2024 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

What is the ones digit of $\begin{align*} & 222,222 - 22,222 - 2,222 \\ &- 222 - 22 - 2? \end{align*}$

$0$

$2$

$4$

$6$

$8$

Answer: B

Concepts:units digit

Difficulty rating: 370

Small Hint:

Only the ones digits affect the ones digit of the result

Big Hint:

Compute the ones digit of $2-2-2-2-2-2$

Video solution:

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Written solution:

Only the ones digits matter. Modulo $10,$ the expression is $\begin{gathered} 2-2-2-2-2-2=-8,\\ -8\equiv2\pmod{10}. \end{gathered}$ Therefore its ones digit is $2.$

Thus, B is the correct answer.

2.

What is the value of this expression in decimal form? $\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}$

$6.4$

$6.504$

$6.54$

$6.9$

$6.94$

Answer: C

Concepts:fraction decimal

Difficulty rating: 450

Small Hint:

Simplify each fraction first

Big Hint:

$\frac{44}{11}=4$ , $\frac{110}{44}=2.5$ , and $\frac{44}{1100}=0.04$

Video solution:

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Written solution:

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.

3.

Four squares of side length $4,$ $7,$ $9,$ and $10$ units are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate shaded and unshaded, as shown in the figure. What is the area of the visible shaded region in square units?

$42$

$45$

$49$

$50$

$52$

Answer: E

Concepts:area decomposition square (geometry)

Difficulty rating: 720

Small Hint:

Use shaded and unshaded nested square areas

Big Hint:

The visible shaded area is $10^2-9^2+7^2-4^2$

Video solution:

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Written solution:

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 $\begin{gathered} 10^2-9^2+7^2-4^2 \\ =100-81+49-16 \\ =52. \end{gathered}$

Thus, E is the correct answer.

4.

When Yunji added all the integers from $1$ to $9,$ she mistakenly left out a number. Her incorrect sum turned out to be a square number. Which number did Yunji leave out?

$5$

$6$

$7$

$8$

$9$

Answer: E

Concepts:triangular number perfect square

Difficulty rating: 770

Small Hint:

The sum from $1$ to $9$ is $45$

Big Hint:

The incorrect sum must be a square below $45$

Video solution:

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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.

5.

Aaliyah rolls two standard $6$ -sided dice. She notices that the product of the two numbers rolled is a multiple of $6.$ Which of the following integers cannot be the sum of the two numbers?

$5$

$6$

$7$

$8$

$9$

Answer: B

Concepts:dice (probability)divisibility casework

Difficulty rating: 960

Small Hint:

A multiple of $6$ needs a factor of $2$ and a factor of $3$

Big Hint:

List the possible sums from dice products divisible by $6$

Video solution:

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Written solution:

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: $\begin{gathered} 5=2+3,\quad 7=1+6, \\ 8=2+6,\quad 9=3+6. \end{gathered}$ There is no way to get sum $6$ while also having a product divisible by $6$ : the pairs with sum $6$ are $(1,5),$ $(2,4),$ $(3,3),$ $(4,2),$ and $(5,1)$ , and none have product divisible by $6$ .

Thus, B is the correct answer.

6.

Sergei skated around an ice rink, gliding along different paths. The marked lines in the figures below show four of the paths labeled P, Q, R, and S. What is the sorted order of the four paths from shortest to longest?

P, Q, R, S

P, R, S, Q

Q, S, P, R

R, P, S, Q

R, S, P, Q

Answer: D

Concepts:arc bounding to limit cases

Difficulty rating: 1070

Small Hint:

Compare the path pieces: arcs are longer than their chords

Big Hint:

The answer choices already agree that Path R is shortest and Path Q is longest

Video solution:

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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.

7.

A $3 \times 7$ rectangle is covered without overlap by $3$ shapes of tiles: $2 \times 2,$ $1 \times 4,$ and $1 \times 1,$ shown below. What is the minimum possible number of $1 \times 1$ tiles used?

