2020 AMC 8 Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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Tiempo restante:

40:00

1.

Luka is making lemonade to sell at a school fundraiser. His recipe requires 44 times as much water as sugar and twice as much sugar as lemon juice. He uses 33 cups of lemon juice. How many cups of water does he need?

66

88

1212

1818

2424

Answer: E
Video solution:
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Written solution:

Since Luka needs twice as much sugar as lemon, he needs 23=62\cdot3=6 cups of sugar. Since Luka also needs 44 times as much water as sugar, he needs 46=244\cdot6=24 cups of water.

Thus, the correct answer is E.

2.

Four friends do yardwork for their neighbors over the weekend, earning $15, $20, $25, and $40 respectively. They decide to split their earnings equally among themselves. In total how many dollars will the friend who earned $40 give to the others?

55

1010

1515

2020

2525

Answer: C
Video solution:
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Written solution:

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

Since they divide this equally, they each get 1004=25\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.

3.

Carrie has a rectangular garden that measures 66 feet by 88 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 44 strawberry plants per square foot, and she harvests an average of 1010 strawberries per plant. How many strawberries can she expect to harvest?

560560

960960

11201120

19201920

38403840

Answer: D
Video solution:
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Written solution:

First, the size of the garden is 6ft8ft=48ft2.6\text{ft} \cdot 8\text{ft} = 48\text{ft}^2.

Next, since there are 44 plants per square foot, Carrie can plant 448=1924\cdot 48=192 plants total.

Finally, since there are 1010 strawberries per plant, Carrie can harvest 10192=192010\cdot 192=1920 strawberries total.

Thus, the correct answer is D.

4.

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?

3535

3737

3939

4343

4949

Answer: B
Video solution:
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Written solution:

Firstly, let's try and find a pattern for the number of dots in the nnth ring. Notice that for the nnth ring, there is always nn dots on each edge of the hexagon. However, this overcounts as each vertex of the hexagon will be counted twice.

Therefore, we can claim that for the nnth ring, the number of dots is equal to 6n6=6(n1).6n-6 = 6(n-1). Note that this pattern does not hold for n=1.n=1.

Therefore, for the first hexagon, we have 11 dot, the 2nd hexagon adds 6(21)=66(2-1)=6 dots, the 3rd hexagon adds 6(31)=126(3-1)=12 dots, as reflected in the diagram. Extending the pattern, we can say that the fourth hexagon adds 6(41)=186(4-1)=18 dots.

Finally, notice that the total number of dots in the hexagon is equal to the sum of all the rings up to the nnth ring. This is the same as saying: 1+i=2n6(i1)1+\sum_{i=2}^n 6(i-1) Therefore, for n=4n=4 we have a total of 1+i=246(n1)=1+6+12+18=37\begin{align*} &1+\sum_{i=2}^4 6(n-1)\\ &= 1+6+12+18 \\&= 37\end{align*}

Thus, the correct answer is B.

5.

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of 55 cups. What percent of the total capacity of the pitcher did each cup receive?

55

1010

1515

2020

2525

Answer: C
Video solution:
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Written solution:

Since we start with 34\dfrac{3}{4} of the pitcher, we have 75%75\% of the pitcher full as 34=75100.\dfrac{3}{4} = \dfrac{75}{100}.

Now, since 55 cups each have the same amount of juice, they each have one-fifth of the 75%.75\%.

This means they have 1575%=(755)%=15%.\dfrac15\cdot 75\%=\left(\dfrac{75}{5}\right)\% = 15\%.

Thus, the correct answer is C.

6.

Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?

Aaron

Darren

Karen

Maren

Sharon

Answer: A
Video solution:
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Written solution:

We know that they must be arranged as such: \text{__, __, __, __, Maren}, with each "__" representing people whose location we don't yet know.

Note that Maren must be there since he sat in the last car.

Since we know Sharon sat in front of Aaron, we can take cases of the possible locations of them both.

Case 1: \text{Sharon, Aaron, __, __, Maren} Since Darren sits in front of Aaron, we can't find a space for him so this case can't work.

