Answer: D

Solution(s):

Cagney can make \(60 \div 20 = 3\) cupcakes per minute, and Lacey can make \(60 \div 30 = 2.\)

In \(5\) minutes, they can make \[ 5(2 + 3) = 5 \cdot 5 = 25. \]

Thus, D is the correct answer.

Answer: E

Solution(s):

Note that the one of the sides remains the same as the original square.

This means that one dimension is \(8.\) The other dimension is the original side cut in half, which is \(4.\)

Thus, E is the correct answer.

Answer: E

Solution(s):

It crawls a distance of \[ |-2 - (-6)| = |-4| = 4 \] when it moves from \(-2\) to \(-6.\) It then travels a distance of \[ |-6 - 5| = |-11| = 11 \] as it moves from \(-6\) to \(5.\)

The total distance is then \[ 4 + 11 = 15. \]

Thus, E is the correct answer.

Answer: C

Solution(s):

Note that both of the angles share the ray \(AB.\) To minimize the desired degree, we want \(D\) to be between \(A\) and \(C.\)

This would make \[ \angle CBD = \angle ABC - \angle ABD = 4^{\circ}. \]

Thus, C is the correct answer.

Answer: B

Solution(s):

We have that there are \(100 \div 2 = 50\) female cats. We then have that \(50 \div 2 = 25\) cats that have kittens.

Since the average number of kittens per litter is \(4,\) the total number of kittens is \[ 4 \cdot 25 = 100. \]

The total number of cats and kittens is then \[ 100 + 100 = 200. \]

Thus, B is the correct answer.

Answer: D

Solution(s):

Let the two numbers be \(x\) and \(y\) such that \[ xy = 9 \text{ and } \dfrac{1}{x} = \dfrac{4}{y}. \]

We get that \[ y = \dfrac{9}{x} \]\[ \dfrac{1}{x} = \dfrac{4x}{9}. \]\[ x = \dfrac{3}{2}, \] since \(x\) is positive.

Then \[ y = 9 \div \dfrac{3}{2} = 6. \]

The desired sum is then \[ 6 + \dfrac{3}{2} = \dfrac{15}{2}. \]

Thus, D is the correct answer.

Answer: C

Solution(s):

WLOG, let the number of marbles in the bag be \(5.\) Since we only care about ratios, we can do this. Then there are \(3\) blue marbles and \(2\) red marbles.

Doubling the red marbles gives us \(4\) of them. Then the fraction of red marbles is \[ \dfrac{4}{4 + 3} = \dfrac{4}{7}. \]

Thus, C is the correct answer.

Answer: D

Solution(s):

Let the three numbers be \(a, b, c\) where \(a \lt b \lt c.\) None of them are equal, since all three sums are different.

Then \[ a + b = 12, a + c = 17, \] and \[ b + c = 19. \]

Adding all three equations together gives us \[ 2(a + b + c) = 48. \]

Then \[ a + b + c = 24, \] from which we can subtract \[ a + c = 17, \] to get \(b = 7.\)

Thus, D is the correct answer.

Answer: D

Solution(s):

The pairs of numbers that sum to \(7\) are \[ (2, 5), (4, 3), \text{ and } (6, 1). \]

There is a \[ \dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9} \] chance that we get any of these pairs.

There are \(3\) pairs, which means that the total probability that the rolls sum to \(7\) is \[ 3 \cdot \dfrac{1}{9} = \dfrac{1}{3}. \]

Thus, D is the correct answer.

Answer: C

Solution(s):

Let \(a\) be the smallest possible sector angle and \(d\) be the difference in the arithmetic sequence.

Then we have that \[ \dfrac{12}{2}(a + a + 11d) = 12a + 66d \] is the sum of the arithmetic sequence.

We have that this sums to \(360,\) so \[ 12a + 66d = 360 \]\[\]\[ 2a + 11d = 60. \]

We want to minimize \(a,\) so we maximize \(d.\) If \(d = 5,\) then \(a\) is not an integer, so \(d = 4\) and \(a = 8.\)

Thus, C is the correct answer.

Answer: D

Solution(s):

Let \(x\) be \(BC.\) Note that \(\triangle CEB\) and \(\triangle CDA\) are similar due to angle-angle (tangent lines are perpendicular to radii).

