Solution:

During the first $5$ weeks, Theresa works for a total of $$8 + 11 + 7 + 12 + 10 = 48$$ hours.

She, however, promised to work for $$10 \cdot 6 = 60$$ hours.

This means that she has to work $$60 - 48 = 12$$ hours during the final work to earn the tickets.

Thus, D is the correct answer.

Solution:

There are $250$ students who preferred spaghetti and $100$ that preferred manicotti.

The ratio is therefore $$\dfrac{250}{100} = \dfrac{5}{2}.$$

Thus, E is the correct answer.

Solution:

We can prime factorize $250$ to get $$250 = 2 \cdot 5^3.$$ From this, we can see that $250$ only has two prime factors: $2$ and $5.$ The sum of these is $7.$

Thus, C is the correct answer.

Solution:

Georgie has $6$ options for which window he enters through. He, however, only has $5$ options for the exit since it must be different from the entrance.

The total number of paths is therefore $6 \cdot 5 = 30.$

Thus, D is the correct answer.

Solution:

The total amount of money that Chandler got from his birthday is $$50 + 35 + 15 = 100$$ dollars. Therefore, he only needs to raise $500 - 100 = 400$ more dollars.

This can be achieved in $400 \div 16 = 25$ weeks through his paper route.

Thus, B is the correct answer

Solution:

The difference in the costs is $41 - 7 = 34$ cents. The approximate percent decrease is therefore $$100 \cdot \dfrac{35}{41} \approx 100 \cdot \dfrac{32}{40}$$$$=100 \cdot \dfrac{4}{5}$$$$= 80 \%.$$

Thus, E is the correct answer.

Solution:

Initially, the total age of everyone in the room is $5 \cdot 30 = 150$ years.

After the person leaves, the total age is $150 - 18 = 132.$ With $4$ people remaining, the average age becomes $\dfrac{132}{4} = 33.$

Thus, D is the correct answer.

Solution:

We know that $$EC = DC - DE$$$$= 6 - 3$$$$= 3.$$ We also know that $$BE = AD = 3.$$ Therefore, the area of $\triangle BEC$ is $$\dfrac{1}{2} \cdot 3 \cdot 3 = \dfrac{9}{2}.$$

Thus, B is the correct answer.

Solution:

Consider the last element in the second row. This has to be a $1$ since $4$ cannot be in that column.

Then, the number in the top right square must be $3$ since it cannot be $2.$

This forces the $2$ to be in the bottom right square.

Thus, B is the correct answer.

Solution:

First, we find that $$\boxed{11} = 1 + 11 = 12$$ since $11$ is prime. Then we need to find $\boxed{12}.$ The factors of $12$ are $$1, 2, 3, 4, 6, 12.$$ Adding these yields $\boxed{12} = 28.$

Thus, D is the correct answer.

Solution:

Note that only Tile $III$ has the number of $0.$ This forces it to be either $C$ or $D.$ This tile also is the only one with $5.$ This makes sure that Tile $III$ is $D.$

The only tile that can match with $1$ on Tile $III$ is Tile $IV.$ Therefore, Tile $IV$ is $C.$

Similarly, we get that Tile $II$ is $A$ and Tile $I$ is $B.$

Thus, the correct answer is D

Solution:

Note that we can split the hexagon into $6$ congruent equilateral triangles as follows.

Since each of them share an edge with an exterior triangle, all the triangles are congruent. Therefore, the ratio of areas is $1:1.$

Thus, A is the correct answer.

Solution:

Let $x$ be the number of elements in each $A$ and $B.$ Also let $y$ be the number of elements in their intersection.

The conditions give us that $$2x - y = 2007$$ and $$y = 1001.$$ Plugging $y$ into the first equation yields $$\begin{align*} 2x - 1001 &= 2007 \\ 2x &= 3008 \\ x &= 1504. \end{align*}$$

Thus, C is the correct answer.

Solution:

Construct $\overline{BD}$ as the altitude from $B$ to $\overline{AC}.$

Then $$60 = \dfrac{1}{2} \cdot BD \cdot 24,$$ which gives us that $BD = 5.$

From this, we apply the Pythagorean Theorem on $\triangle ABD:$ $$AB^2 = 5^2 + 12^2 = 169 = 13^2.$$ This gives us that $AB = 13.$

Thus, C is the correct answer.

Solution:

We know that $b < c$ and $0 < a.$ Adding these two inequalities together yields $$b \lt c + a.$$ This shows that A is impossible, and therefore the right answer.

To ensure that this is correct, we can show that the other options are possible.