Case 2: \text{__, Sharon, Aaron, __, Maren} Since Darren sits in front of Aaron, we know the configuration must be Darren, Sharon, Aaron, __, Maren -- leaving one spot for Karen.

The final configuration is Darren, Sharon, Aaron, Karen, Maren.

Case 3: \text{__, __, Sharon, Aaron, Maren} This case would involve Darren and Karen being next to each other, which contradicts the condition that at least one person is between them.

Since the only valid configuration is Darren, Sharon, Aaron, Karen, Maren, we know Aaron is in the middle.

Thus, the correct answer is A.

7.

How many integers between 20202020 and 24002400 have four distinct digits arranged in increasing order? (For example, 23572357 is one integer.)

99

1010

1515

2121

2828

Answer: C
Video solution:
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Written solution:

Since the integers are between 20202020 and 2400,2400, we know the thousands digit must be 2 and the hundreds digit must be between 00 and 3.3.

Since the digits are increasing, the second digit must be greater than 2,2, so it can only be 3.3.

This means the tens and units digits are different digits each greater than or equal to 4.4.

Suppose we choose 2 distinct digits each greater than or equal to 4.4. There are 66 digits for the first choice and 55 digits for the second choice.

This means there are 3030 combinations.

However, we ignore exactly half of these combinations as each combination has an equal likelihood of being ascending or descending.

This leaves 302=15\frac{30}{2}=15 combinations.

Thus, the correct answer is C.

8.

Ricardo has 20202020 coins, some of which are pennies (11-cent coins) and the rest of which are nickels (55-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?

80628062

80688068

80728072

80768076

80828082

Answer: C
Video solution:
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Written solution:

Let pp be the number of pennies Ricardo has and let nn be the number of nickels he has.

We know that p+n=2020,p+n=2020, p1,p \geq 1, and n1n \geq 1 by the problem statement.

This means 2020n=p    2020n1.\begin{align*}2020 -n &=p \\ \implies 2020 -n &\geq 1.\end{align*} Therefore, 1n2019.1 \leq n \leq 2019. It follows, then, that Ricardo has p+5n=p+n+4n=2020+4n\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 2019,2019, so he can have at most 2020+2019(4)2020+2019(4) cents. The minimum number of nickels he can have is 1,1, so he has at most 2020+1(4)2020+1(4) cents. The difference between the maximum and minimum amount of money he can have is: 2020+2019(4)(2020+1(4))=2018(4)=8072. \begin{gather*} 2020+2019(4) - (2020+1(4)) \\ = 2018(4) = 8072 . \end{gather*}

Thus, the correct answer is C.

9.

Akash's birthday cake is in the form of a 4×4×44 \times 4 \times 4 inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into 6464 smaller cubes, each measuring 1×1×11 \times 1 \times 1 inch, as shown below. How many small pieces will have icing on exactly two sides?

1212

1616

1818

2020

2424

Answer: D
Video solution:
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Written solution:

On the top side, there are 44 cubes that are only on the top. Also, there are 44 cubes that have icing on 33 sides. Therefore, we have 1644=816-4-4=8 cubes on the top that have icing on exactly 22 sides.

For the 44 other sides, we need to find the number of cubes that have icing on exactly two sides, exluding the cubes we counted on the sides. Each face has 66 cubes on the edge that have icing on exactly two sides. However, this would double count when taking into account the other faces, as each cube would be counted for two faces. Therefore, we need to add 642=12\dfrac{6\cdot 4}{2} = 12 cubes.

The final answer is 8+12=20.8+12 = 20.

Thus, the correct answer is D.

10.

Zara has a collection of 44 marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?

66

88

1212

1818

2424

Answer: C
Video solution:
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Written solution:

Let XX and YY represent the Steelie and the Tiger in some order, with XX always coming before Y.Y.

