Solution:

We get that $$\begin{aligned} &\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10} \\ &= \dfrac{5}{10} + \dfrac{2}{10} + \dfrac{1}{10} \\&= \dfrac{4}{5}. \end{aligned}$$

Then $\left(\dfrac{4}{5}\right)^{-1} = \dfrac{5}{4}$ and then finally, $$10 \cdot \dfrac{5}{4} = \dfrac{25}{2}.$$

Thus, C is the correct answer.

Solution:

The cat eats $$\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{12}$$ cans of food each full day. After $10$ full days, it has eaten $10\cdot\frac7{12}=\frac{35}{6}$ cans, leaving $\frac16$ of a can.

The cat finishes that remainder during its next morning feeding. The eleventh morning, counting Monday as the first, is Thursday.

Thus, C is the correct answer.

Solution:

In the morning Bridget sells $24$ loaves for $24\cdot\$2.50=\$60$.

She has $24$ loaves left. In the afternoon she sells $\frac23\cdot24=16$ loaves at half price, earning $16\cdot\$1.25=\$20$.

The remaining $8$ loaves sell for $\$8$, so her revenue is $\$60+\$20+\$8=\$88$. Her cost is $48\cdot\$0.75=\$36$.

Her profit is $\$88-\$36=\$52$.

Thus, E is the correct answer.

Solution:

Blue must come before yellow but not next to it, so they sit in positions $(1,3)$ or $(1,4)$ or $(2,4)$.

In each case orange and red fill the two remaining spots with orange before red, which is forced. The three orderings are orange, blue, red, yellow; blue, orange, red, yellow; and blue, orange, yellow, red.

There are $3$ possible orderings.

Thus, B is the correct answer.

Solution:

The median is $90$, because $45\%$ of the students scored below $90$ and $25\%$ scored above $90$.

The mean is $0.10\cdot70$ $+0.35\cdot80$ $+0.30\cdot90$ $+0.25\cdot100=87$.

The difference is $90-87=3$.

Thus, C is the correct answer.

Solution:

We have to multiply $b$ by $\dfrac{d}{a}$ to account for the new number of cows.

We then have to multiply by $\dfrac{e}{c}$ to account for the new time that we have.

This gives us a final answer of $$b \cdot \dfrac{d}{a} \cdot \dfrac{e}{c} = \dfrac{bde}{ac}.$$

Thus, A is the correct answer.

Solution:

Adding the two inequalities together gets us

$$x + y \lt a + b,$$ which shows that (I) is correct.

One cannot subtract inequalities, which means that (II) is not necessarily true.

Consider $x = 1,$ $y = 1,$ $a = 2,$ and $b = 3$ as a counter-example. This would give us $0 \lt -1.$

(III) is also not always true, since $x$ and $y$ might be negative numbers.

Let $x = -3,$ $y = -2,$ $a = 1,$ and $b = 1.$ Then $xy = 6$ and $ab = 1$ which shows that (III) is wrong.

The same thing occurs with (IV) . Using the same values as above, we have $\dfrac{x}{y} = 1.5$ and $\dfrac{a}{b} = 1.$

This shows that (I) is the only true statement.

Thus, B is the correct answer.

Solution:

Note that all of these answer choices are of the form $$\dfrac{n!(n + 1)!}{2} = \dfrac{(n!)^2(n + 1)}{2}.$$ We have that $(n!)^2$ is square, so we need $\dfrac{n + 1}{2}$ to be square as well.

This means that $n + 1$ must be twice a perfect square. The only choice we have is $n + 1 = 18,$ which gives us $n = 17.$

Thus, D is the correct answer.

Solution:

We get that the area of the triangle is $$\dfrac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}.$$ The length of the hypotenuse is $$\sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{48} = 4\sqrt{3}.$$

Dropping the altitude, $h,$ from the vertex to the hypotenuse, we get that $$\dfrac{1}{2} \cdot h \cdot 4\sqrt{3} = 6\sqrt{3}$$$$h = 3.$$

Thus, C is the correct answer.

Solution:

Note that the average of $5$ consecutive numbers starting with $x$ is $$\dfrac{5x + 1 + 2 + 3 + 4}{5} = x + 2.$$

This means that the average of $5$ consecutive integers starting with $a$ is $a + 2,$ which we know is $b.$

Furthermore, the average of $5$ consecutive numbers starting with $b$ is $$b + 2 = a + 4.$$

Thus, B is the correct answer.

Solution:

Let us analyze what these coupons do to an arbitrary price, $x.$

Coupon $1$ changes this price to $.9x.$ Coupon $2$ changes the price to $x - 20.$ Coupon $3$ changes the price to $$x - .18(x - 100) = .82x + 18.$$

We want $$.9x \lt x - 20$$ and $$.9x \lt .82x + 18.$$ Solving both gives us $$200 \lt x \lt 225.$$

The only answer choice that works is $\$ 219.95.$

Thus, C is the correct answer.

Solution:

Note that we can split the hexagon up into $6$ equilateral triangles each with side length $6.$

Recall that the area of an equilateral triangle with side length $s$ is $$\dfrac{s^2 \sqrt{3}}{4}.$$

This means that the area of the hexagon is $$6 \cdot \dfrac{6^2 \sqrt{3}}{4} = 54\sqrt{3}.$$

Since each interior angle of a regular hexagon is $120^{\circ},$ the six sectors form $2$ full circles.

This means that the area of all the sectors is $$2 \cdot 3^2 \pi = 18\pi.$$

The area of the shaded region is then $$54\sqrt{3} - 18\pi.$$

Thus, C is the correct answer.

