Small Hint:

A tip equal to $10\%$ of a bill means the bill is $10$ times the tip

Big Hint:

Recover each bill from its tip, then subtract

Solution:

Mike’s $\$2$ tip is $10\%$ of his bill, so his bill is $2 \times 10 = 20$ dollars. Joe’s $\$2$ tip is $20\%$ of his bill, so his bill is $2 \times 5 = 10$ dollars. The difference is $20 - 10 = 10$ dollars.

Thus, the correct answer is D.

Small Hint:

Evaluate the inner operation $(1 \star 2)$ first

Big Hint:

$(1 \star 2) = \dfrac{1+2}{1-2},$ then apply $\star 3$ to that result

Solution:

First $(1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3.$ Then $(-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0.$

Thus, the correct answer is C.

Small Hint:

Solve $2x + 7 = 3$ for $x$ first

Big Hint:

Substitute that value of $x$ into $bx - 10 = -2$ and solve for $b$

Solution:

From $2x + 7 = 3$ we get $x = -2.$ Substituting, $-2b - 10 = -2,$ so $-2b = 8$ and $b = -4.$

Thus, the correct answer is B.

Small Hint:

Let the width be $w;$ then the length is $2w$

Big Hint:

By the Pythagorean theorem $x^2 = w^2 + (2w)^2;$ solve for $w^2$ and then find the area $2w^2$

Solution:

Let the width be $w,$ so the length is $2w.$ Then $x^2 = w^2 + (2w)^2 = 5w^2,$ giving $w^2 = \dfrac{x^2}{5}.$ The area is $w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.$

Thus, the correct answer is B.

Small Hint:

Every four windows purchased comes with a fifth window free

Big Hint:

Compare buying alone ($7$ windows and $8$ windows) with buying all $15$ together

Solution:

Alone, Dave pays for $6$ windows and receives one free to reach $7,$ costing $\$600;$ Doug pays for $7$ and receives one free to reach $8,$ costing $\$700.$ Separately they pay $\$1300.$ Together they need $15$ windows: buying $12$ yields $3$ free, for $\$1200.$ The savings are $1300 - 1200 = 100$ dollars.

Thus, the correct answer is A.

Small Hint:

An average multiplied by the count gives the total sum

Big Hint:

Add the two totals, then divide by $50$

Solution:

The combined sum is $20 \cdot 30 + 30 \cdot 20 = 600 + 600$ $= 1200.$ The average of all $50$ numbers is $\dfrac{1200}{50} = 24.$

Thus, the correct answer is B.

Small Hint:

Distance equals rate times time; write Josh’s distance in terms of Mike’s

Big Hint:

Josh’s distance is $\frac{4}{5} \cdot 2 = \frac{8}{5}$ of Mike’s, and the two distances add to $13$

Solution:

Let Mike ride $m$ miles. Josh rides $\frac{4}{5}$ the rate for $2$ times the time, so Josh’s distance is $\frac{8}{5}m.$ Together they cover $13,$ so $m + \frac{8}{5}m = \frac{13}{5}m = 13,$ giving $m = 5.$

Thus, the correct answer is B.

Small Hint:

The four corner triangles, such as $\triangle ABH,$ are congruent right triangles

Big Hint:

Each corner triangle is right with hypotenuse $\sqrt{50}$ and one leg $BE = 1;$ the inner square’s side is the other leg minus $1$

Solution:

The triangles $ABH,$ $BCE,$ $CDF,$ and $DAG$ are congruent right triangles. In $\triangle BCE$ the hypotenuse is $BC = \sqrt{50}$ and $BE = 1,$ so $CE = \sqrt{50 - 1} = 7.$ Since $BH = CE = 7$ and $E$ lies on $BH$ with $BE = 1,$ the inner square’s side is $EH = 7 - 1 = 6,$ giving area $6^2 = 36.$

Thus, the correct answer is C.

Small Hint:

Count the distinct arrangements of three X’s and two O’s in a row

Big Hint:

The two O’s occupy $\binom{5}{2}$ equally likely position sets, and only one of them gives XOXOX

Solution:

The three X positions can be any of $\binom{5}{3} = 10$ equally likely choices, and exactly one of them produces XOXOX. So the probability is $\dfrac{1}{10}.$

Thus, the correct answer is B.

Small Hint:

A quadratic has exactly one solution when its discriminant equals $0$

Big Hint:

Combine the linear terms into $(a + 8)x,$ then set $(a+8)^2 - 4 \cdot 4 \cdot 9 = 0$

Solution:

Writing the equation as $4x^2 + (a+8)x + 9 = 0,$ there is one solution exactly when the discriminant $(a+8)^2 - 144 = 0.$ Then $a + 8 = \pm 12,$ so $a = 4$ or $a = -20,$ and their sum is $-16.$

Thus, the correct answer is A.

Small Hint:

Each of the $n^3$ unit cubes has $6$ faces, so there are $6n^3$ faces in all

Big Hint:

Only the original surface is red, contributing $6n^2$ red faces; set $\dfrac{6n^2}{6n^3} = \dfrac{1}{4}$

Solution:

The unit cubes have $6n^3$ faces total, of which the original surface accounts for $6n^2$ red faces. Then $\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4},$ so $n = 4.$

Thus, the correct answer is B.

