Solution:

Harry calculates it correctly as follows: $$\begin{align*} 8 - (2 + 5) &= 8 - 7 \\ &= 1. \end{align*}$$ Terry calculates it incorrectly as follows: $$\begin{align*} 8 - 2 + 5 &= 6 + 5 \\ &= 11. \end{align*}$$ Therefore: $$\begin{align*} H - T &= 1 - 11 \\ &= -10. \end{align*}$$

Thus, A is the correct answer.

Solution:

To use the most number of coins, Paul would have to use only $5$-cent coins. This would require $35 / 5 = 7$ coins.

To use the least number of coins, Tom would use a $25$-cent coin and a $10$-cent coin, for a total of $2$ coins.

Therefore, the difference is $7 - 2 = 5.$

Thus, E is the correct answer.

Solution:

During the first three days, Isabella read $36 \cdot 3 = 108$ pages.

During the next three days, Isabella read $44 \cdot 3 = 132$ pages.

Therefore, she read a total of $108 + 132 + 10 = 250$ pages.

Thus, B is the correct answer.

Solution:

The only for two numbers to add to an odd number is if one is even and the other is odd. The only even prime is $2,$ so $2$ is one of the prime numbers.

Therefore, the other prime is $85 - 2 = 83,$ and the product is $2 \cdot 83 = 166.$

Thus, E is the correct answer.

Solution:

With $\$ 20,$ Margie can buy $20 / 4 = 5$ gallons of gas. With $5$ gallons of gas, Margie can travel $5 \cdot 32 = 160$ miles.

Thus, C is the correct answer.

Solution:

To find the area of each rectangle we multiply $2$ by their respective lengths. This means that we can factor out the $2$ and we are left with the sum of the lengths. Adding together the lengths, we get $91.$

Therefore, the sum of the areas is $2 \cdot 91 = 182.$

Thus, D is the correct answer.

Solution:

Let $x$ be the number of boys in the class. This means that there are $x + 4$ girls. As there are $28$ students, we know that: $$\begin{align*} x + x + 4 &= 28\\ 2x&=24 \\ x&=12. \end{align*}$$ Therefore, the ratio of girls to boys is $16 : 12 = 4 : 3.$

Thus, B is the correct answer.

Solution:

Note that since $11$ people paid the same amount, then the resulting sum is divisible by $11.$ Therefore, $\underline{1} \underline{A} \underline{2}$ must be divisible by $11.$

Remember the divisibility rule for $11$: if we take the difference of the sums of alternating digits, and this difference is divisible by $11,$ the whole number is divisible by $11.$

For $\underline{1} \underline{A} \underline{2},$ the aforementioned difference is $$1 + 2 - A = 3 - A.$$ The only way for this to be divisible by $11$ is if $A = 3.$

Thus, D is the correct answer.

Solution:

Since $\triangle BDC$ is isosceles, we know that $$\angle DBC = \angle DCB = 70^{\circ}.$$ This means that $$\begin{align*} \angle BDC &= 180^{\circ} - 70^{\circ} - 70^{\circ} \\ &= 40^{\circ}. \end{align*}$$ Therefore, $$\begin{align*} \angle ADB &= 180^{\circ} - \angle BDC \\ &= 180^{\circ} - 140^{\circ} \\ &= 140^{\circ}. \end{align*}$$ Thus, D is the correct answer.

Solution:

The seventh AMC 8 would have been administered $6$ years after the first one. Therefore, Samantha took it in $1985 + 6 = 1991.$

This means that Samantha was born $12$ years prior in $1991 - 12 = 1979.$

Thus, A is the correct answer.

Solution:

Allow the X is the location of the dangerous intersection.

Let E represent moving one block east and N represent moving one block north. To avoid the dangerous intersection, Jack must go either EE or NN first.

After these moves, there are only $4$ possible remaining options: $$\begin{align*} &• EEENN\\ &• EENEN\\ &• EENNE\\ &• NNEEE \end{align*}$$

Thus, A is the correct answer.

Solution:

Notice that there are $3! = 6$ total ways that a reader could match the celebrities. However, only one of these is the correct matching. Therefore, the probability that the reader guesses it correctly is $\frac{1}{6}.$

Thus, B is the correct answer.

Solution:

Since $n^2 + m^2$ is even, either $n^2$ and $m^2$ are both odd, or both even.

If they are both odd, then $n$ and $m$ are both odd. If they are both even, then $n$ and $m$ are both even. If $n$ and $m$ are both odd or even, their sum will always be even.

Therefore, $n + m$ is never odd.

Thus, D is the correct answer.

Solution:

The area of $ABCD$ is $$5 \cdot 6 = 30.$$

The area of $$\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*}$$ Therefore: $$\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12. \end{align*}$$ Then using the Pythagorean theorem we get that $$\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13. \end{align*}$$

Thus, B is the correct answer.

Solution:

Note that each of the $12$ arcs splits the circle evenly, so they each cover $360^{\circ} / 12 = 30^{\circ}.$

$\angle AOE$ spans $4$ of these arcs, so $$\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}.$$ Similarly, $$\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}.$$ We also know that both triangles are isosceles since two of their sides are radii. Therefore, $$x = \dfrac{180 - 120}{2} = 30^{\circ}$$ and $$y = \dfrac{180 - 60}{2} = 60^{\circ}.$$ Therefore, $x + y = 90^{\circ}.$

Thus, C is the correct answer.

Solution:

Each time plays $4$ non-conference games, totalling $4 \cdot 8 = 32$ games.

