Solution:

The text charge is $100 \cdot 5 = 500$ cents $= \$5.$ She talked $30$ minutes past the $30$-hour allowance, so the overage is $30 \cdot 10 = 300$ cents $= \$3.$

The total is $\$20 + \$5 + \$3 = \$28.$

Thus, the correct answer is D.

Solution:

Coin $C$ is drawn as a complete, unbroken circle, so nothing covers it and it lies on top.

Reading the remaining overlaps, each coin's uncovered arc shows it sits above the next: $C$ covers $E,$ $E$ covers $D,$ $D$ covers $B,$ and $A$ covers $B$ while lying under the others. This gives the top-to-bottom order $(C, E, D, A, B).$

Thus, the correct answer is E.

Solution:

Fourteen bottles hold $14 \cdot 35 = 490$ milliliters, which is not enough. Fifteen bottles hold $15 \cdot 35 = 525$ milliliters, which suffices.

So Jasmine needs $15$ bottles.

Thus, the correct answer is E.

Solution:

Take the grade sizes in the ratio $4 : 2 : 1$ for third, fourth, and fifth grades. The weighted average is $$\dfrac{4 \cdot 12 + 2 \cdot 15 + 1 \cdot 10}{7} = \dfrac{48 + 30 + 10}{7} = \dfrac{88}{7}.$$

Thus, the correct answer is C.

Solution:

The birds that are not swans make up $100\% - 25\% = 75\%$ of the total, and geese are $30\%$ of the total. The requested fraction is $$\dfrac{30}{75} = \dfrac{2}{5} = 40\%.$$

Thus, the correct answer is C.

Solution:

Let $a$ be the number of two-point shots. The two-point shots score $2a$ points, and the three-point shots score the same $2a$ points. The free throws number $a + 1$ and score $a + 1$ points.

The total is $$2a + 2a + (a + 1) = 5a + 1 = 61,$$ so $a = 12$ and the free throws number $a + 1 = 13.$

Thus, the correct answer is A.

Solution:

Total cents is $1771 = 7 \cdot 11 \cdot 23.$ Writing (students)(pencils each)(cost per pencil) $= 1771,$ the number of students is a divisor of $1771$ that is a majority of $30,$ hence more than $15.$ The only such divisor is $23.$

Then (pencils)(cost) $= 77 = 7 \cdot 11$ with cost $\gt$ pencils $\gt 1,$ forcing $7$ pencils at $11$ cents each.

Thus, the correct answer is B.

Solution:

Since $A + B + C = B + C + D = 30,$ we get $D = A,$ and likewise the sequence repeats with period $3.$ Thus $H,$ the eighth term, equals $B.$

From $A + B + C = 30$ and $C = 5,$ we have $A + H = A + B = 30 - 5 = 25.$

Thus, the correct answer is C.

Solution:

There are $18$ twins and $18$ triplets.

Twin-twin handshakes: each twin shakes $18 - 2 = 16$ other twins, giving $\dfrac{18 \cdot 16}{2} = 144.$

Triplet-triplet handshakes: each triplet shakes $18 - 3 = 15$ other triplets, giving $\dfrac{18 \cdot 15}{2} = 135.$

Twin-triplet handshakes: each twin shakes half the $18$ triplets, giving $18 \cdot 9 = 162$ (each such handshake counted once).

The total is $144 + 135 + 162 = 441.$

Thus, the correct answer is B.

Solution:

For diameter $d,$ area $\lt$ circumference means $\dfrac{\pi d^2}{4} \lt \pi d,$ i.e. $d \lt 4.$ Since $d \ge 2,$ this needs a sum of $2$ or $3.$

A sum of $2$ has probability $\dfrac{1}{36}$ and a sum of $3$ has probability $\dfrac{2}{36},$ totaling $\dfrac{3}{36} = \dfrac{1}{12}.$

Thus, the correct answer is B.

Solution:

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin.

The distance from $C$ to $A$ (and to $B$) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $$2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) - \tfrac{\sqrt2}{2}\sqrt{4 - 2} = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1.$$

Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $$\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$$

Thus, the correct answer is C.

