Answer: B

Solution:

With $b$ bagels she buys $5 - b$ muffins, costing $50(5-b) + 75b = 250 + 25b$ cents. For a whole number of dollars this must be a multiple of $100,$ so $25b$ must end in $50,$ meaning $b$ is even.

Among $b = 2$ and $b = 4,$ only $b = 2$ works: $250 + 50 = 300$ cents $= \$3.$

Thus, the correct answer is B.

Answer: C

Solution:

Losing $3$ cans cost her $5$ rooms, so $3$ cans paint $5$ rooms and each room needs $\dfrac{3}{5}$ of a can.

For $25$ rooms she used $25 \cdot \dfrac{3}{5} = 15$ cans.

Thus, the correct answer is C.

Answer: D

Solution:

Twenty percent less than $60$ is $0.8 \cdot 60 = 48.$

One-third more than $n$ is $\dfrac{4}{3}n,$ so $\dfrac{4}{3}n = 48$ gives $n = 36.$

Thus, the correct answer is D.

Answer: C

Solution:

The parallel sides differ by $25 - 15 = 10,$ so each triangle has legs $\dfrac{10}{2} = 5$ and area $\dfrac{1}{2} \cdot 5^2 = \dfrac{25}{2}.$ The two beds total $25.$

The rectangle measures $25$ by $5,$ so its area is $125,$ and the fraction occupied is $\dfrac{25}{125} = \dfrac{1}{5}.$

Thus, the correct answer is C.

Answer: D

Solution:

Since $128 = 2^7,$ each age is a power of $2.$ The twins share an age $t,$ so Kiana's age is $\dfrac{128}{t^2}.$

Taking $t = 8$ gives Kiana $\dfrac{128}{64} = 2,$ who is younger than the twins. (Smaller twins would make Kiana older, which is not allowed.) The sum is $8 + 8 + 2 = 18.$

Thus, the correct answer is D.

Answer: C

Solution:

The genuinely different groupings give

$(2 \times 3) + (4 \times 5) = 26,$ $\ ((2 \times 3) + 4) \times 5 = 50,$ $\ 2 \times (3 + (4 \times 5)) = 46,$ and $\ 2 \times (3 + 4) \times 5 = 70.$

These are all distinct, so $4$ values can be obtained.

Thus, the correct answer is C.

Answer: B

Solution:

After January through March the price is $1.2 \cdot 0.8 \cdot 1.25 = 1.2$ times the original.

To return to the original, April must multiply by $\dfrac{1}{1.2},$ a decrease of $1 - \dfrac{1}{1.2} = \dfrac{1}{6} \approx 16.7\%.$ To the nearest integer, $x = 17.$

Thus, the correct answer is B.

Answer: E

Solution:

Let $x$ be the bucket's weight and $y$ the weight of a full bucket of water. Then $x + \dfrac{2}{3}y = a$ and $x + \dfrac{1}{2}y = b.$

Subtracting gives $\dfrac{1}{6}y = a - b,$ so $y = 6a - 6b$ and $x = b - \dfrac{1}{2}y = 4b - 3a.$ The full weight is $x + y = 3a - 2b.$

Thus, the correct answer is E.

Answer: A

Solution:

Line $AB$ has equation $x + y = 3,$ which is parallel to $x + y = 7,$ so the area is independent of where $C$ lies on that line.

Take $C = (7, 0).$ Then the base $AC = 4$ lies on the $x$-axis with height $3,$ giving area $\dfrac{1}{2} \cdot 4 \cdot 3 = 6.$

Thus, the correct answer is A.

Answer: A

Solution:

The hours containing a $1$ are $1, 10, 11, 12,$ so $8$ of the $12$ hours display correctly, a fraction $\dfrac{2}{3}.$

A minute is wrong if either digit is $1$: the tens digit gives $10\text{-}19$ ($10$ minutes), and the ones digit adds $01, 21, 31, 41, 51$ ($5$ more), $15$ in all. So $\dfrac{45}{60} = \dfrac{3}{4}$ of minutes are correct.

The fraction of the day is $\dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{2}.$

Thus, the correct answer is A.

Answer: D

Solution:

Each day the birds leave $\dfrac{3}{4}$ of the millet and Millie adds $\dfrac{1}{4}$ quart of new millet, so after $n$ days the millet is $$\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) = 1 - \left(\dfrac{3}{4}\right)^n$$ quart.

