Solution:

Let the number of pennies be $p.$ Then, the number of nickels is $13-p$ and $p-1,$ so $13-p=p-1.$ This means we have $7$ pennies and $6$ nickels.

Therefore, the number of cents is $7+6\cdot 5=37.$

Thus, the correct answer is C .

Solution:

$$\begin{align*} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{align*}$$

Thus, the correct answer is E .

Solution:

Let the path distance be $t.$ Then, we get: $$\begin{align*}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{align*}$$

Thus, the correct answer is E .

Solution:

Let the price for a muffin be $m$ and let the price for a banana be $b.$ Then, $$\begin{align*} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{align*}$$

Thus, the correct answer is B .

Solution:

Let the smaller side of a pane be a distance of $x.$

Then, the side length is $5\cdot 2 + 4x.$ Also, the other direction has pane lengths of $2.5x.$

This means the side length is $3\cdot 2 + 2(2.5x) .$ We solve for $x$ as follows: $$\begin{align*}10+4x &= 6+5x \\ x&= 4\end{align*}$$

Therefore, the side length is $$10+4\cdot 4 = 26.$$

Thus, the correct answer is A .

Solution:

Suppose we buy $6$ balloons. Then, we can buy $3$ at full price and $3$ at a price of $\frac 23$ of a balloon.

Therefore, we can buy it at a price of $5$ balloons. Thus, with the money to buy $30$ balloons, we could buy $30 \cdot \frac 65 = 36$ balloons.

Thus, the correct answer is C .

Solution:

By definition, we know $$A = \dfrac {x+ 100}{100}B$$$$= B + \dfrac x{100}B.$$

This implies, $$\begin{align*} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{align*}$$

Thus, the correct answer is A .

Solution:

This means it travels $\dfrac{b}{18}$ yards in $t$ seconds since a yard is $3$ feet. Then, in one second, it travels $\dfrac{b}{18t}$ yards.

Therefore, in $3$ minutes which is $180$ seconds, it travels $$\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.$$

Thus, the correct answer is E .

Solution:

Observe that: $$\begin{align*} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{align*}$$

Thus, the correct answer is A .

Solution:

From the leftmost column, there is no carry into a sixth digit, so $A+B=D\le 9$.

The units column is $B+A=D$, so it also has no carry. The tens column then gives $C+D=D$, hence $C=0$.

Since $A$ and $B$ are distinct nonzero digits, $D=A+B$ can be any digit from $3$ through $9$. For example, $(A,B)=(1,2),(1,3),(2,3),(2,4),(2,5),(2,6),(2,7)$ give $D=3,4,5,6,7,8,9$.

Thus there are $7$ possible values of $D$, and the correct answer is C .

Solution:

We need to find the smallest possible $n$ such that $$1- \dfrac n{100} < (0.85)^2,$$ $$1- \dfrac n{100} < (0.9)^3,$$ $$1- \dfrac n{100} < (0.75)(0.95).$$ Note that $$0.75\cdot 0.95 = 0.85^2-0.1^2 < 0.85^2,$$ so we don't need to worry about the first condintion since the last condition is true. Then, the second condition yields $$1- \dfrac n{100} < 0.729.$$ $$n > 27.1.$$

Also, we also can see that $$1- \dfrac n{100} < 0.95\cdot 0.75$$$$1- \dfrac n{100} < \dfrac 34 \cdot \dfrac{19}{20}$$$$1- \dfrac n{100} < \dfrac{57}{80}$$$$1- \dfrac n{100} < \dfrac{285}{400}.$$

Therefore, $$n > \dfrac{115}{4} = 28.75.$$ Combining our conditions yields a smallest $n$ of $n=29.$

Thus, the correct answer is C .

Solution:

The fifth-largest divisor is $2,014,000,000$ divided by the fifth smallest divisor.

The prime factorization of $2,014,000,000$ is: $$2^7\cdot 5^6 \cdot 19\cdot 53.$$ This makes the first $5$ smallest divisors $1,2,4,5,8$ Therefore, the fifth smallest divisor is $8,$ and the fifth largest divisor must be: $$\dfrac{2014000000}8 =251750000.$$

Thus, the correct answer is C .

Solution:

Since $AB = BC = AC$ by rotational symmetry, we know it is an equilateral triangle.

