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If we have $8$ hamburgers, we have $\dfrac{1}{3}$ of the $24$ hamburgers. This means we have $\dfrac{1}{3}$ of the meat when we have $3$ pounds. The total amount of meat is therefore $\dfrac{1}{\dfrac{1}{3}} \cdot 3 = 3\cdot3 = 9.$

Thus, the answer is E .

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Since we have $1$ birth every $8$ hours, we have $3$ births every $24$ hours. Therefore, we have $3$ births a day and $1$ death a day. The net change in population every day should be on average $2.$ Since we have $365$ days in a year and $2$ added to the population every day, the net change in population should be around $2\cdot 365 = 730 .$ This is approximately $700.$

Thus, the answer is B .

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Since $10$ hours after $6:57 \text{AM}$ is $4:57 \text{PM} ,$ we can then say $10$ hours and $3$ minutes after sunrise is $5:00 \text{PM} .$ We then have $21$ more minutes until sunset, so sunset is $5:21 \text{PM} .$

Thus, the answer is B .

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Peter ate $1$ full slice, and he ate $\dfrac 12$ of the slice that he split.

Therefore, he ate $\dfrac{3/2}{12}$ of the pizza, which is equivalent to $\dfrac 18.$

Thus, the answer is C .

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First, we can find the height of the object by getting the sum of the heights on the right. Therefore, the height is $1+2+1+6 = 10.$

Next, we can find the height of the object by getting the sum of the heights on the left. Therefore, the height is $1+1+1+2+X = 5+X.$

Since the heights are the same, we know $10=5+X ,$ so $X = 5.$

Thus, the answer is E .

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If we add $2$ inches on each side, we add $4$ inches total on both sides. This means that the dimensions of the outer part of the frame are $12 \times 14.$ The area of this is $12\cdot 14 = 168.$

However, we must take out the area of the inner part of the frame which has area $8 \cdot 10 =80.$

Therefore, the total area is $168-80=88.$

Thus, the answer is E .

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If the average is $95,$ then the sum of the tests is $95\cdot 4 = 380 .$ Since we have the first two tests, the sum of the last two tests is $380$ minus the first two scores.

This makes the sum of the last two scores equal to $380-91-97 = 192.$ Her last two scores therefore have a sum of $192.$

Given the sum of the tests we try to minimize one score, then we must maximize the other test. Therefore, we maximize the fourth test by making it $100.$ This would make the third test equal to $192 -100 = 92.$

Thus, the answer is B .

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Let $p$ be the original price. If everything is half off, we have the new price as $0.5p.$

Having a $20\%$ discount makes it such that we keep $80\%$ of the price, so the price is $0.5p\cdot 0.8 =0.4p.$

This would have $0.6p$ off, so we get a discount of $\dfrac{0.6p}{p} = 0.6,$ which is $60\%.$

Thus, the answer is D .

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Let $f$ be the number of animals with $4$ legs and let $t$ be the number of animals with $2$ legs.

Counting the number of legs yields $2t+4f = 522$ and counting the number of heads yields $t + f = 200.$ This means $4t+4f = 800,$ and subtracting the first equation from the second yields $2t = 278 \implies t = 139.$ This means there are $139$ two-legged birds.

Thus, the answer is C .

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First, we can't have the $0$ in the thousands position. Therefore, we have $3$ spots we can put it. Then, we have $3$ available positions for the $1,$ and then the two $2$s are placed. This makes it such that we have $3\cdot 3 = 9$ combinations.

Thus, the answer is D .

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Every listed value except $6$ appears once while $6$ appears twice. If any of the other values are chosen, then we have two modes, which means we do not have a unique mode. Otherwise, the only value that shows up more than once is $6$ making that the unique mode. This also means $6$ is the mean. Since there are $7$ elements, the sum of the elements is $6\cdot 7 = 42.$ The sum is also $$3+4+5+6+6+7+x$$ $$= 31+x$$$$= 42,$$ so $x = 11.$

Thus, the answer is D .

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We have to find $13^{2012} \mod 10.$ The following is true:

$$\begin{align*}13^{2012} &\equiv 3^{2012} \mod 10\\&\equiv 81^{503} \mod 10\\&\equiv 1^{503} \mod 10\\&\equiv 1 \mod 10\end{align*}$$

This means $13^{2012}$ has the same units digit as $1,$ so the units digit of $13^{2012}$ is $1.$

Thus, the answer is A .

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Let $c$ be the price of one pencil in cents. Then $c$ divides both $143$ and $187$, and $c\gt1$.

Since $143=11\cdot13$ and $187=11\cdot17$, the greatest common divisor is $11$. Thus each pencil costs $11$ cents.

Sharona paid $187-143=44$ cents more, so she bought $44/11=4$ more pencils. Thus, the answer is C .

