Small Hint:

Find how much Susan spent on rides before subtracting from $50.$

Big Hint:

Twice as much as $\$12$ is $\$24$.

Solution:

Susan spent $2\cdot 12=24$ dollars on rides, so she spent $12+24=36$ dollars total.

She had $50-36=14$ dollars left.

Thus, B is the correct answer.

Small Hint:

Write the digit under each letter in BEST OF LUCK.

Big Hint:

Because the first letter represents $0,$ each letter represents one less than its position in the code.

Solution:

The code BEST OF LUCK represents the digits $0$ through $9$ in order, so $C=8$, $L=6$, $U=7$, and $E=1$.

Therefore, $\text{CLUE}=8671$.

Thus, A is the correct answer.

Small Hint:

Dates one week apart fall on the same weekday.

Big Hint:

Compare February $13$ to February $6,$ then count back to February $1.$

Solution:

Since the $13$th is a Friday, we know that the $6$th is also a Friday. The $1$st is $5$ days before the Friday, making it a Sunday.

Thus, the answer is A .

Small Hint:

Subtract the inner triangle area from the outer triangle area.

Big Hint:

The remaining area is split equally among the three congruent trapezoids.

Solution:

The outer triangle has area $16$, and the inner triangle has area $1$.

So the three congruent trapezoids have total area $16-1=15$, and each trapezoid has area $15\div3=5$.

Thus, C is the correct answer.

Small Hint:

Find the miles ridden by subtracting the odometer readings.

Big Hint:

The total riding time is the $4$ hours plus the $6$ hours.

Solution:

The odometer increased by $1661-1441=220$ miles.

Barney rode for $4+6=10$ hours, so his average speed was $220\div10=22$ miles per hour.

Thus, E is the correct answer.

Small Hint:

Subdivide the central shaded square into unit-size pieces matching the others.

Big Hint:

Count shaded and unshaded equal-area small squares.

Solution:

After subdividing the central shaded square, the figure has $16$ equal small squares.

Of these, $6$ are shaded and $10$ are unshaded, so the desired ratio is $6:10=3:5$.

Thus, D is the correct answer.

Small Hint:

Solve for $M$ using $\frac{M}{45}=\frac{3}{5}$.

Big Hint:

Solve for $N$ using $\frac{60}{N}=\frac{3}{5}$, then add.

Solution:

From $\dfrac{3}{5}=\dfrac{M}{45}$, we get $M=\dfrac{3\cdot45}{5}=27$.

From $\dfrac{3}{5}=\dfrac{60}{N}$, we get $3N=300$, so $N=100$.

Therefore, $M+N=27+100=127$.

Thus, E is the correct answer.

Small Hint:

Read the four monthly bar heights from the chart.

Big Hint:

Average means total sales divided by $4$ months.

Solution:

The four monthly sales are $100,$ $60,$ $40,$ and $120$ dollars.

Their average is $\dfrac{100+60+40+120}{4}$ $=\dfrac{320}{4}$ $=80$ dollars.

Thus, D is the correct answer.

Small Hint:

After a $15\%$ loss, $85\%$ of the original investment remains.

Big Hint:

Apply the $20\%$ gain to the reduced amount, not to the original amount.

Solution:

The $15\%$ loss changes the investment from $\$100$ to $0.85\cdot\$100=\$85$.

The $20\%$ gain then changes it to $1.2\cdot\$85=\$102$.

The investment went from $\$100$ to $\$102$, which is a $2\%$ gain.

Thus, D is the correct answer.

Small Hint:

Convert each room average into a total age.

Big Hint:

The combined average is the combined total age divided by all $10$ people.

Solution:

The sum of the ages in Room A is $$6\cdot 40 = 240.$$

The sum of the ages in Room B is $$4\cdot 25 = 100.$$

The total sum is $$240+100 = 340.$$

The average is therefore $$\dfrac{340}{10} =34.$$

Thus, the answer is D .

Small Hint:

Adding dog owners and cat owners counts students with both pets twice.

Big Hint:

Use inclusion-exclusion with the total number of students as the union.

Solution:

The number of people that have both animals is equal to the number of people that own a cat plus the number of people that own a dog minus the number of people that own either.

Therefore, the number of people who own both is $$20+26-39 = 7.$$

Thus, the answer is A .

Small Hint:

List the bounce heights by multiplying by $\frac{2}{3}$ each time.

