Answer: B

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Written solution:

We only need to consider the ones digits of each number (except for the first one so we avoid getting a negative answer): $$22 - 2 - 2 - 2 - 2 - 2 = 12$$ which has a ones digit of $2$.

Thus, B is the correct answer.

Answer: C

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We can simplify the fractions by taking out the common factor $11$: $\dfrac{44}{11}$ simplifies to $4$, $\dfrac{110}{44}$ simplifies to $\dfrac{10}{4} = \dfrac{5}{2} = 2.5$, and $\dfrac{44}{1100}$ simplifies to $\dfrac{4}{100} = 0.04$. Therefore, we have $4 + 2.5 + 0.04 = 6.54.$

Thus, C is the correct answer.

Answer: E

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The visible shaded region is the part inside the $10$ by $10$ square but outside the $9$ by $9$ square, together with the part inside the $7$ by $7$ square but outside the $4$ by $4$ square. Its area is $$10^2-9^2+7^2-4^2=100-81+49-16=52.$$

Thus, E is the correct answer.

Answer: E

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Written solution:

To find the number that Yunji left out, we need to find the sum of the integers from $1$ to $9$ and find its difference with the largest perfect square below the sum. We can calculate the sum of the integers from $1$ to $9$ as follows: $$1 + \ldots + 9 = \dfrac{9(9+1)}{2} = 45$$ The largest perfect square less than $45$ would be $36$ and $45 - 36 = 9$.

Thus, E is the correct answer.

Answer: B

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For the product to be a multiple of $6$, the two dice together must supply a factor of $2$ and a factor of $3$. The possible sums among the answer choices can occur as follows: $$5=2+3,\quad 7=1+6,\quad 8=2+6,\quad 9=3+6.$$ There is no way to get sum $6$ while also having a product divisible by $6$: the pairs with sum $6$ are $(1,5),(2,4),(3,3),(4,2),(5,1)$, and none have product divisible by $6$.

Thus, B is the correct answer.

Answer: D

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Written solution:

Path R is shortest because it replaces curved arc portions with straight-line chords. Path Q is longest because it includes the most interior crossing distance while still using the curved ends.

It remains to compare paths P and S. The relevant straight piece in path S is a diagonal across the rink, while the corresponding straight piece in path P is a side of the same right triangle. A diagonal is longer than a side, so path S is longer than path P.

The order from shortest to longest is R, P, S, Q.

Thus, D is the correct answer.

Answer: E

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Written solution:

The $2\times2$ and $1\times4$ tiles each have area $4$. Since the rectangle has area $21$, the number of $1\times1$ tiles must be congruent to $1\pmod4$. Among the choices, only $1$ and $5$ are possible by area.

It is impossible to use just one $1\times1$ tile. If the larger tiles covered the other $20$ cells, then two rows would have all $7$ cells covered by larger tiles. But each $2\times2$ tile covers $2$ cells in any row it meets, and each $1\times4$ tile covers $4$ cells in one row, so each row would have an even number of cells covered by larger tiles. A row cannot have $7$ such cells.

The following tiling shows that $5$ unit tiles are possible.

Thus, E is the correct answer.

Answer: D

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Written solution:

After Tuesday, Taye could have $5$ or $4$ dollars. After Wednesday, the possible amounts are $$8,10,7,8,$$ so the distinct amounts are $7,8,10$.

After Thursday, these can become $$10,14,\quad 11,16,\quad 13,20.$$ These are $6$ distinct dollar amounts.

Thus, D is the correct answer.

Answer: E

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We can let $r$ be the number of red marbles that Maria has. Since Maria has half as many red marbles as green, then we know that she has $2r$ green marbles. Moreover, since she has twice as many blue marbles as green, then she will have $2(2r) = 4r$ blue marbles. Adding these together gives us $r + 2r + 4r = 7r,$ and so the answer must be a multiple of $7.$ Among the answer choices, only $28$ is a multiple of $7.$

Thus, E is the correct answer.

Answer: B

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There are $50$ years between $1980$ and $2030$, so we can expect the CO₂ reading to increase by $50 \times 1.515 = 75.75 \approx 76$ ppm by $2030$. Since the CO₂ reading in $1980$ was $338$ ppm, then we will have $338 + 76 = 414$ ppm by $2030$.

Thus, B is the correct answer.

Answer: D

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Consider the base of the triangle to be $\overline{\rm AB}$ which has length $11-5=6$. Given that the area of the triangle is $12$, its height must be of length $\dfrac{2(12)}{6}=4$. Since $y>7$, then $y$ must be $7+4=11$.

Thus, D is the correct answer.

