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We can simplify this as follows.

$$\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}$$

Thus, D is the correct answer.

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We can unfold the cut up paper to achieve the following figure.

Thus, E is the correct answer.

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Using the formula, we can calculate the wind chill to be $$\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$$

Thus, B is the correct answer.

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We can fill in the other numbers to get the complete grid.

From this, we can see that the only prime numbers in the shaded boxes are $19, 23,$ and $47.$

Thus, D is the correct answer.

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Note that $\dfrac{30}{180} = \dfrac{1}{6}.$ This means that a sixth of the fish in the lake are trout, or in other words, the total number of fish is $6$ times the number of trout.

Therefore, there are $6 \cdot 250 = 1500$ fish in the lake.

Thus, B is the correct answer.

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Note that we do not want $0$ as a base, since that would make the expression equal to $0.$

This means that $0$ must be an exponent. The number whose exponent is $0$ will automatically evaluate to $1.$

Therefore, we should minimize this number, so we can put $2^0.$

Now the other term is $2^3$ or $3^2.$ Clearly, $3^2$ is larger, so we should use that.

This gives us a final value of $$2^0 \times 3^2 = 1 \times 9 = 9.$$

Thus, C is the correct answer.

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We can graph the two lines.

From this, we see that only the top left corner of the rectangle intersects either line.

Thus, B is the correct answer.

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Each column represents one round of two matches, so every column must contain exactly two $1$'s and two $0$'s.

Completing each column with the value needed to reach two wins gives Tiyo's record $000101.$

Thus, A is the correct answer.

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The first time that she hits an elevation of $7$ meters is at $2$ seconds.

She then dips below $4$ meters after $4$ seconds. This adds $4 - 2 = 2$ seconds to the total answer.

Malaika then goes above $4$ meters at $6$ seconds. She hits $7$ meters again at $10$ seconds.

This adds $10 - 6 = 4$ more seconds to the total. She finally dips below $7$ meters for the last time at $12$ seconds.

She then falls below $4$ meters at $14$ seconds, finally adding $14 - 12 = 2$ seconds to the total time.

The desired answer is therefore $$2 + 4 + 2 = 8$$ Thus, B is the correct answer.

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Harold left $1 - \frac{1}{4} = \frac{3}{4}$ of the pie for his friends.

The moose ate $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$ of the pie, leaving $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the pie.

Finally, the porcupine ate $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ of the pie. This leaves $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the pie.

Thus, D is the correct answer.

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We can round the distance up to $300,000,000$ miles. We can also approximate $6.5$ months as $$30 \times 6.5 = 195 \approx 200$$ days. This means that the Rover traveled $$300,000,000 \div 200 = 1,500,000$$ miles per day. Dividing this by $24$ to get the speed in miles per hour yields $62,500,$ which is close to $60,000.$

Thus, C is the correct answer.

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Let each smallest circle have radius $1.$

This means that there are $3$ shaded unit circles, which total to $3 \cdot 1^2 \pi = 3 \pi$ area.

There is also a shaded circle with radius $4$ with two unshaded circles of radius $2$ inside.

This gives us an extra shaded area of $$\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$$

Adding these values together yields $$8\pi + 3\pi = 11\pi.$$ The area of the large unshaded circle is $$6^2\pi = 36\pi.$$ Therefore, the desired fraction is $$\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.$$

Thus, B is the correct answer.

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The $3$rd water station is located $\frac{3}{8}$ of the way along the race (the water stations split the race up into $8$ equal spaces).

The first repair station is located $\frac{1}{3}$ of the way along the race. The distance between the two is $$\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$$ the length of the race. We know that this equals $2,$ which means the race is $24 \cdot 2 = 48$ miles long.

Thus, D is the correct answer.

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Let $a,b,c$ be the numbers of $5$-, $10$-, and $25$-cent stamps, and let $n=a+b+c.$ Dividing the value equation by $5$ gives $$a+2b+5c=142,$$ so $n+b+4c=142.$

Suppose $n\ge56.$ Then $b+4c\le86.$ Also $$a=142-2b-5c\le20,$$ so $2b+5c\ge122.$ Combining these inequalities gives $c\le16,$ but then $2b+5c\le2(20)+5(16)=120,$ a contradiction. Thus at most $55$ stamps can be used.

The bound is attainable with $a=19, b=19, c=17:$ $$19(5)+19(10)+17(25)=710.$$ Hence the greatest possible number of stamps is $19+19+17=55.$

Thus, E is the correct answer.

