Solution:

We can evaluate it as follows. $$\begin{align*} &(2^0-1+5^2+0)^{-1} \times 5 \\&=\dfrac{1}{1 - 1 + 25} \times 5 \\ &= \dfrac{5}{25}\\ &= \dfrac{1}{5} \end{align*}$$

Thus, C is the correct answer.

Solution:

Let $x$ be the number of triangular tiles and $y$ be the number of square tiles. We then have that $$x + y = 25$$ and $$3x + 4y = 84.$$ Multiplying the first equation by $3$ gives us $$3x + 3y = 75$$ and subtracting this from the other equation gives us $y = 9.$

Thus, D is the correct answer.

Solution:

Let us try to find a pattern between the number of toothpicks needed for the staircases.

For a $1$-step staircase, we would only need $4$ toothpicks (just a square).

For a $2$-step staircase, we would need $10$ toothpicks according to the diagram.

Similarly, we would need $18$ toothpicks for a $3$-step staircase.

A $2$-step staircase needs $10 - 4 = 6$ more toothpicks than a $1$-step staircase. A $3$-step staircase needs $18 - 10 = 8$ more toothpicks than a $2$-step staircase.

Following this pattern, we can see that a $4$-step staircase will need $18 + 10 = 28$ toothpicks, and a $5$-step staircase will need $28 + 12 = 40$ toothpicks.

This means that Ann would need to add $40 - 18 = 22$ more toothpicks.

Thus, D is the correct answer.

Solution:

Let $m$ be the number of candy eggs that Mia had. Then $Sofia$ had $2m$ eggs and $Pablo$ had $6m$ eggs.

The total number of eggs is then $$m + 2m + 6m = 9m.$$ For all of them to have the same number of eggs, they each must have $9m \div 3 = 3m$ eggs.

Sofia needs $3m - 2m = m$ more eggs. This means Pablo must give $\dfrac{m}{6m} = \dfrac{1}{6}$ of his eggs to Sofia.

Thus, B is the correct answer.

Solution:

The total for the first $14$ graded tests was $14\cdot 80=1120$.

After Payton's test was included, the total became $15\cdot 81=1215$. Therefore Payton's score was $1215-1120=95$.

Thus, E is the correct answer.

Solution:

Let $x$ and $y$ be the two numbers. Then we have that $$x + y = 5(x - y).$$ Note that we are assuming $x \gt y.$ This gives us $$x + y = 5x - 5y$$$$6y = 4x.$$

Dividing through yields $$\dfrac{x}{y} = \dfrac{6}{4} = \dfrac{3}{2}.$$

Thus, B is the correct answer.

Solution:

Recall that the $n$ th term of an arithmetic sequence is $a + d(n - 1),$ where $a$ is the first term and $d$ is the common difference.

For us, $a = 13$ and $d = 3.$ Plugging these in, we get that $$\begin{align*} 73 &= 13 + 3(n - 1) \\ 20&= n - 1 \\ n &= 21. \end{align*}$$ Thus, B is the correct answer.

Solution:

Let $p$ and $c$ be Pete's and Claire's current ages respectively.

Then we have that $$p - 2 = 3(c - 2)$$ and $$p - 4 = 4(c - 4).$$ Simplifying both equations gives us $$p = 3c - 4$$ and $$p = 4c - 12.$$ Setting them equal, we have $$3c - 4 = 4c - 12$$$$c = 8.$$

This means that $$p = 3 \cdot 8 - 4 = 20.$$

Now, we need to find the number of years ($y$) until $$20 + y = 2(8 + y),$$ which gives us $$20 + y = 16 + 2y$$$$y = 4.$$ Thus, B is the correct answer.

Solution:

Let $r_1$ and $h_1$ be the radius and height of the first cylinder and similarly define $r_2$ and $h_2$ for the second cylinder.

We know that $$r_2 = \dfrac{11}{10}r_1$$ and $$\pi r_1^2h_1 = \pi r_2^2h_2.$$

Substituting and simplifying gives us $$r_1^2h_1 = \dfrac{121}{100}r_1^2h_2,$$ which tells us that $$h_1 = \dfrac{121}{100}h_2.$$

Thus, D is the correct answer.

Solution:

The initial step would be to choose a letter that could work and build from there.

For instance, if we begin with $a,$ the only letters that can be placed next to it are $c$ or $d.$

From here, we cannot proceed further since the combinations $acbd$ and $acdb$ are not allowed due to the presence of $cb$ and $cd,$ respectively.

The same issue arises if we start with $d,$ as a $b$ would have to be placed in the middle and end up being adjacent to either an $a$ or a $c.$

On the other hand, if we start with a $b,$ the next letter would have to be $d,$ and after that, an $a$ and a $c$ can be placed, making this arrangement possible.

