Answer: A

Small Hint:

The average rate is the total rainfall divided by the total number of hours in the month

Big Hint:

July has $31$ days, and each day has $24$ hours

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 $\dfrac{366}{31 \times 24} = \dfrac{61}{124}$ inches per hour.

Thus, the correct answer is A .

Answer: A

Small Hint:

The reciprocal of a positive number is largest when the number itself is smallest

Big Hint:

Compare the five numbers and find the smallest one

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 .

Answer: C

Small Hint:

To make the sum as small as possible, choose the three smallest numbers

Big Hint:

The three smallest numbers are $-3,$ $-1,$ and $7$

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 .

Answer: C

Small Hint:

Round each factor: $1.8$ is near $2$ and $40.3 + 0.07$ is near $40$

Big Hint:

Estimate $2 \times 40$ and adjust slightly downward since $1.8 \lt 2$

Solution:

The second factor $40.3 + 0.07 = 40.37$ is about $40,$ and $1.8$ is about $2.$ A quick estimate is $2(40) - 0.2(40) = 80 - 8 = 72,$ so the product is closest to $74.$

Thus, the correct answer is C .

Answer: D

Small Hint:

Convert $1000$ minutes into hours and minutes by dividing by $60$

Big Hint:

$1000$ minutes is $16$ hours and $40$ minutes; add that to noon

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 .

Answer: E

Small Hint:

First simplify the denominator $1 - \dfrac23$

Big Hint:

Dividing by $\dfrac13$ is the same as multiplying by $3$

Solution:

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

Thus, the correct answer is E .

Answer: B

Small Hint:

Find perfect squares just above $8$ and just below $80$

Big Hint:

$\sqrt{8}$ is between $2$ and $3,$ and $\sqrt{80}$ is between $8$ and $9$

Solution:

Since $2 \lt \sqrt{8} \lt 3$ and $8 \lt \sqrt{80} \lt 9,$ 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 .

Answer: E

Small Hint:

The units digit of the product is $6,$ and it comes from $B \times 2$

Big Hint:

$B \times 2$ ends in $6$ when $B = 3$ or $B = 8;$ the product is large, so test which value works

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 .

Answer: E

Small Hint:

List the routes by the first step: from $M$ you can go to $A$ or to $B$

Big Hint:

Follow the arrows carefully; from $B$ you can reach $A,$ $C,$ or $N$ directly

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 .

Answer: B

Small Hint:

The picture is centered, so the two gaps on the sides are equal

Big Hint:

Subtract the picture width from the wall width, then split the remainder between the two sides

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 .

Answer: A

Small Hint:

The operation $A * B$ gives the average of $A$ and $B$

Big Hint:

First compute $3 * 5,$ then combine that result with $8$

Solution:

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

Thus, the correct answer is A .

Answer: D

Small Hint:

A student with the same grade on both tests lies on the diagonal running from the top-left to the bottom-right of the table

Big Hint:

Add the diagonal entries, then divide by $30$ and convert to a percent

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 .

Answer: C

Small Hint:

All the angles are right angles, so the horizontal edges and the vertical edges can each be regrouped

Big Hint:

The rightward and leftward horizontal edges each span the full width $8,$ and the upward and downward vertical edges each span the full height $6$

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 .

Answer: C

Small Hint:

A quotient is largest when the numerator is as large as possible and the denominator as small as possible

Big Hint:

Use the largest $b$ and the smallest $a$

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 .

Answer: B

Small Hint:

A $50\%$ discount leaves half the original price

Big Hint:

Then take $80\%$ of that sale price to apply the extra $20\%$ off

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 .

Answer: A

Small Hint:

The fall sales of $4$ million are $25\%$ of the yearly total, so find the yearly total first

Big Hint:

Subtract the spring, summer, and fall sales from the yearly total

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 .

Answer: E

Small Hint:

Factor: $o^2 + no = o(o + n)$

Big Hint:

Since $o$ is odd, the parity of $o(o + n)$ matches the parity of $o + n$

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 .

Answer: B

Small Hint:

To use the fewest posts, let the $60$ meter side lie along the wall, so the fenced length is as short as possible

Big Hint:

The fenced path is $36 + 60 + 36 = 132$ meters; place a post every $12$ meters, counting both ends

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 .

Answer: D

Small Hint:

The gas used during the trip is only what was added after the start: $12 + 20$ gallons

Big Hint:

Divide the total miles driven by the gallons used during the trip

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 .

Answer: D

Small Hint:

Round the numbers: $304 \approx 300,$ $29.7 \approx 30,$ and $399 \approx 400$

Big Hint:

Simplify $\dfrac{300^5}{30 \cdot 400^4} = 10 \left(\dfrac34\right)^4$

Solution:

Estimating,

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

Thus, the correct answer is D .

Answer: E

Small Hint:

A topless box has $5$ faces, so the four shaded squares plus the one added square must fold into a cube missing its top

Big Hint:

Try each lettered square as the fifth face; a placement fails when folding would force two faces into the same spot

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 .

Answer: C

Small Hint:

If a student earlier in the chain gets an A, everyone after them in the chain must also get an A

Big Hint:

For exactly two A’s, the two must be the last students in the chain

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 .

Answer: B

Small Hint:

The large circle has radius $2,$ since $AC$ equals two small diameters

Big Hint:

By symmetry the shaded region is half the difference between the large circle’s area and the two small circles’ area

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 .

Answer: B

Small Hint:

Fix Al’s group; then find the chance that Bob lands in that same group

Big Hint:

After Al’s group is fixed, Bob’s chance is $\frac{199}{599}$ and then Carol’s is $\frac{198}{598}$

Solution:

Whatever group Al is in, Bob joins that same group with probability $\dfrac{199}{599}.$ If Bob does, Carol joins them with probability $\dfrac{198}{598}.$

Thus all three share a group with probability $\dfrac{199}{599} \cdot \dfrac{198}{598} \approx 0.110,$ closest to $\dfrac19.$

Thus, the correct answer is B .