Answer: B

Solution(s):

Susan spent \(2 \cdot $12 = $24\) on rides. This means she spent a total of \($12 + $24 = $36\) at the carnival.

This means that she has \($50 - $36 = $14\) left to spend.

Thus, the answer is B.

Answer: A

Solution(s):

The letter \(C\) is in the \(9\)th position, so it would be the letter \(8.\) The letter \(L\) is in the \(7\)th position, so it would be the letter \(6.\) The letter \(U\) is in the \(8\)th position, so it would be the letter \(7.\) The letter \(E\) is in the \(2\)nd position, so it would be the letter \(1.\)

Therefore, when putting together the word \(\text{CLUE},\) we get \(8671.\)

Thus, the answer is A.

Answer: A

Solution(s):

Since the \(13\)th is a Friday, we know that the \(6\)th is also a Friday. The \(1\)st is \(5\) days before the Friday, making it a Sunday.

Thus, the answer is A.

Answer: C

Solution(s):

Since the area of the larger triangle is \(16\) and the area of the smaller triangle is \(1.\) Thus, the area of the other \(3\) trapezoids is \(15.\) Since the \(3\) trapezoids have a combined area of \(15,\) each of their areas is \(\dfrac{15}{3} =5.\)

Thus, the answer is C.

Answer: E

Solution(s):

The total distance traveled is \(1661-1441 = 220\) miles. He also travelled \(10\) hours. Thus, the average speed is \(\dfrac{220}{10} = 22\) miles per hour.

Thus, the answer is E.

Answer: D

Solution(s):

The total area of the entire square is \(16\) since it is a \(4 \times 4\) square. The area of the colored region is \(6,\) making the remaining area \(16-6 = 10.\) Thus, the ratio is \(6:10 = 3:5.\)

Thus, the answer is D.

Answer: E

Solution(s):

\(\dfrac{3}{5}=\dfrac{M}{45}\) means \[M = \dfrac{3\cdot 45}{5}=27.\]

\(\dfrac{3}{5}=\dfrac{60}{N}\) means \[N = \dfrac{5\cdot 60}{3}=100.\]

Therefore, \(M +N = 127.\)

Thus, the answer is E.

Answer: D

Solution(s):

The total sales are \[100+60+40+120 = 320.\]

The average sales is then \(\dfrac{320}{4} =80.\)

Thus, the answer is D.

Answer: D

Solution(s):

The \(15\%\) loss means the investment went from \(\$100\) to \(0.85\cdot \$100 = \$85.\)

The \(20\%\) gain means the investment went from \(\$85\) to \(\$85 \cdot 1.2 = $102.\) The total investment went from \(\$100\) to \($102,\) making a gain of \(2\%.\)

Thus, the answer is D.

Answer: D

Solution(s):

The sum of the ages in Room A is \[6\cdot 40 = 240.\]

The sum of the ages in Room B is \[4\cdot 25 = 100.\]

The total sum is \[240+100 = 340.\]

The average is therefore \[\dfrac{340}{10} =34.\]

Thus, the answer is D.

Answer: A

Solution(s):

The number of people that have both animals is equal to the number of people that own a cat plus the number of people that own a dog minus the number of people that own either.

Therefore, the number of people who own both is \[20+26-39 = 7.\]

Thus, the answer is A.

Answer: C

Solution(s):

Since the height of a bounce decreases by \(\frac 23\) each bounce, the height of each bounce is \(3\cdot \left(\dfrac 23\right)^n.\)

After \(4\) bounces, the ball bounces \[3\cdot \left(\dfrac 23\right)^4 = \dfrac{16}{27},\] which is greater than \(\frac 12.\)

After \(5\) bounces, the ball bounces \[3\cdot \left(\dfrac 23\right)^5 = \dfrac{32}{81},\] which is less than \(\frac 12.\)

Therefore, we are under \(\frac 12\) after \(5\) bounces.

Thus, the answer is C.

