Small Hint:

Compute the even and odd sums separately

Big Hint:

Convert both fractions to a common denominator after simplifying

Solution:

$$\begin{aligned} &\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6} \\ \ &= \dfrac {12}9 - \dfrac 9{12} \\ &= \dfrac 43 - \dfrac 34 \\ &= \dfrac 7{12} \end{aligned}$$

Thus, the correct answer is C.

Small Hint:

Convert each garden side from steps to feet

Big Hint:

Multiply area by half a pound per square foot

Solution:

The dimensions of the garden are $2\cdot 15 = 30$ feet by $2\cdot 20=40$ feet. Thus, the square footage is $30\cdot 40 = 1200.$

Therefore, there are $$1200\cdot \dfrac 12 =600$$ pounds of potatoes.

Thus, the correct answer is A .

Small Hint:

The high and low temperatures are equally spaced around the average

Big Hint:

The low temperature is $8$ degrees below the average

Solution:

Let $l$ represent the low temperature. Then the high temperature is $l+16.$

The average satisfies $$\dfrac{l+l+16}2 = l+8 = 3.$$ Therefore, $l = -5.$

Thus, the correct answer is C.

Small Hint:

Count backward by using $201$ as position $1$

Big Hint:

A number $x$ is in backward position $202-x$

Solution:

When counting backward, $201$ is the first number, $200$ is the second, and in general $x$ is the $202-x$th number.

Thus $53$ is the $202-53=149$th number counted.

Thus, the correct answer is D .

Small Hint:

Factor the expression as $a(2-b)$

Big Hint:

Make $2-b$ as negative as possible while making $a$ large

Solution:

The expression is $2a-ab=a(2-b)$.

To make it as small as possible, choose $b$ as large as possible so that $2-b$ is most negative, and choose $a$ as large as possible. Since $a,b<6$, take $a=b=5$.

The minimum value is $5(2-5)=-15$.

Thus, the correct answer is B .

Small Hint:

Use total age, not the average of the two averages

Big Hint:

Compute $33\cdot11+55\cdot33$ and divide by $88$

Solution:

The sum of the ages of all the fifth-graders and their parents is: $$33\cdot 11+55\cdot 33 = 66\cdot 33.$$

Then, as the average is the sum divided by the number of people, the average age must be: $$\dfrac{66\cdot 33}{88} = \dfrac 34 \cdot 33 = 24.75 .$$

Thus, the correct answer is C .

Small Hint:

The only allowed non-isosceles triangle from six equally spaced points is a $30$-$60$-$90$ triangle

Big Hint:

Use the diameter as the hypotenuse

Solution:

Six equally spaced points on the circle form a regular hexagon. A triangle using three of them is not equilateral or isosceles only when its angles are $30^\circ,60^\circ,90^\circ$.

The hypotenuse is a diameter of the unit circle, so it has length $2$. The legs of the $30$-$60$-$90$ triangle are $1$ and $\sqrt3$.

The area is $\frac12\cdot1\cdot\sqrt3=\frac{\sqrt3}{2}$.

Thus, the correct answer is B .

Small Hint:

Let both cars drive a convenient common distance

Big Hint:

Combined miles per gallon is total miles divided by total gallons

Solution:

Let each car drive $40$ miles. Together they drive $80$ miles.

Ray uses $1$ gallon, while Tom uses $4$ gallons, so together they use $5$ gallons.

The combined rate is $80\div5=16$ miles per gallon.

Thus, the correct answer is B .

Small Hint:

Put each prime-power factor wholly into one of the three numbers

Big Hint:

Pairwise relatively prime means no prime can appear in two different numbers

Solution:

Only one number is a multiple of $2,$ only one is a multiple of $3,$ and only one is a multiple of $5.$

Since each positive integer is greater than $1,$ each must be a multiple of one of these primes.

Now $$27000=2^3\cdot 3^3\cdot 5^3.$$ Therefore, the numbers must be $2^3,3^3,5^3,$ and their sum is $160.$

Thus, the correct answer is D.

Small Hint:

Let $x$ be the number of three-point attempts

Big Hint:

Translate made-shot percentages into total points in terms of $x$

Solution:

Let $t$ be the number of three-point attempts.

The number of made three-point shots is $0.4 t,$ giving $1.2t$ points.

The number of two-point attempts is $1.5t,$ so the number made is $0.75t,$ giving $1.5t$ points.

Thus the total number of points satisfies $$2.7t=54,$$ so $t=20.$

Thus, the correct answer is C.

Small Hint:

Move all terms to one side and complete two squares

Big Hint:

A sum of squares equals zero only when both squares are zero

Solution:

This can be rewritten as $$x^2-10x+y^2+6y+34=0.$$

Completing the square yields $$x^2-10x+25+y^2+6y+9$$ $$=(x-5)^2+(y+3)^2$$ $$=0.$$ Since both squared terms are nonnegative and their sum is $0,$ both must be $0$. Thus $$(x-5)^2 = (y+3)^2=0.$$

Therefore, $x=5,y=-3,$ so $x+y=2.$

Thus, the correct answer is B.

Small Hint:

A regular pentagon has five equal sides and five equal diagonals

Big Hint:

After one segment is chosen, count same-length choices among the remaining nine

Solution:

A regular pentagon has $5$ sides of one length and $5$ diagonals of another length.

After the first segment is chosen, there are $9$ segments left, and exactly $4$ of them have the same length as the first chosen segment.

Therefore the probability is $\frac49$.

Thus, the correct answer is B .