To place XX and YY we can have the following cases:

\quadX,X, __, Y,Y, __ ;

\quadX,X, __, __, YY;

\quad__, X,X, __, Y,.Y,.

leaving 3 configurations, with "__" being the reserved spots for the other marbles. With any other configuration, we have XX and YY next to each other or YY before X.X.

Now, each configuration can have either the Steelie be XX and the Tiger be YY or Steelie be YY and the Tiger be X.X. This doubles the number of configurations we have, making 6.6.

Also, each configuration can have either the Bumblebee be the first spot and Aggie be the second spot or vice versa. This again doubles the number of configurations we have, making 12.12.

Thus, the correct answer is C.

11.

After school, Maya and Naomi headed to the beach, 66 miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?

66

1212

1818

2020

2424

Answer: E
Video solution:
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Written solution:

Naomi traveled 66 miles in 1010 minutes. This ratio is 6 miles10 minutes=36 miles60 minutes=36 mileshour \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 66 miles in 3030 minutes. This ratio is 6 miles30 minutes=12 miles60 minutes=12 mileshour \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 3612=24.36-12=24.

Thus, the correct answer is C.

12.

For a positive integer n,n, the factorial notation n!n! represents the product of the integers from nn to 1.1. For example:

6!=6×5×4×3×2×1 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1

What value of NN satisfies the following equation?

5!×9!=12×N! 5! \times 9! = 12 \times N!

1010

1111

1212

1313

1414

Answer: A
Video solution:
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Written solution:

Note first that n!=n(n1)(n2)1=n((n1)(n2)1)=n(n1)!\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: 5!9!=543219!=1209!=12(109!)\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 12N!=12(109!),12 \cdot N! = 12(10\cdot 9!), we know N!=109!.N! = 10\cdot 9!.

Using our note from above, we know that 109!=10!10\cdot 9!= 10! so N=10.N=10.

Thus, the correct answer is A.

13.

Jamal has a drawer containing 66 green socks, 1818 purple socks, and 1212 orange socks. After adding more purple socks, Jamal noticed that there is now a 60%60\% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

66

99

1212

1818

2424

Answer: B
Video solution:
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Written solution:

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

Also, since there are 3636 total socks to begin with, we have s+36s+36 socks after adding the socks.

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

Solving for ss yields: s+18=0.6s+21.60.4s=3.6s=9.\begin{align*}s+18 &= 0.6s+21.6 \\ 0.4s &= 3.6\\ s &= 9.\end{align*}

Therefore, 99 socks are added.

Thus, the correct answer is B.

14.

There are 2020 cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all 2020 cities?

65,00065,000

75,00075,000

85,00085,000

95,00095,000

105,000105,000

Answer: D
Video solution:
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Written solution:

Looking at the horizontal dashed line, the average population is around 4750.4750.

Since there are 2020 cities, and the average population is total populationnumber of cities,\frac{\text{total population}}{\text{number of cities}}, we know: total population20=4750.\dfrac{\text{total population}}{20} = 4750.

Multiplication yields that the total population is 95,000.95,000.

Thus, the correct answer is D.

15.

Suppose 15%15\% of xx equals 20%20\% of y.y. What percentage of xx is y?y?

55

3535

7575

13313133 \frac{1}{3}

300300

Answer: C
Video solution:
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Written solution:

If a number mm is pp percent of nn then p100m=n.\frac{p}{100}m=n.

This means that 0.15x=0.2y.0.15x = 0.2y.

Dividing each side by 0.20.2 gives that 0.75x=y.0.75x=y.

Therefore, we know that xx is 75%75\% of y.y.

Thus, the correct answer is C.

16.

Suppose Each of the points A,B,C,D,E,A,B,C,D,E, and FF in the figure below represents a different digit from 11 to 6.6. Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is 47.47. What is the digit represented by B?

11

22

33

44

55

Answer: E
Video solution:
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Written solution:

Each number is added once per line it is on. Every point is on 22 lines except for BB which is on 3.3.

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

Since A,B,C,D,E,FA,B,C,D,E,F are unique digits from 11 to 6,6, each digit is represented exactly once, making A+B+C+D+E+F=21.\begin{align*} &A + B + C + D + E + F \\ &= 21. \end{align*}

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

Thus, the correct answer is E.