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Each of the $N$ teams plays $N-1$ games. However, $2$ teams play each game, so multiplying $N$ and $N-1$ would be twice the number of games. Therefore, we know $N(N-1) = 42.$ This leads to $$N^2 -N +0.25 = 42.25$$ which implies $$(N-0.5)^2 = 6.5^2.$$ In turn, this suggests: $$N-0.5 = 6.5$$$$N = 7$$

Thus, the answer is B .

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Let the number be $x.$ Since it leaves a remainder of $2$ when divided by $3,4,5,6,$ we know $x-2$ is a multiple of $3,4,5, \text{ and }6.$ This means $x-2$ is a multiple of $lcm(3,4,5,6)$ which is $60.$ Therefore, $x-2$ must be a multiple of $60.$ The next number such that this occurs is when $x-2 = 60 \implies x = 62 .$

Thus, the answer is D .

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To construct two five digit numbers, first digits of the numbers from the left must be as great as possible. Therefore, the leftmost unused number must be either of the greatest two numbers. This means the first digit must be either of $9,8,$ the second digit must be either of $7,6,$ the third digit must be either of $5,4,$ the fourth digit must be either of $3,2$ and the last digit must be either of $1,0.$

The only one of the given numbers that satisfy this is $87431.$

Thus, the answer is C .

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Since all the $10$ squares have integer side length, they each must have a side length greater than or equal to $1.$ This means the total area must be over $10\cdot 1 = 10.$ Therefore, the square can't have a side length less than or equal to $3$ or else it would have an area less than $9.$

We can make a configuration with side length $4$ however with the following configuration.

Thus, the answer is B .

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The number is composite, not prime, and has no prime factor below $50$. The smallest possible prime factors are therefore $53$, $59$, $61$, and so on.

A square such as $53^2$ is not allowed, and $53^3$ is much larger than $53\cdot59$. The smallest allowed nonsquare composite is $53\cdot59=3127$.

Thus, the answer is A .

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Let $r,g,b$ be the number of red marbles, green marbles, and blue marbles respectively. We then know $$r+g+b -r=6,$$$$r+g+b-g = 8,$$$$r+g+b-b = 4$$ by the statements given. Adding these equations yields $3(r+g+b) -(r+g+b) = 18.$ This would mean $2(r+g+b) = 18,$ so $r+g+b = 9.$ Therefore, the sum of all of the marbles is $9.$

Thus, the answer is C .

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We can start with $19 \lt 21 \lt 23.$

Then, $\dfrac{1}{19} \gt \dfrac{1}{21} \gt \dfrac{1}{23}$ since we take the reciprocal of positive numbers.

Then, multiplying by the negative constant $-14$ yields $-\dfrac{14}{19} \lt -\dfrac{14}{21} \lt -\dfrac{14}{23}$ since that switches the direction of the inequalities.

Adding one to each of them then yields $\dfrac{5}{19} \lt \dfrac{7}{21} \lt \dfrac{9}{23} .$

Thus, the answer is B .

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The total surface area is $6\cdot 10\cdot 10 = 600.$ Therefore, $600-300 = 300$ square feet aren't covered. This would be $\dfrac{300}{6} = 50$ square feet per face.

Thus, the answer is D .

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In a sorted set of nine distinct integers, the median is the fifth number.

The median cannot be below $3$, since then $3,4,6,9,14$ would already give at least five larger elements. It cannot be above $9$, since $2,3,4,6,9$ would already give at least five smaller elements.

Each integer from $3$ through $9$ can be made the median by choosing the three missing integers appropriately. Therefore there are $7$ possible medians. Thus, the answer is D .

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Let the side length of the triangle be $s.$ This means the perimeter is $3s.$ Therefore, the side length for the hexagon is $\frac{3s}{6} = \frac s2.$

A hexagon can be made of $6$ equilateral triangles with side length $\dfrac{s}{2}$ as shown above. Each triangle is the original triangle scaled down by $\dfrac{1}{2},$ so the area is scaled down by $(\dfrac{1}{2})^2 = \dfrac{1}{4} .$ Therefore, the area of each of these triangles is $4 \cdot \dfrac{1}{4} = 1.$ Since there are $6$ of them, the area is $6\cdot 1 = 6.$

Thus, the answer is C .

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The area of the original circle is $\pi\cdot2^2=4\pi$.

Join the four quarter-circle endpoints to form a square. The square has diagonals $4$ and $4$, so its area is $\frac12\cdot4\cdot4=8$.

The part inside the circle but outside this square has area $4\pi-8$. Those four pieces are congruent to the pieces inside the square but outside the star.

Thus the star area is $8-(4\pi-8)=16-4\pi$. The desired ratio is $\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}$. Thus, the answer is A .

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Since all the triangles can be made from each other by rotating them around, they are all congruent. Therefore, we can place the $a$ as we have. The total area of the triangles is $5-4 = 1,$ so we have $4$ congruent triangles with a combined area of $1.$ This means the area of each triangle is $\dfrac{1}{4}.$ The area of each triangle is also $\frac {ab}2,$ so $\frac{ab}2 = \dfrac{1}{4}.$ This means $ab = \dfrac{1}{2} .$

Thus, the correct answer is C .