Solution:

If Jane buys $b$ bagels, she buys $5-b$ muffins, for a total of $$50(5-b)+75b = 250+25b$$ cents. This is a whole number of dollars when $250+25b$ is a multiple of $100,$ that is, when $25b\equiv50\pmod{100},$ or $b\equiv2\pmod4.$

The only value with $0\le b\le5$ is $b=2.$

Thus, the correct answer is B.

Solution:

The least common denominator of the small fractions is $12,$ so multiply top and bottom by $12:$ $$\dfrac{\frac13-\frac14}{\frac12-\frac13}=\dfrac{4-3}{6-4}=\dfrac12.$$

Thus, the correct answer is C.

Solution:

The lost $3$ cans would have painted $30-25=5$ rooms, so each room takes $\dfrac35$ of a can.

For $25$ rooms she used $\dfrac35\cdot25=15$ cans.

Thus, the correct answer is C.

Solution:

The two parallel sides differ by $25-15=10,$ split evenly between the two triangles, so each isosceles right triangle has legs of length $5$ and area $\dfrac12\cdot5^2=\dfrac{25}{2}.$

Together the beds cover $25$ square meters. The rectangle has length $25$ and width $5,$ so area $125.$ The fraction is $\dfrac{25}{125}=\dfrac15.$

Thus, the correct answer is C.

Solution:

Twenty percent less than $60$ is $\dfrac45\cdot60=48.$

If $n$ is the unknown number, one-third more than $n$ is $\dfrac43 n,$ so $$\dfrac43 n=48$$$$n=36.$$

Thus, the correct answer is D.

Solution:

Since $128=2^7,$ every age is a power of $2.$ Writing the twins' common age as $t$ and Kiana's as $k,$ we need $t^2k=128$ with $k\lt t.$

Taking $t=8$ gives $k=2,$ which works. The sum is $8+8+2=18.$

Thus, the correct answer is D.

Solution:

The three operations can be ordered in $3!=6$ ways, but performing the addition first or last leaves the two multiplications interchangeable, so at most four values arise.

Indeed $$(2\times3)+(4\times5)=26,\quad (2\times3+4)\times5=50,$$$$2\times(3+4\times5)=46,\quad 2\times(3+4)\times5=70$$ are four distinct values.

Thus, the correct answer is C.

Solution:

Let $p$ be the starting price. After March the price is $$(1.2)(0.8)(1.25)p=1.2p.$$ The April drop must return it to $p,$ so it removes $0.2p,$ a fraction $$x=100\cdot\dfrac{0.2p}{1.2p}=\dfrac{100}{6}\approx16.7.$$

To the nearest integer, $x=17.$

Thus, the correct answer is B.

Solution:

Since $\triangle ABC$ is isosceles with $AB=BC,$ we have $\angle A=\angle C.$ With $\angle A=\dfrac52\angle B,$ the angle sum gives $$\dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ,$$ so $\angle B=30^\circ$ and $\angle ACB=75^\circ.$

By vertical angles $\angle DCE=75^\circ.$ Since $CD=CE,$ triangle $CDE$ is isosceles, so $$2\angle D+75^\circ=180^\circ,$$ giving $\angle D=52.5^\circ.$

Thus, the correct answer is A.

Solution:

The standing stump has height $x,$ and the snapped piece of length $5-x$ is the hypotenuse of a right triangle with legs $x$ and $1.$ By the Pythagorean Theorem, $$x^2+1^2=(5-x)^2=x^2-10x+25,$$ so $10x=24$ and $x=2.4.$

Thus, the correct answer is E.

Solution:

A $7$-digit palindrome has the form with the middle digit used once and the outer three digits each used twice. Only $5$ appears an odd number of times, so $5$ must be the middle digit.

The remaining digits $2,3,5$ fill the first three positions in some order and mirror to the last three. There are $3!=6$ such orderings.

Thus, the correct answer is A.

Solution:

A positive-area triangle uses two points on one line as its base and one point on the other line as its apex. The height is always the fixed distance between the lines, so the area depends only on the base length.

Bases on the first line can be $1, 2,$ or $3;$ a base on the second line is $1.$ So the distinct base lengths are $1, 2, 3,$ giving three possible areas.

Thus, the correct answer is A.

Solution:

The pentagon has perimeter $3+4+6+3+7=23.$ One full roll advances the contact point by $23,$ and $2009=23\cdot87+8.$

After $87$ rolls, vertex $A$ sits at $x=23\cdot87=2001$ and $B$ at $2004.$ Rolling further, $C$ touches at $2004+4=2008$ and $D$ at $2008+6=2014.$

Since $2009$ lies between $2008$ and $2014,$ side $\overline{CD}$ touches that point.

Thus, the correct answer is C.

Solution:

Each quart adds $\dfrac14$ quart of millet, and the birds leave $\dfrac34$ of the standing millet. On day $n$ the millet present is $$\dfrac14\left(1+\dfrac34+\cdots+\Big(\dfrac34\Big)^{n-1}\right)=1-\Big(\dfrac34\Big)^{n}.$$

The other seeds always total $\dfrac34$ quart. Millet exceeds half when $1-\Big(\dfrac34\Big)^n\gt\dfrac34,$ i.e. $\Big(\dfrac34\Big)^n\lt\dfrac14.$

Since $\Big(\dfrac34\Big)^4=\dfrac{81}{256}\gt\dfrac14$ but $\Big(\dfrac34\Big)^5=\dfrac{243}{1024}\lt\dfrac14,$ this first happens on day $5,$ which is Friday.

