Solution:

Simply solving directly: $$\begin{align*} &\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} \\ &= \dfrac{12}9 - \dfrac 9{12} \\ &= \dfrac {16}{12} - \dfrac 9{12} \\ &=\dfrac 7{12}.\end{align*}$$

Thus, the correct answer is C .

Solution:

Her current average is $$\dfrac{90+80+70+60+85}5 = 77,$$ and the sum of her scores is $385.$ The desired average is then $80,$ so the sum of scores required is $80\cdot 6.$ Therefore, the answer is $$80\cdot 6-385 = 95.$$

Thus, the correct answer is E .

Solution:

The smallest possible dimensions are $1.5\times 2.5,$ so the area is $$1.5\cdot 2.5=3.75.$$

Thus, the correct answer is A .

Solution:

The amount they each would have to pay is $\dfrac{A+B}2,$ and LeRoy paid $A.$ Thus, he has to pay $$\dfrac{A+B}2-A = \dfrac{B-A}2$$ more.

Thus, the correct answer is C .

Solution:

The number $161$ is equal to $7\cdot 23.$ There are no other pairs of numbers that multiply to $161$ besides $1\cdot 161,$ so $23$ is the only two digit factor. Thus, $23$ is the number reversed, so he meant to get $32\cdot 7$ which is $224.$

Thus, the correct answer is E .

Solution:

Work backward. Before giving $4$ candies to his sister, Casper had $12$ candies, because he ended the second day with $8$.

Those $12$ candies were $\dfrac23$ of what he had after the first day, so after the first day he had $18$ candies.

Before giving $2$ candies to his brother, he had $20$ candies. This was $\dfrac23$ of his original amount, so he began with $30$ candies.

Thus, A is the correct answer.

Solution:

The two angles add to $\frac 65 \cdot 90 = 108.$ This makes the other angle $180-108=72.$ Then, if the larger of the two angles is $x$ then the smaller of them is $x-30$ so their sum is $2x-30=108,$ making $x=69.$

This means no angle is larger than $72,$ making the largest equal to $72.$

Thus, the correct answer is B .

Solution:

The statement says that if the weather was at least $80^\circ F$ and sunny, then the beach was crowded.

Because the beach was not crowded, the two conditions could not both have been true. Therefore the temperature was cooler than $80^\circ F$, or it was not sunny, or both.

Thus, B is the correct answer.

Solution:

By angle angle similarity, we have $BDE \sim BCA .$

Then, since the ratio of the areas is $\frac 13,$ the ratio of the sidelengths is $\frac{1}{\sqrt 3}.$

As such, $$\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3},$$ making $$BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3} .$$

Thus, the correct answer is D .

Solution:

The largest number is $10^{10}.$ The rest of the number have a sum of $$S=\sum_{i=0}^9 10^i.$$ Then, $10S = \sum_{i=0}^9 10^{i+1},$ making $9S = 10^{10}-1.$ This means that $\dfrac{10^{10}}{9S}$ is close to one, so the ratio between $10^{10}$ and the sum is close to $9.$

Thus, the correct answer is B .

Solution:

It isn't necessarily true for $n\geq 6$ as we could have $5$ people born in the first $4$ months and $4$ people born in the subsequent months.

However, one month must be greater than or equal to $5$ as the average of the number of people born in each month is $\frac{52}{12}$ which is greater than $4,$ and some month must be above average.

Thus, the correct answer is D .

Solution:

Let the inner semicircle radius be $r$. The straight parts of the inside and outside paths have the same total length, so only the semicircular ends change the distance.

The two inner semicircles have total length $2\pi r$, while the two outer semicircles have total length $2\pi(r+6)$. The outside path is therefore $12\pi$ meters longer.

Keiko takes $36$ more seconds to walk $12\pi$ more meters, so her speed is $12\pi/36=\dfrac\pi3$ meters per second.

Thus, A is the correct answer.

Solution:

There is $0$ probability that our number is $0,$ so we need to just find the probability that the product isn't less than $0.$ The number product is less than zero if one of the numbers is less than $0$ and one of them is greater than $0.$

First, there are $2$ ways to choose the designated lower number. Then, the probability that the designated lower number is less than $0$ is $\frac 23$ and the probability that the designated higher number is greater than $0$ is $\frac 13.$

This makes the probability that the product is less than $0$ equal to $$2 \cdot \dfrac 23 \cdot \dfrac 13 = \dfrac 49.$$ As such, the probability that the product is greater than $0$ equal to $\dfrac 59.$

Thus, the correct answer is D .

Solution:

Let the side lengths be $l,w.$ We wish to find $2(l+w).$ From the Pythagorean Theorem, we get $$25 = \sqrt{l^2+w^2}$$$$l^2+w^2 = 625$$ We also know $lw = 168.$

As such $$\begin{align*}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2.\end{align*}$$ This makes $l+w = 31,$ and as such, our answer is $31\cdot 2=62.$

Thus, the correct answer is C .

Solution:

In text 1, the left hand side equals $\dfrac{x+y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they aren't equal.

In text 2, the left hand side equals $x+ \dfrac{y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they are equal.

In text 3, the left hand side equals $\dfrac{2x+y+z}4$ and the right hand side equals $\dfrac{2x+y+z}4,$ so they are equal.

Thus, the correct answer is E .

Solution:

Let the side length be $1.$ Then, the area of the center is $1.$

Then, we must find the area of the octagon. It can be found as a square with $4$ isosceles right triangles taken out. The side length of this square is $$1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .$$ It has an area of $$(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.$$

Then, the side length of the right triangles is $\dfrac {1}{\sqrt 2} ,$ making the area of one equal to $$\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .$$ This makes them have a total combined area of $1,$ so the area of the octagon is $2+ 2 \sqrt 2.$

Thus, the ratio is $$\begin{align*}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{align*}$$

Thus, the correct answer is A .

