Solution:

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

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 day. This means that it will take the cat $$6 \div \dfrac{7}{12} = \dfrac{72}{7}$$ days to finish all the foods. Note that this value is between $10$ and $11.$

This means that the cat will finish eating the food in $11$ days, which is $10$ days after Monday.

$10$ days after Monday is the same as $3$ days after Monday, and as such, the answer is Thursday.

Thus, C is the correct answer.

Solution:

In the morning, she sells $48 \div 2 = 24.$ From this, she makes 24 \cdot $ 2.50 = $60. In the afternoon, she has $24$ loaves left, $24 \cdot \dfrac{2}{3} = 16$ of which she sells for 16 \cdot \dfrac{$ 2.50}{2} = $20. Finally, she sells the remaining $24 - 16 = 8$ loaves for a dollar each, for a total of $ 8.

Her total revenue for the day is $ 60 + $ 20 + $ 8 = $ 88. The cost of all the loaves is 48 \cdot $ 0.75 = $ 36. Her total profits are then $88 - $36 = $52.

Thus, E is the correct answer.

Solution:

There are only two choices for the position of the yellow house, the third or fourth spot.

Case $1:$ the yellow house is in the third spot

This forces the blue house to be the first, and this makes the orange house second and the red house fourth.

Case $2:$ the yellow house is the last house

If the blue house is first, then the orange house is second, and the red house is third.

If the blue house is second, then the orange house is first, and the red house is second.

This gives us $3$ possible orderings for the houses that satisfy all the conditions.

Thus, B is the correct answer.

Solution:

We can assign the number of students as $20$ since this value will not affect the answer.

Then we have that $2$ students got $70$ points, $7$ students got $80,$ $6$ got $90,$ and $5$ got $100.$

We get that the mean is $$\dfrac{2 \cdot 70 + 7 \cdot 70 + 6 \cdot 90 + 5 \cdot 100}{20}$$ $$= \dfrac{1740}{20}$$$$= 87.$$ We also get that the median is $90,$ since $9$ students got below a $90,$ and the $10$ th and $11$ th students got $90.$

The difference between the two 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 $s$ $$\dfrac{s^2 \sqrt{3}}{4}.$$

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

We also have that $\triangle EAF$ is isosceles, which means that we can rearrange the triangle by splitting it down the middle and recombining it into an equilateral triangle.

This means that the area of the three triangles is then $$3 \cdot \dfrac{1^2 \sqrt{3}}{4} = \dfrac{3\sqrt{3}}{4}.$$

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 $z.$ We also have that that the distance from $A$ to the origin is $z$ since it is the median to the midpoint of the hypotenuse.

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. Let the distance he still needs to be drive be $d.$

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: $$\begin{align*} \dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) &= \dfrac 32\\ \dfrac x{35} - \dfrac{x-35+50}{50} &= \dfrac 32\\ 10x - 7(x+15)&= 525\\ 10x - 7x-105&= 525\\ 3x&= 630\\ x&=210 \end{align*}$$ 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 $AB = 2 \cdot DF,$ we have that $BX = 2 \cdot AX.$

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 have 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:

Let the points be $$A = (a, 0,)$$ $$B = (b, 1),$$ $$C = (c, 4),$$ and $$D = (d, 5).$$

Note that the difference in $y$-coordinates of $A$ and $C$ is $4.$

As the angles of a square are right, we have that the difference in $x$-coordinates of $A$ and $B$ must be $4.$

Using the distance formula, we get that $$AB = \sqrt{4^2 + 1^2} = \sqrt{17}.$$ Squaring this tells us that the square's area is $17.$

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

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.

Solution:

To see if any pattern exists, we can test out small values of $k.$

We have that $$8 \cdot 8 = 64,$$ $$8 \cdot 88 = 704,$$ $$8 \cdot 888 = 7104,$$ $$8 \cdot 8888 = 71104.$$

From this, it is pretty safe to guess that for every increment of $k,$ there is an extra $1$ added to the product. If you know how to use induction, you can prove that this pattern holds, but that's not necessary to solve the problem.

This means that for any $k \geq 3,$ the sum of the digits in the product is $$7 + 4 + 0 + k - 2 = k + 9.$$

Finally, we get $$k + 9 = 1000$$$$k = 991.$$

Thus, D is the correct answer.

