Since one side is $12$ feet, it would be $\dfrac{12}{3} = 4$ yards.

Since another side is $9$ feet, it would be $\dfrac{9}{3} = 3$ yards.

Since the dimensions are $4 \text{ yards} \times 3 \text{ yards},$ the area is equal to $$4\cdot3 = 12.$$

Thus, the correct answer is A .

First notice that there are $8$ equally sized triangles that can be created with $O$ and any two consecutive points. Therefore, they each take up $\dfrac {1}{8}$ of the total area of the octagon.

The shaded area has three complete triangles and half of the triangle $ABO.$ Therefore, the shaded area is $$\dfrac{3.5}{8} = \dfrac{7}{16}$$ of the total area of the octagon.

Thus, the correct answer is D .

Jack travels at a rate of $4$ miles per $60$ minutes. Therefore, it takes him $\dfrac{60}{4} = 15$ minutes to get to the pool.

Jill travels at a rate of $10$ miles per $60$ minutes. Therefore it takes her $\dfrac{60}{10} = 6$ minutes to get to the pool.

Therefore, the difference in their times is $15-6 = 9$ minutes.

Thus, the correct answer is D .

There are $2! = 2$ ways to place the two boys at the two ends. There are $3! = 6$ ways to arrange the three girls in the middle seats.

Thus the total number of arrangements is $2\cdot6=12$.

Thus, E is the correct answer.

When considering all $12$ games, $40$ -- from the $12$th game -- will be the lowest score. Therefore, compared to the range of just the first $11$ games, the range of all $12$ games would increase from $73-42 = 31$ to $73-40=33.$

Thus, the correct answer is A .

Drop the altitude from $B$ to $AC$, meeting $AC$ at $X$. Since $AB=BC$, point $X$ is the midpoint of $AC$, so $AX=21$.

In right triangle $ABX$, the altitude is $$\begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20. \end{aligned}$$

The area of $\triangle ABC$ is $\dfrac12\cdot42\cdot20=420$.

Thus, B is the correct answer.

The product is odd only when both chips are odd. Each box has two odd chips, $1$ and $3$, out of three chips, so the probability of an odd product is $(\dfrac23)^2=\dfrac49$.

The probability of an even product is the complement, $1-\dfrac49=\dfrac59$.

Thus, E is the correct answer.

Let the third side length be $s$. The triangle inequality gives $s<5+19=24$, so the perimeter satisfies $$5+19+s<48.$$

Perimeters can be made arbitrarily close to $48$ from below, so the smallest whole number larger than the perimeter of any such triangle is $48$.

Thus, D is the correct answer.

We want to find $$1+3 + \cdots + 39.$$ This sum can be rewritten as $$\begin{align*}&(1+39) + (3+ 37) \\&+\cdots +(19+21),\end{align*}$$ which we can see has $10$ terms. Further notice that each term is equal to $40.$ Therefore, the sum is $$10\cdot40=400.$$

Thus, the correct answer is D .

First, there are $9$ digits to choose for the thousands digit since $0$ can't be chosen.

Then, after that, there are $9$ ways to choose the hundreds digit, $8$ ways to choose the tens digit, and $7$ ways to choose the ones digit. Therefore, we get $9\cdot9\cdot8\cdot7 = 4536$ ways to choose such an integer.

Thus, the correct answer is B .

There are $5$ choices for the first symbol, $21$ choices for the second, $20$ choices for the third because it must be a different non-vowel, and $10$ choices for the final digit.

Thus there are $5\cdot21\cdot20\cdot10=21000$ possible plates. Exactly one of these is AMC8, so the probability is $\dfrac{1}{21000}$.

Thus, B is the correct answer.

A cube has $12$ edges. For any edge, there are $3$ other edges parallel to it.

This counts each pair twice, once from each edge in the pair, so the number of pairs of parallel edges is $\dfrac{12\cdot3}{2}=18$.

Thus, C is the correct answer.

The original set has sum $1+2+\cdots+11=66$. After removing two numbers, $9$ numbers remain and must have mean $6$, so their sum must be $9\cdot6=54$.

Therefore the two removed numbers must have sum $66-54=12$. The possible two-element subsets are $\{1,11\},$ $\{2,10\},$ $\{3,9\},$ $\{4,8\},$ $\{5,7\}$, so there are $5$ choices.

Thus, D is the correct answer.