The same methodology holds true for starting with a $c.$ This gives us a total of $2$ rearrangements.

Thus, C is the correct answer.

Solution:

Let the side lengths be $4x$ and $3x.$ Then the diagonal has length $$\sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x.$$

The area of the rectangle is $$4x \cdot 3x = 12x^2.$$ Then we get that $$kd^2 = 12x^2$$$$k = \dfrac{12x^2}{25x^2} = \dfrac{12}{25}.$$

Thus, C is the correct answer.

Solution:

Substitute $x=\sqrt{\pi}$. Then $x^2=\pi$ and $x^4=\pi^2$, so the equation becomes $$y^2-2\pi y+\pi^2=1.$$

Hence $(y-\pi)^2=1$, giving the two possible values $y=\pi+1$ and $y=\pi-1$. Their distance is $2$.

Thus, C is the correct answer.

Solution:

Let the number of $5$-cent coins be $x$ and the number of $10$-cent coins be $12 - x.$

Then we have that any multiple of $5$ between $5$ and $$5x + 10(12 - x) = 120 - 5x$$ can be achieved by a combination of coins.

There are $24 - x$ such multiples of $5,$ which means that $x = 7$ to get $17$ possible different values.

The number of $10$-cent coins is therefore $12 - 7 = 5.$

Thus, C is the correct answer.

Solution:

The disk of radius $10$ rolls externally around the clock face of radius $20$. If the point of tangency moves through central angle $\theta$ around the clock, the disk rotates through $\frac{20+10}{10}\theta=3\theta$ relative to its original direction.

The arrow next points upward when $3\theta$ is a positive multiple of $2\pi$. The first time this happens is $\theta=\frac{2\pi}{3}$, which is one-third of the way around the clock from 12 o'clock, namely 4 o'clock.

Thus, C is the correct answer.

Solution:

The condition is $$\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}.$$ Cross-multiplying gives $10y(x+1)=11x(y+1)$, or $xy+11x-10y=0$.

Factoring by grouping after subtracting $110$ gives $$(x-10)(y+11)=-110.$$ Since $x,y$ are positive, the useful negative factor pairs are $(-1,110)$, $(-2,55)$, and $(-5,22)$, producing $(x,y)=(9,99),(8,44),(5,11)$.

Only $\frac{5}{11}$ has relatively prime numerator and denominator, so exactly one fraction works.

Thus, B is the correct answer.

Solution:

Adding the two equations gives us $$x^2 + y^2 - 4x - 4y + 8$$ $$= x + y + 8.$$ We can rearrange this equation to get $$x^2 + y^2 = 5(x + y).$$ We can then subtract them to get $$x^2 - y^2 - 4x + 4y = y - x.$$ Once again rearranging, we can find $$x^2 - y^2 = 3(x - y).$$ We have that $x \neq y,$ which means that we can divide both sides by $x - y.$ This gives us $$x + y = 3$$$$x^2 + y^2 = 15.$$

Thus, B is the correct answer.

Solution:

Since one of the sides of the equilateral triangle is a vertical line, the line of symmetry perpendicular to this side must be horizontal.

This means that the slope of the third side must be opposite the slope of the second side, which would be $-\dfrac{\sqrt{3}}{3}.$

To find the perimeter, we only need to find the length of one of the sides of the triangle.

We can plug in $x = 1$ into the two other equations to get the two vertices on the vertical line.

The two $y$-values are then $$1 + \dfrac{\sqrt{3}}{3} \text{ and } -\dfrac{\sqrt{3}}{3}.$$ The distance between the two is $1 + \dfrac{2\sqrt{3}}{3},$ which makes the perimeter $$3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$$ Thus, D is the correct answer.

Solution:

Note that $1000$ converted to hexadecimal is $3E8.$ Now we need to count the number of numbers that have only numerical digits in their hexadecimal.

The first digit can be $0, 1, 2$ or $3.$ The second and third digits can be any number from $0 - 9.$ This gives us $$4 \cdot 10 \cdot 10 = 400$$ numbers. This, however, includes $0,$ which is not a positive integer so we have to subtract one.

The sum of the digits in $399$ is $21.$

Thus, E is the correct answer.

Solution:

Since $\triangle ABC$ is isosceles right with area $12.5$, its legs have length $5$. The trisectors make $\angle ACD=30^\circ$ and $\angle BCE=30^\circ$, so $\triangle ACD$ and $\triangle BCE$ have equal area.

In the diagram, let $AF=DF=h$. Then $CF=5-h$, and the $30^\circ$ angle gives $$\frac{CF}{DF}=\sqrt{3}.$$ Thus $5-h=h\sqrt{3}$, so $h=\frac{5}{1+\sqrt{3}}=\frac{5\sqrt{3}-5}{2}$.

