Solution:

Over the common denominator $24$, the fractions are $\dfrac{8}{24}, \dfrac{6}{24}, \dfrac{9}{24}, \dfrac{10}{24},$ and $\dfrac{7}{24}.$

The largest numerator is $10$, so $\dfrac{5}{12} = \dfrac{10}{24}$ is the largest.

Thus, the correct answer is D .

Solution:

The numerators sum to $1 + 2 + \cdots + 9 + 55 = 45 + 55 = 100.$

So the total is $\dfrac{100}{10} = 10.$

Thus, the correct answer is D .

Solution:

Eight hours after 7:25 A.M. is 3:25 P.M.

Adding the $45$-minute lunch break gives an ending time of 4:10 P.M.

Thus, the correct answer is C .

Solution:

A clockwise turn of $120°$ sends each shape to the next position clockwise: the triangle at the top moves to the lower right, the square at the lower right moves to the lower left, and the circle at the lower left moves to the top.

The result therefore has a circle on top, a square at the lower left, and a triangle at the lower right, with each shape keeping its original form.

Thus, the correct answer is B .

Solution:

One mile is $8$ furlongs, and each furlong is $40$ rods, so one mile is $8 \times 40 = 320$ rods.

Thus, the correct answer is B .

Solution:

Any six consecutive whole numbers include at least one multiple of $5$ and at least one even number, so their product is a multiple of $5 \times 2 = 10.$

A multiple of $10$ always ends in $0$.

Thus, the correct answer is A .

Solution:

In triangle $ABE,$ $\angle ABE = 180° - (60° + 40°) = 80°.$

Since $A, B, C$ are collinear and $D$ lies on segment $EB,$ the angle $\angle DBC = 180° - 80° = 100°.$

In triangle $BDC,$ $\angle BDC = 180° - (100° + 30°) = 50°.$

Thus, the correct answer is B .

Solution:

Since the digits sum to $25$ and the maximum is $27$, at least one digit is $9$. The possible digit sets are $\{9, 9, 7\}$ and $\{9, 8, 8\}.$

Each set has $3$ distinct arrangements ($997, 979, 799$ and $988, 898, 889$), giving $6$ numbers in all.

Thus, the correct answer is C .

Solution:

The $20\%$ discount lowers the price to $\$80,$ and the $\$5$ coupon reduces it to $\$75.$

Adding $8\%$ tax gives $1.08 \times \$75 = \$81.00.$

Thus, the correct answer is A .

Solution:

The expression is an integer when $N+2$ divides $36.$ The divisors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36.$

Because $N \gt 0,$ we need $N + 2 \gt 2,$ leaving the divisors $3, 4, 6, 9, 12, 18, 36$: that is $7$ values.

Thus, the correct answer is A .

Solution:

Since $48$ girls attended and $20$ were from Jones, $48 - 20 = 28$ girls were from Clay.

Clay had $60$ students total, so the number of Clay boys is $60 - 28 = 32.$

Thus, the correct answer is B .

Solution:

In square $II,$ $1$ of the $4$ equal cells is shaded, so $\tfrac14$ of it is shaded.

Square $I$ breaks into $8$ equal triangles with $2$ shaded, and square $III$ breaks into $16$ equal triangles with $4$ shaded; each of these equals $\tfrac{2}{8} = \tfrac{4}{16} = \tfrac14.$

Since every figure has exactly $\tfrac14$ shaded, the shaded areas are all equal.

Thus, the correct answer is A .

Solution:

The number halfway between two values is their average: $\dfrac{\frac16 + \frac14}{2}.$

Since $\dfrac16 + \dfrac14 = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12},$ the average is $\dfrac{5}{12} \div 2 = \dfrac{5}{24}.$

Thus, the correct answer is C .

Solution:

Two children play at every moment for $90$ minutes, so the total playing time is $2 \times 90 = 180$ child-minutes.

Split equally among $5$ children, each plays $\dfrac{180}{5} = 36$ minutes.

Thus, the correct answer is E .

Solution:

The pattern repeats every $4$ points, so the arrows leaving a point depend only on its remainder upon division by $4.$

Because $425 = 4(106) + 1$ and $427 = 4(106) + 3,$ the path from $425$ to $427$ looks just like the path from point $1$ to point $3$: from $1$ it goes up to $2,$ then right to $3.$

Thus, the correct answer is A .

