Small Hint:

Add the widths before dividing by the number of books

Big Hint:

Compute $\left(6+\dfrac12+1+2.5+10\right)\div5$

Solution:

Adding the five widths gives $6+\dfrac12+1+2.5+10=20.$ The average is $20\div5=4.$

Thus, D is the correct answer.

Small Hint:

Let each small square have side length $s$

Big Hint:

The rectangle is $4s$ long and one small square shorter than the large square

Solution:

WLOG, let the side lengths of the squares be $1.$

This means that the length of the rectangle is $4.$ We also have that the width must be $4 - 1 = 3.$

The desired ratio is then $\dfrac{4}{3}.$

Thus, B is the correct answer.

Small Hint:

The total number of marbles stays fixed

Big Hint:

The final amounts are in the ratio $2:1$

Solution:

Let $x$ be the number of marbles that Eric ends up with. Then Tyrone ends up with $2x.$

The total number of marbles is $97 + 11 = 108,$ so $$3x = 108$$$$x = 36.$$

Then, Tyrone ends up with $36 \cdot 2 = 72$ marbles. This means he has to give away $97 - 72 = 25$ marbles.

Thus, D is the correct answer.

Small Hint:

Determine the minimum number of discs before averaging

Big Hint:

$7$ discs hold too little, while $8$ discs suffice

Solution:

Note that $7 \cdot 56 = 392$ and $8 \cdot 56 = 448,$ which means that the minimum number of discs needed is $8.$

Then the minutes of reading that each disc contains is $$412 \div 8 = 51.5.$$

Thus, B is the correct answer.

Small Hint:

Use the circumference to find the radius first

Big Hint:

From $2\pi r=24\pi$, the desired $k$ is $r^2$

Solution:

Recall that the formula for the circumference of a circle is $2 \pi r.$ We then have that $$24\pi = 2\pi r$$$$r = 12.$$

The area of a circle is $\pi r^2,$ so we have that $$k \pi = \pi 12^2$$$$k = 144.$$

Thus, E is the correct answer.

Small Hint:

Evaluate the inner $\spadesuit$ expression first

Big Hint:

$\spadesuit(2,2)=2-\dfrac12$, then use that as the second input

Solution:

Evaluating the inner expression, we get $$\spadesuit (2, 2) = 2 - \dfrac{1}{2} = \dfrac{3}{2}.$$ Then we have $$\spadesuit \left(2, \dfrac{3}{2}\right) = 2 - \dfrac{1}{\frac{3}{2}} = \dfrac{4}{3}.$$

Thus, C is the correct answer.

Small Hint:

The northeast and southeast vertical components cancel

Big Hint:

The net displacement has north component $1$ and east component $\sqrt2$

Solution:

From the diagram, we see that the distance traveled is the hypotenuse of a right triangle.

One of the legs is just $1$ from running due north. The other leg is $$\sqrt{1^2 + 1^2} = \sqrt2.$$

The final distance is then $$\sqrt{\sqrt2^2 + 1^2} = \sqrt3.$$

Thus, C is the correct answer.

Small Hint:

Compute the total hours worked and average hourly pay

Big Hint:

Divide the average hourly pay by $\$0.50$ to get the average age

Solution:

Tony worked $2\cdot50=100$ hours, so his average hourly pay was $\$630\div100=\$6.30.$ Because his hourly pay is $\$0.50$ times his age in full years, his average age on the days he worked was $6.30\div0.50=12.6$ years. During a six-month period his age can change by at most one, so he must have worked some days at age $12$ and some at age $13.$ Thus he was $13$ at the end of the period.

Thus, D is the correct answer.

Small Hint:

Bound the four-digit palindrome between $1000$ and $1031$

Big Hint:

Find the only palindrome in that interval, then subtract $32$

Solution:

Note that $x$ is at most $999.$ This means that $x + 32$ has a maximum of $1031.$

Similarly, we have that the minimum value of $x + 32$ is $1000.$

The only palindrome in this range is $1001,$ so this is what $x + 32$ equals.

Then $$x + 32 = 1001$$$$x = 969.$$

The sum of the digits is then $$9 + 6 + 9 = 24.$$

Thus, E is the correct answer.

Small Hint:

Each non-leap year shifts the weekday forward by $1$

Big Hint:

Since May $27$ is after leap day, leap years shift it forward by $2$

Solution:

Note that on a normal year, we have that $$365 = 52 \cdot 7 + 1,$$ which means that for a specific day, it moves to the day after the next year.

On a leap year, the day of the week moves forward two since there is an extra day.

Then in $2009,$ this day falls on a Wednesday. In $2010,$ it falls on a Thursday.

Similarly, in $2011,$ it falls on a Friday. In $2012,$ however, since it is a leap year, it falls on a Sunday.

Now, for the next three years, the day moves forward one. Then in $2016,$ it moves forward two, landing on a Friday.

Finally, in $2017,$ the day of the week is a Saturday.

Thus, E is the correct answer.

Small Hint:

Solve the inequality for the two endpoints of $x$

Big Hint:

The interval length is $\dfrac{b-3}{2}-\dfrac{a-3}{2}$

Solution:

Splitting the inequality into two of them and solving gives us $$a \le 2x + 3$$$$x \ge \dfrac{a - 3}{2}$$ and $$2x + 3 \le b$$$$x \le \dfrac{b - 3}{2}.$$

The range of the solutions is then $$\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10,$$ which then simplifying gives us $$(b - 3) - (a - 3) = 20$$ $$b - a = 20.$$

Thus, D is the correct answer.

