Answer: D

Solution:

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

She promised to work $10\cdot 6=60$ hours in all.

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

Thus, D is the correct answer.

Answer: E

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.

Answer: C

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.

Answer: D

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.

Answer: B

Solution:

The total amount of birthday money Chandler gets is $50+35+15=100$ dollars. Therefore, he still needs $500-100=400$ dollars.

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

Thus, B is the correct answer.

Answer: E

Solution:

The decrease in cost is $41-7=34$ cents per minute.

The percent decrease is $100\cdot\dfrac{34}{41}\%$, which is a little more than $80\%$, so the closest answer is $80$.

Thus, E is the correct answer.

Answer: D

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.

Answer: B

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.

Answer: B

Solution:

The last square in the second row must be $1$, since that row already contains $2$ and $3$, and the last column already contains $4$.

Then the top-right square must be $3$, since the top row already contains $1$ and $2$, and the last column already contains $1$ and $4$.

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

Thus, B is the correct answer.

Answer: D

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.

Answer: D

Solution:

Only Tile $III$ has a $0$ on an edge, so Tile $III$ must be placed in $C$ or $D$, where that edge can be on the outside.

Tile $III$ also has a $5$ on its right edge, and no other tile has a $5$. Therefore Tile $III$ must be placed in $D$, with its $5$ on the outside.

The only tile that can match the $1$ on the left edge of Tile $III$ is Tile $IV$, so Tile $IV$ is placed in $C$.

Thus, D is the correct answer.

Answer: A

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.

Answer: C

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.

Answer: C

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.

Answer: A

Solution:

We know that $b \lt c$ and $0 \lt a.$ 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.

Answer: A

Solution:

The circumferences of circles with radii $1$ through $5$ are $2\pi,4\pi,6\pi,8\pi,10\pi$. Their respective areas are $\pi,4\pi,9\pi,16\pi,25\pi$.

As $C$ increases by equal amounts, $A$ increases more and more quickly, so the points form an increasing quadratic pattern. Only graph A has that shape.

Thus, A is the correct answer.

Answer: C

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.

Answer: D

Solution:

Only the last four digits can affect the thousands digit and units digit of the product. The two numbers end in $0303$ and $0505$, so it is enough to compute $303\cdot 505$.

Since $303\cdot 505=153015$, the last four digits of the full product are $3015$.

This gives $A=3$ and $B=5$, so $A+B=8$.

Thus, D is the correct answer.

Answer: C

Solution:

Let the consecutive positive integers be $x$ and $x+1$. Their sum is $2x+1$, which is less than $100$.

The difference of their squares is $$(x+1)^2-x^2=(x+1+x)(x+1-x)=2x+1.$$

So the difference must be an odd number less than $100$. Among the choices, $79$ is the only possibility, and it occurs for $39$ and $40$.

Thus, C is the correct answer.

Answer: A

Solution:

Let $x$ be the number of games the Unicorns had played before district play. Then they had won $0.45x$ games before district play and $0.45x+6$ games after district play.

In total, they played $x+8$ games. The problem statement tells us that $$0.45x+6=\dfrac{x+8}{2}.$$

Solving gives $0.45x+6=0.5x+4$, so $2=0.05x$, and $x=40$. Therefore, the Unicorns played $40+8=48$ games in all.

Thus, A is the correct answer.

Answer: D

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.

Answer: C

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.

Answer: B

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.

Answer: C

Solution:

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

The only triples of distinct digits from $1,2,3,4$ whose sum is divisible by $3$ are $(1,2,3)$ and $(2,3,4)$.

This means the constructed number is a multiple of $3$ exactly when either $1$ or $4$ is left in the bag.

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

Thus, C is the correct answer.

Answer: B

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.