Solution:

We can find the areas of all the individual pieces and then add them up together.

The area of the center equilateral triangle is $$\dfrac{1^2 \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}.$$

We have that the areas of all the squares is $$3 \cdot 1^2 = 3.$$

We also have that $$\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}$$$$= 120^{\circ}.$$

Also, $AE=AF=1$ and $\angle EAF=120^\circ$. Dropping the altitude from $A$ shows that $EF=\sqrt3$ and the altitude is $\frac12$, so $[EAF]=\frac{\sqrt3}{4}$. The other two outer triangles have the same area. Thus their combined area is $\frac{3\sqrt3}{4}$.

The total area is then $$\dfrac{\sqrt{3}}{4} + \dfrac{3\sqrt{3}}{4} + 3 = 3 + \sqrt{3}.$$

Thus, C is the correct answer.

Solution:

We have that the $y$-intercepts are an equal distance from the origin since their values sum to $0.$

Let this distance be $z.$ Because the two given lines are perpendicular, $\triangle APQ$ is right at $A$. The origin is the midpoint of its hypotenuse $PQ$, so it is equidistant from $P$, $Q$, and $A$. Hence the distance from $A$ to the origin is also $z$.

We then know that $$z = \sqrt{6^2 + 8^2} = 10$$ by the distance formula. We know the altitude from $A$ to $\overline{PQ}$ is $6$ (it is just the $x$-value of $A$).

We also know that $PQ = 2 \cdot 10 = 20,$ which tells us that the area $$[APQ] = \dfrac{1}{2} \cdot 6 \cdot 20 = 60.$$

Thus, D is the correct answer.

Solution:

Note that David drives at $50$ miles per hour after one hour.

Then, if the airport is $x$ miles from David's house, we know that: $$\dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) = \dfrac 32$$ We solve this equation as follows: $$\begin{aligned} \dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) &= \dfrac 32\\ \dfrac x{35} - \dfrac{x-35+50}{50} &= \dfrac 32\\ 10x - 7(x+15)&= 525\\ 10x - 7x-105&= 525\\ 3x&= 630\\ x&=210 \end{aligned}$$ Therefore, the airport is $x=210$ miles from David's house.

Thus, C is the correct answer.

Solution:

We can find the area of the shaded region by finding the area of $\triangle DHC$ and subtracting out the two unshaded triangles.

Extend $\overline{DH}$ so that it hits $B.$ Let the intersection of $\overline{DB}$ and $\overline{AF}$ be $X.$

We have that $\triangle DXF\sim\triangle BXA.$ Since $AB=2\cdot DF$, corresponding sides give $BX=2\cdot DX$.

This means that $DX = \dfrac{1}{3} \cdot DB,$ which means that the altitude of $\triangle DXF$ is $\dfrac{1}{3}$ the height of the rectangle.

The area of $\triangle DXF$ is then $$\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{6}.$$

The area of both unshaded triangles is then $2 \cdot \dfrac{1}{6} = \dfrac{1}{3}.$ The area of $\triangle DHC$ is $$\dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2}.$$

The area of the shaded region is then $\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}.$

Thus, E is the correct answer.

Solution:

Note that if one die is the sum of the other two dice, then it is strictly greater than the other two dice.

There are $3$ ways to choose which of the dice is the sum of the other two, which makes it the greatest.

This die cannot be $1,$ since there is no way to sum two positive integers to get $1.$

There is a $\dfrac{1}{6}$ chance that this die is any of the other numbers.

There is $1$ way to get a sum of $2,$ $2$ ways for $3,$ $3$ for $4,$ $4$ for $5,$ and $5$ for $6.$

We take these numbers of ways out of a total of $6^2 = 36$ possibilities. The desired probability is then $$3 \cdot \dfrac{1}{6} \cdot \dfrac{1 + 2 + 3 + 4 + 5}{36} = \dfrac{5}{24}.$$

Thus, D is the correct answer.

Solution:

Opposite vertices of a square have the same average $y$-coordinate. Thus the opposite pairs must have $y$-coordinates $(0,5)$ and $(1,4)$, the two pairs with equal sum. In particular, a vertex with $y$-coordinate $0$ is adjacent to vertices with $y$-coordinates $1$ and $4$.

Let one vertex be $A=(0,0)$, and let its adjacent vertex with $y$-coordinate $1$ be $B=(x,1)$ with $x>0$.

Rotating the side vector $(x,1)$ by $90^\circ$ gives the next side vector $(-1,x)$, so another vertex has y-coordinate $x$. The remaining y-coordinates are $4$ and $5$, hence $x=4$.

The side length squared is $AB^2=x^2+1^2=4^2+1=17$, which is the area of the square.

Thus, B is the correct answer.

Solution:

The distance between $X$ and $Y$ with respect to the $z$-axis is $$1 + 2 + 3 + 4 = 10.$$

Both the distances along the $x$ and $y$-axes are $4.$

Then $$XY = \sqrt{4^2 + 4^2 + 10^2} = 2\sqrt{33}.$$

Using coordinates $X=(0,0,10)$ and $Y=(4,4,0)$, the line meets the top and bottom of the side-$3$ cube at $(\frac65,\frac65,7)$ and $(\frac{12}5,\frac{12}5,4)$. Both points lie inside those square faces, so the portion inside this cube really does have vertical change $3$.

Let the desired length be $x.$ Then using similar triangles, we have that $$\dfrac{x}{3} = \dfrac{2\sqrt{33}}{10}$$$$x = \dfrac{3\sqrt{33}}{5}.$$

Thus, A is the correct answer.