Then \[ \dfrac{x}{3} = \dfrac{8 + x}{5}. \] Cross-multiplying gives us \[ 5x = 24 + 3x \]\[ x = 12. \]

Thus, D is the correct answer.

Answer: A

Solution(s):

On a typical year with \(365\) days, moving one year into the future moves the day of the week forward one since \[ 365 = 52 \cdot 7 + 1. \]

On a leap year however, the day of the week gets moved forward twice since there is an extra day.

Fifty of the years between \(2012\) and \(1812\) are multiples of \(4,\) but we have to discard \(1900,\) which leaves us with \(49\) leap years.

Therefore, moving back \(200\) years means we have to go back \[ 200 + 49 = 249 = 35 \cdot 7 + 4 \] days, which means we move back \(4\) days in the week.

This takes us back to Friday, which is the day that corresponds to February \(7, 1812.\)

Thus, A is the correct answer.

Answer: C

Solution(s):

Let the order of the numbers be \[ a, b, c, d, e. \]

Then the iterative average is \[ \dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}\]\[ = \dfrac{a + b + 2c + 4d + 8e}{16}. \]

To minimize this, we make the order \[ 5, 4, 3, 2, 1, \] which gives us a sum of \[ \dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.\]

To maximize it, we have to reverse this order to get an average of \[ \dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}. \]

The difference between these is \[ \dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}. \]

Thus, C is the correct answer.

Answer: B

Solution(s):

Note that there are \(15\) rows with \(15\) black tiles and \(16\) rows with \(16\) black tiles.

This can seen by observing that the first row has \(16\) black tiles, and all the other rows alternate with \(15\) and \(16\) tiles.

Then, due to the alternating pattern, there will be a total of \(16\) rows with \(16\) tiles, and the other rows have \(15\) tiles.

The total number of black squares is then \[ 15^2 + 16^2 = 481. \]

Thus, B is the correct answer.

Answer: B

Solution(s):

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let \(A\) be the origin and \(B = (1, 0).\) Then the slope of the line through \(A\) is \(-\dfrac{1}{2},\) which makes the equation of the line \[ y = -\dfrac{1}{2}x. \] The slope of the line through \(B\) is \(2.\) The \(y\)-intercept is \(-2.\) This makes the equation of this line \[ y = 2x - 2. \]

Equating the equations, we get \[ 2x - 2 = -\dfrac{1}{2}x \]\[ x = \dfrac{4}{5}. \]

This makes the \(y\)-coordinate of \(C\) \[ -\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}. \]

The area of triangle \(ABC\) is then \[ \dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}. \]

Thus, B is the correct answer.

Answer: C

Solution(s):

Let us find the amount of time that it takes for the runner running at \(4.8\) meters per second to lap the second fastest person.

We must have that \[ 4.8x - 4.4x = 500 \]\[ x = 1250, \] where \(x\) is the amount of time it takes for the faster runner to lap the other.

Note that \(4.4 \cdot 1250 = 5500,\) which means that these two runners always intersect at the starting line.

We now have to find the least time, \(t,\) such that \(t\) is a multiple of \(1250\) and the fastest runner ends up at the starting line.

Every \(1250\) seconds, the fastest runner runs \(1250 \cdot 5 = 6250\) meters. Then in \(2500\) seconds, the fastest runner runs \(12500\) meters, which is a whole number of laps.

Thus, C is the correct answer.

Answer: C

Solution(s):

Recall that we can factor \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2). \] Canceling out this factor gives us that \[ \dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}. \]

Cross-multiplying and rearranging gives us \[ 70a^2 - 149ab + 70b^2 = 0. \] Since \(b \neq 0,\) we can divide through by \(b^2\) to get \[ 70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0. \]

Applying the quadratic formula and noting that \(a \gt b\) gives us that \(\dfrac{a}{b} = \dfrac{10}{7}.\)

Since \(a\) and \(b\) are relatively prime, we have that \(a = 10\) and \(b = 7.\) Their difference is \(10 - 7 = 3.\)

Thus, C is the correct answer.