$1$

$2$

$3$

$4$

$5$

Answer: E

Concepts:tiling parity extremal argument

Difficulty rating: 1310

Small Hint:

The area left for non-unit tiles must be a multiple of $4$

Big Hint:

If only one unit tile were used, two rows would need an odd number of cells covered by larger tiles

Video solution:

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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.

8.

On Monday Taye has $\$2.$ Every day, he either gains $\$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?

$3$

$4$

$5$

$6$

$7$

Answer: D

Concepts:tree diagram systematic listing

Difficulty rating: 1030

Small Hint:

After each day, branch into adding $3$ or doubling

Big Hint:

List the amounts after Tuesday, then Wednesday, then Thursday

Video solution:

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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.

9.

All of the marbles in Maria’s collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria’s collection?

$24$

$25$

$26$

$27$

$28$

Answer: E

Concepts:ratio and proportion divisibility

Difficulty rating: 870

Small Hint:

If red is $r$ , then green is $2r$

Big Hint:

Blue is twice green, so the total is a multiple of $7$

Video solution:

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Written solution:

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.

10.

In January $1980$ the Mauna Loa Observatory recorded carbon dioxide CO_$2$ levels of $338$ ppm (parts per million). Over the years the average CO_$2$ reading has increased by about $1.515$ ppm each year. What is the expected CO_$2$ level in ppm in January $2030?$ Round your answer to the nearest integer.

$399$

$414$

$420$

$444$

$459$

Answer: B

Concepts:arithmetic sequence decimal estimation

Difficulty rating: 960

Small Hint:

There are $50$ years from January $1980$ to January $2030$

Big Hint:

Estimate $50\cdot1.515$ , then add the $1980$ level

Video solution:

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Written solution:

There are $50$ years between $1980$ and $2030$ , so we can expect the CO_$2$ reading to increase by $50 \times 1.515 = 75.75 \approx 76$ ppm by $2030$ . Since the CO_$2$ reading in $1980$ was $338$ ppm, then we will have $338 + 76 = 414$ ppm by $2030$ .

Thus, B is the correct answer.

11.

The coordinates of $\bigtriangleup ABC$ are $A(5, 7),$ $B(11, 7),$ and $C(3, y),$ with $y > 7.$ The area of $\bigtriangleup ABC$ is $12.$ What is the value of $y?$

$8$

$9$

$10$

$11$

$12$

Answer: D

Concepts:coordinate geometry triangle area linear equation

Difficulty rating: 960

Small Hint:

Use $AB$ as the base of the triangle

Big Hint:

The height from $C$ to $AB$ is $y-7$

Video solution:

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Written solution:

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.

12.

Rohan keeps a total of $90$ guppies in $4$ fish tanks.

• There is $1$ more guppy in the $2$ nd tank than in the $1$ st tank.

• There are $2$ more guppies in the $3$ rd tank than in the $2$ nd tank.

• There are $3$ more guppies in the $4$ th tank than in the $3$ rd tank.

How many guppies are in the $4$ th tank?

$20$

$21$

$23$

$24$

$26$

Answer: E

Concepts:linear equation

Difficulty rating: 980

Small Hint:

Write each tank amount in terms of the first tank

Big Hint:

The fourth tank has $6$ more guppies than the first

Video solution:

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Written solution:

Let $x$ be the number of guppies in the $1$ st tank. Hence, there are $x+1$ guppies in the $2$ nd tank, $x+3$ guppies in the $3$ rd tank, and $x+6$ guppies in the $4$ th 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 $4$ th tank and not the $1$ st. There are $x+6=20+6=26$ guppies in the $4$ th tank.

Thus, E is the correct answer.

13.

Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)

$4$

$5$

$6$

$8$

$12$

Answer: B

Concepts:Catalan Number systematic listing

Difficulty rating: 1310

Small Hint:

The sequence must have three up hops and three down hops

Big Hint:

Buzz can never have more down hops than up hops at any point

Video solution:

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Written solution:

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.

