Solution:

Including you, there are $6$ people sharing the fundraising equally. Each must raise $$\dfrac{1500}{6} = 250$$ dollars.

Thus, the correct answer is A.

Solution:

The inner values are $$\diamond(1,2,3) = \dfrac{1}{2-3} = -1, \quad \diamond(2,3,1) = \dfrac{2}{3-1} = 1, \quad \diamond(3,1,2) = \dfrac{3}{1-2} = -3.$$

Therefore $$\diamond(-1, 1, -3) = \dfrac{-1}{1 - (-3)} = -\dfrac{1}{4}.$$

Thus, the correct answer is B.

Solution:

Since $$20$ equals $2000$ cents, the local tax is $$0.0145 \times 2000 = 29$$ cents per hour.

Thus, the correct answer is E.

Solution:

Since $|x - 1|$ and $|x - 2|$ are the distances from $x$ to $1$ and $2,$ the point $x$ is equidistant from $1$ and $2.$

That midpoint is $$x = \dfrac{1 + 2}{2} = \dfrac{3}{2}.$$

Thus, the correct answer is D.

Solution:

The number of three-point sets is $$\binom{9}{3} = 84.$$

The collinear triples are the $3$ rows, the $3$ columns, and the $2$ main diagonals, for a total of $8.$

The probability is therefore $$\dfrac{8}{84} = \dfrac{2}{21}.$$

Thus, the correct answer is C.

Solution:

Bertha has $30 - 6 = 24$ granddaughters, none of whom have daughters.

These granddaughters belong to $24 / 6 = 4$ of Bertha's daughters. So exactly $4$ women have daughters, and the number with no daughters is $$30 - 4 = 26.$$

Thus, the correct answer is E.

Solution:

There are five layers, each one shorter and narrower than the one below. The total number of oranges is $$5\cdot 8 + 4\cdot 7 + 3\cdot 6 + 2\cdot 5 + 1\cdot 4 = 40 + 28 + 18 + 10 + 4 = 100.$$

Thus, the correct answer is C.

Solution:

After the first three rounds the counts go from $(15, 14, 13)$ to $(14, 13, 12).$ In general, every three rounds each player loses exactly one token.

After $36$ rounds the counts are $(3, 2, 1).$ On the $37$th round the leader gives away three tokens and drops to $0,$ ending the game.

Thus, the correct answer is B.

Solution:

Let $[ABD]$ be the area shared by both large triangles. Then $[ABE] = [ADE] + [ABD]$ and $[ABC] = [BDC] + [ABD].$

Subtracting, $$[ADE] - [BDC] = [ABE] - [ABC].$$ Since $\angle EAB$ and $\angle ABC$ are right angles, $$[ABE] = \tfrac12(4)(8) = 16, \qquad [ABC] = \tfrac12(4)(6) = 12.$$

The difference is $16 - 12 = 4.$

Thus, the correct answer is B.

Solution:

The two coins match when both show $0, 1, 2,$ or $3$ heads. Coin $A$ has weights $1, 3, 3, 1$ out of $8$ and coin $B$ has weights $1, 4, 6, 4, 1$ out of $16.$

The probability is $$\dfrac{1\cdot 1 + 3\cdot 4 + 3\cdot 6 + 1\cdot 4}{8 \cdot 16} = \dfrac{35}{128}.$$

Thus, the correct answer is D.

Solution:

Keeping $\pi r^2 h$ constant while multiplying the radius by $1.25$ requires the height to be multiplied by $$\dfrac{1}{1.25^2} = \dfrac{1}{1.5625} = 0.64.$$

So the height becomes $64\%$ of the original, a decrease of $36\%.$

Thus, the correct answer is C.

Solution:

Each of the $8$ condiments is independently in or out, giving $2^8 = 256$ condiment combinations.

For each of these there are $3$ choices of patty count, so the number of hamburgers is $$3 \times 256 = 768.$$

Thus, the correct answer is C.

Solution:

The number of dancing pairs is $12 \cdot 3 = 36,$ counting from the men's side. Each woman was in exactly $2$ pairs, so the number of women is $$\dfrac{36}{2} = 18.$$

Thus, the correct answer is D.

Solution:

With $n$ coins the total value is $20n$ cents. Adding a quarter gives $$20n + 25 = 21(n + 1),$$ so $n = 4.$

Four coins worth a total of $80$ cents must be three quarters and one nickel. Hence the number of dimes is $0.$

Thus, the correct answer is A.

Solution:

Write $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}.$ Here $\dfrac{y}{x} \lt 0,$ so the expression is largest when $\left|\dfrac{y}{x}\right|$ is smallest.

That happens with $y = 2$ and $x = -4,$ giving $$1 + \dfrac{2}{-4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}.$$

Thus, the correct answer is D.