Answer: E

Solution(s):

Note that the region enclosed by the curve but outside the hexagon consists of \(3\) sectors with angle \(240^{\circ}.\)

This means that together they form \(2\) whole circles with radius \(1.\) Now to find the area of the region inside both the hexagon and the curve.

This area can be found by finding the area of the hexagon and subtracting out the areas of the sectors outside the curve.

There are three \(\dfrac{1}{3}\) circles that together form a whole circle. The area of the hexagon can be given by \[ \dfrac{3}{2} \cdot 2^2 \cdot \sqrt{3} = 6\sqrt{3}. \]

The desired area is then \[ 2\pi + (6\sqrt{3} - \pi) = \pi + 6\sqrt{3}. \]

Thus, E is the correct answer.

Answer: D

Solution(s):

Let Paula work at a rate of \(x \%\) per hour and the helpers combined work at a rate of \(y.\) Let \(l\) be the duration of the lunch break.

Then we have the following equations. \[ \begin{gather*} (8 - l)(x + y) = 50 \\ (6.2 - l)y = 24 \\ (11.2 - l)x = 36. \end{gather*} \]

Adding the second and third equations together gives us \[ 6.2y + 11.2x - L(x + y) = 50. \] We can then subtract the first equation from this to get \[ -1.8y + 3.2x = 0 \]\[ h = \dfrac{16}{9}p. \]

We can now substitute this into the second equation, which gives us \[ (6.2 - l)\dfrac{16}{9}x = 24 \]\[ \] \[ (6.2 - l)p = \dfrac{27}{2}. \] Finally, subtracting this from the third equation gets us \[ 5p = 26 - \dfrac{27}{2} \]\[ p = \dfrac{5}{2}. \]

Plugging in this value gives us \(l = \dfrac{4}{5},\) which is the same as \(48\) minutes.

Thus, D is the correct answer.

Answer: A

Solution(s):

Since the center square does not get affected by the rotation, we have that it must be black.

Now, let us analyze the corners. Clearly, if they are all black, that works. If only one is white, that also works.

If two are white, they must diagonal from each other, since otherwise a white will move onto a white, keeping it white.

There is no way for three or all four of them to be white, since that ensures a white square will move onto a white square.

This gives us \[ 1 + 4 + \binom{4}{2} = 7 \] possible colorings for the corners. Similarly, we have that there are also \(7\) working colorings for the edges.

This gives us a total probability of \[ \dfrac{7 \cdot 7}{2^9} = \dfrac{49}{512}. \]

Thus, A is the correct answer.

Answer: C

Solution(s):

Note that \(EF = \dfrac{1}{2}AD\) since it is a midsegment of \(\triangle ABD.\) Similarly, \(HG = \dfrac{1}{2}AD\) and \(FG = \dfrac{1}{2}BC.\)

We also have that \(\overline{EF}\) and \(\overline{HG}\) are perpendicular to the \(xy\)-plane, which means that they are perpendicular to \(\overline{FG}\) and \(\overline{EH}.\)

This tells us that \(EFGH\) is rectangle since \(EF = HG.\) We have \(EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.\)

We also have that \[ FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}. \]

The area of \(EFGH\) is then \[ EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}. \]

Thus, C is the correct answer.

Answer: A

Solution(s):

Recall that the sum of the first \(m\) odd numbers is \(m^2\) and the first \(n\) even numbers is \(n^2 + n.\)

We have that \[ m^2 = n^2 + n + 212. \]

We can view this equation as a quadratic in terms of \(n\) as follows. \[ n^2 + n + (212 - m^2) = 0. \]

We can apply the quadratic formula to get \[ n = \dfrac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}. \]

We have that \(m\) and \(n\) are integers, so \[ 1 - 4(212 - m^2) = 4m^2 - 847 \] must be a square number. Also, since we add \(-1\) and divide by \(2,\) this must be odd.

