Solution(s):

There are $11+8+9 = 28$ students.

Therefore, the answer is C.

Solution(s):

Given our definition of $a*b,$ we have $$5 *10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}.$$

Thus, the answer is D.

Solution(s):

The highest price is $17$ and the lowest price is $10.$ This means the percent is $$(\dfrac{17}{10}-1) * 100 = 70.$$

Thus, the answer is C.

Solution(s):

The list reordered is $0,0,1,2,3,3,3,4.$

The median is the mean of the middle two numbers, which would be $\dfrac{2+3}2 = 2.5.$ The mode is $3$ since $3$ appears the most. The mean is $$\dfrac{0+0+1+2+3+3+3+4}{8}$$$$= \dfrac{16}{8} = 2 .$$

Their sum is $2.5+3+2 = 7.5.$

Thus, their answer is C.

Solution(s):

We know the height of the ceiling is $2.4 \cdot 100 = 240$ cm. Subtracting out all the given values, we get $$240 - 150 - 46 - 10 = 34.$$ Therefore, the height of the stool is 34 cm. Thus, B is the correct answer.

Solution(s):

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn't on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, lets look at each of the shapes. An equilateral triangle has $3$ lines that intersect a corner or a midpoint, so it has at most $3$ symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangles only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn't bisect the angle. An isosceles trapezoid only has $1$ symmetry line, that goes through the midpoints of opposite sides. A square has $4$ symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E.

Solution(s):

Freddie needs $4$ pennies as this is the only way to pay $4$ cents. Next, he needs $1$ nickel or $5$ pennies to make a $9$ cents after having the $4$ starting pennies. Here, we have $5$ coins. Using $1$ nickel minimizes the coins needed. Now, to make $24$ cents, we need $15$ more cents. This can be done with a dime and nickel. Now, we have $7$ coints Now, for the last $75$ cents, we can use $3$ quarters to make the rest. This leaves us with $10$ coins.

Therefore, the answer is B.

Solution(s):

Let $d$ be how far Emily is ahead of Emerson. Emily sees Emerson if $-\dfrac 12 \leq d \leq \frac 12.$ Suppose at $t= 0,$ where $t$ is in hours, that $d = -\frac 12.$ Then, $d = (12-8)t - \frac 12 .$ Since we must find where $d = \frac 12,$ we find the time where $$\frac 12 = 4t - \frac 12 \implies 1 = 4t$$$$\implies t = 0.25.$$ Since $0.25$ hours passed, we know that $60*0.25 = 15$ minutes passed.

Therefore, the answer is D.

Solution(s):

There were a total of $25+40+10 = 75$ problems.

On the first test, he solved $0.8*25 = 20$ problems.

On the second test, he solved $0.9*40 = 36$ problems.

On the third test, he solved $0.7*10 = 7$ problems.

Therefore, he solved a total of $20+36+7=63.$ This means the fraction he solved is $\dfrac{63}{75} = 84\%.$

Therefore, the answer is D.

Solution(s):

Each circle has $\dfrac{1}{6}$ the diameter of the large circle, so it has $(\frac 16)^2 = \frac 1{36}$ of the total area.

Since there are $24$ pepperoni, they take up $24 * \dfrac 1{36} = \frac 23$ of the area.

Therefore, the answer is B.

Solution(s):

Let $b,s$ be the heights of bigger and smaller trees respectively. Then, $b= s+16$ and $s = 0.75b.$ If we substitute, we get $$b = 0.75b + 16 \implies 0.25b = 16$$$$\implies b = 64.$$

Thus, the answer is B.

Solution(s):

If $r$ is the number of red balls, then $$r = 0.8*500 = 400.$$ Therefore, if $b$ is the number of blue balls, then $$b = 500-400 = 100.$$

If there are $75$ red balls after removing balls, then there are $25$ blue balls. This means the total number of balls is $\dfrac{100}{.25} = 400.$ This means the total number of balls decreased by $500-400 = 100.$

Thus, the answer is D.

Solution(s):

Let $s$ be the smallest length. Then, all the side lengths are $s,s+1,s+2.$ This would make the perimeter equal to $3s+3.$ Since $s = 0.3(3s+3),$ then $s = 0.9s + 0.9 .$ This makes $0.1s = 0.9,$ so $s = 9$ which makes the longest side length $s + 2 = 11.$

Thus, the answer is E.

Solution(s):

First, the primes $2,5$ are factors of $2010$ since it is a multiple of $10.$ Dividing $2010$ by $10$ is $201.$ Then, $3$ is a factor of $201$ since the digit sum of $201$ is a multiple of $3.$ Dividing this by $3$ yields the prime number $67.$ This means the prime factors are $2,3,5,67,$ which makes their sum $77.$

Thus, the correct answer is C.

Solution(s):

Since we have percentages for every color except green, the percent of green is $100\%$ minus the sum of the other colors. This would make the percent of green equal to $$100\%-30\%-20\%-$$$$15\%-10\% = 25\%$$

Since we know $30$ jelly beans is $25\%,$ we know that the total number of jelly beans is $\dfrac { 30}{.25} = 120.$

This means there are $.2*120 = 24$ brown jelly beans to start and $.3*120 = 36$ blue jelly beans. If half of the blue jelly beans are turned to brown, then $\dfrac{36}{2} = 18$ more brown jelly beans are added. Therefore, we have $24+18= 42$ brown jelly beans.