Answer: E

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Let $x$ be the number of guppies in the 1st tank. Hence, there are $x+1$ guppies in the 2nd tank, $x+3$ guppies in the 3rd tank, and $x+6$ guppies in the 4th tank. We then use the fact that there are a total of 90 guppies in the 4 tanks to find $x$: $$x + x + 1 + x + 3 + x + 6 = 90$$ $$4x+10=90$$ $$x=20.$$

Note that we are not yet done since we are asked for the number of guppies in the 4th tank and not the 1st. There are $x+6=20+6=26$ guppies in the 4th tank.

Thus, E is the correct answer.

Answer: B

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We can deduce from the choices that it is possible to exhaust all possible cases for this problem. Note that all sequences must start with up $(U)$ and end with down $(D)$, and that it should not be possible to go down more times than Buzz has gone up so far. Keeping this in mind, we can arrive at the following possible cases: $$UUUDDD$$ $$UUDUDD$$ $$UUDDUD$$ $$UDUDUD$$ $$UDUUDD$$ which is a total of five possible sequences.

Thus, B is the correct answer.

Answer: A

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A systematic way of tracking the shortest overall distance to $Z$ is to consider the shortest distance to get to each town from $A$. For instance, the shortest distance to get to town $X$ from $A$ is $5$ km, trivially.

Then, for town $M$, going to town $X$ first will be shorter compared to going directly from $A$, so the shortest path to town $M$ has a length of $7$ km.

For town $Y$, it will take us $15$ km if we come from town $X$ and only $13$ km coming from $M,$ so $13$ km is the length of shortest path to $Y$ from $A$.

Doing the same for town $C$ will give us $18$ km as the shortest distance by coming from town $Y$.

Finally, for town $Z$, we can either come from town $Y,$ $C,$ or $M$. The total distance if we come from each three towns respectively would be $30,$ $28,$ and $32$. Hence, $28$ km is the shortest distance from $A$ to $Z$.

Thus, A is the correct answer.

Answer: C

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Firstly, note that $$\text{FLYFLY} = 1001(\text{FLY})$$ and, similarly, $$\text{BUGBUG}=1001(\text{BUG})$$ so the equation can be simplified to $8 \cdot \text{FLY} = \text{BUG}.$

For $\text{BUG}$ to remain three digits, $F$ must be $1.$ Moreover, $L$ must also be less than $3$ to avoid carrying over $2$ to the hundreds digit and making the product $4$ digits. Since we need $\text{FLY}$ to be the greatest number, $L$ must be $2.$

To identify the possible values for $Y,$ we note that so far we have $8(12)=96$, so we must avoid carrying $4$ to the tens digit to keep the resulting product three digits. Hence, $Y\le5$. We can try $4$ and verify that the resulting product has unique digits that haven't been used yet: $8(124) = 992$, which does not have unique digits. Trying $3,$ we get $8(123) = 984$, which satisfies our criteria.

Hence, $$\text{FLY} + \text{BUG} = 123 + 984 = 1107.$$

Thus, C is the correct answer.

Answer: D

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Written solution:

There are $27$ multiples of $3$ from $1$ through $81$. A row or column has product divisible by $3$ exactly when it contains at least one of these multiples.

Suppose $r$ rows and $c$ columns have products divisible by $3$. Every multiple of $3$ must lie in one of those $r$ rows and also in one of those $c$ columns, or else it would create another marked row or column. Thus the $27$ multiples must fit in the $rc$ intersection cells. If $r+c\le10$, then $rc\le25$, which is too small. So at least $11$ rows and columns are needed.

This can be done by placing $25$ multiples of $3$ in a $5\times5$ block, then placing the remaining $2$ multiples in a sixth column within two of those same rows. Then exactly $5$ rows and $6$ columns are marked, for a total of $11$.

Thus, D is the correct answer.

Answer: E

Solution:

Count ordered placements by first choosing the square for the white king. If the white king is in the center, it attacks every other square, so there are $0$ choices for the black king.

If the white king is in a corner, it attacks $3$ squares, so the black king has $9-1-3=5$ safe squares. There are $4$ corner choices, giving $4\cdot5=20$ placements.

If the white king is on an edge but not a corner, it attacks $5$ squares, so the black king has $9-1-5=3$ safe squares. There are $4$ such edge choices, giving $4\cdot3=12$ placements.

The total number of placements is $20+12=32$.

Thus, E is the correct answer.

