Answer: C

Small Hint:

Add the even numbers and odd numbers first

Big Hint:

The expression becomes $\frac{12}{9}-\frac{9}{12}$

Solution:

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

Thus, the correct answer is C .

Answer: E

Small Hint:

Find the current total score and target average

Big Hint:

The sixth score must bring the total to $6\cdot80$

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 .

Answer: A

Small Hint:

Use the smallest possible length and width

Big Hint:

The reported $2$ by $3$ tile can be as small as $1.5$ by $2.5$

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 .

Answer: C

Small Hint:

Compare how much more Bernardo paid than LeRoy

Big Hint:

Only half of the difference has to be transferred

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 .

Answer: E

Small Hint:

Factor $161$

Big Hint:

The only two-digit factor tells you the reversed value of $a$

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 .

Answer: A

Small Hint:

Work backward from the final $8$ candies

Big Hint:

Before giving away $4$, Casper had $12$, which was two thirds of the previous amount

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.

Answer: B

Small Hint:

Convert $\dfrac65$ of a right angle to degrees

Big Hint:

Compare the third angle with the two angles whose sum and difference are known

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 .

Answer: B

Small Hint:

Use the contrapositive of “hot and sunny implies crowded”

Big Hint:

Not crowded means at least one of the two conditions failed

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.

Answer: D

Small Hint:

Use similarity between $\triangle BDE$ and $\triangle BCA$

Big Hint:

Area ratio $1:3$ gives side-length ratio $1:\sqrt3$

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 .

Answer: B

Small Hint:

The denominator is a geometric sum

Big Hint:

$1+10+\cdots+10^9$ $=\frac{10^{10}-1}{9}$

Solution:

The largest number is $10^{10}.$ The other ten numbers have sum $$S=\sum_{i=0}^9 10^i=\dfrac{10^{10}-1}{9}.$$ Therefore their ratio is $$\dfrac{10^{10}}{S}=9\cdot\dfrac{10^{10}}{10^{10}-1},$$ which is just slightly greater than $9$ and hence closest to $9.$

Thus, the correct answer is B .

Answer: D

Small Hint:

Use the pigeonhole principle with $12$ months

Big Hint:

Try distributing $52$ birthdays as evenly as possible

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 .

Answer: A

Small Hint:

The straight portions cancel when comparing outside and inside paths

Big Hint:

Only the two semicircular ends add extra distance

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 $\frac{12\pi}{36}=\dfrac\pi3$ meters per second.

Thus, A is the correct answer.

Answer: D

Small Hint:

The product is positive when both numbers have the same sign

Big Hint:

The interval has $20$ negative units and $10$ positive units

Solution:

A selected number is negative with probability $\frac{20}{30}=\frac{2}{3}$ and positive with probability $\frac{10}{30}=\frac{1}{3}.$ (Selecting exactly $0$ has probability $0.$) The product is positive exactly when both numbers have the same sign, so the probability is $$\left(\dfrac23\right)^2+\left(\dfrac13\right)^2=\dfrac49+\dfrac19=\dfrac59.$$

Thus, the correct answer is D .

Answer: C

Small Hint:

Let the side lengths be $x$ and $y$

Big Hint:

Use $(x+y)^2=x^2+y^2+2xy$

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

Thus, the correct answer is C .

Answer: E

Small Hint:

Expand each side using $a@b=\frac{a+b}{2}$

Big Hint:

Test each of I, II, and III symbolically

Solution:

In statement I, the left-hand side equals $\dfrac{x+y+z}2$ and the right-hand side equals $x+\dfrac{y+z}2,$ so they are not always equal.

In statement II, both sides equal $x+\dfrac{y+z}2,$ so it holds.

In statement III, both sides equal $\dfrac{2x+y+z}4,$ so it also holds.

Thus, the correct answer is E .

Answer: A

Small Hint:

Let each side of the center square be $1$

Big Hint:

Find the octagon area by adding the center square, four rectangles, and four corner triangles

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 $$\begin{aligned}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{aligned}$$

Thus, the correct answer is A .

Answer: C

Small Hint:

Use the fact that an angle subtending a diameter is right

Big Hint:

Parallel chords create equal base angles in an isosceles trapezoid

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.

Answer: E

Small Hint:

Use $AB\parallel CD$ to transfer one angle to $\angle CDM$

Big Hint:

Triangle $CMD$ becomes isosceles

Solution:

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

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

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

Answer: A

Small Hint:

Square both sides after checking the radicals are defined

Big Hint:

Write the equation in terms of $|x|$

Solution:

Because $x^2=|x|^2,$ squaring the equation gives $$\begin{aligned} 5|x|+8 &= |x|^2-16,\\ |x|^2-5|x|-24&=0,\\ (|x|-8)(|x|+3)&=0. \end{aligned}$$ The algebraic possibilities for $|x|$ are $8$ and $-3,$ but an absolute value cannot be negative. Hence $|x|=8,$ so the roots are $8$ and $-8$, with product $-64.$

Thus, the correct answer is A .

Answer: C

Small Hint:

Use perpendicular bisectors to locate points closer to $B$

Big Hint:

The desired region is the central pentagon cut out by two congruent small triangles

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.