Small Hint:

Find the triangular number just before $53$

Big Hint:

The next spoken list starts again at $1$

Solution:

Through the turn ending with $n$, the total number of numbers spoken is $$1+2+\cdots+n=\frac{n(n+1)}2.$$

After the turn ending with $9$, exactly $\frac{9\cdot10}{2}=45$ numbers have been spoken. The next turn is the list $1,2,\ldots,10$, so its eighth entry is the $53$rd number spoken. That entry is $8$.

Thus, the correct answer is E .

Small Hint:

Expand both custom-operation expressions

Big Hint:

Factor the resulting equation into line factors

Solution:

The condition is $x\clubsuit y=y\clubsuit x$, so $x^2y-xy^2=y^2x-yx^2$.

Combining like terms gives $2x^2y-2xy^2=2xy(x-y)=0$.

Thus $x=0$, $y=0$, or $x=y$. These are three lines.

Thus, the correct answer is E .

Small Hint:

Compare the side length of the large equilateral triangle with the small equilateral triangles inside the hexagon

Big Hint:

Areas scale as the square of side length

Solution:

Let $s$ be the side length of the equilateral triangle and $A$ its area. Then $a=3s.$

A regular hexagon of side length $s$ consists of $6$ equilateral triangles of side length $s,$ so its area is $6A.$

Therefore, a regular hexagon of area $A$ has side length $\dfrac{s}{\sqrt 6} .$ Hence $$b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .$$

It follows that $$\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.$$

Thus, the correct answer is B.

Small Hint:

Use the centroid ratio on both medians

Big Hint:

The $1.5,2,2.5$ triangle is right

Solution:

Since $PE:PD:DE=1.5:2:2.5$ $=3:4:5$, triangle $DPE$ is right at $P$. Thus medians $AD$ and $CE$ are perpendicular.

The centroid divides each median in a $2:1$ ratio, so $CE=3\cdot PE=4.5$ and $AD=3\cdot PD=6$.

Quadrilateral $AEDC$ has perpendicular diagonals $AD$ and $CE$, so its area is $\frac12(AD)(CE)=\frac12\cdot6\cdot4.5=13.5$.

Thus, the correct answer is B .

Small Hint:

Let $m$ and $n$ be the numbers of the two exchange types

Big Hint:

No more moves means fewer than $2$ red and fewer than $3$ blue tokens

Solution:

Suppose Alex makes $m$ exchanges at the red-token booth and $n$ exchanges at the blue-token booth.

He then has $75-2m+n$ red tokens and $75+m-3n$ blue tokens. At the end he must have fewer than $2$ red tokens and fewer than $3$ blue tokens.

Solving these terminal possibilities gives only two candidate final token counts: $(1,2)$, which comes from $(m,n)=(59,44)$, or $(0,0)$, which comes from $(m,n)=(60,45)$.

The final count $(0,0)$ is impossible, because the last exchange would always create either one blue token or one red token.

The final count $(1,2)$ is attainable. Starting from $(75,75)$ red and blue tokens, make $25$ blue-booth exchanges, then $50$ red-booth exchanges, then $16$ blue-booth exchanges, then $8$ red-booth exchanges, then $3$ blue-booth exchanges, and finally $1$ red-booth exchange. The red-blue counts become $$\begin{aligned} &(100,0),(0,50),(16,2),\\ &(0,10),(3,1),(1,2). \end{aligned}$$

Therefore Alex ends with $59+44=103$ silver tokens, and the correct answer is E .

Small Hint:

Count numbers from $1001$ to $1999$ by their units digit

Big Hint:

Handle the small range from $2000$ to $2012$ separately

Solution:

Once the first three digits are given, their sum determines the units digit, provided that the sum is at most $9.$

We separate the possibilities according to the thousands digit.

If the thousands digit is $1,$ then the sum of the hundreds and tens digits must be at most $8.$ For each possible sum $s,$ there are $s+1$ choices, because the hundreds digit can range from $0$ to $s.$

Thus, this case gives the ninth triangular number: $$\dfrac{9\cdot 10}2 =45.$$

If the thousands digit is $2,$ the only qualifying number below $2013$ is $2002,$ giving $46$ cases in total.

Thus, the correct answer is D.

Small Hint:

Write the arithmetic sequence as $b-d,b,b+d$

Big Hint:

Use the zero discriminant condition

Solution:

Let the common difference be $d$, so $c=b-d$ and $a=b+d$.

A quadratic with exactly one real root has discriminant $0$, so $b^2-4ac=0$. Substituting gives $b^2=4(b+d)(b-d)=4b^2-4d^2$, hence $4d^2=3b^2$.

Since $b,d\ge0$, $2d=\sqrt3 b$. The double root is $\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3$.

Thus, the correct answer is D .

Small Hint:

The numerator must include a factor of $61$

Big Hint:

Any unwanted prime below $61$, especially $59$, must be canceled from the denominator

Solution:

The prime factorization is $2013=3\cdot11\cdot61$, so the numerator must contain a factor of $61$. Hence $a_1\ge61$.

But $61!$ also contains the prime factor $59$, which is not in $2013$, so the denominator must contain a factor of $59$. Hence $b_1\ge59$.

The lower bound $a_1+b_1\ge120$ is attainable because $2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}$.

Thus $|a_1-b_1|=61-59=2$, and the correct answer is B .

Small Hint:

Write the seventh term in terms of the first two terms

Big Hint:

Use divisibility by $8$ and $5$ to force the smallest distinct starts

Solution:

A sequence starting with $u,v$ has seventh term $5u+8v$.