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If they buy $6$ sandwiches, they would spend $6 \times 4.5 = 27$ dollars. This means that they would not be able to buy any more sandwiches.

They would then have $30 - 27 = 3$ dollars left. With this, they could buy $3$ sodas. They would therefore buy a total of $6 + 3 = 9$ items.

Thus, the correct answer is D.

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From the figure, we can see that the longer side has the same length as two of the shorter sides. This makes it $2 \cdot 5 = 10$ feet long.

This tells us that $BC = 10$ feet and $DC = 10 + 5 = 15$ feet. Therefore, the area would be $10 \cdot 15 = 150 \text{ ft.}^2$

Thus, the correct answer is E.

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We can rewrite the fraction as follows: $$\begin{align*} \dfrac{15}{11} &= 1 + \dfrac{4}{11} \\ \dfrac{17}{13} &= 1 + \dfrac{4}{13} \\ \dfrac{19}{15} &= 1 + \dfrac{4}{15}. \end{align*}$$

Recall that if two fractions have the same numerator, then the fraction with the larger denominator is smaller.

Using this fact, we can see that the correct ordering is $$\dfrac{19}{15} \lt \dfrac{17}{13} \lt \dfrac{15}{11}.$$

Thus, the correct answer is E.

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We can split the rhombus up into $4$ triangles, each of which looks like this.

We know that this is a right triangle, since the diagonals of a rhombus are perpendicular, and we know the hypotenuse is $52 \div 4 = 13.$

Using the Pythagorean theorem, we get the other leg to be $$\sqrt{13^2 - 12^2} = \sqrt{25} = 5.$$

This means that the area of the rhombus is $$4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120.$$ Thus, the correct answer is D.

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The hare's path is the one with the horizontal line in the middle, which represent the period where the hare took a nap.

We also know that the time it takes the hare to reach the finish line is longer than the time it took the tortoise.

This means that the end of the hare's path should have a larger $x$-value. This is exactly graph $B.$

Thus, the correct answer is B.

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Note the only lines of symmetry for a square are the two diagonals and the two lines connecting opposite midpoints.

Therefore, $Q$ must be a point on any one of these four lines. Each line consists of $9$ points, for a total of $4 \cdot 9 = 36$ points.

Note that $Q$ cannot be the same as $P,$ so we have to subtract out the $4$ points which coincide with $P,$ for a total of $36 - 4 = 32$ points.

The probability is therefore $\dfrac{32}{80} = \dfrac{2}{5}.$

Thus, the correct answer is C.

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To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a $100$ on her fourth test.

The sum of the $4$ tests is then $$76 + 94 + 87 + 100 = 357.$$ For an average of $81,$ Shauna's test scores must add to $5 \cdot 81 = 405.$ This means that she needs to get a $405 - 357 = 48$ on her last test.

Thus, the correct answer is A.

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Assume that Gilda starts off with $100$ marbles. After giving $20$ marbles to Pedro, she has $100 - 20 = 80$ marbles left.

She then gives $80 \times .1 = 8$ marbles to Ebony, with her having $80 - 8 = 72$ marbles left.

Finally, Gilda gives $72 \times .25 = 18$ marbles to Jimmy. She has a total of $72 - 18 = 54$ marbles left, which is $54\%$ of her original total.

Thus, the correct answer is E.

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Felicia's diameter is twice as much as Alex's which means that her radius is also twice as much.

Recall that the formula for the area of a circle is $\pi r^2.$ If the radius is doubled, then the area the is quadrupled.

This means that the base of Felicia's can has $4$ times the area as the base of Alex's can.

We also see that Alex's can is twice as tall as Felicia's can. The formula for the volume of a can is base times height.

Using this formula, we see that Felicia's can will have twice the volume of Alex's can since $4 \div 2 = 2.$

Thus, the correct answer is B.