Solution:

Every $5\times5,$ $4\times4,$ and $3\times3$ square contains the center cell, and there are $$1^2 + 2^2 + 3^2 = 14$$ of them.

Among the smaller squares, $4$ of the $2\times2$ squares and $1$ of the $1\times1$ squares cover the center, giving $$14 + 4 + 1 = 19.$$

Thus, the correct answer is D.

Solution:

Before the first meeting the two together cover half the track. Between the first and second meetings they together cover a full track, which is twice as far, so Brenda runs $2 \cdot 100 = 200$ meters in that stretch.

Sally runs $150$ meters in the same stretch, so the full track length is $$200 + 150 = 350.$$

Thus, the correct answer is C.

Solution:

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d,$ so the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $$(11 + d)^2 = 9(29 + 2d),$$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The third terms $29 + 2d$ are $49$ and $1.$ The smallest is $1.$

Thus, the correct answer is A.

Solution:

Unrolling the stripe flattens it into a parallelogram. Its base (the horizontal width) is $3$ feet and its height spans the full $80$ feet of the silo.

The area is therefore $$3 \times 80 = 240$$ square feet.

Thus, the correct answer is C.

Solution:

Let the square have side $1,$ and by symmetry let $ED = DF = x,$ so $AE = 1 - x.$

Since $\triangle BEF$ is equilateral, $EF^2 = EB^2,$ giving $$2x^2 = 1 + (1 - x)^2,$$ which simplifies to $x^2 = 2(1 - x).$

The right triangles have areas $[DEF] = \tfrac12 x^2$ and $[ABE] = \tfrac12(1 - x),$ so $$\dfrac{[DEF]}{[ABE]} = \dfrac{x^2}{1 - x} = \dfrac{2(1 - x)}{1 - x} = 2.$$

Thus, the correct answer is D.

Solution:

Let $\theta$ be the acute angle. The shaded region has three parts: two acute sectors of the unit disk with total area $\theta,$ two obtuse sectors of the ring between radii $1$ and $2$ with total area $3(\pi - \theta),$ and two acute sectors of the ring between radii $2$ and $3$ with total area $5\theta.$

Adding these gives a shaded area of $$\theta + 3(\pi - \theta) + 5\theta = 3\pi + 3\theta.$$

The shaded region is $\dfrac{8}{13}$ of the unshaded region, so it is $\dfrac{8}{21}$ of the total area $9\pi.$ Then $$3\pi + 3\theta = \dfrac{8}{21}(9\pi) = \dfrac{24\pi}{7},$$ which gives $\theta = \dfrac{\pi}{7}.$

Thus, the correct answer is B.

Solution:

Let $F$ be the point where $CE$ touches the semicircle and let $x = AE.$ Since tangents from a point are equal, $CF = CB = 2$ and $EF = EA = x,$ so $CE = 2 + x.$

In right triangle $CDE,$ we have $DE = 2 - x$ and $DC = 2,$ so $$(2 - x)^2 + 2^2 = (2 + x)^2.$$ This gives $x = \dfrac{1}{2},$ hence $CE = 2 + \dfrac{1}{2} = \dfrac{5}{2}.$

Thus, the correct answer is D.

Solution:

Because circle $A$ passes through $D$'s center and is internally tangent to $D,$ circle $D$ has radius $2.$ Place $D$'s center at the origin and $A$'s center at $(-1, 0).$

Let circle $B$ have radius $r$ and center $(x, r),$ using the symmetry of $B$ and $C$ about the horizontal axis. Tangency gives $$(x + 1)^2 + r^2 = (1 + r)^2, \qquad x^2 + r^2 = (2 - r)^2.$$

Subtracting yields $x = 3r - 2.$ Substituting into the second equation gives $9r^2 - 8r = 0,$ so $r = \dfrac{8}{9}.$

Thus, the correct answer is D.

Solution:

Applying the rule repeatedly, $$a_{2^1} = 2^0, \quad a_{2^2} = 2^1, \quad a_{2^3} = 2^{1+2}, \quad a_{2^4} = 2^{1+2+3}, \ldots$$ so in general $a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{n(n-1)/2}.$

For $n = 100,$ the exponent is $\dfrac{100 \cdot 99}{2} = 4950,$ so $a_{2^{100}} = 2^{4950}.$

Thus, the correct answer is D.

Solution:

The three small centers form an equilateral triangle of side $2,$ each $1$ unit above the plane. Its centroid $D$ is at distance $\dfrac{2\sqrt{3}}{3}$ from each vertex.

The large sphere's center $E$ sits directly above $D,$ and the distance between $E$ and a small center is $1 + 2 = 3.$ Thus $$DE = \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} = \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}.$$

Adding the $1$ unit from the plane to $D$ and the $2$ units from $E$ to the top of the large sphere gives $$1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$$

Thus, the correct answer is B.