17.

How many factors of 20202020 have more than 33 factors? (As an example, 1212 has 66 factors, namely 1,1, 2,2, 3,3, 4,4, 6,6, and 12.12.)

66

77

88

99

1010

Answer: B
Video solution:
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Written solution:

Let's begin by firstly simply factoring 20202020: 2020=12020=21010=4505=5404=10202=20101\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 20202020 can be classified by the number of their factors:

\quad - 11 has one factor

\quad - 2,5,2, 5, and 101101 has two factors

\quad - 44 has three (distinct) factors

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

Thus, the correct answer is B.

18.

Rectangle ABCDABCD is inscribed in a semicircle with diameter FE,\overline{FE}, as shown in the figure. Let DA=16,DA=16, and let FD=AE=9.FD=AE=9. What is the area of ABCD?ABCD?

240240

248248

256256

264264

272272

Answer: A
Video solution:
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Written solution:

Since FEFE is the diameter of the semicircle, we know the length of the diameter is 34,34, and so the radius is 17.17. Let OO be the center of the diameter.

The length from OFOF therefore is 17.17.

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

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

Finally, since ABCDABCD is a rectangle, we know ODC\angle ODC is a right angle. This means we can find DCDC by the Pythagorean Theorem. We know OD2+DC2=OC282+DC2=172DC=15.\begin{align*} OD^2+DC^2&=OC^2 \\ 8^2+DC^2 &= 17^2 \\ DC &= 15. \end{align*}

As such, the area of the rectangle is DCDA=1516=240.DC\cdot DA = 15\cdot 16=240.

Thus, the correct answer is A.

19.

A number is called flippy if its digits alternate between two distinct digits. For example, 20202020 and 3737337373 are flippy, but 38833883 and 123123123123 are not. How many five-digit flippy numbers are divisible by 15?15?

33

44

55

66

88

Answer: B
Video solution:
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Written solution:

For a number to be divisible by 1515 the number must be divisible by 33 and by 5.5.

First, to ensure the number is divisible by 55 it must end in 00 or in 5.5.

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

This would make our number either 0d0d00d0d0 or 5d5d5.5d5d5.

We can ignore the numbers in the form 0d0d00d0d0 as they would reduce to a 44 digit number.

As such, we know our number is 5d5d55d5d5 for some digit d.d.

To ensure our number is also a multiple of 33 the sum of the digits must be a multiple of 3.3.

The sum of our digits is 5+d+5+d+5=15+2d.5+d+5+d+5 = 15+2d. Since 1515 is a multiple of 3,3, all that is required is that dd is a multiple of 3.3.

This means dd can be 0,3,60,3,6 or 9.9.

Therefore, we have 44 solutions.

Thus, the correct answer is B.

20.

A scientist walking through a forest recorded as integers the heights of 55 trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?

\renewcommand{\arraystretch}{1.5} \begin{array}{|c|c|} \hline \text{Tree 1} & \text{__ meters} \\ \text{Tree 2} & 11 \text{ meters} \\ \text{Tree 3} & \text{__ meters} \\ \text{ Tree 4 }& \text{__ meters} \\ \text{Tree 5} & \text{__ meters} \\ \hline \text{Average height} & \text{__.2 meters} \\ \hline \end{array}

22.222.2

24.224.2

33.233.2

35.235.2

37.237.2

Answer: B
Video solution:
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Written solution:

We know that tree 2 is 1111 meters tall, and since the trees on either side of any given true were said to be either double the height or half the height, we can conclude that tree 11 and three 33 must both be 2222 meters tall, as if they were half that of tree 2,2, they would not be integers.