Answer: C

Solution(s):

Let the weights be \(a,b,c.\) We know \[\begin{cases}a+b = 122\\a+c = 125\\b+c = 127.\end{cases}\] Adding all of this yields \[2(a+b+c) =\]\[ 122+125+127=374.\] This makes \[a+b+c = 187.\]

Thus, the answer is C.

Answer: C

Solution(s):

There are \(2\) rows and \(2\) columns to put B in, so there are \(4\) places for it. After that, there is \(1\) column and \(1\) row available for C, so the number of combinations is \[2\cdot 2\cdot 1\cdot 1 = 4.\]

Thus, the answer is C.

Answer: B

Solution(s):

The sum of her first \(8\) scores is \(37.\) Since the average of the first \(9\) scores is an integer, the sum of the first \(9\) scores is a multiple of \(9.\)

Since the score is less than \(10,\) the sum of the scores after \(9\) games is between \(37\) and \(47,\) and is a multiple of \(9,\) making the sum \(45.\) Thus, the score of the \(9\)th game is \(45-37=8.\)

The sum of Theresa's first \(9\) scores is \(45.\) Since the average of the first \(10\) scores is an integer, the sum of the first \(10\) scores is a multiple of \(10.\)

Since the score is less than \(10,\) the sum of the scores after \(10\) games is between \(45\) and \(55,\) and is a multiple of \(10,\) making the sum \(50.\) Thus, the score of the \(10\)th game is \(50-45=5.\)

Therefore, their product is \(8\cdot 5=40.\)

Thus, the answer is B.

Answer: D

Solution(s):

There are \(7\) unit cubes, so the volume is \(7.\)

If we look at the perspective of the inner cube, we can see that there is one cube connected to each side. Furthermore, there are \(5\) exposed on each of the outer cubes. This makes the surface area \(5\cdot 6=30.\)

This makes the ratio of volume to surface area \(7:30.\)

Thus, the answer is D.

Answer: D

Solution(s):

Let the length and width be \(l,w\) respectively. The rectangle would have an area of \[l+r+l+r = 2(l+r)=50.\] Thus, \(l+r =25,\) so \(r=25-l.\) The area is \[lr = l(25-l) \]\[= 156.25 - (l-12.5)^2.\] This makes the area largest when \(l\) is as close as possible to \(12.5\) and the area is the smallest when \(l\) is as far from \(12.5.\) Therefore, the largest area is when \(l=12,13\) and the smallest area is when \(l=1,24.\)

This makes the larger area equal to \[156.25-(13-12.5)^2 \]\[= 156.25-0.25 =156\] and the smaller area equal to \[156.25-(24-12.5)^2 \]\[= 156.25-132.25 =24.\]

Thus, the difference is \(156-24=132.\)

Thus, the answer is D.

Answer: E

Solution(s):

The circumference of a circle is \(\pi d = 2r\pi ,\) so going a quarter of the way around is \(\dfrac {\pi r }2.\)

He goes a quarter of the way around the large circle, so this part is \(10 \pi \) meters. He then goes from the larger circle to the smaller circle, which is \(20-10=10\) meters.

He goes a quarter of the way around the smaller circle, so this part is \(5 \pi \) meters. He then goes through the diameter of the smaller circle, which is \(10\cdot 2=20\) meters.

He then goes a quarter of the way around the smaller circle, so this part is \(5 \pi \) meters. He finally goes from the smaller circle to the larger circle, which is \(20-10=10\) meters.

The total length is \[10\pi + 10+5\pi + 20 + 5\pi + 10 \]\[=20\pi + 40.\]

Thus, the answer is E.

Answer: B

Solution(s):

Each dot has \(2\) dots that are one unit away from it.

Therefore, regardless of the choice of the first dot, \(2\) of the other \(7\) dots would be within one unit, so the probability is \(\dfrac 27.\)

Thus, the answer is B.

