Solution:

Since Luka needs twice as much sugar as lemon, he needs $2\cdot3=6$ cups of sugar. Since Luka also needs $4$ times as much water as sugar, he needs $4\cdot6=24$ cups of water.

Thus, the correct answer is E.

Solution:

First, the total amount of money that they make is $\$15+\$20+\$25+\$40=\$100.$

Since they divide this equally, they each get $\frac{100}{4}=25$ dollars.

This means that the person who earned $\$40$ earned $\$15$ more than what he will end up with, so he gives $\$15$ to the others.

Thus, the correct answer is C.

Solution:

First, the size of the garden is $6\text{ft} \cdot 8\text{ft} = 48\text{ft}^2.$

Next, since there are $4$ plants per square foot, Carrie can plant $4\cdot 48=192$ plants total.

Finally, since there are $10$ strawberries per plant, Carrie can harvest $10\cdot 192=1920$ strawberries total.

Thus, the correct answer is D.

Solution:

The first three hexagons contain $1,$ $7,$ and $19$ dots. Each new band adds $6$ more dots than the previous band.

The fourth hexagon adds $18$ dots around the third hexagon, so it contains $19+18=37$ dots.

Thus, the correct answer is B.

Solution:

Since we start with $\dfrac{3}{4}$ of the pitcher, we have $75\%$ of the pitcher full as $$\dfrac{3}{4} = \dfrac{75}{100}.$$

Now, since $5$ cups each have the same amount of juice, they each have one-fifth of the $75\%.$

This means they have $$\dfrac{1}{5}\cdot 75\%=\left(\dfrac{75}{5}\right)\% = 15\%.$$

Thus, the correct answer is C.

Solution:

Maren is in the last car. Since Aaron is directly behind Sharon and Darren is in front of Aaron, the possible placements of Darren, Sharon, and Aaron before Maren are limited.

If Sharon and Aaron were in cars $1$ and $2,$ Darren could not be in front of Aaron. If they were in cars $3$ and $4,$ then Darren and Karen would have to occupy cars $1$ and $2,$ which are adjacent.

Thus Sharon and Aaron must be in cars $2$ and $3.$ Darren is then in car $1,$ Karen is in car $4,$ and Aaron is in the middle car.

Thus, the correct answer is A.

Solution:

The thousands digit must be $2.$ Since the digits are distinct and increasing, the hundreds digit must be $3.$

The last two digits must be chosen from $4,5,6,7,8,9.$ Once the two digits are chosen, their order is forced.

There are $\binom{6}{2}=15$ such integers.

Thus, the correct answer is C.

Solution:

Let $p$ be the number of pennies Ricardo has and let $n$ be the number of nickels he has.

We know that $$p+n=2020,$$ $$p \geq 1,$$ and $$n \geq 1$$ by the problem statement.

This means $$\begin{align*}2020 -n &=p \\ \implies 2020 -n &\geq 1.\end{align*}$$ Therefore, $1 \leq n \leq 2019.$ It follows, then, that Ricardo has $$\begin{align*}p+5n &= p+n + 4n\\ &= 2020 + 4n \end{align*}$$ cents.

Therefore, to maximize the money he has, we maximize the number of nickels he has, and minimizing the money he has involves minimizing the number of nickels he has. The maximum number of nickels he can have is $2019,$ so he can have at most $2020+2019(4)$ cents. The minimum number of nickels he can have is $1,$ so he has at least $2020+1(4)$ cents. The difference between the maximum and minimum amount of money he can have is: $$\begin{align*} &2020+2019(4) - (2020+1(4)) \\ &= 2018(4) = 8072 . \end{align*}$$

Thus, the correct answer is C.

Solution:

In the top layer, the non-corner edge pieces have icing on exactly two sides. There are $4$ edges with $2$ such pieces each, for $8$ pieces.

In each of the other $3$ layers, the only pieces with exactly two iced sides are the $4$ corner pieces. This adds $3\cdot4=12$ pieces.

The total is $8+12=20.$

Thus, the correct answer is D.

Solution:

There are $4!=24$ total arrangements of the marbles.

If the Steelie and Tiger are adjacent, treat them as one block. Then there are $3!$ ways to arrange the block with the other two marbles, and $2$ orders inside the block, for $3!\cdot2=12$ adjacent arrangements.

Therefore, $24-12=12$ arrangements keep the Steelie and Tiger separated.

Thus, the correct answer is C.