We don't want to double count the conference game, so we can just count all the games every team played out home. Each team played $7$ home games, for a total of $8 \cdot 7 = 56$ home games.

Therefore, the total number of games in a season is $$32 + 56 = 88.$$

Thus, B is the correct answer.

Solution:

If George normally walks $1$ mile at $3$ miles per hours, it will take him $\frac{1}{3}$ hours to walk to school.

Today, he walked $\frac{1}{2}$ miles at $2$ miles per hours, which means he walked for $$\dfrac{1}{2} \div 2 = \dfrac{1}{4}$$ hours.

This means he has $$\dfrac{1}{3} - \dfrac{1}{4} = \dfrac{1}{12}$$ hours left to run to school. In order to run $\frac{1}{2}$ mile in $\frac{1}{12}$ hours, he must run at a speed of $$\dfrac{1}{2} \div \dfrac{1}{12} = 6$$ miles per hour.

Thus, B is the correct answer.

Solution:

We can count the number of ways for each outcome to occur, and whichever has the most is the answer.

A: There is only one way for the children to all be boys.

B: There is only one way for the children to all be girls.

C: This scenario has $6$ possibilities: $$\begin{align*} \text{BBGG}\\ \text{BGBG}\\ \text{BGGB}\\ \text{GBBG}\\ \text{GBGB}\\ \text{GGBB} \end{align*}$$

D: This scenario has $8$ possibilities: $$\begin{align*} \text{BBBG}\\ \text{BBGB}\\ \text{BGBB}\\ \text{GBBB} \end{align*}$$ (swapping G and B yields the other $4$ possibilities).

E: Clearly, this is not the case.

Thus, D is the correct answer.

Solution:

Note that the interior cube has none of its faces showing, which means that this must be a white cube to minimize the white surface area. We should also place the other $5$ white cubes in the middle of every space to minimize the number of white faces showing.

This results in a white surface area of $5 \text{ in}^2.$ The total surface area is $6 \text{ in} \cdot (3 \text{ in})^2 = 54 \text{ in}^2.$

Therefore, the fraction of white surface area is $\dfrac{5}{54}.$

Thus, A is the correct answer.

Solution:

The three sectors inside the rectangle are all quarter-circles, with areas $\frac{\pi}{4},$ $\pi,$ and $\frac{9\pi}{4},$ and therefore their total area is $\frac{7\pi}{2}.$

We can use $\frac{22}{7}$ as an approximation for $\pi.$ Substituting this value in, we get $\frac{7}{2} \cdot \frac{22}{7} = 11.$ The area of the rectangle is $3 \cdot 5 = 15,$ so the area outside the circles is (approximately) $15 - 11 = 4.$

Thus, B is the correct answer.

Solution:

For a number to be divisible by $3,$ the sum of the digits has to be divisible by $3.$ The sum of the digits of $\underline{74A52B1}$ is $19 + A + B,$ so this has to be divisible by $3.$

Similarly for $\underline{326AB4C},$ we get that $3$ divides $15 + A + B + C,$ which means that $3$ divides $A + B + C.$

From the first condition, we get that $A + B$ is $1$ less than a multiple of $3.$ This means that $C$ must be $1$ more than a multiple of $3.$

Therefore, the only way that all these conditions can be satisfied is if $C=1.$

Thus, A is the correct answer.

Solution:

We can represent the number as $10a + b,$ where $a$ and $b$ are both digits. Then the condition implies that $$ab + a + b = 10a + b.$$ Simplifying this, we get $ab = 9a,$ which shows that $b = 9$ since $a$ cannot equal $0.$

Thus, E is the correct answer.

Solution:

Since there are no more than $31$ days in a month, the sum of any two girls' uniform numbers must be less than or equal to $31.$ The only possible such pairs of $2$-digit primes are $$11 + 13 = 24$$$$11 + 17 = 28,$$ $$11 + 19 = 30$$$$13 + 17 = 30.$$ This shows that the desired dates are $24, 28,$ and $30.$ Caitlin's birthday should be the latest, which means that her uniform number must be the smallest.

Therefore, Caitlin's number is $11.$

Thus, A is the correct answer.

Solution:

We can order the number of cans bought by customer from least to greatest. In this ordering, the median is the average of the $50$th and $51$st customer.

To maximize this, we want to minimize the first $49$ purchases. To minimize them, we can set them all equal to $1.$ If the $50$th number is $4,$ then there would be $51$ purchases greater than or equal to $4.$

This would cause the total number of cans sold to be at least $49 + 51 \cdot 4 = 253,$ which is too large. This forces the $50$th purchase to be $3$ cans, which lowers the total by $1$ to the desired amount. This means that the $51$st person bought $4$ cans, causing the median to be $$(3 + 4) / 2 = 3.5.$$

Thus, C is the correct answer.

Solution:

If we compare a straight portion to a semi-circular path, we can see that the semi-circular path is $\frac{\pi}{2}$ longer than the straight path due to the equation for circumference.

The diameter of each semi-circular path is $80$ feet, which evenly divides a mile, showing that there are an integer number of semi-circular paths. Therefore, the amount of time it takes Robert to ride along the semi-circular path is directly proportional to the length of the straight path.

The straight path would take Robert $\frac{1}{5}$ hours to ride. This means that it would take him $$\dfrac{1}{5} \cdot \dfrac{\pi}{2} = \dfrac{\pi}{10}$$ hours to ride along the semi-circular path.

Thus, B is the correct answer.