Small Hint:

Compute inside the parentheses first for Harry.

Big Hint:

Compare Harry’s result with Terry’s left-to-right calculation.

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.

Small Hint:

Use the smallest coins to maximize the number of coins.

Big Hint:

Use a quarter and a dime to minimize the number of coins.

Solution:

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

Small Hint:

Convert each average into a total number of pages.

Big Hint:

Add the first three days, next three days, and last day.

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.

Small Hint:

An odd sum of two primes must include the only even prime.

Big Hint:

Subtract that even prime from $85$.

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.

Small Hint:

First find how many gallons $\$20$ buys.

Big Hint:

Multiply the gallons by $32$ miles per gallon.

Solution:

With $\$20$, Margie can buy $\frac{20}{4}=5$ gallons of gas.

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

Thus, C is the correct answer.

Small Hint:

Each rectangle has area $2$ times its listed length.

Big Hint:

Add the lengths first, then multiply by $2$.

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.

Small Hint:

Let the number of boys be $b$.

Big Hint:

Then the number of girls is $b+4$, and the total is $28$.

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.

Small Hint:

The total paid by $11$ members must be divisible by $11$.

Big Hint:

For a three-digit number, use the alternating-sum test for divisibility by $11$.

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.

Small Hint:

Since $BD=DC$, triangle $BCD$ is isosceles.

Big Hint:

Find $\angle BDC$, then use the straight angle at $D$.

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.

Small Hint:

The seventh AMC $8$ was six years after the first AMC $8.$

Big Hint:

Subtract Samantha’s age from that year.

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.

Small Hint:

Any shortest path uses three east moves and two north moves.

Big Hint:

Count or list only the paths that avoid the intersection after one east and one north move.

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.

Small Hint:

There are $3!$ ways to match the babies to celebrities.

Big Hint:

Only one matching is completely correct.

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.

Small Hint:

A square has the same parity as its base integer.

Big Hint:

For a sum of two squares to be even, the two integers have the same parity.

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.

Small Hint:

Set the triangle area equal to the rectangle area.

Big Hint:

After finding $CE$, use the $5$-$12$-$13$ right triangle.

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.

Small Hint:

Each adjacent pair of letters cuts off $30^\circ$ at the center.

Big Hint:

The two relevant triangles are isosceles because their sides are radii.

Solution:

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

$\angle AOE$ 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.

Small Hint:

Count non-conference games separately from conference games.

Big Hint:

For conference games, count each team’s home games against the other seven teams.

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.

Small Hint:

Find the usual time for one mile at $3$ mph.

Big Hint:

Subtract the time already spent walking the first half mile.

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.

Small Hint:

List outcomes by the number of boys and girls.

Big Hint:

The most likely category is the one with the most arrangements.

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.

Small Hint:

Hide one white cube in the center of the $3\times3\times3$ cube.

Big Hint:

Put the remaining white cubes in face centers to expose only one white face each.

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.

Small Hint:

Inside the rectangle, each circle contributes a quarter-circle sector.

Big Hint:

Subtract the total sector area from the rectangle area and approximate $\pi$.

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.

Small Hint:

Use the divisibility-by-$3$ test on both numbers.

Big Hint:

Compare what the first number says about $A+B$ with what the second number needs.

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.

Small Hint:

Let the tens digit be $a$ and the units digit be $b$.

Big Hint:

Translate the condition into $ab+a+b=10a+b$.

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.