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If half a mile is the same as $10$ blocks, then one block is $\dfrac{1}{2} \div 10 = \dfrac{1}{20}$ miles.

Starting from the detour, Visvam has to walk $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ miles.

Normally, from this spot Visvam would take $5$ minutes to walk to school. Now he has to travel $\dfrac{7}{20}$ miles in $5$ minutes.

Note that $5$ minutes is $\dfrac{5}{60} = \dfrac{1}{12}$ hours. This means his speed must be $$\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$$ mph.

Thus, B is the correct answer.

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Since $20 = 3 \cdot 6 + 2,$ the bottom $2$ letters in each column will occur one more time than the third letter.

This means that the third letter in each column will occur $6$ times, whereas the other $2$ will appear $7$ times.

Using the same analysis, we can see that $R$ and $P$ appear in the third row $7$ times, whereas $Q$ only appears $6$ times.

Therefore, in $20 - 7 = 13$ columns, $P$ and $R$ appear $7$ times, for a total of $13 \cdot 7 = 91$ times.

They also appear $6$ times in $7$ columns, adding $6 \cdot 7 = 42$ appearances to the total. This gives us a total of $91 + 42 = 133$ appearances for $P$ and $R.$

$Q$ will therefore appear $$\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}$$ times.

Thus, C is the correct answer.

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Begin by observing that when folded, the sides labelled $2,$$3,$$4,$ and $5$ form the bottom half of the octahedron. As such, the remaining four faces must make up the top half of the octahedron.

From here, we have narrowed down our possibilities to $1,$$6,$ and $7.$ We can see that $6$ will be the number to the left of the shaded region $Q.$ This also gives us that $7$ is to the left of $6.$

Therefore, we know that the only remaining face, $1,$ must be to the right of the shaded region $Q.$

Thus, A is the correct answer.

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Let $r$ be the number of right jumps and $l$ be the number of left jumps. We need $$5r-3l=2023.$$ Modulo $5$, this gives $-3l\equiv3\pmod5$, so $l\equiv4\pmod5$.

To minimize the total number of jumps, use the smallest possible $l$, namely $l=4$. Then $5r-12=2023$, so $r=407$.

The fewest number of jumps is $407+4=411$.

Thus, D is the correct answer.

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Since the inner triangle's side length is $\frac{2}{3}$ the side length of the outer triangle, its area is $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$ the area of the outer triangle.

This means that the three trapezoids are $1 - \frac{4}{9} = \frac{5}{9}$ the area of the outer triangle.

Therefore, one trapezoid is $\frac{5}{9} \div 3 = \frac{5}{27}$ the area of the outer triangle.

This makes the ratio of the areas of one trapezoid and the inner triangle $$\dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}.$$

Thus, C is the correct answer.

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The original range is $28-3=25,$ so the new range must be $50.$ To keep the median $8,$ one added number $x$ must be less than $8$ and the other, $y,$ must be greater than $8.$

If $x\ge3,$ the minimum remains $3,$ so the new maximum must be $53.$ The value $x=3$ would change the mode, so $x\le7.$ Hence $x+y\le7+53=60.$

If $x<3$ and $y>28,$ then $y-x=50,$ so $x+y=2x+50\le54.$ If $y\le28,$ the maximum remains $28,$ forcing $x=-22$ and giving an even smaller sum. Therefore no case exceeds $60.$

The values $x=7$ and $y=53$ preserve the mode and median and give range $50,$ so the maximum sum is $60.$

Thus, D is the correct answer.

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The sum of all the numbers is $\dfrac{9 \cdot 10}{2} = 45.$ This means that the sum of each group is $45 \div 3 = 15.$

Consider the group with $9$ in it. The other two numbers must add to $6.$ Therefore, the other cards in this group are $1$ and $5$ or $2$ and $4.$

Case $1:$ One group is $1, 5, 9$

Consider the group with $8$ in it. The other numbers must add to $7.$ The only option is $3$ and $4$ with the remaining cards.

The other group is then $2,6,$ and $7.$ This adds to $15,$ so this case contributes one possibility.

Case $2:$ One group is $2, 4, 9$

Consider the group with $8$ in it. As above, the other numbers have to add to $7.$ The only option is $1$ and $6.$

The final group is $3, 5,$ and $7,$ which adds to $15.$ This is another configuration.

We have gone through all the cases, which revealed that there are only $2$ possible groupings.

Thus, C is the correct answer.

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