Let \[ x = \sqrt{4m^2 - 947}. \]

Squaring and rearranging gives us \[ 4m^2 - x^2 = 847 \]\[ \] \[ (2m + x)(2m - x)\]\[ = 847 = 7 \cdot 11^2. \]

Note that \[ n = \dfrac{-1 + x}{2}. \]

Given the values for \(2m + x\) and \(2m - x,\) we have that the difference between the two is \(2x.\)

The possible pairs of values are \[ 847 \cdot 1, 121 \cdot 7, 77 \cdot 11. \]

These pairs contribute the following values for \(x\) respectively: \( 423, 57, 33. \) Then the possible values for \(n\) are \(211, 28,\) and \(16.\)

Thus, A is the correct answer.

Answer: B

Solution(s):

We case on the value of friends that each person has. This value ranges from \(1\) to \(4\) since all of them are not friends.

Note that the cases for \(1\) and \(2\) friends correspond with the case for \(4\) and \(3\) friends, since choosing who are friends determines who are not friends.

Case 1: everyone has \(1\) friend

This means that the \(6\) people must split up into \(3\) pairs where the people in each pair are friends.

There are \(5\) choices for the friend for the first person. This leaves \(4\) people remaining.

There are then \(3\) choices for the friend of the next person. The remaining \(2\) people are then forced to be friends.

Therefore, there are \(3 \cdot 5 = 15\) possibilities for this case.

Case 2: everyone has \(2\) friends

There are two possibilities for this case. There could be two triples where everyone in a triple is friends with each other.

For this possibility, there are \(\binom{6}{3} = 20\) ways to choose the people in the first pair. We have to divide by \(2\) since we can swap the pairs. This gives us \(20 \div 2 = 10\) configurations.

The second possibility is if the friends form a hexagon where the person at each vertex is friends with the adjacent people.

The first person can be placed anywhere on the hexagon. There are \(\binom{5}{2} = 10\) ways to choose the people adjacent to this person.

The final \(3\) people can placed in \(3! = 6\) ways in the remaining spots. This case then has a total number of \[ 10 + 10 \cdot 6 = 70 \] configurations.

The total number of arrangements is then have \[ 2(15 + 70) = 170. \]

Thus, B is the correct answer.

Answer: E

Solution(s):

Adding together the equations gives us \[ 2(a^2 + b^2 + c^2) - 2(ab + ac + bc)\]\[ = 14. \]

We can group terms and factor this to get \[ (a - b)^2 + (a - c)^2 + (b - c)^2\]\[ = 14. \]

Note that every term on the left hand side is a positive square integer. The only triple of squares that add to \(14\) is \(9, 4,\) and \(1.\)

We have that \(a - c\) is the biggest difference among the three pairs. Therefore, \(a - c = 3.\)

We cannot discern which of the other terms we can match with the other squares. Let us try \(a - b = 1\) and \(b - c = 2.\)

Plugging in these values into the first equation gives us \[ a^2 - (a - 1)^2 - (a - 3)^2 \] \[ + a(a - 1) = 2011. \] Simplifying yields \(7a = 2021.\) Since \(2021\) is not divisible by \(7,\) we have that \(a - b = 2\) and \(b - c = 1.\)

Plugging these in again and solving gives us \(a = 253.\)

Thus, E is the correct answer.

Answer: D

Solution(s):

This problem lends itself to geometric probability since we can view the interval as a range on an axis.

WLOG, let \[ n \geq x \geq y \geq z \geq 0. \]

Then we have that the points \((x, y, z)\) which satisfy this restriction form a tetrahedron.

The height of this tetrahedron is \(n,\) and the base has an area of \(\dfrac{n^2}{2}.\) This makes the volume \[ \dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}. \]

Now we have to apply the restrictions from the problem statement. We need to find the region where \[ |x - y|, |x - z|, |y - z| \geq 1. \]

From our ordering condition that we imposed, these inequalities reduce to \[ x - y \geq 1 \text{ and } y - z \geq 1. \]

These two restrictions form another tetrahedron as shown below.

Note that in the new tetrahedron, all the dimensions have been reduced by \(2.\) This makes the height \(n - 2\) and the base \(\dfrac{(n - 2)^2}{2}.\)

The area is then \[ \dfrac{1}{2} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)\]\[ = \dfrac{(n - 2)^3}{6}. \]

The desired probability is then \[ \dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}. \]

Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than \(\dfrac{1}{2}\) is \(10.\)

Thus, D is the correct answer.