The non-millet seeds always total $\dfrac{3}{4}$ quart, so millet exceeds half when $1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4},$ that is, $\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}.$

Since $\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4}$ and $\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4},$ this first happens on day $5,$ which is Friday.

Thus, the correct answer is D.

Answer: E

Solution:

The eighth term divided by the fifth term is $r^3 = \dfrac{8!}{7!} = 8,$ so $r = 2.$

The fifth term is $a r^4 = 7!,$ so $a = \dfrac{7!}{16} = \dfrac{5040}{16} = 315.$

Thus, the correct answer is E.

Answer: D

Solution:

Let $D$ be the foot of the altitude from $A.$ Then $BD = \sqrt{13^2 - 12^2} = 5$ and $DC = \sqrt{15^2 - 12^2} = 9.$

If $D$ lies between $B$ and $C,$ then $BC = 5 + 9 = 14$; if the triangle is obtuse, $BC = 9 - 5 = 4.$ The sum is $14 + 4 = 18.$

Thus, the correct answer is D.

Answer: C

Solution:

The five squares have total area $5,$ so each region must have area $\dfrac{5}{2}.$

The line from $(a, 0)$ to $(3, 3)$ together with the axes bounds a triangle of base $3 - a$ and height $3$; the region on the lower-right side of the line is this triangle with one unit square removed. Setting $$\dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2}$$ gives $3(3 - a) = 7,$ so $a = \dfrac{2}{3}.$

Thus, the correct answer is C.

Answer: B

Solution:

Each equation gives $x = \dfrac{\log(7/3)}{\log(1 + f(r))},$ which is largest when the positive quantity $f(r)$ is smallest.

For $0 \lt r \lt 3,$ among $r,\ \dfrac{r}{10},\ 2r,\ \sqrt{r},\ \dfrac{1}{r},$ the smallest is $\dfrac{r}{10}$: it is below $r$ and below $\sqrt{r}$ since $r \lt 3 \lt 100.$ So equation (B) has the largest solution.

Thus, the correct answer is B.

Answer: B

Solution:

Draw the line through $D$ parallel to $AB,$ meeting $BC$ at $E,$ so $ABED$ is a parallelogram with $BE = AD.$ Then $\angle BDE = \angle DBA = 23^\circ,$ and since $\angle BDC = 46^\circ,$ segment $DE$ bisects $\angle BDC.$

By the angle bisector theorem in $\triangle BDC,$ $\dfrac{EC}{BE} = \dfrac{DC}{DB},$ so $$CD = DB \cdot \dfrac{BC - AD}{AD} = 1 \cdot \left(\dfrac{9}{5} - 1\right) = \dfrac{4}{5}.$$

Thus, the correct answer is B.

Answer: B

Solution:

Each of the $6$ faces has $2$ equally likely stripe orientations, for $2^6 = 64$ configurations.

An encircling stripe runs around one of the $3$ pairs of opposite faces. Fixing such a band, the four faces it passes through must be aligned, with probability $\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16},$ while the two remaining faces are free. The $3$ possible bands are disjoint events, so the probability is $3 \cdot \dfrac{1}{16} = \dfrac{3}{16}.$

Thus, the correct answer is B.

Answer: C

Solution:

The picture covers the arc within $\dfrac{1}{8}$ lap of the start on each side. At $600$ s Rachel has run $6\tfrac{2}{3}$ laps, $30$ s short of the line; a quarter lap takes her $22.5$ s, so she is in view between $30 - 11.25 = 18.75$ s and $30 + 11.25 = 41.25$ s of the $10$th minute.

At $600$ s Robert is $40$ s from the line; a quarter lap takes $20$ s, so he is in view between $30$ and $50$ s. Both appear between $30$ and $41.25$ s, a window of $11.25$ s out of $60,$ giving probability $\dfrac{11.25}{60} = \dfrac{3}{16}.$

Thus, the correct answer is C.

Answer: E

Solution:

Write $$f(n) = n^4 + 40n^2 + 400 - 400n^2 = (n^2 + 20)^2 - (20n)^2 = (n^2 + 20n + 20)(n^2 - 20n + 20).$$

For $f(n)$ to be prime the smaller factor must be $1$: solving $n^2 - 20n + 20 = 1$ gives $(n - 1)(n - 19) = 0,$ so $n = 1$ or $n = 19.$

Then $f(1) = 41$ and $f(19) = 761$ are both prime, summing to $802.$

Thus, the correct answer is E.