This gives us only four possibilities for the remaining two trees, of which we simply test each case for an average that ends in ".2.2":

\quad - Tree 44 is 4444 meters; \quadTree 55 is 8888 meters:

\quad\quad 55+44+885=37.4\dfrac{55+44+88}{5}=37.4

\quad - Tree 44 is 4444 meters; \quadTree 55 is 2222 meters:

\quad\quad 55+44+225=24.2\dfrac{55+44+22}{5}=24.2

\quad - Tree 44 is 1111 meters; \quadTree 55 is 2222 meters:

\quad\quad 55+11+225=17.6\dfrac{55+11+22}{5}=17.6

\quad - Tree 44 is 1111 meters; \quadTree 55 is 5.55.5 meters:

\quad\quadThis means that Tree 55 is not an integer, as therefore this case is invalid and discarded.

The only one of these cases that has an average that ends in ".2.2" is the 11,2211,22 case. As such, the average height is 24.224.2 meters.

Thus, the correct answer is B.

21.

A game board consists of 6464 squares that alternate between blank and colored. The figure below shows square PP in the bottom row and square QQ in the top row. A marker is placed at P.P. A step consists of moving the marker onto one of the adjoining blank squares in the row above. How many 77-step paths are there from PP to Q?Q? (The figure shows a sample path.)

2828

3030

3232

3333

3535

Answer: A
Video solution:
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Written solution:

Every move must go either down or up and either left or right. Since the final position is 77 moves up and we move 77 times, every move must go up. Since after nn moves, we are on the nnth row, we can use the nnth row to construct the ways to get to each position from the previous moves. Each move can come from the down-left or down-right direction, so we get the ways to get to a point by adding the number of paths from those two directions.

We have then constructed the ways to get to each point from PP assuming we always go up. The circled point is Q,Q, so we have 2828 ways to get to Q.Q.

Thus, the correct answer is A.

22.

When a positive integer NN is fed into a machine, the output is a number calculated according to the rule shown below.N{N2,N is even3N+1,N is odd N\mapsto \begin{cases}\dfrac N2, \text{$N$ is even} \\ 3N+1, \text{$N$ is odd} \end{cases}

For example, starting with an input of N=7,N=7, the machine will output 37+1=22.3 \cdot 7 +1 = 22. Then if the output is repeatedly inserted into the machine five more times, the final output is 26.26. 7221134175226 \begin{gather*} 7 \to 22 \to 11 \to 34 \\ \to 17 \to 52 \to 26 \end{gather*} When the same 66-step process is applied to a different starting value of N,N, the final output is 1.1. What is the sum of all such integers N?N? \begin{gather*} N \to \text{ __ } \to \text{ __ } \to \text{ __ } \\ \to \text{ __ } \to \text{ __ } \to 1 \end{gather*}

7373

7474

7575

8282

8383

Answer: E
Video solution:
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Written solution:

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

This would take NN and yield either 2N2N if 2N2N is even (which it always is) or N13\frac{N-1}{3} if N13\frac{N-1}{3} is an odd integer. Note that N13\frac{N-1}{3} is an integer only if N1mod3.N \equiv 1 \mod 3. Also, if NN is even, then N1N-1 is odd. That would mean N13\frac{N-1}{3} would be odd. Therefore, our inverter machine yields 2N2N and also N13\frac{N-1}{3} if N1mod3N \equiv 1 \mod 3 and NN is even.

Now, we must see what the inverting machine can yield after 66 moves:

1)1) We can only get 2.2.

2)2) From 2,2, we can only get 4.4.

3)3) From 4,4, we can get 11 and 8.8.

4)4) From 1,1, we can get only 22; from 8,8, we can only get 16.16.

5)5) From 2,2, we can get only 44; from 16,16, we can get 55 and 32.32.

6)6) From 4,4, we can get 11 and 8,8,; 5,5, we can get 1010; from 3232 we can get 64.64.

Move 66 can yield 1,8,10,1,8,10, and 6464 and their sum is 83.83.

Thus, the correct answer is E.

23.

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?

120120

150150

180180

210210

240240

Answer: B
Video solution:
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Written solution:

First, we can calculate the number of way to just give out 55 awards, without making sure every student has at least 1.1. This is 35=243.3^5=243.

