Solution:

Since $\dfrac{128}{12} = 10\dfrac{2}{3},$ ten cans hold only $120$ ounces, which is not enough.

Therefore $11$ cans are needed.

Thus, the correct answer is E.

Solution:

To get a multiple of $5$ cents, the number of pennies must be a multiple of $5.$ With only four coins, that means using no pennies, but then the four coins are each worth at least $5$ cents, for a total of at least $20$ cents.

So $15$ cents cannot be made. The others can: $25 = 10 + 3(5),$ $35 = 3(10) + 5,$ $45 = 25 + 10 + 2(5),$ and $55 = 25 + 3(10).$

Thus, the correct answer is A.

Solution:

Working outward, $$1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}} = 1 + \cfrac{1}{1 + \dfrac{1}{2}} = 1 + \dfrac{1}{\frac{3}{2}} = 1 + \dfrac{2}{3} = \dfrac{5}{3}.$$

Thus, the correct answer is C.

Solution:

The swim takes $\dfrac{1/4}{2} = \dfrac18$ hour and the run takes $\dfrac{3}{6} = \dfrac12$ hour. This leaves $$2 - \dfrac18 - \dfrac12 = \dfrac{11}{8}$$ hours for the bicycle ride.

His average speed must be $$\dfrac{15}{11/8} = \dfrac{120}{11}$$ miles per hour.

Thus, the correct answer is A.

Solution:

The square of the nine-digit repunit is the palindrome $$111{,}111{,}111^2 = 12{,}345{,}678{,}987{,}654{,}321.$$

Its digits are $1, 2, \ldots, 9, 8, \ldots, 1,$ so the sum is $$2(1 + 2 + \cdots + 8) + 9 = 2 \cdot 36 + 9 = 81.$$

Thus, the correct answer is E.

Solution:

The inscribed circle rests on the diameter and is tangent to the arc, so the semicircle has radius $4.$ Its area is $$\dfrac12 \pi (4)^2 = 8\pi.$$

The circle's area is $\pi(2)^2 = 4\pi,$ so the shaded area is $8\pi - 4\pi = 4\pi.$

The shaded fraction is $\dfrac{4\pi}{8\pi} = \dfrac12.$

Thus, the correct answer is A.

Solution:

Let whole milk be $x\%$ fat. Since $2$ is $40\%$ less than $x,$ we have $$0.6x = 2,$$ so $$x = \dfrac{2}{0.6} = \dfrac{10}{3}.$$

Thus, the correct answer is C.

Solution:

The senior ticket costs $\$6,$ which is $\dfrac34$ of the full price, so a full ticket costs $\dfrac43 \cdot 6 = \$8,$ and a child ticket costs $\dfrac12 \cdot 8 = \$4.$

The total is $$2(6 + 8 + 4) = \$36.$$

Thus, the correct answer is B.

Solution:

Let the common ratio be $r.$ Then $a r^2 = 2009 = 7^2 \cdot 41.$

Since $r$ must be an integer greater than $1,$ the only possibility is $r = 7,$ giving $a = 41$ and the sequence $41, 287, 2009.$

Thus, the correct answer is B.

Solution:

For the altitude from the right angle to the hypotenuse, $$BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$$ so $BD = 2\sqrt{3}.$

The hypotenuse is $AC = 3 + 4 = 7,$ so the area is $$\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.$$

Thus, the correct answer is B.

Solution:

Let the cube have side length $x.$ The new solid has volume $$x(x+1)(x-1) = x^3 - x.$$

Setting this equal to $x^3 - 5$ gives $x^3 - x = x^3 - 5,$ so $x = 5.$

The cube's volume is $5^3 = 125.$

Thus, the correct answer is D.

Solution:

In $\triangle BCD,$ the triangle inequality gives $5 + BD \gt 17,$ so $BD \gt 12.$

In $\triangle ABD,$ it gives $5 + 9 \gt BD,$ so $BD \lt 14.$

The only integer with $12 \lt BD \lt 14$ is $BD = 13.$

Thus, the correct answer is C.

Solution:

Since $12 = 2^2 \cdot 3,$ $$12^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = P^{2n} Q^m.$$

Thus, the correct answer is E.

Solution:

Let each rectangle have shorter side $x$ and longer side $y.$ The outer square has side length $y + x$ and the inner square has side length $y - x.$

Since the area ratio is $4,$ the side ratio is $2,$ so $$y + x = 2(y - x),$$ which gives $y = 3x.$

The ratio of longer to shorter side is $\dfrac{y}{x} = 3.$

Thus, the correct answer is A.

Solution:

Going from $F_{n-1}$ to $F_n,$ the new outside square carries $4(n-1)$ diamonds. Starting from the single diamond of $F_1,$ $$F_n = 1 + 4\big(1 + 2 + \cdots + (n-1)\big) = 1 + 4 \cdot \dfrac{(n-1)n}{2} = 1 + 2n(n-1).$$

Therefore $$F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761.$$

Thus, the correct answer is E.