Solution:

Measure everything relative to the water. In that frame the raft is stationary at the point where the boat started, and the boat moves at its constant speed $v$ relative to the water, both downstream and upstream.

The boat leaves the raft, travels away for some time, then returns to it at the same relative speed, so it spends equal times going and returning. Hence the outbound leg to $B$ takes half of $9,$ which is $4.5$ hours.

Thus, the correct answer is D.

Solution:

Let $I$ be the incenter. Because $\overline{BI}$ bisects $\angle B$ and $MN \parallel BC,$ alternate angles give $\angle MIB = \angle IBC = \angle MBI,$ so $\triangle MBI$ is isosceles with $MB = MI.$ Similarly $NC = NI.$

Therefore the perimeter of $\triangle AMN$ is $$AM + MN + NA = AM + (MI + IN) + NA = AM + MB + NC + NA = AB + AC = 12 + 18 = 30.$$

Thus, the correct answer is B.

Solution:

Substituting $x = a,$ $y = b,$ the point is above the parabola when $b \gt a^3 - ab,$ i.e. $b(a + 1) \gt a^3.$

For $a = 1:$ $b \gt \tfrac12,$ all $9$ values work. For $a = 2:$ $b \gt \tfrac83,$ so $b \ge 3,$ giving $7.$ For $a = 3:$ $b \gt \tfrac{27}{4} = 6.75,$ so $b \ge 7,$ giving $3.$ For $a \ge 4,$ no $b \le 9$ works.

The count is $9 + 7 + 3 = 19$ out of $81,$ so the probability is $\dfrac{19}{81}.$

Thus, the correct answer is E.

Solution:

Let the base have side $s,$ centered at the origin, with apex at height $6.$ Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from $\left(\tfrac{s}{2}, 0\right)$ to $(0, 6).$

This line is $\tfrac{2}{s}x + \tfrac16 y = 1.$ The hemisphere is tangent to the face, so the distance from the origin to this line is the radius $2:$ $$\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.$$

Then $\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,$ so $\tfrac{4}{s^2} = \tfrac{2}{9}$ and $s^2 = 18,$ giving $s = 3\sqrt2.$

Thus, the correct answer is A.

Solution:

The diagonals connect the vertices in the order $A - C - E - B - D - A,$ which is a $5$-cycle. The condition is exactly that this cycle is properly colored.

The number of proper $k$-colorings of a cycle of length $n$ is $(k-1)^n + (-1)^n (k-1).$ With $n = 5$ and $k = 6,$ $$5^5 + (-1)^5 \cdot 5 = 3125 - 5 = 3120.$$

Thus, the correct answer is C.

Solution:

The centers are separated by the sums of radii: $3,$ $4,$ and $5,$ a right triangle with the right angle at the radius-$1$ center. Place that center at $(0,0),$ the radius-$2$ center at $(3,0),$ and the radius-$3$ center at $(0,4).$

The tangency points lie on the segments at distances equal to the radii: $(1, 0),$ $(0, 1),$ and on the hypotenuse at $(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).$

By the shoelace formula the area is $$\tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65.$$

Thus, the correct answer is D.

Solution:

The identity $|x+y| + |x-y| = 2\max(|x|, |y|)$ turns the condition into $\max(|x|, |y|) = 1,$ the boundary of the square with $|x| \le 1$ and $|y| \le 1.$

On this region $x^2 - 6x + y^2$ increases as $x$ decreases and as $y^2$ increases, so the maximum is at $x = -1,$ $y = \pm 1:$ $$1 + 6 + 1 = 8.$$

Thus, the correct answer is D.

Solution:

Let $m = \lfloor \log_2 (N - 1) \rfloor,$ so the elite count is $2^{m+1} - N = 19,$ giving $N = 2^{m+1} - 19.$

Consistency requires $2^m \le N - 1 = 2^{m+1} - 20,$ i.e. $2^m \ge 20,$ so $m \ge 5.$

The two smallest choices are $m = 5$ giving $N = 64 - 19 = 45,$ and $m = 6$ giving $N = 128 - 19 = 109.$ Their sum is $154.$

Thus, the correct answer is C.

