There are $60$ minutes in an hour, so the total time is $60 \cdot 11 + 5 = 665$ minutes.

Thus, C is the correct answer.

From the diagram, we can see that the base of $\triangle AMC$ is $4$ and the altitude is $6.$ The area is therefore $\dfrac{1}{2} \cdot 4 \cdot 6 = 12.$

Thus, A is the correct answer.

From the average, we can calculate the sum of the scores to be $4 \cdot 70 = 280.$ This means that the remaining score is $$280 - 70 - 80 - 90 = 40.$$

Thus, A is the correct answer.

To better compare the rates, we can change his speed into minutes per mile.

As a boy he ran $15$ miles in $3 \cdot 60 + 30 = 210$ minutes, which means that he ran at a pace of $210 / 15 = 14$ minutes per mile.

As an adult, he can walk $10$ miles in $4 \cdot 60 = 240$ minutes, which means he walks at a pace of $240 / 10 = 24$ minutes per mile.

Subtracting the two, we get that he takes $10$ more minutes to walk a mile as an adult.

Thus, B is the correct answer.

The two-digit numbers that leave a remainder of $1$ when divided by $9$ are: $$\begin{align*}&10, 19, 28, 37, 46,\\& 55, 64, 73, 82, 91.\end{align*}$$ The two-digit numbers that leave a remainder of $3$ when divided by $10$ are: $$\begin{align*}&13, 23, 33, 43, 53,\\& 63, 73, 83, 93.\end{align*}$$ Among these numbers, $73$ is the only common number. The remainder of $73$ when divided by $11$ is $7.$

Thus, E is the correct answer.

Since there are $19$ people, each with one corresponding name length, the middle length will be the tenth one. Counting from the left side, the tenth value that we arrive upon is $4.$

Thus, B is the correct answer.

Since any number with an even exponent is a perfect square, we can eliminate A, C, and E. Also, a square number to any power remains a square number, so that rules out D.

Thus, B is the correct answer.

We can group the sum as follows: $$\begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2). \end{gathered}$$ Note that each pair evaluates to $2$ and there are $25$ pairs. Therefore, the total sum is $2 \cdot 25 = 50.$

Thus, C is the correct answer.

We can prime factorize $2016$ as $2^5 \cdot 3^2 \cdot 7.$ This shows that the prime divisors of $2016$ are $2, 3,$ and $7.$ The sum of these is $12,$ so B is the correct answer.

We can simplify the equation as follows: $$\begin{align*} 1 &= 2 * (5 * x) \\ &= 2 * (3 \cdot 5 - x) \\ &= 2 * (15 - x) \\ &= 3 \cdot 2 - (15 - x) \\ &= x - 9. \end{align*}$$ Solving yields $x = 10.$

Thus, D is the correct answer.

Let $ab$ be the two-digit number in question. Then, it follows that the number obtained by reversing its digits is $ba.$ Therefore, in order for $ab$ to satisfy the property in the question: $$\begin{align*}10(a + b) + a + b &=132\\11(a + b)&=132\\a+b&=12.\end{align*}$$ The only possible solutions $(a,b)$ to this equation, where $a,b$ are both one digit, are: $$(3,9), (4,8), (5,7), (6,6),$$$$(7,5), (8,4),(9,3).$$ As such, there are $7$ solutions.

Thus, B is the correct answer.

To more easily compare, we can convert the fractions to have the same denominator:$$\dfrac{3}{4} = \dfrac{9}{12}$$ $$\dfrac{2}{3} = \dfrac{8}{12}$$ This shows that the ratio of girls to boys is $9 : 8,$ which means that the fraction of girls on the field trip is $\dfrac{9}{17}.$

Thus, B is the correct answer.

The only way for the product to be $0$ is if one of the numbers chosen is $0.$ If the first number chosen is $0,$ then there are $5$ options for the second number.

Similarly, there are $5$ combinations if $0$ was chosen second.

Therefore, there are $10$ total pairs where the product is $0.$ The total number of pairs is $6 \cdot 5 = 30,$ so the probability is $$\dfrac{10}{30} = \dfrac{1}{3}.$$

Thus, D is the correct answer.

If Karl drove $350$ miles, then he used $350 / 35$ gallons of gas.

When he bought more gas, he added $8$ gallons to $14 - 10 = 4$ gallons, attaining a total of $12$ gallons.