Answer: C

Solution:

Each of the $100$ edges is cut once near each endpoint, so $R$ has $2 \cdot 100 = 200$ vertices.

The cut at vertex $V_k$ creates a small polygon whose number of edges equals the degree of $V_k$; summed over all vertices this is $200,$ the total number of edge-endpoints. The middle portion of each original edge also survives, adding $100$ edges. So $R$ has $200 + 100 = 300$ edges.

Thus, the correct answer is C.

Answer: A

Solution:

Let $S_n$ be the number of valid reseatings of $n$ women. The rightmost woman either keeps her seat, leaving $S_{n-1}$ ways for the rest, or swaps with her left neighbor — the only other way to fill the end seat — leaving $S_{n-2}$ ways.

Thus $S_n = S_{n-1} + S_{n-2}$ with $S_1 = 1$ and $S_2 = 2,$ giving the Fibonacci values $1, 2, 3, 5, 8, 13, 21, 34, 55, 89.$ So $S_{10} = 89.$

Thus, the correct answer is A.

Answer: C

Solution:

Let $B = (b, b)$ and $D = (d, kd)$ with $b, d, k$ positive integers and $k \gt 1.$ The area is $(k - 1)bd = 1{,}000{,}000 = 2^6 \cdot 5^6.$

Each parallelogram corresponds to an ordered triple $(k - 1, b, d)$ of positive integers with product $2^6 \cdot 5^6.$ The six $2$'s distribute among the three factors in $\binom{6 + 2}{2} = 28$ ways, and likewise the six $5$'s in $28$ ways, giving $28^2 = 784.$

Thus, the correct answer is C.

Answer: D

Solution:

Expanding, $\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy) = \dfrac{3}{4}(x - y) + \dfrac{3}{4}(x + y)i.$ Both parts lie in $[-1, 1]$ iff $|x - y| \le \dfrac{4}{3}$ and $|x + y| \le \dfrac{4}{3}.$

Within the square $S$ (area $4$) these fail only in four corner triangles. Near $(1, 1),$ the line $x + y = \dfrac{4}{3}$ cuts off a right triangle with legs $\dfrac{2}{3},$ area $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}.$

The four corners remove $4 \cdot \dfrac{2}{9} = \dfrac{8}{9},$ leaving $4 - \dfrac{8}{9} = \dfrac{28}{9}.$ The probability is $\dfrac{28/9}{4} = \dfrac{7}{9}.$

Thus, the correct answer is D.

Answer: B

Solution:

On $[0, \pi],\ \cos^{-1}(\cos x) = x.$ Since $\sin^{-1}$ takes values in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ any solution requires $x \in [0, \tfrac{\pi}{2}],$ where the equation becomes $\sin 6x = \sin x.$

As $x$ goes from $0$ to $\tfrac{\pi}{2},$ $\sin 6x$ runs $0 \to 1 \to -1 \to 1 \to 0$ (peaks at $\tfrac{\pi}{12}, \tfrac{5\pi}{12},$ trough at $\tfrac{\pi}{4}$), while $\sin x$ increases from $0$ to $1.$

Besides $x = 0,$ the graphs cross once in each of $[\tfrac{\pi}{12}, \tfrac{\pi}{4}],$ $[\tfrac{\pi}{4}, \tfrac{5\pi}{12}],$ and $[\tfrac{5\pi}{12}, \tfrac{\pi}{2}],$ for $4$ solutions in all.

Thus, the correct answer is B.

Answer: E

Solution:

$G$ consists of four $5 \times 5$ blocks $G_1, \ldots, G_4,$ one in each quadrant. Any square of side $\ge 6$ uses exactly one vertex in each block, since two points in one block are less than $6$ apart while points in different blocks are at least $6$ apart.

Sliding each block inward by $(\pm 5, \pm 5)$ superimposes them on one $5 \times 5$ grid $G'$ (points with $|x|, |y| \le 2$). Each such square maps to either a single point of $G'$ or a square in $G'.$ So the count equals the number of points of $G'$ plus $4$ times the number of squares with vertices in $G'.$

A $5 \times 5$ grid has $25$ points and $50$ squares of all tilts, so the total is $25 + 4 \cdot 50 = 225.$

Thus, the correct answer is E.