The high exceeds the low by $16,$ so the low is $8$ below the average. Since the average is $3^\circ,$ the low temperature is $3 - 8 = -5^\circ.$ Thus, the correct answer is C.

The garden is $2\cdot 15 = 30$ feet by $2\cdot 20 = 40$ feet, an area of $1200$ square feet. At half a pound per square foot, Mr. Green expects $\tfrac12\cdot 1200 = 600$ pounds. Thus, the correct answer is A.

Counting down from $201,$ the value $x$ is the $(202-x)$th number. So $53$ is the $(202-53) = 149$th number. Thus, the correct answer is D.

If each drives $D$ miles, together they cover $2D$ miles using $\dfrac{D}{40}+\dfrac{D}{10} = \dfrac{D}{8}$ gallons. The combined rate is $\dfrac{2D}{D/8} = 16$ miles per gallon. Thus, the correct answer is B.

The parents' ages sum to $55\cdot 33$ and the fifth-graders' to $33\cdot 11,$ a total of $33\cdot 66.$ Dividing by $88$ people gives $\dfrac{33\cdot 66}{88} = 24.75.$ Thus, the correct answer is C.

Rearranging gives $x^2 - 10x + 25 + y^2 + 6y + 9$ $= 0,$ that is $(x-5)^2 + (y+3)^2 = 0.$ Hence $x = 5$ and $y = -3,$ so $x + y = 2.$ Thus, the correct answer is B.

After the turn that counts up to $n,$ exactly $1 + 2 + \cdots + n = \tfrac12 n(n+1)$ numbers have been said. For $n = 9$ that is $45.$ The next turn starts $1, 2, \ldots,$ so the $53$rd number is the $8$th number of that turn, namely $8.$ Thus, the correct answer is E.

Solving $3x - 2y = 1$ with $y = 1$ gives $B = (1, 1).$ The distance from $A = (-1, -2)$ to the line $y = 1$ is $3,$ so $\tfrac12\cdot BC\cdot 3 = 3$ gives $BC = 2.$ Then $C = (3, 1)$ or $C = (-1, 1);$ the latter makes $\ell_3$ vertical, so $C = (3, 1)$ and the slope is $\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34.$ Thus, the correct answer is B.

Since $12! = 2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11,$ the largest perfect square dividing it is $2^{10}\cdot 3^4\cdot 5^2,$ whose square root is $2^5\cdot 3^2\cdot 5.$ The exponents sum to $5 + 2 + 1 = 8.$ Thus, the correct answer is C.

After $m$ red-booth and $n$ blue-booth exchanges, Alex has $75 - (2m - n)$ red tokens, $75 - (3n - m)$ blue tokens, and $m + n$ silver tokens. Exchanges are impossible exactly when $2m - n \ge 74$ and $3n - m \ge 73.$ Equality holds at $(m, n) = (59, 44),$ giving $59 + 44 = 103$ silver tokens. Thus, the correct answer is E.

Take east, north, up as $x, y, z.$ After $7$ feet bee $A$ is at $(2, 3, 2)$ and bee $B$ is at $(-3, -4, 0),$ a distance $\sqrt{78} \lt 10.$ On the next foot bee $A$ moves east to $(3, 3, 2)$ and bee $B$ moves west to $(-4, -4, 0),$ a distance $\sqrt{102} \gt 10.$ So they pass through $10$ feet apart while $A$ heads east and $B$ heads west. Thus, the correct answer is A.

City $E$ (roads $AE, DE$) is a detour on an $A$–$D$ trip, and city $C$ (roads $BC, CD$) is a detour on a $B$–$D$ trip. Replace them to get a graph on $A, B, D$ with two $A$–$D$ connections, two $B$–$D$ connections, and one $A$–$B$ road. The trails from $A$ to $B$ using each once are of $4$ types: $ABDADB,$ $ADABDB,$ $ADBADB,$ and $ADBDAB.$ Each detour (through $E,$ through $C$) can be taken on either passage, so each type gives $4$ actual routes, for $4\cdot 4 = 16$ routes. Thus, the correct answer is D.

The angles of a triangle form an arithmetic progression exactly when the middle one is $60^\circ.$ With $\angle DBA = x$ and $\angle ADB = y,$ the four angles of $ABCD$ are $x, y, 180 - y, 180 - x,$ which must itself be an arithmetic progression. Combined with a $60^\circ$ angle in the triangles, this forces either $x = 60$ or $x + y = 120.$ Working through the cases, the possible angle sets are $(60, 80, 100, 120)$ and $(45, 75, 105, 135).$ The two largest angles sum to at most $105 + 135 = 240.$ Thus, the correct answer is D.

A sequence starting $a_1, a_2$ has seventh term $5a_1 + 8a_2.$ For the two sequences, $5a_1 + 8a_2 = 5b_1 + 8b_2,$ so $5(b_1 - a_1) = 8(a_2 - b_2).$ Since $\gcd(5, 8) = 1,$ we need $8 \mid b_1 - a_1$ and $5 \mid a_2 - b_2.$ Taking $a_1 \lt b_1$ with nondecreasing terms gives $a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13.$ Choosing $a_1 = 0,$ $b_1 = b_2 = 8,$ $a_2 = 13$ yields $N = 5\cdot 0 + 8\cdot 13 = 104.$ Thus, the correct answer is C.

Since $2013 = 3\cdot 11\cdot 61,$ the numerator needs a factorial at least $61!$ to supply the prime $61,$ so $a_1 \ge 61.$ But $61!$ also has a factor of $59,$ which $2013$ does not, so the denominator needs $b_1 \ge 59.$ Thus $a_1 + b_1 \ge 120,$ attained by $a_1 = 61,$ $b_1 = 59$ via $2013 = \dfrac{61!\,11!\,3!}{59!\,10!\,5!}.$ Then $|a_1 - b_1| = 2.$ Thus, the correct answer is B.

An equiangular pentagon has all interior angles $108^\circ,$ so each point of the star is an isosceles triangle with base angles $72^\circ$ and apex $36^\circ.$ By the equal base angles, each point contributes two sides that are the same fixed multiple $c$ of the pentagon side it rests on. Summing over the five points, the star perimeter equals $2c\cdot(\text{pentagon perimeter}) = 2c,$ independent of the individual side lengths. So $s$ is constant, and the difference between its maximum and minimum values is $0.$ Thus, the correct answer is A.

From the equations, $a + b = 2 - c$ and $a^2 + b^2 = 12 - c^2.$ Real numbers $a, b$ with a given sum and sum of squares exist iff $(a + b)^2 \le 2(a^2 + b^2),$ i.e. $(2 - c)^2 \le 2(12 - c^2).$ This simplifies to $(3c - 10)(c + 2) \le 0,$ so $-2 \le c \le \tfrac{10}{3}.$ The difference is $\tfrac{10}{3} - (-2) = \tfrac{16}{3}.$ Thus, the correct answer is D.

Work modulo $5.$ With $2013 \equiv 3$ coins, Jenna wins either way: going first she takes $3$ to leave a multiple of $5,$ then answers Barbara's $2$ with $3$ and $4$ with $1$ to keep multiples of $5,$ eventually taking the last coin; going second she keeps the count $\equiv 3 \pmod 5$ until Barbara is stuck at $3$ coins, must remove $2,$ and leaves Jenna the last coin. With $2014 \equiv 4$ coins, whoever goes first wins: Jenna first reduces to the $2013$ case, while Barbara first takes $4$ and then keeps multiples of $5.$ This is choice B. Thus, the correct answer is B.