Solution:

Since $EB$ is a diameter, $\angle EAB=90^\circ$. The ratio $\angle AEB:\angle ABE=4:5$ then gives $\angle AEB=40^\circ$ and $\angle ABE=50^\circ$.

Because $AB\parallel ED$, $\angle DEB=50^\circ$. Since $EB\parallel DC$, quadrilateral $EBCD$ is an isosceles trapezoid, so $\angle BCD=\angle CDE$.

Angles $DEB$ and $CDE$ are supplementary, so $\angle CDE=180^\circ-50^\circ=130^\circ$. Hence $\angle BCD=130^\circ$.

Thus, C is the correct answer.

Solution:

The angles $\angle AMD$ and $\angle MDC$ are equal since $AB \mid \mid DC.$

As such, $\angle MDC = \angle DMC ,$ making $MDC$ isosceles and $MC = DC = 6.$

As we can see, $\sin (CMB) = \frac 12,$ making $\angle CMB = 30^\circ .$

Therefore, $\angle AMC = 150^\circ .$ Since $\angle AMD$ is half of that, $\angle AMD = 75^\circ .$

Thus, the correct answer is E .

Solution:

The equation is equal to $$\sqrt{5 | x | + 8} = \sqrt{|x|^2 - 16}.$$ Solving, we get that: $$\begin{align*} 5|x|+8 &= |x|^2-16 \\ |x|^2-5|x|-24&=0 \\ (|x|-8)(|x|+3)&=0 \end{align*}$$ This makes $|x|=-3,8,$ making $|x|=8$ the only possible value. Thus, $x =8,-8$ with a product of $-64.$

Thus, the correct answer is A .

Solution:

The points closer to $B$ than to another vertex are bounded by the perpendicular bisectors of $BA$, $BC$, and $BD$. The bisector of $BD$ is diagonal $AC$, so the desired region lies in $\triangle ABC$.

Triangle $ABC$ has area $\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3$. The perpendicular bisectors of $BA$ and $BC$ cut off two congruent $30$-$60$-$90$ triangles, each with area $\dfrac{\sqrt3}{6}$.

Therefore the desired area is $\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}$.

Thus, C is the correct answer.

Solution:

The largest difference is $9$, so $w-z=9$. For either middle number $n$, the two differences $w-n$ and $n-z$ must add to $9$.

The available pairs of differences that add to $9$ are $3+6$ and $4+5$, and the remaining difference between the two middle numbers is $1$.

One possible set is ${w,w-5,w-6,w-9}$, giving $4w-20=44$ and $w=16$. The other is ${w,w-3,w-4,w-9}$, giving $4w-16=44$ and $w=15$.

The sum of the possible values of $w$ is $16+15=31$.

Thus, B is the correct answer.

Solution:

Let the cube have side length $x$. Take a vertical diagonal cross-section of the pyramid through opposite vertices of the square base.

This cross-section is an isosceles right triangle with hypotenuse $\sqrt2$. The cube appears as a rectangle of height $x$ and width $\sqrt2x$, leaving two congruent right isosceles triangles of leg $x$.

Thus $\sqrt2=\sqrt2x+2x$, so $x=\sqrt2-1$. The cube volume is $x^3=(\sqrt2-1)^3=5\sqrt2-7$.

Thus, A is the correct answer.

Solution:

Because $2011\equiv11\pmod{1000}$, it is enough to find $11^{2011}=(10+1)^{2011}$ modulo $1000$.

All terms with $10^3$ or higher are divisible by $1000$, so only the first three terms matter: $$1+2011\cdot10+\binom{2011}{2}10^2.$$

Modulo $1000$, this is $1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100\equiv1+110+500=611$.

The hundreds digit is therefore $6$.

Thus, D is the correct answer.

Solution:

Shift the graph down by $2$. The problem is equivalent to finding the smallest slope $m\gt\dfrac12$ for which $y=mx$ passes through a lattice point with $0\lt x\le100$.

For a fixed integer $x$, the smallest integer $y$ with $y/x\gt1/2$ is $x/2+1$ when $x$ is even, and $(x+1)/2$ when $x$ is odd.

Thus the candidate slopes are $\dfrac12+\dfrac1x$ for even $x$, minimized at $x=100$ as $\dfrac{51}{100}$, and $\dfrac12+\dfrac1{2x}$ for odd $x$, minimized at $x=99$ as $\dfrac{50}{99}$.

The smaller of these is $\dfrac{50}{99}$, so every $m$ with $\dfrac12\lt m\lt\dfrac{50}{99}$ avoids such lattice points, and this upper endpoint is best possible.

Thus, B is the correct answer.

Solution:

For a triangle with side lengths $a=BC$, $b=CA$, and $c=AB$, equal tangents from the same vertex give the next side lengths $$\dfrac{b+c-a}{2},\quad \dfrac{a+c-b}{2},\quad \dfrac{a+b-c}{2}.$$

If the current side lengths are $s-1,s,s+1$, then the next side lengths are $\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1$. Thus the same form persists while the middle side halves each time.

For $T_n$, the middle side is $2012/2^{n-1}$. A triangle of the form $s-1,s,s+1$ exists exactly when $s>2$.

The last valid triangle has $2012/2^{n-1}>2$, but the next one does not. This gives $n=10$, with middle side $2012/2^9=503/128$.

The perimeter is $3\cdot\dfrac{503}{128}=\dfrac{1509}{128}$.

Thus, D is the correct answer.