Then, one-fourth of $AB$ can be found as a the leg of a right triangle with hypotenuse $1$ and is opposite to the $60^{\circ}$ angle, making it $$\sin(60^\circ) = \dfrac{\sqrt 3}2.$$

As such, $$AB = 4\cdot \dfrac{\sqrt 3}2 = 2\sqrt 3.$$

Then, since it is an equilateral triangle, it has area $$\dfrac{s^2 \sqrt 3}4 = \dfrac{12\sqrt 3}4 = 3 \sqrt 3.$$

Thus, the correct answer is B .

Solution:

We know that the difference of the numbers $cba$ and $abc$ is equal to: $$100c + 10b+a - 100a - 10b-c$$$$= 99(c-a)$$ We know that this number also must be a multiple of $55.$ As $\gcd(55,99)$ is $11,$ we know that $c-a$ is a multiple of $5,$ and $c > a.$

This makes $a = 1, b = 0, c = 6$ the only possible value with $a+ b+c \leq 7$ as every other combination has $a+b+c > 7.$ As such, $a^2+b^2+c^2 = 37.$

Thus, the correct answer is D .

Solution:

The area of $EFB$ is equal to $$\dfrac{EF\cdot AD}2 .$$

Similarly, The area of $ABCD$ is equal to $$AB\cdot AD .$$

Thus, their ratio is $$\dfrac{EF}{2\cdot AD} = \dfrac{EF}{4\cdot AB}.$$

Then, $$\begin{align*} EF &= AF - AE \\&= AB \tan(\angle ADE) \\&\quad - AB \tan(\angle ADF) \\&= AB( \tan(60^\circ) - \tan(30^ \circ)) \\&= AB\left(\sqrt 3 - \dfrac {\sqrt{3}}3\right) \\&= AB \left(\dfrac{2\sqrt 3}3\right)\end{align*}$$ This makes our result $$\dfrac{EF}{4\cdot AB} = \dfrac{AB(\frac{2 \sqrt 3}3)}{4\cdot AB} = \dfrac {\sqrt 3}6.$$

Thus, the correct answer is A .

Solution:

There are $6^4$ equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in $6$ ways, the different value in $5$ ways, and the position of the different die in $4$ ways. This gives $6\cdot5\cdot4=120$ outcomes.

If all four dice match, there are $6$ outcomes.

The probability is $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$.

Thus, the correct answer is B .

Solution:

Factor out the obvious power of $2$: $10^{1002}-4^{501}=2^{1002}(5^{1002}-1)$.

Since $5^{1002}-1=(5^{501}-1)(5^{501}+1)$, and $501$ is odd, $5^{501}-1$ is divisible by $4$ but not by $8$, while $5^{501}+1$ is divisible by $2$ but not by $4$.

Thus $5^{1002}-1$ contributes exactly $2^3$, so the whole expression is divisible by $2^{1005}$ but not $2^{1006}$.

Thus, the correct answer is D .

Solution:

The list has total sum $11\cdot10=110$. To maximize the largest entry, minimize the sum of the other ten entries.

In nondecreasing order, the sixth entry is $9$, and $8$ must be the unique mode. If $8$ appears twice, the least possible first ten entries sum to $1+2+3+8+8+9+10+11+12+13=77$, giving largest entry $33$.

If $8$ appears three times, the least possible first ten entries are $1,1,8,8,8,9,9,10,10,11$, with sum $75$, giving largest entry $35$.

If $8$ appears four or five times, the least possible sum of the first ten entries is at least $80$, so the largest entry is at most $30$.

Therefore the largest possible entry is $35$, and the correct answer is E .

Solution:

First, without loss of generality, we could choose some point on the outer circle. Then, the second point can be chosen in a region on the other circle.

This region is such that it has a line that intersects the circle, so the edge of the region is such that the chord is perpendicular with the inner circle.

If we look at the angle at the center, we can see that it has 2 right triangles where the adjacent side is $1$ and the hypotenuse is $2,$ making $$\cos \left(\dfrac \theta 2\right) = \dfrac 12.$$

Thus, $\dfrac \theta 2 = 60^\circ,$ making $\theta = 120 ^\circ .$

Therefore, the probability is $$\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$$

Thus, the correct answer is D .

