Small Hint:

Convert both side lengths from feet to yards first.

Big Hint:

Then multiply the two yard measurements to get square yards.

Video solution:

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Written 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 $4$ yards by $3$ yards, the area is $$4\cdot3=12$$ square yards.

Thus, the correct answer is A .

Small Hint:

Draw segments from $O$ to every vertex of the octagon.

Big Hint:

The shaded region contains three full central triangles and half of another.

Video solution:

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Written 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 .

Small Hint:

Find how many minutes each person takes to travel one mile.

Big Hint:

Subtract Jill’s travel time from Jack’s travel time.

Video solution:

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Written 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 .

Small Hint:

Place the two boys at the two ends first.

Big Hint:

Then arrange the three girls in the middle seats.

Video solution:

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Written solution:

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.

Small Hint:

Adding $40$ creates a new lowest score.

Big Hint:

Check which statistic depends on the difference between the largest and smallest scores.

Video solution:

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Written solution:

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 .

Small Hint:

Drop the altitude from $B$ to $AC.$

Big Hint:

The altitude bisects $AC,$ giving a right triangle with hypotenuse $29.$

Video solution:

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Written solution:

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.

Small Hint:

It may be easier to count the complement: an odd product.

Big Hint:

The product is odd only if both drawn chips are odd.

Video solution:

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Written solution:

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.

Small Hint:

Use the triangle inequality to bound the third side.

Big Hint:

The perimeter can get arbitrarily close to twice $5+19,$ but not equal it.

Video solution:

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Written solution:

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.

Small Hint:

The daily sales are the first $20$ positive odd numbers.

Big Hint:

Pair the first and last terms, second and next-to-last terms, and so on.

Video solution:

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Written solution:

We want to find $$1+3 + \cdots + 39.$$ Pairing terms from the ends gives $$\begin{aligned} &(1+39)+(3+37)+\cdots\\ &\qquad +(19+21). \end{aligned}$$ There are $10$ pairs, and each pair sums to $40.$ Therefore, the sum is $$10\cdot40=400.$$

Thus, the correct answer is D .

Small Hint:

Choose the thousands digit first.

Big Hint:

After that, each later digit must avoid all previously chosen digits.

Video solution:

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Written 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 .

Small Hint:

Count all license plates allowed by the rules.

Big Hint:

Only one of those plates is exactly AMC8.

Video solution:

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Written solution:

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 $5\cdot21\cdot20\cdot10=21{,}000$ possible plates. Exactly one of these is AMC8, so the probability is $\dfrac{1}{21{,}000}.$

Thus, B is the correct answer.

Small Hint:

Group cube edges by direction.

Big Hint:

Each edge is parallel to three other edges, but pair-counting double counts.

Video solution:

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Written solution:

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.

Small Hint:

Find the sum the remaining nine numbers must have.

Big Hint:

So the two removed numbers must have a fixed sum.

Video solution:

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Written solution:

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.

Small Hint:

Write the four consecutive odd integers algebraically.

Big Hint:

Their sum is always a multiple of $8.$

Video solution:

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Written solution:

Let the four consecutive odd integers be $2k+1,$ $2k+3,$ $2k+5,$ and $2k+7.$ 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.

Small Hint:

First find how many students voted for at least one issue.

Big Hint:

Use inclusion-exclusion on the two yes-vote counts.

Video solution:

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Written 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 .

Small Hint:

Let $s$ and $n$ be the numbers of sixth and ninth graders.

Big Hint:

The paired sixth graders and paired ninth graders are the same number of pairs.

Video solution:

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Written solution:

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.

Small Hint:

Convert $20$ minutes and $12$ minutes to hours.

Big Hint:

Set the two distance expressions equal.

Video solution:

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Written solution:

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 $\dfrac{27}{3}=9$ miles.

Thus, D is the correct answer.

Small Hint:

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

Big Hint:

Apply that fact to the top row, bottom row, and then the middle column.

Video solution:

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Written solution:

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.

Small Hint:

Put the triangle inside a small rectangle aligned to the grid.

Big Hint:

Subtract the three corner right triangles from the rectangle.

Video solution:

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Written solution:

The total area of the grid is $6\cdot5=30.$ To find the area of the triangle, place it inside the $4$-by-$3$ rectangle shown below.

The rectangle has area $4\cdot3=12.$ The three corner triangles have areas $\dfrac12(4)(2)=4,$ $\dfrac12(1)(3)=\dfrac32,$ and $\dfrac12(3)(1)=\dfrac32.$ Therefore, $$\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5. \end{aligned}$$

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

Thus, the correct answer is A .

Small Hint:

Let the numbers of $\$1,$ $\$3,$ and $\$4$ pairs be variables.

Big Hint:

Subtract the pair-count equation from the cost equation.

Video solution:

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Written solution:

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.

Small Hint:

Find $JB$ and $BC$ from the two square areas.

Big Hint:

In the diagram, $BK$ is perpendicular to $BC.$

Video solution:

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Written solution:

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 $BC=4\sqrt2.$

At $B,$ the equilateral triangle, square, and equiangular hexagon contribute angles of $60^\circ,$ $90^\circ,$ and $120^\circ,$ respectively. Hence $$\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ. \end{aligned}$$ Therefore $$[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.$$