct a perpendicular line segment from the point to the line, to find the reflected coordinates of the point, we simply double the distance along that line segment.

The line segment from $(1000,2012)$ to $y=2000$ is of length $12,$ so the reflected point is along this same line segment, but a distance of $12$ on the other side of the horizontal line. This yields $(1000,1988).$

Thus, the correct answer is B .

Solution:

As we know that when Ringo's marbles are divided by $6,$ we have a remainder of $4,$ we conclude that he has $6x+4$ marbles for some $x.$

Using the same logic, we can also conclude that Paul has $6y+3$ marbles for some $y.$

Therefore, the total number of marbles is $$(6x+4)+(6y+3)$$$$= 6(x+y+1)+1$$ which, when divided by $6,$ only leaves $1$ left over.

Thus, the correct answer is A .

Solution:

Suppose the original price is $x.$ Then, the tax is $0.1$ and the tip is $0.15x.$ This makes the total payment equal to: $$\begin{align*}x+0.1x+0.15x&=1.25x\\&=27.5\end{align*}$$ Therefore, $x=22.$

Thus, the correct answer is D .

Solution:

The value when $x$ is rounded up is greater than $x.$

The value when $-y$ is rounded up is greater than $-y.$

Therefore, we add two numbers which are greater than their corresponding parts in $x-y,$ making it greater.

Thus, the correct answer is A .

Solution:

Let the number of acorns they each hid be $x.$ Then, the number of holes from the chipmunk is $\dfrac x3$ and the number of holes from the squirrel is $\dfrac x4 .$

This means $$\dfrac x3 - \dfrac x4 = \dfrac{x}{12}= 4$$ Therefore, $x=48.$

Thus, the correct answer is D .

Solution:

Suppose we have $x=2+k$ as a solution. Then, $x=2-k$ would also be a solution as $$((2+k)-2)^2 = ((2-k)-2)^2$$ The sum of these two solutions would be $4.$ Thus, the sum of all integer solutions to the above equation is four times the number of positive $k$'s that work.

To find the number of $k$'s, we need to find the number of positive solutions to: $$1 < k^2 < 25,$$ which would be $3,$ as $k=2,3,4.$

Therefore, there are a total of $4\cdot 3=12$ solutions.

Thus, the correct answer is B .

Solution:

The first two integers have even sum $26$, so they can both be odd. The next two integers have sum $41-26=15$, which is odd, so one of them must be even and one odd.

The last two integers have sum $57-41=16$, so they can both be odd. Therefore at least one integer must be even, and one is attainable.

Thus, A is the correct answer.

Solution:

By cross multiplying, we can see that $MN = 36.$ Thus, we can make $M$ any factor of $36$ and then determine $N$ from it.

Since $36=2^2\cdot 3^2,$ we have $(2+1)(2+1)=9$ possible choices for $M,$ each of which also determine a unique $N.$

Thus, the correct answer is D .

Solution:

We know that there must be cake served on Friday, and as such, on Saturday, we cannot have cake. Therefore, we have $3$ choices for Saturday's menu.

Furthermore, for each of the $5$ previous days, we could go backwards and have $3$ choices on each day, making the total number of choices $3^5\cdot 3 = 729.$

Thus, the correct answer is A .

Solution:

We know $AB$ and $BC$ are perpendicular, so $$AB^2 + BC^2 = (10\sqrt 2)^2 = 200.$$ Also, as $\angle BAC = 45^\circ,$ we know that $\triangle ABC$ is an isosceles right triangle, so $AB = BC ,$ making $2AB^2 = 200.$ Thus, $AB = BC = 10.$

As such, we know that $$BD= BC+CD = 30.$$ Thus, by the Pythagorean Theorem, we have that $$\begin{align*}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{align*}$$ Thus, since $$31^2 < AD^2 < 32^2,$$ we have $$31 \lt AD\lt 32$$

Thus, the correct answer is B .

