Solution:

Checking each pair: $(-4)(-9) = 36,$ $(-3)(-12) = 36,$ $\dfrac12 \times (-72) = -36,$ $(1)(36) = 36,$ and $\dfrac32 \times 24 = 36.$

Only $\dfrac12 \times (-72) = -36$ fails to equal $36.$

Thus, the correct answer is C .

Solution:

Since $49 = 7 \times 7$ and $84 = 7 \times 12,$ the fraction reduces to $\dfrac{7}{12}.$

The sum of numerator and denominator is $7 + 12 = 19.$

Thus, the correct answer is C .

Solution:

Factoring: $39 = 3 \times 13,$ $51 = 3 \times 17,$ $77 = 7 \times 11,$ $91 = 7 \times 13,$ and $121 = 11 \times 11.$

The largest prime factor among these is $17,$ which is a factor of $51.$

Thus, the correct answer is B .

Solution:

Regroup as $(1000 \times 10) \times 0.1993 \times 1993 = 10000 \times 0.1993 \times 1993.$

Since $10000 \times 0.1993 = 1993,$ the product is $1993 \times 1993 = (1993)^2.$

Thus, the correct answer is E .

Solution:

The two shaded regions are each one quarter of the circle, and the unshaded region is one half. So the three quantities are in the ratio $\tfrac14 : \tfrac14 : \tfrac12,$ or $1 : 1 : 2.$

A matching bar graph must have the two shaded bars equal in height and the unshaded bar exactly twice as tall. Only one bar graph has two equal shaded bars with the white bar double their height.

Thus, the correct answer is C .

Solution:

Feeding $15$ children uses $15 \div 5 = 3$ cans, leaving $5 - 3 = 2$ cans.

Those $2$ cans feed $2 \times 3 = 6$ adults.

Thus, the correct answer is B .

Solution:

Adding three equal terms, $3^3 + 3^3 + 3^3 = 3 \times 3^3 = 3^4 = 81.$

Thus, the correct answer is A .

Solution:

She takes half a pill every two days, so one pill lasts $4$ days. Then $60$ pills last $60 \times 4 = 240$ days.

At about $30$ days per month, that is roughly $240 \div 30 = 8$ months.

Thus, the correct answer is D .

Solution:

From the table, $2 * 4 = 3$ and $1 * 3 = 3.$

Then $(2 * 4) * (1 * 3) = 3 * 3 = 4.$

Thus, the correct answer is D .

Solution:

The price changes month to month are: January $\$2.50 \to \$2.00$ (drop $\$0.50$), February $\$2.00 \to \$4.00$ (rise), March $\$4.00 \to \$1.50$ (drop $\$2.50$), April $\$1.50 \to \$4.50$ (rise), May $\$4.50 \to \$3.00$ (drop $\$1.50$), and June $\$3.00 \to \$1.00$ (drop $\$2.00$).

The largest drop is $\$2.50,$ which occurred during March.

Thus, the correct answer is B .

Solution:

Since $81$ students took the test, the median is the $41$st score counting up from the lowest.

Adding the bar heights from the left gives running totals $1, 3, 7, 12, 18, 28, 42, \ldots$ The total first passes $41$ at the interval labeled $70,$ which contains the $29$th through $42$nd scores. So the $41$st score lies in the interval labeled $70.$

Thus, the correct answer is C .

Solution:

The six arrangements give $5 \times 4 + 6 - 3 = 23,$ $5 \times 4 - 6 + 3 = 17,$ $5 + 4 \times 6 - 3 = 26,$ $5 - 4 \times 6 + 3 = -16,$ $5 + 4 - 6 \times 3 = -9,$ and $5 - 4 + 6 \times 3 = 19.$

The only value among the choices is $19.$

Thus, the correct answer is E .

Solution:

The full sign has area $5 \times 15 = 75$ square units. Counting the shaded unit squares in each letter gives $H = 11,$ $E = 11,$ $L = 7,$ and $P = 10,$ for a shaded total of $11 + 11 + 7 + 10 = 39.$

The unshaded area is $75 - 39 = 36.$

Thus, the correct answer is D .

