Answer: A

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.

Answer: E

Solution:

To use the largest number of coins, Paul would use only $5$-cent coins, which gives $35/5=7$ coins.

To use the smallest number of coins, Paul would use a $25$-cent coin and a $10$-cent coin, for a total of $2$ coins. The difference is $7-2=5$.

Thus, E is the correct answer.

Answer: B

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.

Answer: E

Solution:

The sum of two odd primes is even. Since $85$ is odd, one of the primes must be the only even prime, $2$.

The other prime is $85-2=83$, so the product is $2\cdot83=166$.

Thus, E is the correct answer.

Answer: C

Solution:

With $\$20$, Margie can buy $20/4=5$ gallons of gas.

At $32$ miles per gallon, she can drive $5\cdot32=160$ miles.

Thus, C is the correct answer.

Answer: D

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.

Answer: B

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.

Answer: D

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.

Answer: D

Solution:

Since $BD=DC$, triangle $BDC$ is isosceles, so $\angle DBC=\angle BCD=70^\circ$.

Thus $$\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ.\end{aligned}$$ Because $A,D,C$ are collinear, $\angle ADB$ and $\angle BDC$ form a straight angle, so $\angle ADB=180^\circ-40^\circ=140^\circ$.

Thus, D is the correct answer.

Answer: A

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.

Answer: A

Solution:

Let E represent traveling one block east and N represent traveling one block north. A shortest route uses three E moves and two N moves. To avoid the dangerous intersection one block east and one block north of Jack's house, the first two moves must be either EE or NN.

The possible routes are $EEENN$, $EENEN$, $EENNE$, and $NNEEE$, for $4$ routes.

Thus, A is the correct answer.

Answer: B

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.

Answer: D

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.

Answer: B

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.

Answer: C

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.

Answer: B

Solution:

Each team plays $4$ non-conference games, for $8\cdot4=32$ games involving Middle School Eight teams and non-conference opponents.

Within the conference, each of the $8$ teams has $7$ home games against the other teams, so there are $8\cdot7=56$ conference games. The total is $32+56=88$.

Thus, B is the correct answer.

Answer: B

Solution:

If George normally walks $1$ mile at $3$ miles per hour, it takes him $\dfrac{1}{3}$ hour, or $20$ minutes, to get to school.

Today he walked the first $\dfrac{1}{2}$ mile at $2$ miles per hour, taking $\dfrac{1}{2}\div2=\dfrac{1}{4}$ hour, or $15$ minutes. He has $5$ minutes, which is $\dfrac{1}{12}$ hour, left to cover $\dfrac{1}{2}$ mile.

His required speed is $\dfrac{1}{2}\div\dfrac{1}{12}=6$ miles per hour.

Thus, B is the correct answer.

Answer: D

Solution:

There are $2^4=16$ equally likely birth-order outcomes. The counts by category are: all boys, $1$; all girls, $1$; two boys and two girls, $\binom{4}{2}=6$; and three of one gender and one of the other, $2\binom{4}{1}=8$.

The largest count is $8$, so the most likely outcome is three children of one gender and one of the other.

Thus, D is the correct answer.

Answer: A

Solution:

To minimize visible white area, place one white cube in the center of the large cube, where no faces are visible. Place each of the other five white cubes at the center of a face of the large cube, where each contributes only one visible white square.

The visible white surface area is therefore $5$ square inches. The total surface area of the $3$-inch cube is $6\cdot3^2=54$ square inches, so the fraction that is white is $\dfrac{5}{54}$.

Thus, A is the correct answer.

Answer: B

Solution:

The parts of the three circles inside the rectangle are quarter-circles with radii $1$, $2$, and $3$. Their total area is $$\dfrac{\pi}{4}+\pi+\dfrac{9\pi}{4}=\dfrac{7\pi}{2}.$$

Using $\pi\approx\dfrac{22}{7}$, this total is about $11$. The rectangle has area $3\cdot5=15$, so the desired area is about $15-11=4$.

Thus, B is the correct answer.

Answer: A

Solution:

For $\underline{74A52B1}$ to be divisible by $3$, the digit sum $19+A+B$ must be a multiple of $3$. Hence $A+B$ is $1$ less than a multiple of $3$.

For $\underline{326AB4C}$ to be divisible by $3$, the digit sum $15+A+B+C$ must be a multiple of $3$. Since $15$ is already divisible by $3$, $A+B+C$ must be divisible by $3$.

Therefore $C$ must be $1$ more than a multiple of $3$. Among the answer choices, only $1$ has that form.

Thus, A is the correct answer.

Answer: E

Solution:

Let the number be $10a+b$, where $a$ is the tens digit and $b$ is the units digit. The condition gives $$ab+a+b=10a+b.$$

Canceling $b$ from both sides gives $ab+a=10a$, so $ab=9a$. Since $a\neq0$, we get $b=9$.

Thus, E is the correct answer.

Answer: A

Solution:

Any pair-sum must be a date in the month, so it is at most $31$. The possible sums of two two-digit primes that are at most $31$ are $11+13=24$, $11+17=28$, $11+19=30$, and $13+17=30$.

The three dates are earlier, today, and later, so they must be distinct and increasing. The only possible set of three distinct dates is $24,28,30$, coming from the numbers $11$, $13$, and $17$.

Bethany's date is Ashley plus Caitlin, Ashley's date is Bethany plus Caitlin, and Caitlin's later date is Ashley plus Bethany. This means Caitlin's number is the smallest of the three, namely $11$.

Thus, A is the correct answer.

Answer: C

Solution:

Order the $100$ purchases from least to greatest. The median is the average of the $50$th and $51$st entries. To maximize it, make the first $49$ entries as small as possible, so set them all equal to $1$.

If the $50$th entry were at least $4$, then the total would be at least $49+51\cdot4=253$, too many cans. So the $50$th entry is at most $3$.

This maximum is attainable: use $49$ entries equal to $1$, the $50$th entry equal to $3$, and the final $50$ entries equal to $4$. The total is $49+3+50\cdot4=252$, and the median is $(3+4)/2=3.5$.

Thus, C is the correct answer.

Answer: B

Solution:

Each semicircle advances Robert $40$ feet along the highway, so a one-mile stretch requires $5280/40=132$ semicircles.

Each semicircle has diameter $40$ feet, so its arc length is $20\pi$ feet. The total distance Robert rides is $132\cdot20\pi=2640\pi$ feet, which is $\dfrac{2640\pi}{5280}=\dfrac{\pi}{2}$ miles.

At $5$ miles per hour, the time is $\dfrac{\pi}{2}\div5=\dfrac{\pi}{10}$ hours.

Thus, B is the correct answer.