Solution:

Let $c$ represent Clea's speed walking down the non-operational escalator. Similarly, let $e$ represent Clea's speed standing still on the operational escalator. Then, as speed is equal to distance over time, we know that $c = \dfrac d{60}$ and $$c+e = \dfrac d{24}.$$ Therefore, $$e = \dfrac d{24}-\dfrac d{60} = \dfrac d{40},$$ meaning that Clea takes $40$ to descend the escalator by simply standing still.

Thus, the correct answer is B .

Solution:

This rhombus is created by placing two congruent equilateral triangles. Let the side length of it be $s.$ Then, the area of one of them is $\dfrac{s^2 \sqrt 3}4,$ making the total area $\dfrac{s^2 \sqrt 3}2.$

The side length of the larger equilateral triangle is $2 \sqrt 3.$ The height of it is $3$ since the height is equal to $2 \sqrt 3 \sin(60^\circ) .$

Half of the square is $\sqrt 3,$ so the height of the smaller triangle is $3 - \sqrt 3 .$ Thus, the ratio between $s$ and $2 \sqrt 3$ is $\dfrac {3-\sqrt 3}3 .$

As such, $$s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2 .$$

Therefore, the combined area is $$\begin{align*} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{align*}$$

Thus, the correct answer is D .

Solution:

They would have to share $\binom 62 = 15$ wins.

This means we cannot have a $6$ way tie as that would be $2.5$ wins per team.

If we had a $5$ way tie, each team could have $3$ wins, which is possible if one team loses all of its games, and out of the $5$ winning teams, they each split their games.

If we label the teams from $1$ to $5,$ and designate team $6$ to lose all their games. To get a $5$ way tie, we could have each team $1\le x \le 5$ beat team $x+1 \mod 5$ and team $x+2 \mod 5,$ as well as team $6.$

Thus, the correct answer is D .

Solution:

Connect the centers of the three circles. This forms an equilateral triangle of side $4$, with area $4\sqrt3$.

The included part of each circle is a $300^\circ$ sector, whose area is $\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}$.

The total area is $3\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3$.

Thus, A is the correct answer.

Solution:

Each sector forms a cone with slant height $12$. The smaller sector has angle $120^\circ$, so its arc length is $\dfrac13\cdot2\pi\cdot12=8\pi$, giving base radius $4$. Its cone height is $\sqrt{12^2-4^2}=8\sqrt2$.

The larger sector has arc length $16\pi$, giving base radius $8$. Its cone height is $\sqrt{12^2-8^2}=4\sqrt5$.

The volume ratio is $$\dfrac{\frac13\pi\cdot4^2\cdot8\sqrt2}{\frac13\pi\cdot8^2\cdot4\sqrt5}=\dfrac{\sqrt{10}}{10}.$$

Thus, C is the correct answer.

Solution:

Among $500$ people, about $1$ person has the disease and tests positive. Of the remaining $499$ people, about $2\%$ test falsely positive, which is about $10$ people.

So among about $11$ positive tests, only about $1$ is a true positive. The probability is closest to $\dfrac1{11}$.

Thus, C is the correct answer.

Solution:

The polygon $BFDG$ is a trapezoid with bases $DG$ and $BF$ and height $6.$ Also, since $G$ is the midpoint between $A$ and $D,$ we have $GD= 15.$

We can see that $EBF \sim EAD ,$ so $$\dfrac{BF}{AD} = \dfrac{EB}{EA}$$ $$\dfrac{BF}{30} = \dfrac 28$$$$BF = 7.5$$

This makes the area of $BFDG$ equal to $\dfrac{6(15+7.5)}2 = \dfrac{135}2.$

Thus, the correct answer is C .

Solution:

Suppose $x$ was our initial number. Then, it becomes $2x$ when given to Silvia, and $2x+50$ when given to Bernardo. Repeatedly doing this can yield that it eventually becomes $16x+700$ when given to Silvia and $16x+750$ when given to Bernardo. Any more iterations makes the number greater than $1000.$

The number given to Silvia must be below $1000$ and the number Silvia makes is greater than $1000,$ so $$16x+750 > 1000 > 16x+700 .$$

Therefore, $300 > 16x > 250 .$ This makes $18 \geq x \geq 16,$ so $N=16.$ As such, the sum of the digits of $N$ is $1+6=7.$

Thus, the correct answer is A .

