Answer: C

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 .

Answer: E

Solution:

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

Thus, the correct answer is E .

Answer: E

Solution:

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

Thus, the correct answer is E .

Answer: B

Solution:

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

Thus, the correct answer is B .

Answer: A

Solution:

Let each pane have height $5x$ and width $2x$. Vertically, the window contains two pane heights and three horizontal borders, so its height is $10x+6$. Horizontally, it contains four pane widths and five vertical borders, so its width is $8x+10$.

Because the window is square, $10x+6=8x+10$, so $x=2$.

Therefore, the side length is $10(2)+6=26$ inches.

Thus, the correct answer is A .

Answer: C

Solution:

Each pair of balloons costs $1+\frac23=\frac53$ times the regular price of one balloon.

Thus, with the money for $30$ regular-price balloons, Orvin can buy $30\cdot\frac65=36$ balloons.

Thus, the correct answer is C .

Answer: A

Solution:

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

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

Thus, the correct answer is A .

Answer: E

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 .

Answer: A

Solution:

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

Thus, the correct answer is A .

Answer: C

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 .

Answer: C

Solution:

The three successive-discount plans leave the customer paying $$\begin{aligned} 0.85^2&=0.7225,\\ 0.9^3&=0.729,\\ 0.75(0.95)&=0.7125 \end{aligned}$$ times the listed price. Their effective discounts are therefore $27.75\%$, $27.1\%$, and $28.75\%$.

A single $n\%$ discount must exceed all three, so $n>28.75$. The smallest positive integer that works is $29$.

Thus, the correct answer is C .

Answer: 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 .

Answer: B

Solution:

Follow the $60^\circ$ grid formed by the unit hexagons and place $A=(0,0)$. The marked vertices may then be written as $B=(3,\sqrt3)$ and $C=(3,-\sqrt3)$.

Thus $BC=2\sqrt3$, and $$\begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3. \end{aligned}$$ Hence $\triangle ABC$ is equilateral, with area $$\frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3.$$

Thus, the correct answer is B .

Answer: D

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 .

Answer: A

Solution:

Let $AD=h.$ Since $DC=AB=2h,$ the area of rectangle $ABCD$ is $2h^2.$

The rays $DE$ and $DF$ make angles of $60^\circ$ and $30^\circ,$ respectively, with $DC.$

Therefore, $$AE=h\tan30^\circ=\dfrac{h}{\sqrt3}$$ and $$AF=h\tan60^\circ=h\sqrt3.$$

Hence $$\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}$$ The area of $\triangle DEF$ is $$\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.$$

The desired ratio is $$\dfrac{h^2\sqrt3/3}{2h^2}=\dfrac{\sqrt3}{6}.$$ Thus, the correct answer is A.

Answer: B

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 .

Answer: D

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 .

Answer: E

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 .

Answer: D

Solution:

Fix the first endpoint $A$ on the outer circle. Draw the two chords from $A$ that are tangent to the inner circle, and call their other endpoints $B$ and $C$. A chord from $A$ meets the inner circle exactly when its second endpoint lies on the minor arc $BC$.

If $O$ is the common center and $D$ is a tangency point, then $\triangle AOD$ is right, with $OA=2$ and $OD=1$. Hence $\angle OAD=30^\circ$. The two tangent chords therefore make a $60^\circ$ angle at $A$, so the intercepted minor arc $BC$ measures $120^\circ$.

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

Thus, the correct answer is D .

Answer: C

Solution:

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

The product is negative exactly when its two factors have opposite signs. Since $x^2-50<x^2-1$, this requires $x^2-50<0<x^2-1$. Thus $1<x^2<50$, or $2\le |x|\le7$. There are $6$ positive and $6$ negative integer solutions, for a total of $12$.

Thus, the correct answer is C .

Answer: B