If his tank was half full when he arrived, he used $12 - 7 = 5$ gallons, which equates to $5 \cdot 35 = 175$ miles.

Therefore, he travelled a total distance of $$350 + 175 = 525.$$ miles.

Thus, A is the correct answer.

We can factor this expression using difference of squares.

$$\begin{align*} (13^2 + 11^2)&(13^2 - 11^2) \\ &= 290 \cdot 48 \\ &= 32 \cdot 145 \cdot 3 \end{align*}$$

This shows that $32$ is the largest power of $2$ that divides the expression.

Thus, C is the correct answer.

Since Annie is $25\%$ faster than Bonnie, for every lap Bonnie finishes, Annie completes $1 \dfrac{1}{4}$ laps. Therefore, Annie gains a quarter lap every time Bonnie finishes a lap.

With this in mind, for Annie to completely lap Bonnie, Bonnie must finish $4$ laps, which means that Annie finished $5$ laps.

Thus, D is the correct answer.

The total number of passwords with no conditions is $10^4.$ The condition removes $10$ possible passwords since the first $3$ are determined, and the last one can be anything. Therefore, the number of acceptable passwords is $$10,000 - 10 = 9990.$$

Thus, D is the correct answer.

Note that each race eliminates $5$ people. For there to be a winner, $215$ must be eliminated. Therefore, $215 / 5 = 43$ races are required to eliminate this number of people.

Thus, C is the correct answer.

The average of these numbers is $10,000 / 25 = 400.$ The largest number is $12$ even numbers away, which means that it equals $400 + 12 \cdot 2 = 424.$

Thus, E is the correct answer.

We know that $b$ has to divide both $12$ and $15,$ so it must equal either $1$ or $3.$

If $b = 1,$ then $a = 12$ and $c = 15,$ making their least common multiple $60.$ If $b = 3,$ then we may take $a = 4$ and $c = 5.$ The least common multiple in this scenario is $20.$

Thus, A is the correct answer.

The $3$ reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are $10$ equally likely ways to choose the two positions of the green chips, and $4$ of them have a green chip in the last position.

Therefore, the desired probability is $$\dfrac{4}{10} = \dfrac{2}{5}.$$

Thus, B is the correct answer.

Define $I$ to be the midpoint of $\overline{AD}$ and $G$ to be the midpoint of $\overline{EF}.$ Also define $H$ to be the intersection of $\overline{CF}$ and $\overline{BE}.$

The area of $\triangle BCE$ equals $$\dfrac{1}{2} \cdot 1 \cdot 4 = 2.$$ By symmetry, we can see that $\triangle BCH$ and $\triangle EFH$ are similar. Since their bases are in a $1 : 3$ ratio, so are their altitudes. This means that $3IH = HG,$ which implies that $IH = 1.$

Therefore, the area of $$\begin{align*}\triangle BCH &= \dfrac{1}{2} \cdot 1 \cdot 1\\ &= \dfrac{1}{2}.\end{align*}$$ This implies that the area of $$\begin{align*}\triangle ECH &= 2 - \dfrac{1}{2} \\&= \dfrac{3}{2}.\end{align*}$$ Since the figure is symmetric, the total area of the bat wings is $2 \cdot \dfrac{3}{2} = 3.$

Thus, C is the correct answer.

We know that $AE = EB = AB$ since they are all radii of congruent circles, so they form an equilateral triangle, which means that $\angle AEB = 60^{\circ}.$

Also, since $\overline{DB}$ and $\overline{AC}$ are diameters, $$\angle DEB = \angle AEC = 90^{\circ}.$$ Therefore, $$\begin{align*}\angle CED = \angle DEB &+ \angle AEC\\ &- \angle AEB,\end{align*}$$ which equals $120^{\circ}.$

Thus, C is the correct answer.

Since $QRS$ is divisible by $5,$ we know that $S = 5.$

Since $PQR$ is divisible by $4,$ $QR$ equals either $12, 24,$ or $32.$

If $QR=12,$ then $R=2,$ so $RST=25T$ cannot be divisible by $3$ using the remaining digits. If $QR=32,$ then $R=2$ again, giving the same obstacle. Thus $QR=24,$ and then $RST=45T.$ Among the remaining digits, only $T=3$ makes $453$ divisible by $3$.

Therefore, $PQRST = 12453.$

Thus, A is the correct answer.