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There are $5$ more participants, which means the mean is increased by $5 \div 5 = 1.$

Right now, we can see that median $20.$ If $16$ gets replaced with $21,$ then the median becomes $21.$

This shows that the median is also increased by $1.$

Thus, the correct answer is B.

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The number of kids that are taking both classes is $$70 + 54 - 93 = 31.$$ Subtracting this from the number of kids taking a math class will give us the number of kids taking only a math class.

The desired number is $70 - 31 = 39.$

Thus, the correct answer is D.

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Using the first and third cubes, we can see that black is opposite purple (both have red and green connected).

Similarly, with the first and second cubes, we see that green and white are opposites.

This means that yellow must be opposite red.

Thus, the correct answer is A.

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Note that $2$-digit palindromes have the tens and units digits the same. This means that they are all multiples of $11.$

If a $3$-digit number is the sum of $3$ $2$-digit palindromes, then it itself must also be a multiple of $11.$

The smallest $3$-digit multiple of $11$ that is not a palindrome is $110.$ This can be achieved by adding $11 + 22 + 77.$

Therefore, $N = 110.$ The sum of its digits is $1 + 1 + 0 = 2.$

Thus, the correct answer is A.

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Let us go through the answer choices.

If we she starts on Monday, her next day is Thursday, and the day after is Sunday.

If we start with Tuesday, the next day is Friday. The day after is Monday, which as we now from above will hit Sunday in $2$ more days.

If we start with Wednesday, the next day is Saturday. The day after is Tuesday, which we know hits Sunday only after $4$ more iterations.

Therefore, if we start with Wednesday, we loop through $6$ iterations with hitting Sunday.

Thus, the correct answer is C.

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If the probability is $\dfrac{2}{5},$ that means $\dfrac{2}{5}$ of the people wearing caps are also wearing sunglasses.

This means that $35 \cdot \dfrac{2}{5} = 14$ people are wearing both caps and sunglasses.

The probability of a person wearing sunglasses also wearing a cap is $\dfrac{14}{50} = \dfrac{7}{25}.$

Thus, the correct answer is B.

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For the first $15$ miles, Qiang drove for $\dfrac{15}{30} = \dfrac{1}{2}$ an hour.

Let $x$ be the distance that Qiang must drive for to satisfy the condition. It will take him $\dfrac{x}{55}$ hours to drive this distance.

The total trip is now $15 + x$ miles. We know the average speed is $50$ miles per hour, so this will take him $\dfrac{x + 15}{50}$ hours.

Setting the two times equal to each other, we get $$\dfrac{1}{2} + \dfrac{x}{55} = \dfrac{x + 15}{50}.$$

Cross-multiplying and simplifying yields $$15 + x = 25 + \dfrac{10x}{11} \Rightarrow x = 110.$$

Thus, the correct answer is D.

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We can regroup all the number as follows. $$\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}$$

From this representation, we can see that all the middle terms cancel leaving only $$\dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.$$

Thus, the correct answer is B.

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The only two ways that the sum can be even is if both rolls are even or both are odd.

The probability that a roll is even is $\dfrac{2}{6} = \dfrac{1}{3},$ and the probability that it is odd is $\dfrac{4}{6} = \dfrac{2}{3}.$

Therefore, the desired probability is $$\dfrac{1}{3} \cdot \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{5}{9}.$$

Thus, the correct answer is C.

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We can assume that the top $3$ teams won every game against every team not amongst themselves.

They play $3 \cdot 2 = 6$ games in total, getting a total of $6 \cdot 3 = 18$ points.

Now, amongst the top $3,$ each team plays each other time twice. To even out the scores, we can let one team win one game and let the other team win the other game.

This means that for every pair of the top $3$ teams, each team in the pair gets $3$ points. There are $3$ such pairs, with each team appearing in $2$ pairs.

This means that each team will get an extra $3 \cdot 2 = 6$ points. Therefore, their maximum score is $18 + 6 = 24.$

Thus, the correct answer is C.