Therefore $$[ACD]=\frac12\cdot 5\cdot h=\frac{25\sqrt{3}-25}{4}.$$ Subtracting the two congruent corner triangles from $\triangle ABC$, $$[CDE]=\frac{25}{2}-2\cdot\frac{25\sqrt{3}-25}{4}=\frac{50-25\sqrt{3}}{2}.$$

Thus, D is the correct answer.

Solution:

Let the side lengths be positive integers $x$ and $y$. Then $$A+P=xy+2x+2y=(x+2)(y+2)-4.$$ Hence $A+P+4$ must factor into two integers both at least $3$.

The answer choices plus $4$ are $104,106,108,110,112$. All except $106$ have a factorization with both factors at least $3$: $$104=4\cdot26, \quad 108=9\cdot12, \quad 110=10\cdot11, \quad 112=7\cdot16.$$ But $106=2\cdot53$, so it cannot equal $(x+2)(y+2)$.

Thus, B is the correct answer.

Solution:

We claim that triangles $ABC$ and $ABD$ are perpendicular to each other.

We can show this be dropping the altitudes from $C$ to $AB$ and from $D$ to $AB$ in each triangle.

Since $AC = AD$ and $BC = BD,$ we have that the feet of these altitudes will coincide at point $P.$

Then we have that $$CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}.$$ We then have that $CD = CP\sqrt{2},$ which shows that $CPD$ is an isosceles right triangle.

This proves the above claim. Finally, the volume of the tetrahedron is $$\dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}.$$

Thus, C is the correct answer.

Solution:

Count the possible sets of people who stand. For $0$ and $1$ people standing, there are $1$ and $8$ possibilities.

For $2$ people standing, choose any pair and subtract the $8$ adjacent pairs: $\binom82-8=20$.

For $3$ people standing, first choose one standing person. Among the remaining five non-neighbor seats, $10$ pairs are possible, but $4$ of those pairs are adjacent, leaving $6$. This counts each final set three times, so there are $8\cdot6/3=16$ possibilities.

For $4$ people standing, the only possibilities are the two alternating sets. Thus the number of favorable coin-flip outcomes is $1+8+20+16+2=47$. Since all $2^8=256$ outcomes are equally likely, the probability is $\frac{47}{256}$.

Thus, A is the correct answer.

Solution:

Let the zeroes be $r$ and $s.$ Using Vieta's formulas, we have that $a = r + s$ and $2a = rs.$

Then we get that $$rs = 2(r + s),$$ which rearranges to $$rs - 2r - 2s = 0$$ $$rs - 2r - 2s + 4 = 4$$ $$(r - 2)(s - 2) = 4.$$

The only possible pairs $(r - 2, s - 2)$ that work are $$(1, 4), (-1, -4), (4, 1), (-4, -1),$$ $$(2, 2), (-2, -2).$$ For any of these pairs, we have that $$a = r - 2 + s - 2 + 4.$$ We want all the such unique values of $a.$ We get that they are $$-1, 0, 8, 9.$$ The sum of these values is $-1 + 8 + 9 = 16.$

Thus, C is the correct answer.

Solution:

Let $BC=x$ and $CD=AD=y$. Dropping the altitude from $A$ to $CD$ gives a right triangle with legs $x$ and $y-2$, and hypotenuse $y$. Therefore $$(y-2)^2+x^2=y^2,$$ so $x^2=4(y-1)$.

Since $x$ is an integer, write $x=2k$. Then $y=k^2+1$, and the perimeter is $$p=2+x+y+y=2k^2+2k+4.$$

We need $2k^2+2k+4<2015$, or $k^2+k<1005.5$. This holds for $k=1,2,\ldots,31$, while $k=32$ is too large. Thus there are $31$ possible perimeters.

Thus, B is the correct answer.

Solution:

Fix one of the two points. The second point is on the same side with probability $\frac14$, on an adjacent side with probability $\frac12$, and on the opposite side with probability $\frac14$.

On the same side, two coordinates $a,b\in[0,1]$ are at distance at least $\frac12$ when $|a-b|\ge\frac12$. This region consists of two right triangles with total area $\frac14$.

On adjacent sides, the distance has the form $\sqrt{a^2+b^2}$. The failing region is a quarter circle of radius $\frac12$, so the success probability is $1-\frac{\pi}{16}$.

On opposite sides, the distance is always at least $1$, so the success probability is $1$. Therefore the desired probability is $$\frac14\cdot\frac14+\frac12\left(1-\frac{\pi}{16}\right)+\frac14=\frac{26-\pi}{32}.$$ Hence $a+b+c=26+1+32=59$.

Thus, A is the correct answer.