Solution:

After Carlos takes $70\%,$ the remaining portion is $30\%.$

Maria takes one third of this, namely $\tfrac13 \cdot 30\% = 10\%,$ leaving $30\% - 10\% = 20\%.$

Thus, C is the correct answer.

Solution:

Split each letter into its segments. The $A$ is a diagonal of length $2\sqrt{2},$ a vertical of length $2,$ and a crossbar of length $1.$

The $M$ has two verticals of length $2$ and two diagonals of length $\sqrt{2}$ each. The $C$ is three sides of length $2.$

The straight pieces total $2 + 1 + 2 + 2 + 2 + 2 + 2 = 13,$ and the diagonal pieces total $2\sqrt2 + \sqrt2 + \sqrt2 = 4\sqrt2.$

The sum is $13 + 4\sqrt2.$

Thus, C is the correct answer.

Solution:

In $2$ hours she drives $120$ miles, earning $120 \cdot \$0.50 = \$60.$

She uses $120 \div 30 = 4$ gallons, costing $4 \cdot \$2.00 = \$8.$

Her net earnings are $\$60 - \$8 = \$52,$ so her rate is $\$52 \div 2 = \$26$ per hour.

Thus, E is the correct answer.

Solution:

To be divisible by $5$ the last digit is $0$ or $5,$ and to be even it must be $0.$ So the units digit is fixed.

The leading digit is a nonzero even digit: $2, 4, 6, 8$ give $4$ choices. Each of the two middle digits is any even digit $0, 2, 4, 6, 8,$ giving $5$ choices each.

The total is $4 \cdot 5 \cdot 5 \cdot 1 = 100.$

Thus, B is the correct answer.

Solution:

The sum of the $25$ integers is $\dfrac{(-10 + 14) \cdot 25}{2} = 50.$

The five rows each have the same sum and together account for the total, so each row sums to $50 \div 5 = 10.$

Thus, C is the correct answer.

Solution:

For both symmetries, the lines must be the vertical and horizontal center lines of the $5$-by-$4$ grid. Every shaded square then forces the squares obtained by reflecting it across each line.

The top square lies off-center, so its reflection group has $4$ squares, requiring $3$ more. The middle square sits on the central column, so its group has $2$ squares, requiring $1$ more. The bottom-right square again has a group of $4,$ requiring $3$ more.

The least number of additional squares is $3 + 1 + 3 = 7.$

Thus, D is the correct answer.

Solution:

The side lengths are $1, 2, \ldots, 7.$ The four side faces of cube $k$ contribute $4k^2,$ so the vertical faces total $4(1^2 + 2^2 + \cdots + 7^2) = 4 \cdot 140 = 560.$

Viewed from directly above, every upward-facing horizontal patch projects onto the $7 \times 7$ base without overlap, giving $49.$ Viewed from below, the same is true, giving another $49.$

The total surface area is $560 + 49 + 49 = 658.$

Thus, B is the correct answer.

Solution:

The median is the average of the $2020$th and $2021$st smallest values.

The perfect squares that are at most $2020$ are $1^2, \ldots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$), so there are $44$ of them.

Among the list, the numbers $\le 1976$ are the $1976$ integers $1, \ldots, 1976$ together with those $44$ squares, totaling $1976 + 44 = 2020.$

Thus the $2020$th value is $1976$ and the $2021$st value is $1977,$ making the median $\dfrac{1976 + 1977}{2} = 1976.5.$

Thus, C is the correct answer.

Solution:

On $[0, 2\pi],$ the graph of $\cos\left(\tfrac{x}{2}\right)$ is a single arc decreasing from $1$ down to $-1.$

The function $\tan(2x)$ has period $\tfrac{\pi}{2}$ with vertical asymptotes at $x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}.$ These split the interval into five stretches, and on each stretch $\tan(2x)$ runs through all real values.

Since the cosine arc is bounded, each of the five branches meets it exactly once, giving $5$ solutions.

Thus, E is the correct answer.

Solution:

Since $\log_{16} n = \tfrac12 \log_4 n,$ set $y = \log_4 n.$ The equation becomes $\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y.$

Multiplying by $2$ gives $\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y,$ so $\left(\tfrac{y}{2}\right)^2 = y,$ which yields $y = 4.$

Then $\log_4 n = 4,$ so $n = 4^4 = 256,$ and the digit sum is $2 + 5 + 6 = 13.$

Thus, E is the correct answer.

Solution:

Let $P(x, y)$ be the probability of ending on a vertical side. On a vertical side $P = 1,$ on a horizontal side $P = 0,$ and at an interior point $P$ is the average of its four neighbors.