Answer: B

Solution(s):

Since the number of boys who passed and the number of girls who passed are the same, we can assign them the same variable. Let this number be \(p.\)

The number of boys in the class is \(\dfrac 32 p,\) and the number of girls in the class is \(\dfrac 43 p.\) Thus, the total number of people is \(\dfrac 43p + \dfrac 32p = \dfrac {17}6p.\)

The total number of people must be a multiple of the numerator of this fraction, so the number of people must be a multiple of \(17,\) making the number of people \(17.\)

Thus, the answer is B.

Answer: C

Solution(s):

The total volume is \(Ah\) where \(A\) is the area of the base and \(h\) is the height.

To find \(A,\) we take the area of the circle. It has a diameter of \(8\)cm, so it has a radius of area \(4\)cm. Thus, the area is \(r^2\pi = 16\pi.\) Since \(h=6\)cm, the volume of the whole thing is \(16\pi \cdot 6 = 96\pi.\)

Since the wedge is half of the volume of the cylinder, the volume of the wedge is \(48\pi.\)

To estimate this, we can find that \[49\pi \approx 49 \cdot \dfrac{22}{7} =154,\] so \[48\pi \approx 154 -\pi \approx 154-3 = 151.\]

Thus, the answer is C.

Answer: A

Solution(s):

Let \(x = \dfrac n3.\) We know \(x\) and \(9\dfrac n3 = 9x\) are both three digit numbers.

Thus, \[100 \leq x, 9x \leq 999,\] so \[100 \leq x \leq 111.\] Every integer in this range has \(x\) that creates a valid \(n,\) so there are \(12\) valid numbers.

Thus, the answer is A.

Answer: C

Solution(s):

Let the side length be \(s.\) Note that the total area is \(s^2.\)

Since \(AF = 2FE,\) we know \[FE = \dfrac{AF}3 = \dfrac s3,\]\[ AF = 2 \dfrac{AE}3 = \dfrac{2s}3 .\]

Since \(CD = 2DE,\) we know \[CD = \dfrac{CE}3 = \dfrac s3,\]\[ AF = 2 \dfrac{AE}3 = \dfrac {2s}{3}.\]

This makes the area of \[ABF = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3,\] the area of \[BCD = \dfrac{s\cdot \dfrac{2s}3}{2} = \dfrac {s^2}3,\] and the area of \[FED = \dfrac {\dfrac s3 \cdot \dfrac s3 }2 = \dfrac {s^2} {18}.\] Thus, the area of \[BFD = s^2 - \dfrac {s^2}3-\dfrac {s^2}3-\dfrac {s^2}{18}\]\[ = \dfrac {5s^2}{18}.\]

Thus, the answer is C.

Answer: C

Solution(s):

If the rolled number was \(1,\) then the tile can be \(1,4,9,\) yielding \(3\) combinations.

If the rolled number was \(2,\) then the tile can be \(2,8,\) yielding \(2\) combinations.

If the rolled number was \(3,\) then the tile can be \(3,\) yielding \(1\) combination.

If the rolled number was \(4,\) then the tile can be \(1,4,9,\) yielding \(3\) combinations.

If the rolled number was \(5,\) then the tile can be \(5,\) yielding \(1\) combination.

If the rolled number was \(6,\) then the tile can be \(6,\) yielding \(1\) combination.

The total number of combinations is \[3+2+1+3+1+1=11.\] There are \(60\) combinations each with equal likelihood, so the probability is \(\dfrac{11}{60} .\)

Thus, the answer is C.

Answer: A

Solution(s):

The largest circle is of radius \(12,\) so the entire design has area of \(12^2 \pi = 144 \pi.\)

Each dark region is the area of its circle minus the area of the previous circle. The largest dark area would be of area \[(10^2-8^2) \pi = 36 \pi,\] the next area would be \[(6^2-4^2) \pi = 20 \pi ,\] and the smallest area would be \[(2^2-0^2) \pi = 4 \pi .\] Therefore, the combined dark area would be \(60 \pi.\)

This fraction would be \( \dfrac{60 \pi }{144 \pi} = \dfrac 5{12} ,\) which is approximately \(42 \%.\)

Thus, the answer is A.