Solution:

Naomi traveled $6$ miles in $10$ minutes. This ratio is $$\begin{align*}\dfrac{6 \text{ miles}}{10 \text{ minutes}} &= \dfrac{36 \text{ miles}}{60 \text{ minutes}} \\&= 36 \dfrac{\text{ miles}}{\text{hour}}\end{align*}$$

Maya traveled $6$ miles in $30$ minutes. This ratio is $$\begin{align*}\dfrac{6 \text{ miles}}{30 \text{ minutes}} &= \dfrac{12 \text{ miles}}{60 \text{ minutes}} \\&= 12 \dfrac{\text{ miles}}{\text{hour}}\end{align*}$$

This difference is $36-12=24.$

Thus, the correct answer is E.

Solution:

Note first that $$\begin{align*} n! &= n \cdot (n - 1) \cdot (n - 2) \cdots 1 \\ &= n((n - 1) \cdot (n - 2) \cdots 1) \\ &= n(n - 1)! \end{align*}$$

With that in mind, further observe that: $$\begin{align*} 5! \cdot 9! &= 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 9! \\ &= 120 \cdot 9! \\ &= 12(10 \cdot 9!) \end{align*}$$

Since $12 \cdot N! = 12(10\cdot 9!),$ we know $N! = 10\cdot 9!.$

Using our note from above, we know that $10\cdot 9!= 10!$ so $N=10.$

Thus, the correct answer is A.

Solution:

Suppose Jamal adds $s$ purple socks. Then, there will be $s+18$ purple socks.

Also, since there are $36$ total socks to begin with, we have $s+36$ socks after adding the socks.

Since we have a $60\%$ chance of choosing a purple sock afterwards, we know $$\dfrac{s+18}{s+36}=0.6.$$

Solving for $s$ yields: $$\begin{align*}s+18 &= 0.6s+21.6 \\ 0.4s &= 3.6\\ s &= 9.\end{align*}$$

Therefore, $9$ socks are added.

Thus, the correct answer is B.

Solution:

Looking at the horizontal dashed line, the average population is around $4750.$

Since there are $20$ cities, and the average population is $\frac{\text{total population}}{\text{number of cities}},$ we know: $$\dfrac{\text{total population}}{20} = 4750.$$

Multiplication yields that the total population is $95,000.$

Thus, the correct answer is D.

Solution:

The statement gives $0.15x=0.20y.$

Solving for $y$ gives $y=\dfrac{0.15}{0.20}x=\dfrac{3}{4}x.$

Therefore, $y$ is $75\%$ of $x.$

Thus, the correct answer is C.

Solution:

Each number is added once per line it is on. Every point is on $2$ lines except for $B$ which is on $3.$

This means $$\begin{align*} 2A + 3B &+ 2C + 2D \\&+ 2E + 2F = 47, \end{align*}$$ so $$\begin{align*} &2(A + B + C + D + E + F) \\ &= 47-B. \end{align*}$$

Since $A,B,C,D,E,F$ are unique digits from $1$ to $6,$ each digit is represented exactly once, making $$\begin{align*} &A + B + C + D + E + F \\ &= 21. \end{align*}$$

With $$\begin{align*} &2(A + B + C + D + E + F) \\ &+ B = 47, \end{align*}$$ we know $B + 2(21)=47,$ so $B=5.$

Thus, the correct answer is E.

Solution:

Let's begin by firstly simply factoring $2020$: $$\begin{align*}2020 &= 1\cdot 2020\\ &=2\cdot 1010\\ &=4\cdot 505\\ &=5\cdot 404\\ &=10\cdot 202\\ &= 20\cdot 101 \end{align*}$$

These twelve factors of $2020$ can be classified by the number of their factors:

$\quad$ - $1$ has one factor

$\quad$ - $2, 5,$and $101$ has two factors

$\quad$ - $4$ has three (distinct) factors

Thus, all the remaining seven numbers must have more than three factors.

Thus, the correct answer is B.

Solution:

Since $FE$ is the diameter of the semicircle, we know the length of the diameter is $34,$ and so the radius is $17.$ Let $O$ be the center of the diameter.

The length from $OF$ therefore is $17.$

Since $D$ is on $OF,$ we know $$\begin{align*} OD + FD &= OF\\ OD + 9 &= 17 \\ OD &= 8. \end{align*}$$

Also, since we have a semicircle, we know $OC = 17.$

Finally, since $ABCD$ is a rectangle, we know $\angle ODC$ is a right angle. This means we can find $DC$ by the Pythagorean Theorem. We know $$\begin{align*} OD^2+DC^2&=OC^2 \\ 8^2+DC^2 &= 17^2 \\ DC &= 15. \end{align*}$$

Thus, the area of the rectangle is $DC\cdot DA = 15\cdot 16=240.$

Thus, the correct answer is A.