Answer: A

Solution:

Let $\theta$ be the measure of $\angle BOC.$

One component of the shaded region is the area of the circle with radius $2$ minus the area of the circle with radius $1.$ This part has area $4\pi-\pi=3\pi.$ The remaining area is a sector of the biggest circle minus the area of the circle with radius $2$. This has area $\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi).$ Hence, the total area of the shaded region is $3\pi + \dfrac{\theta}{360}(5\pi).$

Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of $\pi + \dfrac{360-\theta}{360}(5\pi)$

Lastly, we equate the area of both regions and solve for $\theta:$ $$3\pi + \dfrac{\theta}{360}(5\pi) = \pi + \dfrac{360-\theta}{360}(5\pi)$$ $$2\pi = \dfrac{360-\theta-\theta}{360}(5\pi)$$ $$\dfrac{2}{5} = 1 - \dfrac{2\theta}{360}$$ $$2\theta = \dfrac{3}{5}(360)$$ $$\theta = 108.$$

Thus, A is the correct answer.

Answer: C

Solution:

Jordan has $\dfrac{3}{5} \times 15 = 9$ pairs of red sneakers and $6$ pairs of white sneakers. Moreover, $\dfrac{2}{3} \times 15 = 10$ are high-top and $5$ are low-top. If we want to minimize the number of red high-top sneakers, then we can set all $6$ white sneakers to be high-top, leaving $10-6=4$ red sneakers as high-top. Hence, the fraction of red high-top sneakers would be $\dfrac{4}{15}$.

Thus, C is the correct answer.

Answer: D

Solution:

We first note that we can only form equilateral triangles if we go through the diagonals of the square faces, otherwise at least one angle of the triangle will be different. Afterwards, it is easy to exhaust all possible equilateral triangles that can be formed: $\triangle PVT,$ $\triangle PRT,$ and $\triangle PRV.$

Thus, D is the correct answer.

Answer: E

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We can let $g$ be the number of green frogs and $y$ be the number of yellow frogs. Initially, we have $g=3y$. Then, after some frogs moved, we have the following proportion: $$\dfrac{g+5-3}{y+3-5} = \dfrac{4}{1}.$$

Substituting $3y$ for $g$ will allow us to determine the number of yellow frogs originally ($y$): $$\dfrac{3y+2}{y-2} = 4$$ $$3y+2 = 4(y-2) = 4y-8$$ $$y = 10$$

Hence, there were $10$ yellow frogs and $3 \times 10 = 30$ green frogs initially. After some frogs moved, we now have $10+3-5 = 8$ yellow frogs and $30+5-3=32$ green frogs, giving us a difference of $32-8=24$ between the number of green and yellow frogs.

Thus, E is the correct answer.

Answer: B

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We have a huge margin of error for this problem so we can freely estimate values. Firstly, we note that the entire roll of tape is 1-inch thick, and given that a single tape is $0.015$ inches thick, then there are $\dfrac{1}{0.015}= \dfrac{2}{0.03} = \dfrac{200}{3}$ layers of tape in the entire roll.

Then, we must identify an estimate for the circumference of one layer of tape. Near the center, one layer of tape will have a circumference of $2\pi$ while the layers near the outer section will have a circumference of about $4\pi$. A good estimate would be to take the average circumference which is $3\pi$. With this, we can estimate the total length for the entire roll: $$\dfrac{200}{3}(3\pi) = 200\pi \approx 200(3) \approx 600.$$

Thus, B is the correct answer.

Answer: C

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For a segment whose endpoint differences are $a$ horizontally and $b$ vertically, the segment crosses $a$ vertical grid lines and $b$ horizontal grid lines, but crossings at lattice points are counted twice. Therefore the number of cells whose interiors are intersected is $$a+b-\gcd(a,b).$$

Here the endpoint differences are $3000$ and $5000$, and $\gcd(3000,5000)=1000$. Thus the number of cells colored is $$3000+5000-1000=7000.$$

Thus, C is the correct answer.

Answer: B

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Each mountain is a $45^\circ\text{-}45^\circ\text{-}90^\circ$ triangle. A right isosceles triangle with height $x$ has area $x^2$, since its two perpendicular sides each have length $x\sqrt2$.

The two large mountains have areas $8^2=64$ and $12^2=144$. Their overlap is also a $45^\circ\text{-}45^\circ\text{-}90^\circ$ triangle with height $h$, so its area is $h^2$. Thus $$64+144-h^2=183,$$ so $h^2=25$, and $h=5$.

Thus, B is the correct answer.

Answer: C

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After the first $8$ passengers sit, there are $4$ empty seats among the $12$ seats, so there are $\binom{12}{4}=495$ equally likely sets of empty seats.

Count the complement, where no row has two adjacent empty seats. In a row of three seats, the possible empty-seat patterns with no adjacent empty seats have sizes $0,1,2$, with $1,3,1$ choices respectively. So we need the coefficient of $x^4$ in $$(1+3x+x^2)^4.$$ This coefficient is $$3^4+4\cdot3\cdot3^2+\binom42=81+108+6=195.$$

Therefore the probability that at least one row has adjacent empty seats is $$\frac{495-195}{495}=\frac{20}{33}.$$

Thus, C is the correct answer.