Thus, the correct answer is D.

Solution:

Let $x$ be the bucket's weight and $y$ the weight of a full load of water. Then $$x+\dfrac23 y=a,\qquad x+\dfrac12 y=b.$$

Subtracting gives $\dfrac16 y=a-b,$ so $y=6a-6b,$ and $x=b-\dfrac12 y=4b-3a.$ The full bucket weighs $$x+y=(4b-3a)+(6a-6b)=3a-2b.$$

Thus, the correct answer is E.

Solution:

Let the radius be $r.$ By symmetry $BO$ bisects the $60^\circ$ angle $ABC,$ so $\angle OBC=30^\circ.$ Since $OC\perp BC,$ triangle $BCO$ is a $30$-$60$-$90$ triangle with hypotenuse $BO=2\,OC=2r.$

Then $BD=BO-OD=2r-r=r,$ so $\dfrac{BD}{BO}=\dfrac{r}{2r}=\dfrac12.$

Thus, the correct answer is B.

Solution:

The five unit squares have total area $5,$ so each region must have area $\dfrac52.$

The region to the lower right of the line is a right triangle with legs $3-a$ and $3,$ minus the one unit square it does not cover. Setting its area to $\dfrac52$ gives $$\dfrac{3(3-a)}{2}-1=\dfrac52,$$ so $3(3-a)=7$ and $a=\dfrac23.$

Thus, the correct answer is C.

Solution:

By the Pythagorean Theorem, $AC=\sqrt{8^2+6^2}=10,$ so $AM=5.$ Right triangles $AME$ and $ABC$ share angle $A,$ so they are similar with $$\dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68,$$ giving $ME=\dfrac{15}{4}.$

Then $\text{area}(\triangle AME)=\dfrac12\cdot AM\cdot ME=\dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}.$

Thus, the correct answer is D.

Solution:

Among the hours $1$ through $12,$ exactly $1, 10, 11, 12$ contain a $1,$ so the hour is correct $\dfrac{8}{12}=\dfrac23$ of the time.

A minute is displayed wrong when its tens digit is $1$ (minutes $10$–$19$) or its units digit is $1$ ($01,11,\dots,51$), which is $15$ of the $60$ minutes. So the minute is correct $\dfrac{45}{60}=\dfrac34$ of the time.

The clock is correct $\dfrac23\cdot\dfrac34=\dfrac12$ of the day.

Thus, the correct answer is A.

Solution:

By the Pythagorean Theorem, $AC=\sqrt{1^2+2^2}=\sqrt5.$ The Angle Bisector Theorem gives $$\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5},$$ so $DC=\sqrt5\,BD.$

Since $BD+DC=2,$ we have $BD(1+\sqrt5)=2,$ so $$BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.$$

Thus, the correct answer is B.

Solution:

Any four consecutive powers of $3$ sum to a multiple of $3^0+3^1+3^2+3^3=40,$ which is divisible by $8.$

The terms from $3^2$ to $3^{2009}$ split into such blocks and contribute remainder $0.$ What remains is $3^0+3^1=4.$

Thus, the correct answer is D.

Solution:

Set the top face as a $2\times2$ square. The cut from $M$ toward the far corner creates the top triangle $A$ with legs $1$ and $2,$ so area $1$ and hypotenuse $\sqrt5.$

Triangle $B$ is similar to $A$ but with hypotenuse $2,$ so its area is $\left(\dfrac{2}{\sqrt5}\right)^2\cdot1=\dfrac45.$ Since the cake has height $2,$ the volume is $c=\dfrac45\cdot2=\dfrac85.$

The icing on this piece is its top ($\dfrac45$) plus the full cube side face it borders ($2\times2=4$), so $s=\dfrac45+4=\dfrac{24}{5}.$ Therefore $c+s=\dfrac85+\dfrac{24}{5}=\dfrac{32}{5}.$

Thus, the correct answer is B.

Solution:

The picture spans $\dfrac18$ lap on each side of the start. After $600$ seconds Rachel is $30$ seconds short of the line; running $\dfrac14$ lap in $22.5$ seconds, she is in view between $30-11.25=18.75$ and $30+11.25=41.25$ seconds of the $10$th minute.

After $600$ seconds Robert is $40$ seconds from the line; running $\dfrac14$ lap in $20$ seconds, he is in view between $30$ and $50$ seconds.

Both appear between $30$ and $41.25$ seconds, a window of length $11.25$ out of $60,$ so the probability is $\dfrac{11.25}{60}=\dfrac{3}{16}.$

Thus, the correct answer is C.

Solution:

Adding a mirror image completes the arch into a symmetric closed loop of $18$ trapezoids. Their inner edges form a regular $18$-gon, each interior angle of which is $$\dfrac{(18-2)\cdot180^\circ}{18}=160^\circ.$$

At each inner vertex, two of the trapezoids' larger angles $x$ meet the $160^\circ$ angle around a full turn: $x+x+160^\circ=360^\circ,$ so $x=100.$

Thus, the correct answer is A.

Solution:

Each face's stripe has $2$ orientations, giving $2^6=64$ equally likely configurations.

An encircling stripe runs around one of the $3$ pairs of opposite faces. For a given band, the $4$ faces it crosses must each be oriented to continue it, a probability of $\left(\dfrac12\right)^4=\dfrac{1}{16}.$

The three bands are mutually exclusive, so the probability is $3\cdot\dfrac{1}{16}=\dfrac{3}{16}.$

Thus, the correct answer is B.