Small Hint:

The base of length $2$ spans two radii, so each circular sector has radius $1$

Big Hint:

Rearrange the four equilateral triangles and four circular segments into four $60^\circ$ sectors of radius $1$

Solution:

Since the base $2$ equals two radii, the radius is $1.$ The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four $60^\circ$ sectors of a circle of radius $1.$ Their total area is $4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.$

Thus, the correct answer is B.

Small Hint:

All quantities are positive, so take the $50$th root of each part of the inequality

Big Hint:

The conditions become $130n \gt n^2$ and $n^2 \gt 16,$ which together give $4 \lt n \lt 130$

Solution:

Taking $50$th roots, the condition becomes $130n \gt n^2 \gt 2^4 = 16.$ From $n^2 \gt 16$ we get $n \gt 4,$ and from $130n \gt n^2$ we get $n \lt 130.$ So $n$ ranges over the integers $5, 6, \ldots, 129,$ which is $125$ values.

Thus, the correct answer is E.

Small Hint:

The middle digit is an integer only when the first and last digits share the same parity

Big Hint:

Count first-and-last digit pairs that are both odd or both even, remembering the first digit cannot be $0$

Solution:

The first and last digits must have the same parity so their average is a digit. Both odd gives $5 \cdot 5 = 25$ pairs. Both even, with a nonzero leading digit, gives $4 \cdot 5 = 20$ pairs. Each pair fixes the middle digit, for a total of $25 + 20 = 45$ numbers.

Thus, the correct answer is E.

Small Hint:

Write $3! \cdot 5! \cdot 7!$ as a product of prime powers

Big Hint:

A cube divisor uses each prime to an exponent that is a multiple of $3;$ count the allowed exponents for each prime

Solution:

As a product of primes, $3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.$ A cube divisor uses exponents that are multiples of $3:$ the exponent of $2$ can be $0,$ $3,$ or $6$ ($3$ choices), the exponent of $3$ can be $0$ or $3$ ($2$ choices), and the exponents of $5$ and $7$ must be $0.$ That gives $3 \cdot 2 \cdot 1 \cdot 1 = 6$ cubes.

Thus, the correct answer is E.

Small Hint:

Write the number as $10a + b$ and subtract its digit sum $a + b$

Big Hint:

The result simplifies to $9a,$ whose units digit is $6$ for only one value of $a$

Solution:

If the number is $10a + b,$ then $(10a + b) - (a + b) = 9a.$ The units digit of $9a$ is $6$ only when $a = 4,$ since $9 \cdot 4 = 36.$ The digit $b$ can then be anything from $0$ to $9,$ giving the ten numbers $40$ through $49.$

Thus, the correct answer is D.

Small Hint:

Each of the five numbers is an endpoint of exactly two of the segments

Big Hint:

The five segment sums total $2(3 + 5 + 6 + 7 + 9),$ and the middle term of a five-term arithmetic sequence is its mean

Solution:

Every number is an endpoint of two segments, so the five segment sums total $2(3 + 5 + 6 + 7 + 9) = 60.$ The middle term of a five-term arithmetic sequence equals its mean, which is $\dfrac{60}{5} = 12.$

Thus, the correct answer is D.

Small Hint:

Imagine all five games are played, even after the series is decided, so each five-game sequence is equally likely

Big Hint:

List the sequences in which B wins game $2$ and A wins the series, then see how many also have B winning game $1$

Solution:

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game $2$ and A ends up with the series (three wins) leaves the equally likely sequences

$$\begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA}. \end{gathered}$$

Only in BBAAA does team B win the first game, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is A.

Small Hint:

After the $45^\circ$ rotation, the square’s diagonal (length $\sqrt{2}$) is vertical, with $B$ at the top vertex

Big Hint:

Use a lower edge’s $45^\circ$ slope and its contact with an adjoining top corner to locate the bottom vertex

Solution:

When lowered, the rotated square’s two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ The bottom vertex is centered between those corners, so each corner is horizontally $\tfrac12$ unit from it. A lower edge has slope $1$ in magnitude, so it rises $\tfrac12$ unit on the way to a corner. Therefore the bottom vertex is at height $1-\tfrac12=\tfrac12.$

Point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt2$ higher. Its height is therefore $\tfrac12+\sqrt2.$

Thus, the correct answer is D.

Small Hint:

Every interior angle of an equiangular octagon is $135^\circ;$ enclose the octagon in a larger square

Big Hint:

The four short sides are hypotenuses of isosceles right triangles with legs $\frac{1}{2},$ cut from a square of side $2$

Solution:

Extend the four sides of length $1$ to form a square. Each short side $\dfrac{\sqrt{2}}{2}$ is the hypotenuse of an isosceles right triangle with legs $\dfrac{1}{2},$ and cutting these four corners from a square of side $1 + 2 \cdot \frac{1}{2} = 2$ gives the octagon. Its area is $2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}.$

Thus, the correct answer is A.

Small Hint:

Use $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$

Big Hint:

Then $\dfrac{6n}{\frac{n(n+1)}{2}} = \dfrac{12}{n+1};$ count the $n$ for which $n + 1$ divides $12$

Solution:

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ the quotient is $\dfrac{6n}{\frac{n(n+1)}{2}} = \dfrac{12}{n+1},$ which is an integer exactly when $n + 1$ divides $12.$ The divisors of $12$ that are at least $2$ are $2,$ $3,$ $4,$ $6,$ $12,$ giving $n = 1,$ $2,$ $3,$ $5,$ $11$ — five values.

Thus, the correct answer is B.