Solution:

Note that the lines intersect the $x$-axis when $y = 0.$ This gives us $$0 = ax + 5$$ and $$0 = 3x + b,$$ which when solved gives us $$x = -\dfrac{5}{a}$$ and $$x = -\dfrac{b}{3}.$$

Setting these equal to each other, we have $$\dfrac{5}{a} = \dfrac{b}{3}$$$$ab = 15.$$

We know that $a$ and $b$ are positive, which means that the only pairs of values $(a, b)$ that satisfy the above equation are $$(1, 15),$$$$(3, 5),$$$$(5, 3),$$ $$(15, 1).$$

Plugging these values back into the equations gives us $x$-values of $$x = -5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}.$$ The sum of all these values is $-8.$

Thus, E is the correct answer.

Solution:

To make working with $15^{\circ}$ easier, we can create a $30^{\circ}$ angle and apply the angle bisector theorem.

Let $F$ be the point on $\overline{DC}$ such that $\angle EBF = 15^{\circ}.$ Applying the angle bisector theorem, we get $$\dfrac{BC}{BF} = \dfrac{CE}{EF}.$$

Using the special right triangle properties of a $30-60-90$ triangle, we have that $$CF = \dfrac{10\sqrt{3}}{3}$$and $$BF = \dfrac{20\sqrt3}{3}.$$

We can substitute in some values to get $$\dfrac{10}{\frac{20\sqrt3}{3}} = \dfrac{CE}{EF}$$$$$$$$\dfrac{2\sqrt3}{3}CE = EF.$$

Using $CE + EF = CF,$ we have $$\dfrac{2\sqrt3}{3}CE + CE = \dfrac{10\sqrt3}{3}$$$$$$$$CE = 20 - 10\sqrt3.$$

This means that $DE = 10\sqrt3,$ which gives us $AE = 20$ since $\triangle ADE$ is also a $30-60-90$ triangle.

Thus, E is the correct answer.

Solution:

WLOG, let the width of the rectangle be $1$ and the length be $\sqrt3.$

Draw the line perpendicular to the midpoint of the fold, as shown below.

Note that $QR = \dfrac{2\sqrt3}{3}$ and $$QT = \sqrt{1^2 + \left(\dfrac{\sqrt3}{3}\right)^2}$$$$= \sqrt{1 + \dfrac{1}{3}} = \sqrt{\dfrac{4}{3}}.$$ This tells us $$QT = \dfrac{2\sqrt3}{3} = QR.$$ This means that $\triangle QRT$ is equilateral. Similarly, $\triangle RTS$ is equilateral. This makes the two triangles congruent.

This means that after the rectangle gets folded, this area will be overlapped. The area of the rectangle is $1 \cdot \sqrt3 = \sqrt3.$ The side length of this triangle is $QT = \dfrac{2\sqrt3}{3}.$ The area of it is then $$\left(\dfrac{2\sqrt3}{3}\right)^2 \cdot \dfrac{\sqrt3}{4} = \dfrac{\sqrt3}{3}.$$ The area of the folded figure is then $$\sqrt3 - \dfrac{\sqrt3}{3} = \dfrac{2\sqrt3}{3}.$$ The desired ratio is then $$\dfrac{2\sqrt3}{3} \div \sqrt3 = \dfrac{2}{3}.$$ Thus, C is the correct answer.

Solution:

We can just count the number of skipped numbers and add that on to $500,\!000.$

Note that $$\dfrac{999 \cdot 1000}{2} \lt 500,000$$ and $$500,000 \lt \dfrac{1000 \cdot 1001}{2}.$$

This means that there are $999 - 3 = 996$ skipped blocks of numbers in the sequence. We have to subtract $3$ since we start off by listing $4$ numbers and not $1.$

Then, $$n = \dfrac{996 \cdot 997}{2} = 496,\!506,$$ so the desired answer is $$496,\!506 + 500,\!000 = 996,\!506.$$

Thus, A is the correct answer.

Solution:

Note that between any $2$ consecutive powers of $5,$ there are either $1$ or $2$ powers of $2.$ This is because $$2^2 \lt 5 \lt 2^3.$$

Consider the intervals from $(5^0, 5^1)$ to $(5^{866}, 5^{867}).$

Since $$2^{2013} \lt 5^{867} \lt 2^{2014},$$ we know that these intervals must all together contain $2013$ powers of $2.$

Let $x$ be the number of intervals that contain $2$ powers of $2$ and $y$ contain $3$ powers of $2.$

Then $$\begin{gather*} x + y = 867, \\ 2x + 3y = 2013. \end{gather*}$$ Multiplying the top equation by $2$ and subtracting it from the bottom equation gives us $y = 279.$

Thus, B is the correct answer.