Solution:

Since $a - d = (a-b) + (b-c) + (c-d) = \pm 2 \pm 3 \pm 4,$ the possible absolute values are $$2+3+4 = 9,\quad 2+3-4 = 1,\quad 2-3+4 = 3,\quad -2+3+4 = 5.$$

Their sum is $9 + 1 + 3 + 5 = 18.$

Thus, the correct answer is D.

Solution:

The diagonal is $DB = \sqrt{4^2 + 3^2} = 5.$

Right triangles $EBA,$ $DBC,$ and $BFC$ are all similar to $\triangle DBA.$ From $\triangle EBA,$ $$\dfrac{EB}{AB} = \dfrac{DB}{BC} \implies \dfrac{EB}{4} = \dfrac{5}{3} \implies EB = \dfrac{20}{3}.$$

From $\triangle BFC,$ $$\dfrac{BF}{BC} = \dfrac{DB}{AB} \implies \dfrac{BF}{3} = \dfrac{5}{4} \implies BF = \dfrac{15}{4}.$$

Therefore $$EF = EB + BF = \dfrac{20}{3} + \dfrac{15}{4} = \dfrac{125}{12}.$$

Thus, the correct answer is C.

Solution:

Take $100$ children: $60$ play soccer, and $40\%$ of them, or $24,$ also swim. So $60 - 24 = 36$ soccer players do not swim.

There are $30$ swimmers and $70$ non-swimmers, so the fraction of non-swimmers who play soccer is $$\dfrac{36}{70} \approx 0.514 \approx 51\%.$$

Thus, the correct answer is D.

Solution:

The circumferences are $200\pi$ and $2\pi r,$ so the initial point of tangency returns after $$\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}$$ rolls.

For this to be an integer greater than $1,$ $r$ must be a divisor of $100$ less than $100:$ namely $1, 2, 4, 5, 10, 20, 25,$ and $50.$ That is $8$ values.

Thus, the correct answer is B.

Solution:

Let Lauren's rate be $r$ km/min. Then $r + 3r = 1,$ so $r = \dfrac14.$

In the first $5$ minutes the gap shrinks by $5$ km, leaving $15$ km. Lauren covers this alone at $\dfrac14$ km/min, taking $$\dfrac{15}{1/4} = 60$$ minutes.

The total time is $5 + 60 = 65$ minutes.

Thus, the correct answer is D.

Solution:

Let each small circle have radius $1.$ Their centers form a square of side $2,$ whose diagonal is $2\sqrt2.$

The large circle's diameter is $2 + 2\sqrt2,$ so its radius is $1 + \sqrt2.$

The desired ratio is $$\dfrac{4 \cdot \pi (1)^2}{\pi (1 + \sqrt2)^2} = \dfrac{4}{3 + 2\sqrt2} = 4(3 - 2\sqrt2).$$

Thus, the correct answer is C.

Solution:

Randomly attaching the tiles and then rolling is equivalent to choosing two of the twelve numbers at random and adding them.

Suppose the first top face shows $N.$ For a sum of $7,$ the second must be $7 - N,$ and there are exactly $2$ tiles equal to $7 - N$ among the remaining $11.$

So the probability is $\dfrac{2}{11}.$

Thus, the correct answer is D.

Solution:

Since $[AED] = [BEC],$ adding $[CED]$ to both gives $[ACD] = [BCD].$ These share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$

Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac34,$ so $\dfrac{AE}{EC} = \dfrac34.$

With $AE + EC = AC = 14,$ we get $AE = \dfrac{3}{7} \cdot 14 = 6.$

Thus, the correct answer is E.

Solution:

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the $6$ faces gives $\binom{4}{3} = 4$ triples, so $6 \cdot 4 = 24$ triples lie on a face out of $\binom{8}{3} = 56$ total.

The probability of hitting the interior is $$1 - \dfrac{24}{56} = \dfrac{4}{7}.$$

Thus, the correct answer is C.

Solution:

Write $$I_k = 10^{k+2} + 64 = 2^{k+2} \cdot 5^{k+2} + 2^6.$$

If $k \lt 4,$ the first term has fewer than $6$ factors of $2,$ so $N(k) = k + 2 \lt 6.$

If $k \gt 4,$ the first term has at least $7$ factors of $2$ while the second has exactly $6,$ so their sum has exactly $6:$ $N(k) = 6.$

If $k = 4,$ then $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1 = (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big) = 26 \cdot 601,$ it contributes exactly one more factor of $2.$ Thus $N(4) = 7.$

The maximum value is $N(4) = 7.$

Thus, the correct answer is B.