14.

The one-way routes connecting towns $A,$ $M,$ $C,$ $X,$ $Y,$ and $Z$ are shown in the figure below (not drawn to scale). The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from $A$ to $Z$ in kilometers?

$28$

$29$

$30$

$31$

$32$

Answer: A

Concepts:graph theory

Difficulty rating: 1420

Small Hint:

Track the shortest distance from $A$ to each town

Big Hint:

Once a shortest distance to a town is known, use it to update outgoing routes

Video solution:

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Written solution:

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.

15.

Let the letters $F,$ $L,$ $Y,$ $B,$ $U,$ $G$ represent distinct digits. Suppose $\text{FLYFLY}$ is the greatest number that satisfies the equation $8 \cdot \text{FLYFLY} = \text{BUGBUG}.$ What is the value of $\text{FLY} + \text{BUG}?$

$1089$

$1098$

$1107$

$1116$

$1125$

Answer: C

Concepts:cryptarithm place value optimization

Difficulty rating: 1540

Small Hint:

$\text{FLYFLY}=1001\cdot\text{FLY}$

Big Hint:

The equation reduces to $8\cdot\text{FLY}=\text{BUG}$

Video solution:

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Written solution:

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

Because $\text{BUG}$ has three digits, $\text{FLY}\le124.$ The largest possible three-digit number at most $124$ with distinct digits is $124,$ but $8\cdot124=992$ repeats a digit and also reuses the digit $2.$

The next candidate is $123,$ and $8\cdot123=984.$ The six digits $1,$ $2,$ $3,$ $9,$ $8,$ and $4$ are all distinct, so $123$ is the greatest possible value of $\text{FLY}.$

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

Thus, C is the correct answer.

16.

Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3?$

$8$

$9$

$10$

$11$

$12$

Answer: D

Concepts:divisibility optimization extremal argument

Difficulty rating: 1660

Small Hint:

Only multiples of $3$ make a row or column product divisible by $3$

Big Hint:

Cover the $27$ multiples of $3$ using as few rows and columns as possible

Video solution:

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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.

17.

A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a $3 \times 3$ grid attacks all $8$ other squares, as shown below. Suppose a white king and a black king are placed on different squares of a $3 \times 3$ grid so that they do not attack each other. In how many ways can this be done?

$20$

$24$

$27$

$28$

$32$

Answer: E

Concepts:casework basic counting

Difficulty rating: 1410

Small Hint:

Count ordered placements for the white king first

Big Hint:

A corner allows $5$ safe black-king squares; an edge-center allows $3$

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.

18.

Three concentric circles centered at $O$ have radii of $1,$ $2,$ and $3.$ Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC,$ as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle BOC$ in degrees?

$108$

$120$

$135$

$144$

$150$

Answer: A

Concepts:annulus sector linear equation

Difficulty rating: 1540

Small Hint:

Compute the area of the shaded annulus between radii $1$ and $2$

Big Hint:

Set shaded area equal to unshaded area and solve for the sector angle

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 $\theta:$ $\begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered}$ $2\pi = \dfrac{360-\theta-\theta}{360}(5\pi)$ $\dfrac{2}{5} = 1 - \dfrac{2\theta}{360}$ $2\theta = \dfrac{3}{5}(360)$ $\theta = 108.$

Thus, A is the correct answer.

19.

Jordan owns $15$ pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?

$0$

$\dfrac{1}{5}$

$\dfrac{4}{15}$

$\dfrac{1}{3}$

$\dfrac{2}{5}$

Answer: C

Concepts:inclusion-exclusion optimization fraction

Difficulty rating: 1170

Small Hint:

There are $9$ red pairs and $10$ high-top pairs

Big Hint:

To minimize red high-tops, make as many white pairs high-top as possible

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.

20.