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Recall that $$a^2 - b^2 = (a + b)(a - b).$$ We can do the following rearrangement. $$\begin{gather*} (x^2 - 5)^2 - 4^2 = 0 \\ (x^2 - 5 + 4)(x^2 - 5 - 4) = 0 \\ (x + 1)(x - 1)(x + 3)(x - 3) = 0 \end{gather*}$$

From this we can see that there are $4$ possible values for $x.$

Thus, the correct answer is D.

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First, let us find the vertices of the triangles. The intersections between $y = 5$ and the other two lines are $$(4, 5) \text{ and } (-4, 5).$$

To find the third intersection, we can equate the two equations to get $$1 + x = 1 - x \Rightarrow x = 0.$$

This gives us the point $(0, 1).$

Note that the first two intersection points are mirrored across the $y$-axis. This tells us that the triangle is isosceles.

The base is therefore $2 \cdot 4 = 8,$ and the height is $5 - 1 = 4.$ The area of the triangle is therefore $$\dfrac{1}{2} \cdot 8 \cdot 4 = 16.$$

Thus, the correct answer is E.

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Let $x$ be the percent converted to a real number. Then the percent increase would multiply the original by $1 + x.$

The percent decrease would multiply the new price by $1 - x.$ The final price will the be the original price multiplied by $$(1 + x)(1 - x) = 1 - x^2 = .84.$$

This tells us that $$x^2 = .16 \Rightarrow x = .4.$$

This tells us that the percent would be $40\%.$

Thus, the correct answer is E.

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Let $x$ be the total number of points and $y$ be the desired answer.

Then from the problem statement we get that $$\dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x.$$

Simplifying yields $$y + 15 = \dfrac{13x}{28}.$$

We know that $y \leq 14$ since none of the $7$ team members scored more than $2$ points.

We also know that $x$ must be a multiple of $28.$ If $x = 28,$ then we get that $$y + 15 = 13 \Rightarrow y = -2,$$ which is not allowed.

If $x = 2 \cdot 28 = 56,$ then we have that $$y + 15 = 26 \Rightarrow y = 11,$$ which works.

Thus, the correct answer is B.

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We know that the area of $\triangle BCD$ is twice that of $\triangle ABD$ since its base is twice as long and they have the same heights.

This tells us that the area of $\triangle ABD$ is $360 \div 3 = 120.$ Similarly, we get that the area of $\triangle ABE$ is $120 \div 2 = 60.$

Note that $CD = \dfrac{2}{3} CA.$ This tells us that the altitude of $\triangle BCD$ is $\dfrac{2}{3}$ the altitude of $\triangle ABC.$

We also know that $BD = \dfrac{1}{2}BD,$ which tells us that the altitude of $\triangle BEF$ is $\dfrac{1}{2}$ the altitude of $\triangle BCD.$

Finally, we get that the altitude of $\triangle BEF$ is $\dfrac{2}{3} \cdot \dfrac{1}{2} = \dfrac{1}{3}$ the altitude of $\triangle ABC.$

Note that the altitude of $\triangle ABF$ is the same as the altitude of $\triangle ABC.$

This tells us that the area of $\triangle BEF$ is $\dfrac{1}{3}$ the area of $\triangle ABF$ since they have the same base but different altitudes.

This gives us the following equation, where $x$ equals the area of $\triangle BEF.$

$$\dfrac{x}{x + 60} = \dfrac{1}{3}.$$

Cross-multiplying yields $$3x = x + 60 \Rightarrow x = 30.$$

Thus, the correct answer is B.

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Let us assign everybody $2$ apples. This leaves us with $24 - 3 \cdot 2 = 18$ apples.

Then, we want to solve $$a + b + c = 18,$$ where $a$ is the number of apples that Alice gets and so on.

Note that these are all nonnegative, since we already satisfied the only condition we needed to.

We can use stars and bars to get the number of solutions as $$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = 190.$$

Thus, the correct answer is C.