Next we subtract the number of distributions without everyone having at least 1.1.

This can be counted as the number of distributions with at most 22 people having every award.

We can sort this by two cases. Case 1 has exactly 2 people having at least one award. Case 2 has exactly 1 person having at least one award.

Case 1: If two people have all the awards, there are 252^5 ways to distribute the awards amongst these two people. However, we must subtract 22 as that has only one person owning all the awards, which is outside the definiton of this case. This has 3030 combinations. Finally, we multiply this by 33 as there are 33 ways to choose the group of two people. This case yields 9090 distributions

Case 2: If one people has all the awards, there is 11 way to distribute the awards amongst that person. Finally, we multiply this by 33 as there are 33 ways to choose the person. This case yields 33 distributions.

There are a total of 9393 cases to remove. Therefore, the answer is 24393=150.243-93=150.

Thus, the correct answer is B.

24.

A large square region is paved with n2n^2 gray square tiles, each measuring ss inches on a side. A border dd inches wide surrounds each tile. The figure below shows the case for n=3.n=3. When n=24n=24 , the 576576 gray tiles cover 64%64\% of the area of the large square region. What is the ratio ds\frac{d}{s} for this larger value of n?n?

625\dfrac{6}{25}

14\dfrac{1}{4}

925\dfrac{9}{25}

716\dfrac{7}{16}

916\dfrac{9}{16}

Answer: A
Video solution:
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Written solution:

First, since n=24,n=24, there are n2=242=576n^2=24^2=576 squares.

Since we aren't given an area of the large square and only the ratios within it, we can define the area to be 1.1. Since the total area of the gray is 64100=(45)2,\dfrac{64}{100} = (\dfrac{4}{5})^2, the area for each square is (45)2(1576)=(45)2(124)2=(130)2. \begin{gather*} \left(\dfrac{4}{5}\right)^2 \left(\dfrac{1}{576}\right) \\ = \left(\dfrac{4}{5}\right)^2 \left(\dfrac{1}{24}\right)^2 = \left(\dfrac{1}{30}\right)^2. \end{gather*}

Therefore, the side length of each tile is: (130)2=130.\sqrt{\left(\dfrac{1}{30}\right)^2} = \dfrac{1}{30}.

Now, there are n=24n=24 tiles and n+1=25n+1=25 borders, so, 25d+24s=1.25d+24s=1.

Since s=24,s=24, we have 25d+24(130)=125d=15d=1125.\begin{align*} 25d+24\left(\dfrac{1}{30}\right)&=1 \\ 25d&=\dfrac{1}{5} \\ d&=\dfrac{1}{125}. \end{align*}

Finally, ds=1125d30=30125=625.\begin{align*}\dfrac{d}{s} &= \dfrac{\frac{1}{125}}{\frac{d}{30}}\\ &= \dfrac{30}{125}\\&=\dfrac{6}{25}.\end{align*}

Thus, the correct answer is A.

25.

Rectangles R1R_1 and R2,R_2, and squares S1,S2,S_1,\,S_2,\, and S3,S_3, shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of S2S_2 in units?

651651

655655

656656

662662

666666

Answer: A
Video solution:
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Written solution:

We represent the lengths of each square as s1,s2,s_1,s_2, and s3s_3 respectively. The length of the rectangle is s1+s2+s3s_1+s_2+s_3 as these 33 squares span the entirety of a side of the large rectangle. Therefore, s1+s2+s3=3322.s_1+s_2+s_3=3322.

Also, the height of the large rectangle is the sum of the height of R2R_2 and S3.S_3. Now, note that the sum of height of R2R_2 and s2s_2 is s1,s_1, so height of R2R_2 is equal to s1s2.s_1-s_2. Therefore, the height of the large rectangle is s1s2+s3,s_1-s_2+s_3, which means s1+s2+s3=2020.s_1+s_2+s_3=2020. Subtracting both of our results yields 2s2=33222020=1302.2s_2=3322-2020=1302. This would mean s2=651.s_2=651.

Thus, the correct answer is A.