Thus, the answer is C.

Solution(s):

Let $s$ be the side length of the square and let $r$ be the radius of the circle. Then, since they have the same areas, $s^2 = \pi r^2 .$ This means $(\dfrac sr)^2 = \pi,$ so $\frac sr = \sqrt{ \pi} .$

Thus, the answer is B.

Solution(s):

Since $PQ$ bisects the area, the area under the line is $5.$ Removing the square on the right makes the bottom a triangle of base $5$ with area $4.$ Let the base of this triangle be $PZ.$

The area being $4$ means $$\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4$$$$\implies ZQ = 1.6.$$

Therefore, $QY = QZ-1 = 0.6$ and $XQ = 2-QZ = 0.4.$ This would make $\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.$

Thus, the answer is D.

Solution(s):

Combining the semicircles would make a circle of diameter $d=30.$ This would make the radius equal to $\dfrac{d}{2}.$ Therefore, the combined area of the semicircles is $$(\dfrac{d}{2})^2 * \pi = \dfrac{\pi d^2}{4}.$$

Side $AD = \dfrac 32$ and $AB = \dfrac 32 d.$ The area of the rectangle is therefore $\dfrac 32 d*d = \dfrac 32 d^2.$ The ratio of the area of the rectangle to the area of the semicircles is $\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}.$

Thus, the answer is C.

Solution(s):

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be $$(AC)^2\pi - (CB)^2 \pi$$$$= \pi(AC^2 - CB^2).$$ By the Pythagorean Theorem, we can get $$AC^2 - CB^2 = AB^2.$$ Therefore, we need to find $AB^2 \pi.$

SInce $AB$ is half of $AD,$ we get $AB = 8.$ This makes $AB^2 \pi = 64 \pi.$

Thus, the area is C.

Solution(s):

Since our room has $\dfrac 25$ of the people wearing gloves, the number of people must be a multiple of $5.$ Since our room has $\dfrac 34$ of the people wearing hats, the number of people must be a multiple of $4.$ Therefore, the people in the room must be a multiple of $20.$

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to $\dfrac{2}{5} + \dfrac 34 -$ Fraction of people who wear either. If we wish to minimize the number of who wear both, we maximize the number of people who wear both, to make it $1.$ Therefore, the fraction of people that wear both is $\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.$

Since our number is a (positive) multiple of $20,$ we have the number of people wearing both as $3$ if we choose to have just $20$ people.

Therefore, A is the correct answer.

Solution(s):

The pages left after the third day is $62.$ Before reading the last $18$ pages, she had $80$ pages left. This is $\dfrac 23$ of the pages remaining, so she had $120$ pages left before the third day.

Before reading the $15$ pages, she had $120+15=135$ pages left. This is $\dfrac 34$ of the pages remaining, so she had $180$ pages left before the second day.

Before reading the $12$ pages, she had $180+12=192$ pages left. This is $\dfrac 45$ of the pages remaining, so she had $240$ pages left before the first day, making the book $240$ pages.

Thus, the answer is C.

Solution(s):

Let the units digit be $u,$ and let the tens digit be $t.$ This makes the hundreds digit be $u+2.$ This makes the number equal to $$100(u+2) +10t+u$$$$=201u+200+10t$$ and the reversed number is $$100u +10t+(u+2)$$$$=201u+2+10t.$$ This makes the difference equal to $$(201u+10t+200)-$$$$(201u+10t+2)=198.$$ This makes the units digit $8.$

Therefore, the units digit is E.

Solution(s):

The area of each of the smaller circles have an area of $\pi \dfrac {r^2}{2} .$ Each of them have a radius of $1,$ so their combined area is $\pi \frac{1}{2} + \pi \frac 12 = \pi.$

Next, the radius of the larger circle is equal to the length of $OQ,$ which is equal to $\sqrt{1^2+1^2} = \sqrt 2.$ Its area is $\pi r^2 = \pi(\sqrt{2})^2 = 2\pi.$

This means the ratio is $\dfrac{\pi}{2\pi} = \frac 12.$

Thus, the answer is B.

Solution(s):

First, we get $$2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.$$

Next, we get $$10^8 = (2^8)(5^8) =$$$$(4^4)(5^8) < (5^4)(5^8) = 5^{12}.$$ This means $2^{24} < 10^8 < 5^{12} .$

Thus, the answer is A.

Solution(s):

Lets count first the number of ways to climb a flight of stairs. She must land on the last stair, and for each of the other stairs, she can either step on it or not step on it. This means there are $2^5 = 32$ total combinations. Now, we must exclude the ways that include stepping more than $3$ steps. This would mean we climb $4,5$ or $6$ stairs in one step. To do this, we can do it in the following ways: $$6,1-5,5-1,4-2,2-4,$$$$1-1-4,1-4-1,4-1-1.$$ This means we have $8$ cases to exlude, so we have $32-8 = 24$ total valid cases.

Thus, the answer is E.