Solution:

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 .

Solution:

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.

Solution:

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.

Solution:

Since there are $2$ boys and two places to put them, there are $2! = 2$ ways to place them.

Since there are $3$ boys and three places to put them, there are $3! = 6$ ways to place them.

Therefore, the total number of places to put them is $6\cdot2=12.$

Thus, the correct answer is E.

Solution:

When considering all $12$ games, $40$ -- from the $12^{\text{th}}$ 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 .

Solution:

Let's begin by sketching the triangle:

Define $X$ to be the midpoint of $AC.$ Therefore, $AX=21.$

Then, since $AB=BC,$ we know that $AXB$ is a right triangle. By the Pythagorean Theorem, we know $AX^2+BX^2 = AB^2,$ so we get $$\begin{align*}BX &= \sqrt{AB^2-BX^2} \\&= \sqrt{29^2-21^2} \\&= \sqrt{400} = 20.\end{align*}$$

Then, as the area of a triangle is equal to $\dfrac12 bh$ formula for area, we can see that the area is $\dfrac{42\cdot20}2 = 420.$

Thus, the correct answer is B.

Solution:

First, let us try to find the probability that the product is odd. We know that the product will be odd if both boxes yield odd numbers. As there is a $\dfrac 23$ chance of a box being odd, there is a $$\left(\dfrac 23\right)^2 = \dfrac 49$$ chance of the product being odd.

The probability that a number is even is $1$ minus the probability that it is odd (as they are complements), so the odds of it being even is $$1 - \dfrac 49 = \dfrac 59.$$

Thus, the correct answer is E.

Solution:

Let the third side be $s.$ By the triangle inequality, we know $s < 5+19.$ Therefore, the perimeter of the triangle: $$s+5 +19 < 2(5+19) = 48.$$

As such, the largest triangle we can make under these constraints is a triangle with sides $5,19,23,$ and therefore a perimeter of $47,$ as required. Therefore, the smallest possible number greater than any perimeter is $48.$

Thus, the correct answer is D.

Solution:

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 .

Solution:

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 .

Solution:

First, we find how many license plates are possible.

For the first letter, we have $5$ choices. For the second letter, we have $21$ choices. For the third letter, we have $21-1=20$ choices since we must exlude the letter chosen for the second letter. For the final letter, we have $10$ choices. Overall, there are $5\cdot21\cdot20\cdot10 = 21,000$ choices.

The choice AMC8 is one possible choice of these $21,000$ possibilites, so the probability of it being our choice is $\dfrac{1}{21,000}.$

Thus, the correct answer is B.

Solution:

We have $12$ edges, and each edge has $3$ parallel lines to it. However, this would double count it, so we divide by $2.$ Therefore, there are $\dfrac{3\cdot12}{2} = 18$ set of parallel edges.

Thus, the correct answer is C.

Solution:

Currently, the sum is $66.$ If we remove $2$ elements, then there are $9$ elements remaining. Since their mean is $6,$ the sum of the remaining elements is $54.$ This means the sum of the two elements we take out are $66-54=12.$ There are $5$ two elements sets that work, namely: $$\begin{align*} &\{1,11\}\\ &\{2,10\}\\&\{3,9\}\\&\{4,8\}\\&\{5,7\} \end{align*}$$

Thus, the correct answer is D.

Solution:

Let $2k+1$ be the lowest of the odd integers in question. Then, the next three consecutive odd integers are $2k+3,2k+5,$ and $2k+7.$

It follows that their sum is $8k+16 = 8(k+2).$ This means the integer must a multiple of $8.$ The only answer choice that is not divisible by $8$ is $100.$

Thus, the correct answer is D.

Solution:

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.

Solution:

First, let the number of sixth-grader be $s$ and let the number of ninth-graders be $n.$ Since they are paired in a way that everyone has one buddy, we know $\dfrac{1}{3}n = \dfrac{2}{5}s .$ This means $5n = 6s.$ Using algebra, we get $$11n = 6(s+n)$$$$11s = 5(s+n),$$ so we get $$n = (s+n)\dfrac{6}{11}$$$$s = (s+n)\dfrac{5}{11}.$$ Now, we must determine how many people have a buddy divided by the total number of people. There are $\dfrac{1}{3}n + \dfrac{2}{5}s$ people who have buddies and $s+n$ total people. Therefore: $$\begin{align*}\dfrac{\dfrac{1}{3}n + \dfrac{2}{5}s}{s+n} &=\dfrac{\dfrac{1}{3}\left(\dfrac{6}{11}(s+n)\right)}{s+n} \\&+ \dfrac{\dfrac{2}{5}\left(\dfrac{5}{11}(s+n)\right)}{s+n} \\ &= \dfrac{2}{11} + \dfrac{2}{11} \\&= \dfrac{4}{11}. \end{align*}$$

Thus, the correct answer is B.

Solution:

Let the speed in rush hour be $s$ miles per hour. Since $20$ minutes is $\dfrac 13$ hours, Jeremy's father drives $s(\dfrac 13)$ miles during rush hour. When it's not rush hour, Jeremy's father drives at a speed of $s+18$ miles per hour. Since $12$ minutes is $\dfrac 15$ hours, he drives $(s+18)(\dfrac 15)$ miles when its not rush hour.

As the distance to school is the same, we know $$\begin{align*}s\dfrac{1}{3} &= (s+18) \dfrac {1}{5}\\ s\dfrac{5}{3} &= s+ 18\\ s &= 27.\end{align*}$$ Since the distance to school is $s(\dfrac 13 ),$ we can deduce that the distance to school is $(27)(\dfrac 13) = 9$ miles.