Solution:

Since $f(1) = a + b + c = 0,$ we have $c = -a - b.$ Then $$f(7) = 48a + 6b = 6(8a + b), \qquad f(8) = 63a + 7b = 7(9a + b).$$

From $50 \lt 6(8a + b) \lt 60$ we get $8a + b = 9,$ and from $70 \lt 7(9a + b) \lt 80$ we get $9a + b = 11.$ Subtracting, $a = 2,$ then $b = -7$ and $c = 5.$

So $f(100) = 20000 - 700 + 5 = 19305,$ which lies in $5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4,$ giving $k = 3.$

Thus, the correct answer is C.

Solution:

Each step requires $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}.$ Tracking the domains:

$f_1: (-\infty, 1].$ $f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].$ $f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].$ $f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}$ (only the value $0$ is possible). $f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.$

For $f_6$ we would need $\sqrt{36 - x} = -231,$ impossible, so the domain is empty. Hence $N = 5,$ $c = -231,$ and $N + c = -226.$

Thus, the correct answer is A.

Solution:

For even $n = 2m,$ the $n$-ray partitional points are exactly $\left(\tfrac{i}{m}, \tfrac{j}{m}\right)$ with $1 \le i, j \le m - 1,$ giving $(m-1)^2$ points.

For $n = 100$ ($m = 50$) there are $49^2 = 2401$ points. A point is both $100$- and $60$-ray partitional iff its coordinates are multiples of $\tfrac{1}{10},$ i.e. it is $20$-ray partitional, giving $9^2 = 81$ points.

So the count is $2401 - 81 = 2320.$

Thus, the correct answer is C.

Solution:

Represent $f$ by $M = \begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix}.$ Then $g(g(z)) = z$ says $f$ composed with itself four times is the identity, so $M^4$ is a scalar matrix.

This happens when the ratio of eigenvalues is a fourth root of unity. The order-$4$ case gives $(\operatorname{tr} M)^2 = 2\det M,$ i.e. $(1 + b)^2 = 2(b - a),$ which simplifies to $b^2 = -(1 + 2a).$ The order-$2$ case gives $b = -1.$

Then $|b|^2 = |1 + 2a|,$ and as $a$ runs over $|a| = 1,$ $|1 + 2a|$ ranges over $[1, 3],$ so $|b|$ ranges over $[1, \sqrt3]$ (the value $b = -1$ is included). The difference is $\sqrt3 - 1.$

Thus, the correct answer is C.

Solution:

Because $AB + CD = 14 + 7 = 21 = 9 + 12 = BC + DA,$ a tangential quadrilateral (one with an inscribed circle) with these sides exists. For a tangential quadrilateral the area equals $r \cdot s$ with semiperimeter $s = 21,$ so maximizing $r$ means maximizing the area.

Among tangential quadrilaterals with given sides, the largest area is achieved by the cyclic (bicentric) one, whose area is $$\sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{7 \cdot 12 \cdot 14 \cdot 9} = 42\sqrt6.$$

Then $r = \dfrac{42\sqrt6}{21} = 2\sqrt6.$

Thus, the correct answer is C.

Solution:

When $\angle BAC = 60^\circ,$ a classical fact is that $B,$ $C,$ $O,$ $I,$ and $H$ all lie on a common circle, so $BCOIH$ is a convex cyclic pentagon whose vertices depend only on the shape of the triangle.

Fixing $BC = 1$ and $\angle A = 60^\circ,$ the circumradius is $R = \tfrac{1}{\sqrt3},$ and $O, I, H$ are determined by $\angle CBA = B$ (with $\angle BCA = 120^\circ - B$). Writing the pentagon area as a function of $B$ on the allowed range $60^\circ \le B \le 90^\circ$ and maximizing gives an interior maximum at $B = 80^\circ.$

So the maximizing angle is $\angle CBA = 80^\circ.$

Thus, the correct answer is D.