By left-right symmetry $P(2, 2) = \tfrac12.$ Let $a = P(1, 2),$ $b = P(1, 1) = P(1, 3),$ and $c = P(2, 1) = P(2, 3).$ Then

$a = \tfrac14\left(1 + \tfrac12 + 2b\right),$ $\;b = \tfrac14(1 + c + a),$ and $c = \tfrac14\left(2b + \tfrac12\right).$

Substituting gives $b = \tfrac12,$ hence $a = \tfrac38 + \tfrac12 b = \tfrac58.$

Thus, B is the correct answer.

Solution:

Note $(20, 20)$ satisfies $3x - 5y + 40 = 0,$ so it is on $\ell$ and remains on $k.$ The slope of $\ell$ is $\tfrac{3}{5}.$

Rotating by $45^\circ$ gives slope $\dfrac{\tfrac35 + 1}{1 - \tfrac35} = \dfrac{\tfrac85}{\tfrac25} = 4.$

Line $k$ is $y - 20 = 4(x - 20).$ Setting $y = 0$ gives $-20 = 4(x - 20),$ so $x - 20 = -5$ and $x = 15.$

Thus, B is the correct answer.

Solution:

The left side equals $N$ raised to the exponent $\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc},$ which must equal $\dfrac{25}{36}.$

Trying $a = 2$ leaves $\dfrac{1}{2b} + \dfrac{1}{2bc} = \dfrac{25}{36} - \dfrac12 = \dfrac{7}{36}.$

Then $\dfrac{1}{2b}\left(1 + \dfrac1c\right) = \dfrac{7}{36}.$ Taking $b = 3$ gives $1 + \dfrac1c = \dfrac{7}{6},$ so $c = 6.$

All three are integers greater than $1,$ and $b = 3.$

Thus, B is the correct answer.

Solution:

The four vertices $A, C, E, G$ form a square, since they are every other vertex of the regular octagon.

Taking a unit circumradius, the octagon's area is $2\sqrt2$ and the square $ACEG$ has diagonal equal to the circle's diameter, giving area $2.$

The ratio is $\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}.$

Thus, B is the correct answer.

Solution:

The set $A$ consists of the cube roots of $8:$ $2,$ $-1 + i\sqrt3,$ and $-1 - i\sqrt3.$

Factoring by grouping, $z^3 - 8z^2 - 8z + 64 = z^2(z - 8) - 8(z - 8) = (z - 8)(z^2 - 8),$ so $B = \{8, 2\sqrt2, -2\sqrt2\},$ all real.

The greatest distance is from $-1 \pm i\sqrt3$ to $8:$ $\sqrt{(8 - (-1))^2 + (\sqrt3)^2} = \sqrt{81 + 3} = \sqrt{84} = 2\sqrt{21}.$

Thus, D is the correct answer.

Solution:

By periodicity it suffices to consider one unit cell with a lattice point at each corner. The region within $d$ of a corner consists of four quarter-disks of radius $d,$ forming one full disk of area $\pi d^2.$

Setting $\pi d^2 = \tfrac12$ gives $d = \sqrt{\dfrac{1}{2\pi}} \approx 0.399.$

To the nearest tenth, $d = 0.4.$

Thus, B is the correct answer.

Solution:

Let the vertices have $x$-coordinates $n, n+1, n+2, n+3$ with $y$-coordinates $\ln$ of those values. Applying the shoelace formula and simplifying, the area is $\ln\dfrac{(n+1)(n+2)}{n(n+3)}.$

Setting $\dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{91}{90}$ gives $1 + \dfrac{2}{n^2 + 3n} = \dfrac{91}{90},$ so $n^2 + 3n = 180.$

Then $(n - 12)(n + 15) = 0,$ so $n = 12.$

Thus, D is the correct answer.

Solution:

Place $A = (0,0)$ and $C = (20, 0).$ Since $\angle ACD = 90^\circ,$ $D = (20, 30),$ and $E = (5, 0)$ because $AE = 5.$

Since $\angle ABC = 90^\circ,$ $B$ lies on the circle of radius $10$ centered at $(10, 0).$ Line $DE$ is $(5 + t,\, 2t);$ substituting gives $t^2 - 2t - 15 = 0,$ so $t = 5$ or $t = -3.$

For $E$ to lie between $B$ and $D,$ take $t = -3,$ giving $B = (2, -6),$ a distance $6$ below line $AC.$

Then $[ACD] = \tfrac12 \cdot 20 \cdot 30 = 300$ and $[ABC] = \tfrac12 \cdot 20 \cdot 6 = 60,$ so the total area is $360.$

Thus, D is the correct answer.