Small Hint:

Volume scale is the cube of length scale

Big Hint:

Compare $0.1$ liters to $100000$ liters, then take a cube root

Solution:

The miniature tower holds $$\dfrac{100,000}{0.1} = 1,000,000$$ times less water than the actual tower. Since this is the ratio for volumes, the ratio of heights is $$(1,000,000)^{\frac{1}{3}} = 100.$$ This means that the height of the miniature tower is $$\dfrac{40}{100} = 0.4.$$

Thus, C is the correct answer.

Small Hint:

Subtract the $20$-minute stop from the total time

Big Hint:

After the stop, she drives for $3-\dfrac13-t$ hours

Solution:

Before the stop, Angelina drove $80t$ km.

The stop takes $\frac{1}{3}$ of an hour, so her total driving time is $3-\frac13=\frac83$ hours. After the stop, she drives for $\frac83-t$ hours, covering $100\left(\frac83-t\right)$ km.

The total distance equation is $80t+100\left(\frac83-t\right)=250.$

Thus, A is the correct answer.

Small Hint:

Let $\angle BAE=\angle ACD=x$

Big Hint:

Use $\triangle CFE$ to find $\angle AFC$, then angle-chase $\angle BAC$

Solution:

Let $\angle BAE = \angle ACD = x.$ Note that $\angle CFE = 60^{\circ}$ since $\triangle CFE$ is equilateral.

We then have that $$\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}.$$

Then: $$\begin{aligned} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{aligned}$$

We then get that $$\begin{aligned} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{aligned}$$

Since $AB = 2 \cdot AC$ and $\angle BAC = 60^{\circ},$ we have that $\triangle ABC$ is a $30-60-90$ triangle.

Thus, C is the correct answer.

Small Hint:

Chris and LeRoy cannot have the same species

Big Hint:

If Brian were a toad, Mike’s statement would create a contradiction

Solution:

Chris and LeRoy cannot both be frogs, because then both of their statements would be true. They cannot both be toads either, because then both statements would be false. Thus exactly one of them is a toad.

If Brian were a toad, his statement would make Mike a frog. Brian and the one toad among Chris and LeRoy would then make Mike’s statement true, which is impossible for a frog. Therefore Brian is a frog. His statement is false, so Mike is also a frog. Along with the one frog among Chris and LeRoy, there are $3$ frogs.

Thus, D is the correct answer.

Small Hint:

Use the Angle Bisector Theorem

Big Hint:

$AB:BC=3:8$, and $AC=11$

Solution:

Using the Angle Bisector Theorem, we have that $$\dfrac{AB}{3} = \dfrac{BC}{8}$$$$AB = \dfrac{3}{8} BC.$$

For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$

To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate.

$BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $$16 + 6 + 11 = 33.$$

Thus, B is the correct answer.

Small Hint:

Use inclusion-exclusion for the three rectangular holes

Big Hint:

The three $2\times2\times3$ holes overlap in the central $2\times2\times2$ cube

Solution:

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length $2.$ Then the volume of the cutout region is $$\begin{aligned}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{aligned}$$

We have to subtract out the center region twice since it is included in all $3$ regions.

The remaining volume is then $$3^3 - 20 = 27 - 20 = 7.$$

Thus, A is the correct answer.

Small Hint:

Separate cases according to whether Bernardo picks $9$

Big Hint:

Without a $9$, the two numbers are symmetric except when the chosen sets match

Solution:

There are two cases: Bernardo picks a $9$ or he doesn’t.

Case $1:$ Bernardo picks a $9$

Since a number is fixed, there are $\binom{8}{2} = 28$ ways to choose the other two numbers.

There are a total of $\binom{9}{3} = 84$ ways to pick all three numbers. The probability is then $$\dfrac{28}{84} = \dfrac{1}{3}.$$

Note that if Bernardo picks a $9,$ he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case $2:$ Bernardo doesn’t pick a $9$

There is a $1 - \frac{1}{3} = \frac{2}{3}$ chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a $$\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}$$ chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then $$\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.$$

The total probability of Bernardo getting a higher number is then $$\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.$$

Thus, B is the correct answer.

Small Hint:

Split the hexagon into $\triangle ACE$ and three corner triangles

Big Hint:

Express both areas using $r^2+r+1$ and $r$

Solution:

Note that $\triangle ACE$ is equilateral. Using the Law of Cosines in $\triangle ABC,$ we get $AC^2=r^2+1^2$ $-2r\cos120^\circ$ $=r^2+r+1.$

The area of $\triangle ACE$ is then $$\dfrac{\sqrt3}{4} (r^2 + r + 1).$$

The three corner triangles $\triangle ABC,$ $\triangle CDE,$ and $\triangle EFA$ each have area $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.$

Thus the hexagon has area $\dfrac{\sqrt3}{4}(r^2+r+1)$ $+3\cdot\dfrac{r\sqrt3}{4}$ $=\dfrac{\sqrt3}{4}(r^2+4r+1).$

The condition $[ACE]=70\%\cdot[ABCDEF]$ gives $$r^2+r+1=\dfrac{7}{10}(r^2+4r+1),$$ so $r^2-6r+1=0.$

By Vieta’s formulas, the sum of the possible values of $r$ is $6.$

Thus, E is the correct answer.

Small Hint:

Each move has length $1,\sqrt2,$ or $\sqrt3$

Big Hint:

There are only $4$ space diagonals, so the remaining moves are at most face diagonals

Solution:

Note that all the paths the fly can take have lengths of $1, \sqrt2,$ or $\sqrt3.$

There are only $4$ space diagonals in the cube, so at most $4$ moves can have length $\sqrt3.$ The other $4$ moves have length at most $\sqrt2.$

This upper bound is attainable, for example by alternating space diagonals and face diagonals around the vertices.

The path has length $$4\sqrt2 + 4\sqrt3.$$

Thus, D is the correct answer.