Small Hint:

Follow order of operations before subtracting

Big Hint:

Compute the two parenthesized expressions separately

Video solution:

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Written solution:

We can simplify this as follows.

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

Thus, D is the correct answer.

Small Hint:

Unfold the cut by reflecting it across each fold line

Big Hint:

The single cut appears four times after both folds are undone

Video solution:

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Written solution:

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

Thus, E is the correct answer.

Small Hint:

Substitute $T=36$ and $S=18$

Big Hint:

$0.7\cdot18$ is a little more than $12$

Video solution:

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Written solution:

Using the formula, the wind chill is

$$\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$$

The closest choice is $23.$

Thus, B is the correct answer.

Small Hint:

Continue the spiral until the shaded diagonal is filled

Big Hint:

Check each shaded number for divisibility by small primes

Video solution:

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Written solution:

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.

Small Hint:

The sample has $30$ trout out of $180$ fish

Big Hint:

Trout are $\frac16$ of all fish in the lake

Video solution:

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Written solution:

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.

Small Hint:

Avoid putting $0$ in a base

Big Hint:

Use $0$ as an exponent so one factor becomes $1$

Video solution:

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Written solution:

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 $1.$

Using one of the two copies of $2$ in the factor $2^0$ leaves $2$ and $3$ for the other factor.

The other factor can then be $2^3$ or $3^2.$ Since $3^2$ is larger, we use $3^2.$

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

Thus, C is the correct answer.

Small Hint:

Find the equations of the two lines

Big Hint:

The rectangle has $15\le x\le16$ and $3\le y\le5$

Video solution:

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Written solution:

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.

Small Hint:

Each match contributes one win and one loss

Big Hint:

In each column of the records, two entries must equal $1$ and two must equal $0$

Video solution:

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Written solution:

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

Completing each column to contain two $1$ entries and two $0$ entries gives Tiyo’s record $000101.$

Thus, A is the correct answer.

Small Hint:

Count the time intervals where the graph is between the two heights

Big Hint:

Split the count into the three separate intervals visible on the graph

Video solution:

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Written solution:

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 total is $$2 + 4 + 2 = 8$$ Thus, B is the correct answer.

Small Hint:

Work with the fraction of the original pie remaining

Big Hint:

After Harold, $\frac34$ remains; after the moose, multiply by $\frac23$

Video solution:

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Written solution:

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.

Small Hint:

Approximate $6.5$ months as about $200$ days

Big Hint:

Convert miles per day to miles per hour by dividing by $24$

Video solution:

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Written solution:

We can round the distance to $300{,}000{,}000$ miles and approximate $6.5$ months as $$30 \times 6.5 = 195 \approx 200$$ days.

This gives $$300{,}000{,}000 \div 200 = 1{,}500{,}000$$ miles per day.

Dividing by $24$ gives about $62{,}500$ miles per hour, which is closest to $60{,}000.$

Thus, C is the correct answer.

Small Hint:

Circle areas are proportional to the square of the radius

Big Hint:

Count shaded area in units of the smallest circle’s area

Video solution:

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Written solution:

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 $$\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$$

The total shaded area is $$8\pi + 3\pi = 11\pi.$$ 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.

Small Hint:

Seven water stations divide the race into $8$ equal gaps

Big Hint:

Two repair stations divide the race into $3$ equal gaps

Video solution:

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Written solution:

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 stations is $$\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$$ of the race length. This distance is $2$ miles, so the race is $24 \cdot 2 = 48$ miles long.

Thus, D is the correct answer.

Small Hint:

Use as many low-value stamps as possible

Big Hint:

All $5$- and $10$-cent stamps total $300$ cents, leaving a nonmultiple of $25$

Video solution:

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Written solution:

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.

Small Hint:

One block is $\frac1{20}$ mile

Big Hint:

From the detour point, he must walk $7$ blocks in the usual $5$ minutes

Video solution:

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Written solution:

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, Viswam has to walk $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ miles.

Normally, from this spot Viswam 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 $$\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$$ miles per hour.

Thus, B is the correct answer.

Small Hint:

The pattern repeats every $3$ entries along each row and column

Big Hint:

Since $20=3\cdot6+2$, two letters get one extra appearance in each line

Video solution:

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Written solution:

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.

Across the $20$ columns, $P$ is the letter that appears $6$ times in $7$ columns and appears $7$ times in the other $13$ columns. The same is true of $R.$

Thus $P$ and $R$ each appear $7\cdot6+13\cdot7=42+91=133$ times.

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

Thus, C is the correct answer.

Small Hint:

Track which faces fold around the shaded face $Q$

Big Hint:

Faces $2,$ $3,$ $4,$ $5$ form the opposite half from $Q$

Video solution:

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Written solution:

Begin by observing that when folded, the faces 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 face 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.

Small Hint:

Start with $404$ jumps to the right, which reaches $2020$

Big Hint:

The extra jumps must have net displacement $3$

Video solution:

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Written solution:

Let $r$ be the number of right jumps and $l$ be the number of left jumps. We need $$5r-3l=2023.$$ Reducing modulo $5$ 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.

Small Hint:

Areas of similar triangles scale as the square of side lengths

Big Hint:

The inner triangle has $\left(\frac23\right)^2$ of the large triangle’s area

Video solution:

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Written solution:

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.

Small Hint:

The original range is $25$, so the new range must be $50$

Big Hint:

To keep the median $8$, add one number below $8$ and one above $8$

Video solution:

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Written solution:

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 $60.$