Solution:

Let $x = 2^{17}.$ Then $\dfrac{2^{289} + 1}{2^{17} + 1} = \dfrac{x^{17} + 1}{x + 1} = x^{16} - x^{15} + \cdots - x + 1,$ an alternating sum of the $17$ powers $x^0, x^1, \ldots, x^{16}.$

Pair each subtracted power with the added power just above it: $x^{m+1} - x^m = 2^{17m}(2^{17} - 1) = 2^{17m} + 2^{17m+1} + \cdots + 2^{17m+16},$ a block of $17$ consecutive powers of $2.$

There are $8$ such pairs, together with the leftover $+2^0.$ The blocks occupy disjoint ranges, so the total number of powers is $8 \cdot 17 + 1 = 137.$

Thus, C is the correct answer.

Solution:

Because $T$ is a scalene right triangle, the only isometry carrying $T$ to itself is the identity, so a sequence works exactly when the three transformations compose to the identity.

The five maps are all of the square's symmetry group except the identity and the two diagonal reflections. In an ordered triple the third map must be the inverse of the first two composed, and it is allowed precisely when the product of the first two is again one of the five.

Of the $25$ ordered pairs, $5$ compose to the identity and $8$ compose to a diagonal reflection. The remaining $25 - 13 = 12$ pairs give valid sequences.

Thus, A is the correct answer.

Solution:

Write $n = 2^a 3^b 5^c 7^d \cdots.$ Since $5! = 2^3 \cdot 3 \cdot 5$ has no other primes, $n$ can only involve $2, 3, 5, 7.$ Matching exponents in $\operatorname{lcm}(5!, n) = 5 \cdot \gcd(10!, n):$

For $2:$ $\max(3, a) = \min(8, a),$ so $3 \le a \le 8$ gives $6$ values. For $3:$ $\max(1, b) = \min(4, b),$ so $1 \le b \le 4$ gives $4$ values.

For $5:$ $\max(1, c) = 1 + \min(2, c)$ with $c \ge 1,$ which forces $c = 3,$ giving $1$ value. For $7:$ $\max(0, d) = \min(1, d),$ so $d = 0$ or $1,$ giving $2$ values.

The total is $6 \cdot 4 \cdot 1 \cdot 2 = 48.$

Thus, D is the correct answer.

Solution:

Since $(a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i,$ we have $a_n b_n = \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big) = \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big).$

Therefore the sum is $\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n = \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right).$

This equals $\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right) = \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right) = \tfrac12 \cdot \dfrac{28}{32} = \dfrac{7}{16}.$

Thus, B is the correct answer.

Solution:

Rerolling one die, keeping two dice that sum to $s,$ wins with probability $\tfrac16$ when $s \le 6$ and $0$ otherwise. Rerolling two dice, keeping a die of value $v,$ wins with probability equal to the number of ways two dice sum to $7 - v,$ over $36;$ this is largest when $v$ is smallest. Rerolling all three wins with probability $\tfrac{15}{216}.$

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least $7$ (so rerolling one cannot reach $7$) while the smallest die is $1, 2,$ or $3$ (so keeping it beats rerolling all three).

Counting the ordered rolls with this property gives $42$ out of $216,$ a probability of $\dfrac{42}{216} = \dfrac{7}{36}.$

Thus, A is the correct answer.

Solution:

A point at distances $p, q, r$ from the vertices of an equilateral triangle of side $s$ satisfies $3(p^4 + q^4 + r^4 + s^4) = (p^2 + q^2 + r^2 + s^2)^2.$

With $p^2 = 1,$ $q^2 = 3,$ $r^2 = 4,$ letting $S = s^2$ gives $3(26 + S^2) = (8 + S)^2,$ so $S^2 - 8S + 7 = 0$ and $S = 1$ or $S = 7.$

A triangle of side $1$ cannot contain a point at distance $2$ from a vertex, so $S = 7$ and $s = \sqrt{7}.$

Thus, B is the correct answer.

Solution:

On the interval $x \in [n, n+1)$ the equation becomes $n(x - n) = a x^2,$ i.e. $a x^2 - n x + n^2 = 0,$ whose two roots are $x = \dfrac{n\big(1 \pm \sqrt{1 - 4a}\big)}{2a}.$

For $0 \lt a \lt \tfrac14$ each interval $[n, n+1)$ contributes exactly one root of this quadratic that lies in it (for suitable $n$), and summing the roots over all valid $n$ gives a total that depends only on $a.$

Requiring the sum to be $420$ forces $a = \dfrac{29}{900},$ which is already in lowest terms. Hence $p + q = 29 + 900 = 929.$

Thus, C is the correct answer.