Any three vertices of the cube $PQRSTUVW,$ shown in the figure below, can be connected to form a triangle. (For example, vertices $P,$ $Q,$ and $R$ can be connected to form isosceles $\bigtriangleup PQR.$ ) How many of these triangles are equilateral and contain $P$ as a vertex?

$0$

$1$

$2$

$3$

$6$

Answer: D

Concepts:cube geometry equilateral triangle

Difficulty rating: 1420

Small Hint:

An equilateral triangle through $P$ must use face diagonals from $P$

Big Hint:

Look at the three vertices a face diagonal away from $P$

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.

21.

A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1.$ Then $3$ green frogs moved to the sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1.$ What is the difference between the number of green frogs and yellow frogs now?

$10$

$12$

$16$

$20$

$24$

Answer: E

Concepts:ratio and proportion linear equation substitution

Difficulty rating: 1340

Small Hint:

Let the initial yellow count be $y$

Big Hint:

After the moves, green changes by $+2$ and yellow changes by $-2$

Video solution:

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Written solution:

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.

22.

A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.

$300$

$600$

$1200$

$1500$

$1800$

Answer: B

Concepts:circumference estimation

Difficulty rating: 1580

Small Hint:

The tape occupies an annulus with outer radius $2$ and inner radius $1$

Big Hint:

Area of tape cross-section equals thickness times total length

Video solution:

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Written solution:

The tape occupies an annulus with outer radius $2$ inches and inner radius $1$ inch, so its cross-sectional area is $\pi(2^2-1^2)=3\pi.$

When unrolled, this same cross section is a rectangle of thickness $0.015$ inches and length $L.$ Thus $0.015L=3\pi,$ so $L=200\pi\approx628$ inches. Rounded to the nearest $100$ inches, this is $600$ inches.

Thus, B is the correct answer.

23.

Rodrigo has a very large piece of graph paper. First he draws a line segment connecting point $(0, 4)$ to point $(2, 0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next, Rodrigo draws a line segment connecting point $(2000, 3000)$ to point $(5000, 8000).$ Again he colors the cells whose interiors intersect the segment. How many cells will he color this time?

$6000$

$6500$

$7000$

$7500$

$8000$

Answer: C

Concepts:greatest common divisor lattice point

Difficulty rating: 1810

Small Hint:

For a segment with integer endpoint differences $a$ and $b$ , count grid cells crossed

Big Hint:

Use $a+b-\gcd(a,b)$ for cells whose interiors are intersected

Video solution:

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Written solution:

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.

24.

Jean made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high and the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides of the mountains form $45^\circ$ angles with the ground. The artwork has an area of $183$ square feet. The sides of the mountains meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$

$4$

$5$

$4\sqrt{2}$

$6$

$5\sqrt{2}$

Answer: B

Concepts:special right triangle area inclusion-exclusion

Difficulty rating: 1710

Small Hint:

Each mountain is built from a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle

Big Hint:

Add the two large triangle areas and subtract the overlap

Video solution:

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Written solution:

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.

25.

A small airplane has $4$ rows of seats with $3$ seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be $2$ adjacent seats in the same row for the couple?

$\dfrac{8}{15}$

$\dfrac{32}{55}$

$\dfrac{20}{33}$

$\dfrac{34}{55}$

$\dfrac{8}{11}$

Answer: C

Concepts:basic probability complementary counting generating functions

Difficulty rating: 1950

Small Hint:

Count the four empty seats after the first eight passengers board

Big Hint:

Use the complement: no row has two adjacent empty seats

Video solution:

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Written solution:

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 $0,$ $1,$ and $2,$ with $1,$ $3,$ and $1$ choices respectively. So we need the coefficient of $x^4$ in $(1+3x+x^2)^4.$ This coefficient is $\begin{gathered} 3^4+4\cdot3\cdot3^2+\binom42 \\ =81+108+6 \\ =195. \end{gathered}$

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.