Solution:

For a number to be divisible by $15$ the number must be divisible by $3$ and by $5.$

First, to ensure the number is divisible by $5$ it must end in $0$ or in $5.$

Since the digits alternate between two distinct digits, we call the other digit $d.$

This would make our number either $d0d0d$ or $5d5d5.$ Since the last digit must be $0$ or $5$ and the first digit cannot be $0,$ the number must have the form $5d5d5.$

Thus, we know our number is $5d5d5$ for some digit $d.$

To ensure our number is also a multiple of $3$ the sum of the digits must be a multiple of $3.$

The sum of our digits is $5+d+5+d+5 = 15+2d.$ Since $15$ is a multiple of $3,$ all that is required is that $2d$ is a multiple of $3,$ so $d$ is a multiple of $3.$

This means $d$ can be $0,3,6$ or $9.$

Therefore, we have $4$ solutions.

Thus, the correct answer is B.

Solution:

Tree $2$ is $11$ meters tall. Since all heights are integers and neighboring trees differ by a factor of $2,$ trees $1$ and $3$ must both be $22$ meters tall.

The first three trees total $22+11+22=55$ meters. Tree $4$ is either $11$ or $44$ meters. If tree $4$ were $11,$ then tree $5$ would be $22$ or nonintegral $5.5,$ giving averages ending in $.6$ or invalid.

If tree $4=44,$ then tree $5$ can be $22$ or $88.$ These give averages $24.2$ and $37.4,$ respectively. The only listed average ending in $.2$ is $24.2.$

Thus, the correct answer is B.

Solution:

Each move must go up one row and either left or right. Counting row by row from $P,$ each reachable square gets the sum of the counts from the two adjoining squares below it.

The count at $Q$ is $28,$ so there are $28$ paths.

Thus, the correct answer is A.

Solution:

To see which numbers we can make by inverting it, let's make an inverting machine.

This would take $N$ and yield either $2N$ if $2N$ is even (which it always is) or $\frac{N-1}{3}$ if $\frac{N-1}{3}$ is an odd integer. Note that $\frac{N-1}{3}$ is an integer only if $N \equiv 1 \mod 3.$ Also, if $N$ is even, then $N-1$ is odd. That would mean $\frac{N-1}{3}$ would be odd. Therefore, our inverter machine yields $2N$ and also $\frac{N-1}{3}$ if $N \equiv 1 \mod 3$ and $N$ is even.

Now, we must see what the inverting machine can yield after $6$ moves:

$1)$ We can only get $2.$

$2)$ From $2,$ we can only get $4.$

$3)$ From $4,$ we can get $1$ and $8.$

$4)$ From $1,$ we can get only $2$; from $8,$ we can only get $16.$

$5)$ From $2,$ we can get only $4$; from $16,$ we can get $5$ and $32.$

$6)$ From $4,$ we can get $1$ and $8,$; $5,$ we can get $10$; from $32$ we can get $64.$

Move $6$ can yield $1, 8, 10,$ and $64,$ and their sum is $83.$

Thus, the correct answer is E.

Solution:

There are $3^5=243$ ways to give each of the $5$ distinct awards to one of the $3$ students.

Subtract the distributions in which at least one student receives no award. If a particular student receives none, the awards go to the other two students in $2^5$ ways. This gives $3\cdot2^5$ counts, but the $3$ cases in which one student receives all awards have each been subtracted twice.

By inclusion-exclusion, the desired number is $3^5-3\cdot2^5+3=150.$

Thus, the correct answer is B.

Solution:

For $n=24,$ the shaded tile area is $24^2s^2.$ Each side of the large square consists of $24$ tiles and $25$ borders, so its side length is $24s+25d.$

The shaded tiles cover $64\%=\dfrac{16}{25}$ of the large square, so $$\dfrac{24^2s^2}{(24s+25d)^2}=\dfrac{16}{25}.$$ Taking positive square roots gives $\dfrac{24s}{24s+25d}=\dfrac{4}{5}.$

Thus $120s=96s+100d,$ so $24s=100d$ and $\dfrac{d}{s}=\dfrac{6}{25}.$

Thus, the correct answer is A.

Solution:

We represent the lengths of each square as $s_1,s_2,$ and $s_3$ respectively. The length of the rectangle is $s_1+s_2+s_3$ as these $3$ squares span the entirety of a side of the large rectangle. Therefore, $$s_1+s_2+s_3=3322.$$

Also, the height of the large rectangle is the sum of the height of $R_2$ and $S_3.$ Now, note that the sum of height of $R_2$ and $s_2$ is $s_1,$ so height of $R_2$ is equal to $s_1-s_2.$ Therefore, the height of the large rectangle is $s_1-s_2+s_3,$ which means $$s_1-s_2+s_3=2020.$$ Subtracting both of our results yields $$2s_2=3322-2020=1302.$$ This would mean $$s_2=651.$$

Thus, the correct answer is A.