Thus, the correct answer is D.

Solution:

We can write an arbitrary 5-term arithmetic sequence as: $$a,a+b,a+2b,a+3b,a+4b.$$ In this case, this representation of sequences implies that $a+2b = \dfrac{(a)+(a+4b)}{2},$ so the third team is the average of the first and last terms. Therefore, the middle number on the top row is the average of the top two corners $\dfrac{1+25}{2} = 13,$ and the middle number on the bottom row is the average of the bottom two corners, which would be $\dfrac{17+81}{2} = 49.$ Finally, the middle entry is the average of the third entry in the first and last row, which is $\dfrac{13 + 49}{2} = 31.$

Thus, the correct answer is B.

Solution:

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.

Solution:

Let $a$ be the number of $\$1$ pairs, let $b$ be the number of $\$3$ pairs, and let $c$ be the number of $\$4$ pairs of socks Ralph bought.

We know $$a+b+c = 12$$$$a+3b+4c=24.$$ Therefore, we can subtract the two equations to get that $$2b+3c=12.$$ Since $b > 0,$ we know $2b > 0$ and therefore $12-3c > 0.$ Therefore, $c < 4.$

Also, we know $c > 0.$ Taking $\pmod 2$ of $2b+3c=12$ yields $$c \equiv 0 \pmod 2.$$ Since $c \equiv 0 \pmod 2$ and $0 < c < 4,$ we know $c=2.$ Substituting, we deduce $b=3.$ Finally, with $a+b+c = 12,$ we get $a+2+3 = 12,$ so $a=7.$

This means the number of $1 pairs is $7.$

Thus, the correct answer is D.

Solution:

Since $ABJI$ is a square, each side has a length equal to the square root of the area. This means $JB = \sqrt{18} = 3 \sqrt{2}.$ Since $JBK$ is equiangular, $\angle K = \angle J.$ This means $$BK = JB = 3\sqrt{2}.$$

Similarly, we get $$EF = \sqrt{32} = 4\sqrt{2} .$$

Since $ABCDEF$ is equiangular, we can get that $$BC=EF = 4\sqrt{2}.$$

Therefore, the area of $KBC$ is equal to $$\begin{align*}\dfrac{(BC)(BK)}{2} &=\dfrac{(4\sqrt{2})(3\sqrt{2})}{2} \\&= \dfrac{24}{2} \\&= 12.\end{align*}$$

Thus, the correct answer is C.

Solution:

Let $n$ be the number of students. Since we consider only $12$ days, and each day's row and column pair represents a unique factor pair of $n,$ we have $12$ factors of $n.$

Since $15$ students were in one row, $n$ is a multiple of $15.$

Since $6$ students were in one row on a different day, $n$ is a multiple of $6.$

Since $n$ is a multiple of $6$ and $15,$ it must be a multiple of $lcm(6,15) =30.$ This means $n$ is a multiple of $2\cdot3\cdot5.$ The number of factors is the product of the exponents plus $1,$ so this would have $8$ factors. However, multiplying it by $2$ yields $2^2\cdot3\cdot5=60.$ This yields $12$ total factors as required.

Thus, the correct answer is C.

Solution:

The sum of every slip combined is $35.$ Therefore, the average of the cups would be $7.$ This would make $$A=5$$$$B=6$$$$C=7$$$$D=8$$$$E=9.$$ Since $B$ has one $3,$ the sum of the remaining slips is $3,$ which can only be done with another $3.$

As such, what remains after filling $B$ is: $$\begin{align*}&2, 2, 2, 2.5, 2.5, 3,\\& 3, 3.5, 4, 4.5.\end{align*}$$ Let $E'$ be the sum of $E$ after putting the $2$ slip. This equals $9-2=7.$

Therefore, we need to fill $C,E'$ each to be $7$ and have $A$ have $5$ and $D$ having $8.$ Now, lets try fitting the $3.5$ slip into each of the cups.

When putting it in $A,$ we have $1.5$ left which is invalid. Similarly, we can't put it in $B$ as it is full. When putting it in $C,$ we have $3.5$ left, which cannot work as no other values sum to $3.5.$ Similar to $C,$ putting it in $E$ would have $3.5,$ which we just showed to be invalid. Finally, we can make cup $D$ work by making the cups be $$\begin{align*}&\{2.5,2.5\},\{3,3\},\{3,4\},\\& \{3.5,4.5\},\{2,2,2,3\}. \end{align*}$$

Thus, the correct answer is D.

Solution:

Since there are $3$ other teams in the division and $4$ teams not in the division, a team plays $3N+4M=76$ teams. Also notice that $$76=3N+4M > 10M,$$ since $N > 2M.$ This implies that $76 > 10M,$ and as such, $7 \geq M.$

Under the given conditions, we know $$7 \geq M > 4.$$ Also, taking $$3N+4N=76$$ under modulus $3$ yields $$M \equiv 1 \pmod 3.$$ The only possible $M$ in the range is $7,$ suggesting that he plays $4\cdot7 = 28$ games against non-divisional opponents.

As such, he plays $76-28 = 48$ games against divisional opponents.

Thus, the correct answer is B.

Solution:

The desired square can be drawn like this. It consists a smaller square of size $3 \times 3$ in it, with area $3\cdot 3=9,$ as well as $4$ triangles. As each square has base $3$ and height $1,$ the combined area of the triangles is $4(\dfrac{3\cdot 1}{2}) = 6.$

This makes the total area of this square equal to $15.$

Thus, the correct answer is C.