Small Hint:

Convert the hours to minutes first.

Big Hint:

Add the extra $5$ minutes after multiplying $11$ by $60$.

Video solution:

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Written solution:

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

Thus, C is the correct answer.

Small Hint:

Since $M$ is the midpoint of $\overline{AD}$, find $AM$.

Big Hint:

Use $AM$ as the base and the rectangle’s vertical side length as the height.

Video solution:

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Written solution:

Since $M$ is the midpoint of $\overline{AD},$ the base $AM$ has length $4.$ The altitude from $C$ to $\overline{AM}$ has length $AB=6.$ The area is therefore $\dfrac{1}{2} \cdot 4 \cdot 6 = 12.$

Thus, A is the correct answer.

Small Hint:

First find the total score needed for an average of $70$.

Big Hint:

Subtract the three known scores from the needed total.

Video solution:

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Written solution:

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.

Small Hint:

Compare the two paces in minutes per mile.

Big Hint:

Convert $3$ hours $30$ minutes to minutes before dividing by $15$.

Video solution:

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Written solution:

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 $\frac{210}{15} = 14$ 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 $\frac{240}{10} = 24$ 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.

Small Hint:

The remainder when divided by $10$ tells you the units digit of $N$.

Big Hint:

List the two-digit numbers ending in $3$ and test the condition modulo $9$.

Video solution:

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Written solution:

The second condition says that $N$ ends in $3$, so the possibilities are $13,$ $23,$ $33,$ $43,$ $53,$ $63,$ $73,$ $83,$ $93.$ Of these, only $73$ leaves remainder $1$ when divided by $9$, because its digit sum is $10$. Finally, $73=6\cdot11+7$, so the requested remainder is $7$.

Thus, E is the correct answer.

Small Hint:

With $19$ names, the median is the $10$th value after sorting.

Big Hint:

Count cumulative frequencies from the shortest names upward.

Video solution:

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Written solution:

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.

Small Hint:

Even exponents automatically make perfect squares.

Big Hint:

Also remember that a perfect-square base raised to any positive integer power stays a square.

Video solution:

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Written solution:

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.

Small Hint:

Group the expression into pairs of consecutive terms.

Big Hint:

Count how many pairs of the form $n-(n-2)$ appear.

Video solution:

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Written solution:

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.

Small Hint:

Factor $2016$ into primes.

Big Hint:

The word “distinct” means to count each prime divisor only once.

Video solution:

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Written solution:

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.

Small Hint:

Start by simplifying the inner operation $5\ast x$.

Big Hint:

Apply the definition $a\ast b=3a-b$ twice, from the inside out.

Video solution:

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Written solution:

First, $5\ast x=15-x$. Applying the operation again gives $$\begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9. \end{aligned}$$ Thus $x-9=1$, so $x=10$.

Thus, D is the correct answer.

Small Hint:

Write the two-digit number as $10a+b$.

Big Hint:

Adding the reversed number gives $11(a+b)$.

Video solution:

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Written solution:

Let $a$ be the tens digit and $b$ the units digit. Adding the number and its reversal gives $$\begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132, \end{aligned}$$ so $a+b=12$. The valid digit pairs are $(3,9),$ $(4,8),$ $(5,7),$ $(6,6),$ $(7,5),$ $(8,4),$ $(9,3),$ giving $7$ numbers.

Thus, B is the correct answer.

Small Hint:

Choose a convenient equal number of boys and girls.

Big Hint:

Using $12$ boys and $12$ girls makes both fractions whole numbers.

Video solution:

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Written solution:

Suppose there are $12$ girls and $12$ boys. Then $9$ girls and $8$ boys go on the field trip, so the fraction of field-trip students who are girls is $\dfrac{9}{9+8}=\dfrac9{17}.$

Thus, B is the correct answer.

Small Hint:

A product is $0$ exactly when one of the selected numbers is $0$.

Big Hint:

Count selections containing $0$ out of all ways to choose two different numbers.

Video solution:

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Written solution:

There are $\binom62=15$ ways to select two different numbers. A product is $0$ exactly when one selected number is $0$, and there are $5$ choices for the other number. Thus the probability is $\dfrac5{15}=\dfrac13.$

Thus, D is the correct answer.

Small Hint:

Track gallons of gas rather than miles at first.

Big Hint:

After the $350$-mile drive and the refill, compare the new amount of gas to half a tank.

Video solution:

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Written solution:

If Karl drove $350$ miles, then he used $\frac{350}{35}$ 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 traveled a total distance of $350+175=525$ miles.

Thus, A is the correct answer.

Small Hint:

Factor $13^4-11^4$ as a difference of squares.

Big Hint:

After factoring, count only the powers of $2$ in the product.

Video solution:

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Written solution:

We can factor this expression using difference of squares.

$$\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435. \end{aligned}$$

Because $435$ is odd, $32$ is the largest power of $2$ that divides the expression.

Thus, C is the correct answer.

Small Hint:

A $25\%$ faster pace means Annie runs $\frac{5}{4}$ as far as Bonnie in the same time.

Big Hint:

Annie first passes Bonnie when she has gained one full lap.

Video solution:

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Written solution:

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.

Small Hint:

First count all four-digit passwords when leading zeroes are allowed.

Big Hint:

Then subtract the passwords whose first three digits are forced to be $9,$ $1,$ $1$.

Video solution:

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Written solution:

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.

Small Hint:

Think about how many sprinters are eliminated in each race.

Big Hint:

To leave one champion from $216$ sprinters, all but one sprinter must be eliminated.

Video solution:

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Written solution:

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

Thus, C is the correct answer.

Small Hint:

For an odd number of consecutive terms, the average is the middle term.

Big Hint:

There are $12$ even integers larger than the middle term.

Video solution:

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Written solution:

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

Thus, E is the correct answer.

Small Hint:

Since $b$ appears in both least common multiples, start with possible values of $b$.

Big Hint:

Try to make $b$ absorb the common factor $3$ from $12$ and $15$.

Video solution:

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Written solution:

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.

Small Hint:

The stopping result is determined by which color appears last in the full ordering of the chips.

Big Hint:

Count equally likely positions for the two green chips among the five total chip positions.

Video solution:

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Written solution:

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.

Small Hint:

Use the intersection point of the two diagonals that form the bat wings.

Big Hint:

Similar triangles give the height of the small triangle cut off at the top.

Video solution:

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Written solution:

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 $\triangle BCE$ is $\dfrac12\cdot1\cdot4=2.$ 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, $[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.$ It follows that $[\triangle ECH]=2-\dfrac12=\dfrac32.$ Since the figure is symmetric, the total area of the bat wings is $2\cdot\dfrac32=3.$

Thus, C is the correct answer.