Big Hint:

Compare the fourth and fifth bounce heights with $\frac{1}{2}$.

Solution:

The first bounce rises to $2$ meters. Each later bounce is $\dfrac{2}{3}$ of the previous bounce.

The fourth bounce rises to $2\left(\dfrac{2}{3}\right)^3=\dfrac{16}{27}$, which is greater than $\dfrac{1}{2}$.

The fifth bounce rises to $2\left(\dfrac{2}{3}\right)^4=\dfrac{32}{81}$, which is less than $\dfrac{1}{2}$.

Thus, C is the correct answer.

Small Hint:

Each box appears in exactly two of the three pair weights.

Big Hint:

Add the three pair weights, then divide by $2.$

Solution:

Let the weights be $a,$ $b,$ $c.$ We know $$\begin{cases}a+b = 122\\a+c = 125\\b+c = 127.\end{cases}$$ Adding all of this yields $$2(a+b+c) =$$$$122+125+127=374.$$ This makes $$a+b+c = 187.$$

Thus, the answer is C .

Small Hint:

After the top-left A is fixed, decide where the other two A entries can go.

Big Hint:

For each A-pattern, the top row can place B and C in two orders.

Solution:

The other two A’s must occupy one square in each of the remaining two rows and columns. There are two possible diagonal patterns for the three A’s.

For either A-pattern, the two remaining positions in the top row can be filled as B,C or C,B, and then the rest of the grid is forced.

So there are $2\cdot2=4$ arrangements.

Thus, C is the correct answer.

Small Hint:

First add Theresa’s scores from the first $8$ games.

Big Hint:

The $9$-game total must be a multiple of $9,$ and the $10$-game total must be a multiple of $10.$

Solution:

The sum of her first $8$ scores is $37.$ Since the average of the first $9$ scores is an integer, the sum of the first $9$ scores is a multiple of $9.$

Since the score is less than $10,$ the sum of the scores after $9$ games is between $37$ and $47,$ and is a multiple of $9,$ making the sum $45.$ Thus, the score of the $9$th game is $45-37=8.$

The sum of Theresa’s first $9$ scores is $45.$ Since the average of the first $10$ scores is an integer, the sum of the first $10$ scores is a multiple of $10.$

Since the score is less than $10,$ the sum of the scores after $10$ games is between $45$ and $55,$ and is a multiple of $10,$ making the sum $50.$ Thus, the score of the $10$th game is $50-45=5.$

Therefore, their product is $8\cdot 5=40.$

Thus, the answer is B .

Small Hint:

The volume is just the number of unit cubes.

Big Hint:

For surface area, each of the six outer cubes has $5$ exposed faces.

Solution:

The shape uses $7$ unit cubes, so its volume is $7$ cubic units.

The center cube has no exposed faces. Each of the six outer cubes has $5$ exposed faces, for a total surface area of $6\cdot5=30$ square units.

The ratio of volume to surface area is $7:30$.

Thus, D is the correct answer.

Small Hint:

A perimeter of $50$ means the side lengths have sum $25.$

Big Hint:

For a fixed positive integer sum, the largest product uses numbers as close as possible.

Solution:

If the side lengths are $l$ and $w$, then $2l+2w=50$, so $l+w=25$.

The smallest possible area comes from side lengths $1$ and $24$, giving area $24$.

The largest possible area comes from the closest integer pair with sum $25$, namely $12$ and $13$, giving area $156$.

The difference is $156-24=132$.

Thus, D is the correct answer.

Small Hint:

Break the route into circular arcs and straight radial segments.

Big Hint:

A quarter-circle of radius $r$ has length $\frac{\pi r}{2}$.

Solution:

The circumference of a circle is $\pi d = 2r\pi ,$ so going a quarter of the way around is $\dfrac {\pi r }2.$

He goes a quarter of the way around the large circle, so this part is $10 \pi$ meters. He then goes from the larger circle to the smaller circle, which is $20-10=10$ meters.

He goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He then goes through the diameter of the smaller circle, which is $10\cdot 2=20$ meters.

He then goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He finally goes from the smaller circle to the larger circle, which is $20-10=10$ meters.

The total length is $$10\pi + 10+5\pi + 20 + 5\pi + 10$$$$=20\pi + 40.$$

Thus, the answer is E .

Small Hint:

Count the total number of unordered pairs of points.

Big Hint:

There are exactly $8$ adjacent unit segments around the square.