Solution:

First, note that $$x^4-51x^2+50$$$$= (x^2-50)(x^2-1).$$

If $(x^2-50)(x^2-1) < 0$ means that one of the terms is negative.

Since $$x^2-50 < x^2-1,$$ it must be that x^2-50 < 0, x^2-1 >0. This means $1 < x^2 < 50,$ making $$1< |x| \leq 7,$$ resulting in $12$ solutions.

Thus, the correct answer is C .

Solution:

Let the base of the altitude from $C$ to $AB$ be $E.$ and let the base of the altitude from $D$ to $AB$ be $F.$ Also, let $BC = 10$ since we can assign any value. This yields the following diagram:

Then, let $FB = x$ and the altitude be $h.$ This means $$\begin{align*}AE &= 33-x-EF\\&=33-x-21 \\&= 12-x.\end{align*}$$

This suggests that: $$14^2 = (12-x)^2 + h^2$$ $$10^2 = x^2 + h^2.$$ Subtracting the equations, we get: $$96 = 144 - 24x$$ $$x = 2.$$ Then, we want to find $$\sqrt{ (21+x)^2+h^2}$$$$= \sqrt{21^2 + 42x + (x^2 + h^2) }$$$$= \sqrt{441+42\cdot 2 + 100}$$$$= \sqrt{625}$$$$=25.$$

Thus, the correct answer is B .

Solution:

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $$\sqrt{1^2 + \dfrac 12 ^2} = \dfrac {\sqrt 5}2.$$ Then , we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $$\dfrac{\sqrt 5 -1}2 .$$

Thus, the correct answer is B .

Solution:

Let the top radius be $1$, the bottom radius be $R$, and the inscribed sphere radius be $a$.

In the cross-section, the sphere is tangent to the two bases, so the frustum height is $2a$. The right triangle formed by the side, a radius to the tangency point, and the base radii gives $R=a^2$, as in the official diagram.

The frustum volume is $\frac13\pi(R^2+R+1)(2a)=\frac{2a\pi}{3}(a^4+a^2+1)$.

This is twice the sphere volume, $\frac{8a^3\pi}{3}$. Cancelling gives $a^4-3a^2+1=0$, so $R^2-3R+1=0$.

Thus $R=\frac{3+\sqrt5}{2}$, and the correct answer is E .

Solution:

Single numbers give sums $1$ through $5$, complements give sums $10$ through $14$, and all five numbers give $15$. So an arrangement is good exactly when consecutive blocks can make sums $6$ and $7$.

If sum $6$ is impossible, then $1$ is not adjacent to $5$. By rotating and reflecting, write the arrangement as $1bc5e$. The adjacent pair $bc$ cannot be $\{2,3\}$ or $\{2,4\}$, since $1+2+3=6$ and $2+4=6$. Thus $e=2$, and avoiding the consecutive block $2,1,3$ forces the bad arrangement $14352$.

If sum $7$ is impossible, then $2$ is not adjacent to $5$. Similarly write the arrangement as $2bc5e$. Now $bc$ cannot be $\{3,4\}$ or $\{1,4\}$, so $e=4$. To avoid the consecutive block $4,2,1$, the remaining order must be $b=3,\ c=1$, giving $23154$.

These two arrangements are indeed bad, one missing sum $6$ and the other missing sum $7$. Hence there are $2$ bad arrangements.

Thus, the correct answer is B .

Solution:

Let $p_i$ be the probability that the frog eventually escapes starting from pad $i$. Then $p_0=0$, $p_{10}=1$, and by symmetry $p_5=\frac12$.

For $1\le i\le 4$, $p_i=\frac{i}{10}p_{i-1}+\frac{10-i}{10}p_{i+1}$.

Working downward from $p_5=\frac12$, we get $p_4=\frac25p_3+\frac3{10}$, then $p_3=\frac3{10}p_2+\frac7{10}p_4=\frac5{12}p_2+\frac7{24}$.

Next $p_2=\frac15p_1+\frac45p_3=\frac3{10}p_1+\frac7{30}$. Finally $p_1=\frac9{10}p_2$.

Substituting the expression for $p_2$ gives $p_1=\frac9{10}\left(\frac3{10}p_1+\frac7{30}\right)$, so $p_1=\frac{63}{146}$.

Thus, the correct answer is C .