B and C: $a = 1,$ $b = 2,$ and $c = 4$

D: $a = \dfrac{1}{3},$ $b = \dfrac{1}{2},$ and $c = 1$

E: $a = \dfrac{1}{2},$ $b = 1,$ and $c = 2$

Thus, A is the correct answer.

Solution:

The cirumferences of circles with radii $1$ through $5$ are $$2\pi, 4\pi, 6\pi, 8\pi, 10\pi.$$ Their respective areas are $$\pi, 4\pi, 9\pi, 16\pi, 25\pi.$$ The only graph that shows these points is A, making it the correct answer.

This is the only graph showing a quadratic function.

Thus, A is the correct answer.

Solution:

The amount of yellow tint in the original mixture is $.3 \times 30 = 9$ liters. Adding $5$ liters results in a total of $14$ liters of yellow tint.

The new mixture has a total of $35$ liters, so the percent of yellow tint is $$100 \cdot \dfrac{14}{35} = 100 \cdot \dfrac{2}{5} = 40 \%.$$

Thus, C is the correct answer.

Solution:

We only care about the last $4$ digits, so we can calculate $303 \cdot 505$ to find them (the thousands digit is $0$ for both numbers).

$$\begin{array}{r} &\cdots 303 \\ \times \hspace{-4mm} &\cdots 505 \\ \hline &\cdots 1515 \\ &\cdots 1500 \\ \hline &\cdots 3015 \end{array}$$

This gives us that $A = 3$ and $B = 5.$ Therefore, $A + B = 8.$

Thus, D is the correct answer.

Solution:

Let $x$ and $x + 1$ the two integers, where $2x + 1 \lt 100.$ The difference of the squares is $$(x + 1)^2 - x^2$$$$=(x + 1 + x)(x + 1 - x)$$$$= 2x + 1.$$

From this, we see that the desired difference is less than $100$ and is odd.

Thus, the correct answer is C.

Solution:

Let $x$ be the number of games the Unicorns had played before district play. Then after the entire season, they won a total of $.45x + 6$ games.

In total, they played $x + 8$ games. The problem statement also tell us that $$(x + 8) \div 2 = .45x + 6.$$ Solving, $$\begin{align*} x + 8 &= .9x + 12 \\ .1x &= 4 \\ x &= 40. \end{align*}$$ Therefore, the Unicorns played a total of $40 + 8 = 48$ games.

Thus, A is the correct answer.

Solution:

After drawing the first card, there are $7$ left. $3$ of them have the same color, and $1$ of them has the same letter. Therefore, there are $4$ out of $7$ possibilities that result in a winning pair.

The probability of drawing a pair is then $\dfrac{4}{7}.$

Thus, D is the correct answer.

Solution:

Based on the lengths given in the problem, the lemming is still in the square after it stops.

Since the lemming is still in the square, the sum of the distances to the horizontal sides is $10$ meters and the same for the vertical sides. Therefore, the $4$ distances sum to $20$ meters, making the average $20 \div 4 = 5$ meters.

Thus, C is the correct answer.

Solution:

We can find the area of the shaded part by subtracting the area of the unshaded part from the whole area.

There are $4$ squares with side length $1,$ contributing $1$ each to the unshaded area, for a total of $4.$

There are also $4$ triangles with base $3$ and height $\dfrac{5}{2}.$ This contributes a total area of $$4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15.$$

The total unshaded area is therefore $4 + 15 = 19.$

The total area is $5^2 = 25,$ and subtracting the unshaded area yields $25 - 19 = 6$ as the area of the shaded region.

Thus, B is the correct answer.

Solution:

Recall that a number is divisible if the sum of its digits is divisible by $3.$

The only triples of distinct numbers that satisfy this condition are $(1, 2, 3)$ and $(2, 3, 4).$

This means that the numbers form a three-digit number when either $1$ or $4$ is left in the bag.

Each of these events happens with $\dfrac{1}{4}$ chance, for a total probability of $2 \cdot \dfrac{1}{4} = \dfrac{1}{2}.$

Thus, C is the correct answer.

Solution:

The area of the outer circle is $6^2\pi = 36\pi,$ and the area of the inner circle is $3^2\pi = 9\pi.$ Therefore, the area of the outer ring is $36\pi - 9\pi = 27\pi.$

This means that the probability of hitting an outer segment is $\dfrac{9\pi}{36\pi} = \dfrac{1}{4}.$ Similarly, the probability of hitting an inner segment is $\dfrac{3\pi}{36\pi} = \dfrac{1}{12}.$

Summing over all possibilities, the probability of hitting a $1$ is $$\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}.$$ Similarly, the probability of hitting a $2$ is $$2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}.$$

The only way to get an odd score is to hit one $1$ and one $2.$

The probability of hitting these numbers in either order is $$2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}.$$

Thus, B is the correct answer.