Solution:

A square's perimeter is proportional to its side, so the larger square has side length $3$ times the smaller one.

Area is the side squared, so the larger area is $3^2 = 9$ times the smaller.

Thus, the correct answer is E .

Solution:

The volume of snow is $4 \times 10 \times 3 = 120$ cubic yards.

The amounts shoveled add up to $20 + 19 + 18 + 17 + 16 + 15 + 14 = 119$ after $7$ hours, while $8$ hours would give $132.$ Since $119$ is much closer to $120$ than $132$ is, the $7$-hour answer is closest.

Thus, the correct answer is D .

Solution:

On the way out Mike drives slowly (a shallow slope) then rapidly (a steep slope), so the distance rises slowly and then more steeply. During the hour at the mall the distance stays the same, a flat segment.

On the return he drives rapidly (steep) then slowly (shallow), so the distance falls steeply and then levels off. Only the graph with a slow-then-fast rise, a flat top, and a fast-then-slow fall fits.

Thus, the correct answer is B .

Solution:

Each semicircle is built on a side of length $4,$ so its radius is $2.$ A semicircle bulges out from the middle of each side by that radius.

Each side of $ABCD$ is the original side plus two radii: $4 + 2(2) = 8.$ So the area of $ABCD$ is $8^2 = 64.$

Thus, the correct answer is E .

Solution:

Small numerators and large denominators make small fractions, so use $1$ and $2$ as numerators and $8$ and $9$ as denominators.

Pairing the larger numerator with the larger denominator gives $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},$ which is smaller than $\dfrac19 + \dfrac28 = \dfrac{26}{72}.$ So the minimum sum is $\dfrac{25}{72}.$

Thus, the correct answer is D .

Solution:

In the worst case, a person could draw $3$ red, $3$ white, and $3$ blue, which is $9$ gumballs, without yet having four of any color.

The next (tenth) gumball must match one of these colors, giving four of that color. So $10$ gumballs are needed.

Thus, the correct answer is C .

Solution:

On the first wheel, $P(1) = \dfrac14,$ $P(2) = \dfrac14,$ and $P(3) = \dfrac12.$ On the second wheel, each of $4, 5, 6$ has probability $\dfrac13.$

The sum is even when both numbers are odd or both are even. Both odd: the first is $1$ or $3$ $\left(\dfrac34\right)$ and the second is $5$ $\left(\dfrac13\right),$ giving $\dfrac34 \cdot \dfrac13 = \dfrac14.$ Both even: the first is $2$ $\left(\dfrac14\right)$ and the second is $4$ or $6$ $\left(\dfrac23\right),$ giving $\dfrac14 \cdot \dfrac23 = \dfrac16.$

The total probability is $\dfrac14 + \dfrac16 = \dfrac{5}{12}.$

Thus, the correct answer is D .

Solution:

The hundreds digit of the sum comes from $X$ plus any carry, so if $X = 9$ the sum would spill over into four digits. The largest allowed value is $X = 8.$

To make the sum as large as possible, take $Y = 9:$ then $888 + 98 + 8 = 994.$ Its digits are $9, 9, 4;$ since $9 = Y$ and $4$ is a new digit $Z,$ the sum has the form $YYZ.$

Thus, the correct answer is D .

Solution:

The rule says a green square cannot have a red square on its top or right side, so any green square forces the squares above and to its right to be green as well. The green squares must therefore cluster toward the top-right corner.

The valid colorings are: all four red; only the top-right green; the whole top row green; the whole right column green; all green except the bottom-left; and all four green. That is $6$ colorings.

Thus, the correct answer is B .

Solution:

Small cases show the pattern: $99 \times 44 = 4356$ and $999 \times 444 = 443556.$ In general, a string of $n$ nines times a string of $n$ fours gives $(n-1)$ fours, then a $3,$ then $(n-1)$ fives, then a $6.$

For $n = 94,$ the product has $93$ fours, one $3,$ $93$ fives, and one $6.$ The digit sum is $93(4) + 3 + 93(5) + 6 = 93(9) + 9 = 94(9) = 846.$

Thus, the correct answer is A .