Let the four consecutive odd integers be $2k+1,2k+3,2k+5,2k+7$. Their sum is $$8k+16=8(k+2).$$

So any such sum must be a multiple of $8$. The only answer choice that is not divisible by $8$ is $100$.

Thus, D is the correct answer.

Since $29$ students voted against both, we know that $198-29 = 169$ people voted for at least one.

As we know that $149$ students voted for the first issue, and $119$ students voted for the second issue, and $169$ students that voted for at least one issue, we conclude that the number of students that voted for both is $$149+119-169 = 99.$$

Thus, the correct answer is D .

Let there be $s$ sixth graders and $n$ ninth graders. The number of paired ninth graders equals the number of paired sixth graders, so $$\dfrac13 n=\dfrac25 s.$$

This gives $5n=6s$, so $n:s=6:5$. The total number of students is therefore proportional to $11$ parts.

The paired ninth graders make up $\dfrac13\cdot\dfrac6{11}=\dfrac2{11}$ of all students, and the paired sixth graders make up $\dfrac25\cdot\dfrac5{11}=\dfrac2{11}$ of all students. Altogether, $\dfrac4{11}$ of the students have a buddy.

Thus, B is the correct answer.

Let the rush-hour speed be $s$ miles per hour. The $20$-minute rush-hour trip takes $\dfrac13$ hour, so the distance is $\dfrac{s}{3}$.

Without traffic, the speed is $s+18$ miles per hour and the trip takes $12$ minutes, or $\dfrac15$ hour. The same distance is $\dfrac{s+18}{5}$.

Set the distances equal: $$\dfrac{s}{3}=\dfrac{s+18}{5}.$$ Then $5s=3s+54$, so $s=27$. The distance is $27/3=9$ miles.

Thus, D is the correct answer.

In any five-term arithmetic sequence, the middle term is the average of the first and last terms.

The middle entry of the top row is $\dfrac{1+25}{2}=13$, and the middle entry of the bottom row is $\dfrac{17+81}{2}=49$.

Now apply the same fact to the middle column: $X=\dfrac{13+49}{2}=31$.

Thus, B is the correct answer.

The total area of the grid is $6\cdot5=30.$ In order to find the fraction of this grid that the triangle covers, we must now find the area of the triangle. To do this, we will use the following diagram:

Thus, the area of the triangle $A(\triangle ABC)$ is equal to: $$\begin{align*} &A(PQRB)-A(\triangle PAB)\\&-A(\triangle BCR)-A(\triangle CAQ)\\ &=(3)(4)-\dfrac12 (4)(2)-2 \left(\dfrac32\right)\\ &=12 - 4 - 3\\ &=5. \end{align*}$$

Therefore, the fraction of the area is $\dfrac{5}{30} = \dfrac{1}{6} .$

Thus, the correct answer is A .

Let $a$, $b$, and $c$ be the numbers of $\$1$, $\$3$, and $\$4$ pairs, respectively. Then $$a+b+c=12,$$ $$a+3b+4c=24.$$

Subtracting gives $2b+3c=12$. Since at least one pair of each type was bought, $b>0$ and $c>0$. Also $3c<12$, so $c<4$. Modulo $2$, the equation gives $c$ even, so $c=2$.

Then $2b+6=12$, so $b=3$, and $a=12-3-2=7$.

Thus, D is the correct answer.

The square with area $18$ has side length $JB=\sqrt{18}=3\sqrt2$. Since $\triangle JBK$ is equilateral, $BK=3\sqrt2$.

The square with area $32$ has side length $FE=\sqrt{32}=4\sqrt2$. Since $FE=BC$, we have $BC=4\sqrt2$.

In the equiangular hexagon configuration, $BK\perp BC$. Therefore $$[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.$$

Thus, C is the correct answer.

The possible numbers of students per row are exactly the positive divisors of the total number of students. Since June 1 through June 12 give different arrangements and June 13 gives no new one, the total number of students must have exactly $12$ positive divisors.

The number must be divisible by both $15$ and $6$, hence by $\operatorname{lcm}(15,6)=30=2\cdot3\cdot5$. This number has only $8$ divisors.

The smallest multiple of $30$ with $12$ divisors is $60=2^2\cdot3\cdot5$, which has $(2+1)(1+1)(1+1)=12$ divisors.

Thus, C is the correct answer.

The sum of all the slips is $35$, so the five consecutive integer cup sums must average $7$. Therefore cups $A,B,C,D,E$ must have sums $5,6,7,8,9$, respectively.