Small Hint:

Convert the texting charge and excess-call charge into dollars

Big Hint:

Only the extra half hour is billed by the minute

Solution:

Michelle has to pay $\$20$ for the monthly fee. Her text messages cost $$100\cdot 5=500$$ cents, or $\$5.$ Finally she talked for $0.5$ hours, $30$ minutes, over $30$ hours.

This means her extra charge is $$30\cdot10=300$$ cents, or $\$3.$

Her total cost is $$\$20 + \$5 + \$3 = \$28.$$

Thus, D is the correct answer.

Small Hint:

Compare multiples of $35$ to $500$

Big Hint:

The first multiple of $35$ that reaches $500$ gives the number of bottles

Solution:

The desired amount is $$\dfrac{500}{35} = \dfrac{100}{7} = 14 \dfrac{2}{7}.$$

This means that the smallest number of small bottles she must buy is $15.$

Thus, E is the correct answer.

Small Hint:

Evaluate the inner averages first

Big Hint:

Use $\{1\ 1\ 0\}=\dfrac23$ and $[0\ 1]=\dfrac12$

Solution:

We have that $$\{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}.$$

We also get that $$[0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}.$$

Finally, $$\left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}.$$

Thus, D is the correct answer.

Small Hint:

Most terms in the two sums cancel

Big Hint:

Only the first term of $X$ and the last term of $Y$ do not match

Solution:

Note that the terms $12$ through $100$ are common to both sums. When we subtract, all these terms cancel out.

This means that $$Y - X = 102 - 10 = 92.$$

Thus, A is the correct answer.

Small Hint:

Use a $4:2:1$ ratio for the numbers of third, fourth, and fifth graders

Big Hint:

Compute a weighted average with weights $4,2,1$

Solution:

WLOG, let there be one fifth grader. This then tells us that there are two fourth graders and four third graders.

We can do this, since we are only interested in the average, which is not impacted by the exact number of students.

The total number of minutes the students spend running is $$12 \cdot 4 + 15 \cdot 2 + 10 = 88$$ minutes. The total number of students is $1 + 2 + 4 = 7.$ The average is then $\dfrac{88}{7}.$

Thus, C is the correct answer.

Small Hint:

The union must contain all elements of the larger set

Big Hint:

Make $B$ a subset of $A$ to minimize the union

Solution:

To minimize the number of elements in the union, we want to maximize the overlap between the two sets.

We can then assume that $B$ is contained completely within $A,$ which means that the union is the same as $A,$ which has $20$ elements.

Thus, C is the correct answer.

Small Hint:

Check which equation asks a nonnegative expression to be negative

Big Hint:

An absolute value plus $5$ cannot equal $0$

Solution:

A simplifies to $$x + 7 = 0$$$$x = -7,$$ so it has a solution.

B simplifies to $$|-3x| = -5,$$ which has no solution since absolute value makes everything positive.

Let us make sure that all the other choices have solutions.

C simplifies to $$\sqrt{-x} = 2$$$$-x = 4$$$$x = -4,$$ which is fine.

D simplifies to $$\sqrt{x} = 8$$$$x = 64,$$ which works.

Finally, E simplifies to $$|-3x| = 4$$$$-3x = \pm 4$$ $$x = \pm \dfrac{4}{3},$$ which has a solution as well.

Thus, B is the correct answer.

Small Hint:

Use the percent that are not swans as the denominator

Big Hint:

The desired percent is $\frac{30}{100-25}$

Solution:

WLOG, let there be $100$ birds. Then $75$ birds are not swans. The desired percentage is then $$\dfrac{30}{75} \cdot 100 = 40 \%.$$

Thus, C is the correct answer.

Small Hint:

Find the vertical and horizontal side lengths of the rectangle

Big Hint:

The side lengths are $a+b$ and $c+d$

Solution:

Note that the region is a rectangle with side lengths $$a - (-b) = a + b$$ and $$d - (-c) = c + d.$$ The area is then $$(a + b)(c + d) =$$$$ac + ad + bc + bd.$$

Thus, A is the correct answer.

Small Hint:

Factor $1771$

Big Hint:

The number of students is a majority of $30$, so use the factor greater than $15$

Solution:

Let $p$ be the number of pencils that each student bought, $s$ be the number of students that bought pencils, and $c$ be the cost of a pencil.

We have that $$psc = 1771 = 7 \cdot 11 \cdot 23.$$

We also have the following restrictions: $$30 \geq s \gt 15, p \gt 1, c \gt p.$$

From the above prime factorization, we have that $s = 23$ is the only value that satisfies the conditions.

Finally, we get that $p = 7$ and $c = 11$ are the only remaining values that satisfy the other conditions.

Thus, B is the correct answer.

Small Hint:

Let $EB=x$, so $AB=8x$

Big Hint:

A side of the inner square is the hypotenuse of a $7x$-by-$x$ triangle

Solution:

Let $x = EB.$ Then $AB = 8x.$ Applying the Pythagorean Theorem to a side of $EFGH,$ we get $$\sqrt{(7x)^2 + x^2} = \sqrt{50x^2}$$

The desired ratio is then $$\dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}.$$

Thus, B is the correct answer.

