Small Hint:

Scale from $8$ hamburgers to $24$ hamburgers.

Big Hint:

The hamburger count triples, so the meat amount triples.

Video solution:

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Written solution:

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 .

Small Hint:

Find the net population change in one day.

Big Hint:

A birth every $8$ hours means $3$ births per day.

Video solution:

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Written solution:

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 .

Small Hint:

Add the correct daylight length to the sunrise time.

Big Hint:

From $6:57$ AM to noon is $5$ hours $3$ minutes.

Video solution:

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Written solution:

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 .

Small Hint:

Peter ate one whole slice plus half of another slice.

Big Hint:

Convert $1+\frac{1}{2}$ slices into a fraction of $12$ slices.

Video solution:

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Written solution:

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

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

Thus, the answer is C .

Small Hint:

Add the vertical lengths on each side of the shape.

Big Hint:

The left and right total heights must be equal.

Video solution:

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Written solution:

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 .

Small Hint:

Find the outside dimensions of the framed photograph.

Big Hint:

Subtract the photograph area from the outside rectangle area.

Video solution:

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Written solution:

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 .

Small Hint:

A $95$ average on four tests requires a total of $4\cdot95$.

Big Hint:

Make the fourth test as large as possible to minimize the third score.

Video solution:

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Written solution:

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 .

Small Hint:

After half price, the sale price is $50\%$ of the original.

Big Hint:

The coupon keeps $80\%$ of the sale price.

Video solution:

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Written solution:

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 .

Small Hint:

Start by giving every animal two legs.

Big Hint:

The extra legs determine the number of four-legged mammals.

Video solution:

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Written solution:

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 .

Small Hint:

The first digit cannot be $0$.

Big Hint:

Count arrangements starting with $1$, then arrangements starting with $2$.

Video solution:

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Written solution:

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 .

Small Hint:

The unique mode must be the repeated listed number.

Big Hint:

If the mean is that number, the total sum is fixed.

Video solution:

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Written solution:

Every listed value except $6$ appears once while $6$ appears twice. If $x$ equals any of the other listed values, 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 .

Small Hint:

Only the units digit of the base matters.

Big Hint:

Powers of $3$ have a units-digit cycle of length $4$.

Video solution:

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Written solution:

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 .

Small Hint:

The pencil price in cents divides both $143$ and $187$.

Big Hint:

Use the greatest common divisor of $143$ and $187$.

Video solution:

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Written solution:

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 $\frac{44}{11}=4$ more pencils. Thus, the answer is C .

Small Hint:

List how many games are added when each new team joins.

Big Hint:

With $n$ teams, the number of games is $\frac{n(n-1)}{2}$.

Video solution:

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Written solution:

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 .

Small Hint:

Subtract $2$ from the unknown number.

Big Hint:

The result must be a common multiple of $3$, $4$, $5$, and $6$.

Video solution:

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Written solution:

Let the number be $x.$ Since it leaves a remainder of $2$ when divided by $3$, $4$, $5$, and $6,$ we know $x-2$ is a multiple of $3$, $4$, $5$, and $6.$ This means $x-2$ is a multiple of $\operatorname{lcm}(3,4,5,6)$, 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 .

Small Hint:

Put the largest digits in the largest place values.

Big Hint:

The ten-thousands digits should be $9$ and $8$, then continue in pairs.

Video solution:

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Written solution:

To construct two five-digit numbers, the digits in the largest place values must be as great as possible. Therefore, the leftmost digits must be the greatest two digits. This means the first digit must be either $9$ or $8,$ the second digit must be either $7$ or $6,$ the third digit must be either $5$ or $4,$ the fourth digit must be either $3$ or $2,$ and the last digit must be either $1$ or $0.$

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

Thus, the answer is C .

Small Hint:

A side length of $3$ gives area too small for ten integer squares.

Big Hint:

A $4$ by $4$ square can be cut as shown.

Video solution:

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Written solution:

Since all $10$ squares have positive integer side lengths, each has area at least $1$. Their total area is therefore at least $10$. A square with integer side length at most $3$ has area at most $3^2=9$, so its side length cannot be less than $4$.

The following configuration cuts a $4\times4$ square into ten integer-sided squares, eight of which are unit squares. Therefore side length $4$ is attainable and is the minimum.

Thus, the answer is B .

Small Hint:

The number must be composite but not a square.

Big Hint:

Use the smallest two distinct primes that are at least $50$.

Video solution:

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Written solution:

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 .

Small Hint:

Translate each “all but” statement into a sum of two colors.

Big Hint:

Adding the three equations counts every marble twice.

Video solution:

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Written solution:

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 .

Small Hint:

Compare the fractions to $\frac{1}{3}$.

Big Hint:

Write each fraction as $\frac{1}{3}$ plus or minus a small amount.

Video solution:

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Written solution:

The middle fraction is $\dfrac{7}{21}=\dfrac13$. Also, $\dfrac{5}{19}\lt\dfrac13$ because $3\cdot5=15\lt19$, while $\dfrac{9}{23}\gt\dfrac13$ because $3\cdot9=27\gt23$. Therefore $$\dfrac{5}{19}\lt\dfrac{7}{21}\lt\dfrac{9}{23}.$$

Thus, the answer is B .

Small Hint:

The green paint covers half of the cube surface area.

Big Hint:

The remaining unpainted area is split equally among the six faces.

Video solution:

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Written solution:

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 .