Answer: D

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.

Answer: B

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.

Answer: D

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.

Answer: B

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.

Answer: E

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.

Answer: C

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.

Answer: C

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.

Answer: D

Solution:

Since $2 \cdot .5 = 1,$ we have that Tony makes a dollar per day per full year of his age.

If he is $12$ at the end of the period, then Tony can make a maximum of $$12 \cdot 50 = 600$$ dollars in the period. If he was $13$ at the end, then he could have made $$13 \cdot 50 = 650.$$ By this, we can see that Tony is $13$ the end of the period, since otherwise he would make too much or too little money.

Thus, D is the correct answer.

Answer: E

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.

Answer: E

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.

Answer: D

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.

Answer: C

Solution:

The miniature tower holds $$\dfrac{100,000}{.1} = 1,000,000$$ times less water than the actual tower. Since this is the ratio for volumes, the ratio of heights is $$(1,000,000)^{1 / 3} = 100.$$ This means that the height of the miniature tower is $$\dfrac{40}{100} = .4.$$

Thus, C is the correct answer.

Answer: A

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.

Answer: C

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

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

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.

Answer: D

Solution:

If Brian is a frog, then he must be lying, which means that Mike must be a frog.

If Brian is a toad, then he must be telling the truth, which also means that Mike is a frog.

Therefore, Mike is a frog, which means that Mike is lying. This means that there is at most one toad.

Then, at least one of LeRoy and Chris is a frog. This means the other is telling the truth, which makes them a toad.

This means there is one toad, which makes there be $3$ frogs.

Thus, D is the correct answer.

Answer: B

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.

Answer: A

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 $$\begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{align*}$$

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.

Answer: B

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.

Answer: E

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.

Answer: D

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.

Answer: A

Solution:

Let the roots be positive integers $r\le s\le t.$ By Vieta's formulas, $rst=2010$ and $a=r+s+t.$

Since $2010=2\cdot3\cdot5\cdot67,$ one root must be divisible by $67.$ If that root is larger than $67,$ then it is at least $134,$ which is already worse than the construction below.

Thus take $t=67$ and minimize $r+s$ with $rs=30.$ The factor pair with smallest sum is $5$ and $6,$ so $a=5+6+67=78.$

Thus, A is the correct answer.

Answer: A

Solution:

An interior triangle is formed by three chords that pairwise intersect inside the circle. Such a triangle uses six distinct endpoints on the circle.

Conversely, for any six chosen points in circular order, exactly one set of three chords pairs opposite endpoints so that the three chords intersect pairwise inside the circle.

Therefore the number of triangles is $\binom{8}{6}=\binom{8}{2}=28.$

Thus, A is the correct answer.

Answer: A

Solution:

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it.

The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white.

This happens with a probability of $$\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$$

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$

We have to find the smallest $n$ such that $$\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$$$$$$$$n(n + 1) \gt 2010.$$

Guessing and checking gives us that the smallest $n$ that works is $45.$

Thus, A is the correct answer.

Answer: A

Solution:

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$

There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$

Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Grouping residues modulo $25$ gives $A\equiv1\pmod{25}$ and $\dfrac{B}{5^{21}}\equiv-1\pmod{25}.$

Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25},$ $N=\dfrac{90!}{5^{21}\cdot2^{21}}\equiv-13\equiv12\pmod{25}.$

The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$

Thus, A is the correct answer.

Answer: B

Solution:

We can work backwards starting with $0.$ From this, we can add on $1^2$ to get $1.$

We can again add on $1^2$ to get $2.$ Again, adding on $1^2$ gives us $3.$

If we add on $1^2$ now, we get $4,$ but then $1^2$ is not the greatest square less than or equal to $4.$

Then adding on $2^2$ gives us $7.$ We repeat this process for $8$ steps to get $7223.$

$$\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}$$

Thus, B is the correct answer.