Solution:

Each dot has $2$ dots that are one unit away from it.

Therefore, regardless of the choice of the first dot, $2$ of the other $7$ dots would be within one unit, so the probability is $\dfrac 27.$

Thus, the answer is B .

Small Hint:

Let the common number of passing boys and passing girls be $p$.

Big Hint:

Then total boys and girls are $\frac{3p}{2}$ and $\frac{4p}{3}$, so choose the smallest $p$ that makes both whole.

Solution:

Let $p$ be the common number of boys and girls who passed.

Since $\dfrac{2}{3}$ of the boys passed, the number of boys is $\dfrac{3p}{2}$. Since $\dfrac{3}{4}$ of the girls passed, the number of girls is $\dfrac{4p}{3}$.

The smallest positive $p$ that makes both counts whole is $6$. Then there are $9$ boys and $8$ girls, for a minimum total of $17$ students.

Thus, B is the correct answer.

Small Hint:

The cylinder has radius $4$ cm and length $6$ cm.

Big Hint:

The dashed cut takes half the cylinder.

Solution:

The cylinder has diameter $8$ cm, so its radius is $4$ cm. Its length is $6$ cm.

The whole cylinder has volume $\pi r^2h=\pi\cdot4^2\cdot6=96\pi$.

The dashed cut takes half of the cylinder, so the wedge has volume $48\pi$, which is about $151$ cubic centimeters.

Thus, C is the correct answer.

Small Hint:

Let $x=\frac{n}{3}$. Then $n=3x$ and $3n=9x$.

Big Hint:

Both $x$ and $9x$ must be three-digit whole numbers.

Solution:

Let $x=\dfrac{n}{3}$. Then $n=3x$, so $3n=9x$.

Both $x$ and $9x$ must be three-digit whole numbers. Thus $100\le x$ and $9x\le999$, so $100\le x\le111$.

There are $111-100+1=12$ integer values of $x$, and each gives one positive integer value of $n$.

Thus, A is the correct answer.

Small Hint:

Because only a ratio is needed, choose a convenient side length for the square.

Big Hint:

Take the side length as $3$ so the split lengths are $2$ and $1.$

Solution:

Because the answer is a ratio, choose the side length of the square to be $3$. Then $AF=2$, $FE=1$, $CD=2$, and $DE=1$.

The square has area $9$. The areas of triangles $ABF$ and $BCD$ are each $\dfrac{1}{2}\cdot3\cdot2=3$.

Triangle $DEF$ has area $\dfrac{1}{2}\cdot1\cdot1=\dfrac{1}{2}$.

So $[BFD]=9-3-3-\dfrac{1}{2}=\dfrac{5}{2}$, and the desired ratio is $\dfrac{\frac{5}{2}}{9}=\dfrac{5}{18}$.

Thus, C is the correct answer.

Small Hint:

There are $10\cdot 6$ equally likely tile-die pairs.

Big Hint:

For each die roll, count tile numbers that make the product a square.

Solution:

If the rolled number was $1,$ then the tile can be $1,$ $4,$ $9,$ yielding $3$ combinations.

If the rolled number was $2,$ then the tile can be $2,$ $8,$ yielding $2$ combinations.

If the rolled number was $3,$ then the tile can be $3,$ yielding $1$ combination.

If the rolled number was $4,$ then the tile can be $1,$ $4,$ $9,$ yielding $3$ combinations.

If the rolled number was $5,$ then the tile can be $5,$ yielding $1$ combination.

If the rolled number was $6,$ then the tile can be $6,$ yielding $1$ combination.

The total number of combinations is $$3+2+1+3+1+1=11.$$ There are $60$ combinations each with equal likelihood, so the probability is $\dfrac{11}{60} .$

Thus, the answer is C .

Small Hint:

The total design area is the area of the largest circle.

Big Hint:

The shaded regions are alternating disks or annuli with radii increasing by $2.$

Solution:

The largest circle has radius $12$, so the entire design has area $12^2\pi=144\pi$.

The shaded parts have areas $2^2\pi=4\pi$, $(6^2-4^2)\pi=20\pi$, and $(10^2-8^2)\pi=36\pi$.

The total shaded area is $4\pi+20\pi+36\pi=60\pi$, so the shaded fraction is $\dfrac{60\pi}{144\pi}=\dfrac{5}{12}$, which is about $42\%$.

Thus, A is the correct answer.