Solution:

The average rainfall per hour equals the total rainfall divided by the total number of hours. July has $31$ days, or $31 \times 24$ hours, so the average is $\dfrac{366}{31 \times 24}$ inches per hour.

Thus, the correct answer is A .

Solution:

A large positive number has a small reciprocal, and a small positive number has a large reciprocal. The smallest number listed is $\dfrac13,$ whose reciprocal $3$ is the largest.

Thus, the correct answer is A .

Solution:

The three smallest numbers in the set are $-3, -1,$ and $7.$ Their sum is $-3 + (-1) + 7 = 3.$

Thus, the correct answer is C .

Solution:

The second factor $40.3 + .07 = 40.37$ is about $40,$ and $1.8$ is about $2.$ A quick estimate is $2(40) - .2(40) = 80 - 8 = 72,$ so the product is closest to $74.$

Thus, the correct answer is C .

Solution:

Since $1000$ minutes $= \dfrac{1000}{60}$ hours $= 16\dfrac23$ hours $= 16$ hours $40$ minutes, the contest ended $16$ hours $40$ minutes past noon, which is $4\!:\!40$ a.m.

Thus, the correct answer is D .

Solution:

The denominator is $1 - \dfrac23 = \dfrac13,$ so the expression is $\dfrac{2}{1/3} = 2 \times 3 = 6.$

Thus, the correct answer is E .

Solution:

Since $\sqrt{8} \lt \sqrt{9} = 3$ and $\sqrt{80} \lt \sqrt{81} = 9,$ the whole numbers strictly between $\sqrt{8}$ and $\sqrt{80}$ are $3, 4, 5, 6, 7, 8.$ There are six of them.

Thus, the correct answer is B .

Solution:

The units digit of the product comes from $B \times 2,$ which ends in $6$ only when $B = 3$ or $B = 8.$ Since the product $6396$ exceeds $6000,$ $B$ must be $8:$ indeed $82 \times 78 = 6396.$

Thus, the correct answer is E .

Solution:

Following the arrows, the possible routes are $MADCN,$ $MACN,$ $MBADCN,$ $MBACN,$ $MBCN,$ and $MBN.$ There are six of them.

Thus, the correct answer is E .

Solution:

The picture leaves $19 - 3 = 16$ feet of wall, split equally into two gaps of $8$ feet each. So the nearest edge of the picture is $8$ feet from the end of the wall.

Thus, the correct answer is B .

Solution:

First, $3 * 5 = \dfrac{3 + 5}{2} = 4.$ Then $4 * 8 = \dfrac{4 + 8}{2} = 6.$

Thus, the correct answer is A .

Solution:

A student received the same grade on both tests exactly when counted on the main diagonal. Those entries are $2 + 4 + 5 + 1 + 0 = 12.$

So the fraction is $\dfrac{12}{30} = \dfrac{4}{10} = 40\%.$

Thus, the correct answer is D .

Solution:

Because every angle is a right angle, sliding the edges shows that the horizontal edges together traverse the width twice and the vertical edges traverse the height twice. So the perimeter equals that of the full $8$ by $6$ rectangle, $2(8 + 6) = 28.$

The answer does not depend on exactly where the notch is cut.

Thus, the correct answer is C .

Solution:

The quotient $\dfrac{b}{a}$ is largest with the biggest $b$ and smallest $a,$ giving $\dfrac{1200}{200} = 6.$

Thus, the correct answer is C .

Solution:

The sale price is $50\%$ of $\$180,$ or $\$90.$ The Saturday price takes another $20\%$ off, leaving $80\%$ of $\$90,$ which is $\$72.$

Thus, the correct answer is B .

Solution:

If the fall sales of $4$ million are $25\%$ of the yearly total, then the yearly total is $16$ million.

The winter sales are $16 - (4.5 + 5 + 4) = 2.5$ million.

Thus, the correct answer is A .

Solution:

Factor $o^2 + no = o(o + n).$ Because $o$ is odd, the product is odd exactly when $o + n$ is odd, which happens only when $n$ is even. When $n$ is odd, $o + n$ is even and the product is even.

So the number is odd only if $n$ is even.

Thus, the correct answer is E .

Solution:

The fewest posts are used when the wall serves as the longer $60$ meter side, so the fence covers two $36$ meter sides and one $60$ meter side, a path of $36 + 60 + 36 = 132$ meters.

Placing a post every $12$ meters, including both ends, uses $\dfrac{132}{12} + 1 = 12$ posts.

Thus, the correct answer is B .

Solution:

The trip was $57{,}060 - 56{,}200 = 860$ miles. The initial $6$ gallons only topped off the tank before the trip; the gas actually used during the trip is the $12 + 20 = 32$ gallons later added to replace it.

So the average is $\dfrac{860}{32} \approx 26.9$ miles per gallon.

Thus, the correct answer is D .

Solution:

Estimating, $\dfrac{300^5}{30 \cdot 400^4} = 10 \left(\dfrac{300}{400}\right)^4 = 10 \left(\dfrac34\right)^4 = 10 \cdot \dfrac{81}{256} \approx 3.$

Thus, the correct answer is D .

Solution:

The four shaded squares fold into four faces of an open box, and the added square supplies the fifth face. Picturing the folds, the squares $A, E, H, B, D,$ and $F$ each complete a valid topless box.

The squares $C$ and $G$ do not work, because folding would force four faces to meet at a single corner. That leaves $6$ valid figures.

Thus, the correct answer is E .

Solution:

The statements form a chain: Alan's A forces Beth's, Beth's forces Carlos's, and Carlos's forces Diana's. If Alan got an A, all four would; if Beth got an A, three would.

The only way to have exactly two A's is for them to be the last two in the chain, Carlos and Diana.

Thus, the correct answer is C .

Solution:

The large circle has radius $2,$ so its area is $\pi(2)^2 = 4\pi,$ and each small circle has area $\pi.$

By symmetry the shaded region is half the difference of the areas: $\dfrac12(4\pi - 2\pi) = \pi.$ The ratio of this to one small circle's area $\pi$ is $1.$

Thus, the correct answer is B .

Solution:

Whatever group Al is in, Bob joins that same group with probability about $\dfrac13,$ and Carol joins it with probability about $\dfrac13.$

So all three share a group with probability about $\dfrac13 \times \dfrac13 = \dfrac19.$

Thus, the correct answer is B .

Solution:

For a set of evenly spaced whole numbers, the average is the average of the smallest and largest. The averages are: A: $\dfrac{2 + 100}{2} = 51,$ B: $\dfrac{3 + 99}{2} = 51,$ C: $\dfrac{4 + 100}{2} = 52,$ D: $\dfrac{5 + 100}{2} = 52.5,$ E: $\dfrac{6 + 96}{2} = 51.$

The largest average is $52.5,$ from the multiples of $5.$

Thus, the correct answer is D .