Solution:

Filling the grid so each row and column has $1, 2, 3,$ the top row becomes $1, 3, 2,$ the middle row $3, 2, A,$ and the bottom row $2, 1, B.$ The middle row forces $A = 1,$ and the last column $2, 1, B$ forces $B = 3.$

So $A + B = 1 + 3 = 4.$

Thus, the correct answer is C .

Solution:

The four numbers sum to $4 \times 85 = 340,$ so the remaining three sum to $340 - 97 = 243.$

Their mean is $243 \div 3 = 81.$

Thus, the correct answer is A .

Solution:

Starting inside, $2 + \dfrac13 = \dfrac73,$ so $\dfrac{1}{7/3} = \dfrac37.$

Then $1 + \dfrac37 = \dfrac{10}{7},$ and the whole expression is $\dfrac{1}{10/7} = \dfrac{7}{10}.$

Thus, the correct answer is C .

Solution:

The interior surface is exactly one face of the cardboard after the corners are removed. The sheet has area $20 \times 30 = 600,$ and each removed corner has area $5 \times 5 = 25.$

So the interior surface area is $600 - 4 \times 25 = 500.$

Thus, the correct answer is B .

Solution:

Rectangle $ACDE$ has area $32 \times 20 = 640.$ Triangle $BCD$ has area $\dfrac{16 \times 20}{2} = 160,$ and triangle $DEF$ has area $\dfrac{10 \times 32}{2} = 160.$

The remaining region $ABDF$ has area $640 - (160 + 160) = 320.$

Thus, the correct answer is A .

Solution:

Each number in the first sum is exactly $1800$ more than the matching number in the second sum, and there are $93$ such pairs.

So the difference is $93 \times 1800 = 167{,}400.$

Thus, the correct answer is A .

Solution:

Subtracting $93$ from $10^{93}$ (a $1$ followed by $93$ zeros) gives a number that is $91$ nines followed by $07.$

The digit sum is $91 \times 9 + 0 + 7 = 819 + 7 = 826.$

Thus, the correct answer is D .

Solution:

The new length is $1.2$ times the old and the new width is $1.5$ times the old, so the new area is $1.2 \times 1.5 = 1.8$ times the old area.

That is an increase of $80\%.$

Thus, the correct answer is D .

Solution:

Numbering $1$ through $99$ uses ten $2$'s in the units place and ten in the tens place, a total of twenty $2$'s. Pages $100$ and $101$ use none.

The remaining two $2$'s are used on pages $102$ and $112.$ After that, pages $113$ through $119$ need no $2,$ but $120$ would require another $2,$ so he can number up to $119.$

Thus, the correct answer is D .

Solution:

Since $P$ beats $Q,$ $R,$ and $T,$ and no one beats $P,$ runner $P$ finishes first and so cannot be third.

The clues give the chain $P$ before $T$ before $Q$ before $S.$ So $P, T,$ and $Q$ all finish ahead of $S,$ meaning $S$ is no better than fourth and cannot be third either.

Each of $Q, R, T$ can finish third: for example $P, T, Q, R, S$ puts $Q$ third; $P, R, T, Q, S$ puts $T$ third; and $P, T, R, Q, S$ puts $R$ third. So only $P$ and $S$ cannot be third.

Thus, the correct answer is C .

Solution:

Each row ends at a perfect square, so the row containing $142$ ends at $144 = 12^2,$ and the row above it ends at $121 = 11^2.$

Since the rows are aligned at their right edges, $121$ sits directly above $143,$ and therefore $120$ sits directly above $142.$

Thus, the correct answer is C .

Solution:

Tilt the card $45^\circ$ and center it on a corner where four grid squares meet, as shown. Because the card's diagonal, $\sqrt{1.5^2 + 1.5^2} = \sqrt{4.5} \approx 2.1,$ is longer than $2,$ each of the four corners of the card reaches past a grid line into the next square.

The card covers the central $2 \times 2$ block of $4$ squares and pokes into $2$ more squares on each of its four sides, giving $4 + 4 \times 2 = 12$ squares.

This is also the most possible. The card is only $1.5$ inches wide, so its overall width and height are each at most $2.1$ inches; it therefore lies within a $4 \times 4$ block of $16$ squares. Its four pointed corners are the only parts that reach the edge of that block, so it can never reach the four corner squares of the block, leaving at most $12.$ Since $12$ is achievable, the maximum is $12,$ which falls in the range "$12$ or more."

Thus, the correct answer is E .