Small Hint:

Let $x$ be the number of two-point shots

Big Hint:

The points from two-point shots equal the points from three-point shots

Solution:

Let $x$ be the number of successful two-point shots. Then we have that $$2x + 2x + (x + 1) = 61,$$ which simplifies to $$5x + 1 = 61$$$$x = 12.$$

The number of successful free throws is then $12 + 1 = 13.$

Thus, A is the correct answer.

Small Hint:

The units digit must be $2$ or $8$

Big Hint:

Case on whether the hundreds digit is $2$ or $5$

Solution:

Since the hundreds digit can only be a $2$ or $5,$ we can case on this value.

Case $1:$ hundreds digit is $2$

The only option for the units digit is $8,$ since the number must be even. This leaves $4$ options for the tens digit.

This gives us $1 \cdot 4 = 4$ numbers for this case.

Case $2:$ hundreds digit is $5$

Similarly to above, $2$ and $8$ are the only options for the units digit, leaving $4$ options for the tens digit.

This gives us $2 \cdot 4 = 8$ numbers for this case.

The total number of integers is then $4 + 8 = 12.$

Thus, A is the correct answer.

Small Hint:

Compare $\pi(\frac{d}{2})^2$ with $\pi d$

Big Hint:

The inequality reduces to $d<4$

Solution:

For the area to be less than the circumference, we must have $$\pi r^2 \lt 2\pi r$$$$r \lt 2.$$

This means the diameter must be less than $4.$ There are three possible rolls that satisfy this: $$(1, 1), (1, 2), (2, 1).$$

The probability is then $$\dfrac{3}{36} = \dfrac{1}{12}.$$

Thus, B is the correct answer.

Small Hint:

Let $x$ be the gasoline-only distance

Big Hint:

The miles per gallon equation is $\frac{x+40}{0.02x}=55$

Solution:

Let $x$ be the distance the car drove solely on gasoline. We have that $$\dfrac{40 + x}{0.02x} = 55.$$ Cross-multiplying and simplifying gives $$40 = 0.1x$$$$x = 400.$$ The total length of the trip is then $400 + 40 = 440.$

Thus, C is the correct answer.

Small Hint:

Square the whole expression

Big Hint:

The product $(9-6\sqrt2)(9+6\sqrt2)$ is small

Solution:

Since we have square roots, we can try to change the inside of each radical to be a perfect square.

Note that we can rewrite the expression as $$\begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3}. \end{aligned}$$

Factoring and simplifying gives us $$\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}$$ $$= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6.$$

Thus, B is the correct answer.

Small Hint:

Compare two consecutive three-term sums

Big Hint:

The sequence repeats every three terms, so $A=D=G$ and $C=F$

Solution:

From the condition about the sequence, we get that $$A + B + C = 30$$$$B = 25 - A.$$ Similarly, we get $$B + C + D = 30$$$$D = A.$$

Propagating these values through the sequence and repeating the condition for every consecutive triple, we get that $$E = 25 - A, F = 5, G = A,$$ and finally, $H = 25 - A.$

The desired sum is then $$A + 25 - A = 25.$$

Thus, C is the correct answer.

Small Hint:

Use the unit squares formed by the centers and intersection points

Big Hint:

Two quarter-circle subtractions plus a semicircle leave a simple area

Solution:

The area of this region is the area of circle $C$ minus the area of the overlapping regions with $A$ and $B.$

From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle.

This area is then $$\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.$$

There are four of these that we must subtract, which leaves us with a final answer of $$\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.$$

Thus, C is the correct answer.

Small Hint:

Let the $1991$ population be $p^2$

Big Hint:

Use $(q-p)(q+p)=141$ and test the factor pairs

Solution:

Let the population in $1991$ be $p^2.$ Then let the population in $2001$ be $q^2 + 9.$

Using these values, we have $$p^2 + 150 = q^2 + 9$$ $$q^2 - p^2 = 141.$$

Factoring, we get $$(q - p)(q + p) = 141.$$

As $p$ and $q$ are integers, we have that the only possible values for $q - p$ and $q + p$ are $(1, 141)$ and $(3, 47).$

Trying the first pair, we have $$q-p=1$$ and $$q+p=141,$$ which adding together and dividing gives us $q = 71$ and $p = 70.$

We have that $p^2 + 300$ is not a square number, which means that this pair is the wrong one.

Trying the other pair and using the same strategy gives us $q = 25$ and $p = 22.$

Now, $p^2 + 300 = 784,$ which is a perfect square. The percent increase in population is then $$\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%.$$

Thus, E is the correct answer.

Small Hint:

A chord of length $r$ subtends $60^\circ$

Big Hint:

Fix one chord and look for the arc positions from which the second chord crosses it

Solution:

A chord of length $r$ in a circle of radius $r$ subtends a $60^\circ$ arc. Fix the first chord, with endpoints at angles $0^\circ$ and $60^\circ.$ If the second chord starts at angle $\theta,$ its other endpoint is $60^\circ$ clockwise from there.

The endpoints of the two chords alternate exactly when $\theta$ lies in either of the two $60^\circ$ arcs immediately adjacent to the fixed chord’s endpoints. Thus the favorable starting positions occupy $120^\circ$ of the circle.

The desired probability is then $$\dfrac{2}{6} = \dfrac{1}{3}.$$

Thus, D is the correct answer.