Solution:

Four of the six distances are $a$, so three of the points form an equilateral triangle of side $a$. Call these points $A,B,C$.

The fourth point $D$ is distance $a$ from one of these points, say $A$, and distance $2a$ from another, say $B$. Since $BD$ is a diameter of the circle centered at $A$ with radius $a$, triangle $BCD$ is right.

Thus $b^2=(2a)^2-a^2=3a^2$, so $b/a=\sqrt3$.

Thus, A is the correct answer.

Solution:

Suppose we have $a_1.$ Then we can either add $a_1-1$ or $a_1+1.$ Then, when every we add some number, we must have it in a connected interval to the numbers before.

Thus, our last interval would be $[1,10].$ If we construct a list backwards, we need to take either the lowest or highest numbers in the list. We can do this $9$ times to get $a_{10},$ then $a_9$ and continually until we get $a_2.$ Then, $a_1$ is given. For each index, we choose an upper or lower, so there are $2$ choices. Thus, the total is $2^9=512.$

Thus, the correct answer is B .

Solution:

The discarded tetrahedron has a right isosceles triangle of leg $1$ as one base and height $1$, so its volume is $\dfrac13\cdot\dfrac12\cdot1=\dfrac16$.

The cut face is an equilateral triangle of side $\sqrt2$, so its area is $\dfrac{\sqrt3}{4}(\sqrt2)^2=\dfrac{\sqrt3}{2}$. If $h$ is the height from the opposite vertex to this cut face, then $\dfrac13\cdot\dfrac{\sqrt3}{2}\cdot h=\dfrac16$, so $h=\dfrac{\sqrt3}{3}$.

The full cube diagonal has length $\sqrt3$. After the tetrahedron is removed and the cut face is placed down, the height is $\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{2\sqrt3}{3}$.

Thus, D is the correct answer.

Solution:

There are two cases: Each pair has exactly one liked song in common, or some pair has $2$ liked songs in common. This is because each pair must have at least $1$ liked song in common, and any more pairs than in the cases would result in $5$ songs.

Case 1: Each pair has exactly one liked song in common

There are $4$ ways to choose the song that one pair likes, $3$ ways to choose the song that the second pair likes, and $2$ ways to choose the song the third pair likes if we choose some order for them. Then, for the last song, one of them could like it which has $3$ cases or none of them likes it which is another case. Thus, the number of solutions in this case is $$4\cdot 3\cdot 2\cdot (3+1)=96.$$

Case 2: Some pair has $2$ liked songs in common

There are $3$ ways to choose the pair that has $2$ liked songs in common. Then, there are $\binom 43 = 6$ ways to choose which songs they like. Finally, there are $\binom 21 =2$ ways to figure out who likes the last song. Thus, the number of solutions in this case is $$3\cdot 6\cdot 2=36$$

The total amount is then $96+36 = 132.$

Thus, the correct answer is B .

Solution:

Classify a path by the set $S$ of backward arrows it uses. If $S=\varnothing$, the path is determined by choosing one forward arrow in each column, giving $2\cdot2\cdot4\cdot4\cdot4\cdot2\cdot2=2^{10}$ paths.

If $S$ uses only the left backward arrow, there are $2^8$ paths, and by symmetry the same for only the right backward arrow. If it uses both outer backward arrows but not the middle one, there are $2^6$ paths.

If $S$ uses only the middle backward arrow, there are $2^9$ paths. If it uses the middle arrow and exactly one outer backward arrow, there are $2^7$ paths for each choice of outer arrow. If it uses all three backward arrows, there are $2^5$ paths.

The total is $2^{10}+2\cdot2^8+2^6+2^9+2\cdot2^7+2^5=2400$.

Thus, E is the correct answer.