2022 AMC 8

Complete problem set with solutions and individual problem pages

Problem 1 Easy

The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?

  • A.

    10

  • B.

    12

  • C.

    13

  • D.

    14

  • E.

    15

Answer:A

Solution 1 (A quick solution)

We can see that there are 4 whole squares, since the area of each square will be 1, 4 \times 1 = 4. Next, there are 12 half squares, and 2 half squares is 1 whole square, so 12/2 = 6 whole squares. The area of this will be 6 \times 1 = 6. Finally, we can add the 2 numbers: 4 + 6 = \boxed{(A)\ 10}.

 

Solution 2

Draw the following four lines as shown:

We see these lines split the figure into five squares with side length \sqrt{2}. Thus, the area is 5 \cdot \left(\sqrt{2}\right)^2 = 5 \cdot 2 = \boxed{(A)\ 10}.

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Problem 2 Easy

Consider these two operations:

a \blacklozenge b = a^2 - b^2

a \bigstar b = (a - b)^2

What is the output of (5 \, \blacklozenge \, 3) \, \bigstar \, 6?

  • A.

    -20

  • B.

    4

  • C.

    16

  • D.

    100

  • E.

    220

Answer:D

We can find a general solution to any ((a \, \blacklozenge \, b) \, \bigstar \, c).

~~~~((a \, \blacklozenge \, b) \, \bigstar \, c)

=((a^2-b^2) \, \bigstar \, c)

=(a^2-b^2-c)^2

=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2

=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2

=625+81-225-300-225+108+36

=\boxed{\textbf{(D) } 100}

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Problem 3 Easy

When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:E

Solution 1

The positive divisors of 100 are

1,2,4,5,10,20,25,50,100.

It is clear that 10\leq c\leq50, so we apply casework to c:

- If c=10, then (a,b,c)=(2,5,10).

- If c=20, then (a,b,c)=(1,5,20).

- If c=25, then (a,b,c)=(1,4,25).

- If c=50, then (a,b,c)=(1,2,50).

Together, the numbers a,b, and c can be chosen in \boxed{\textbf{(E) } 4} ways.

 

Solution 2

The positive divisors of 100 are

1,2,4,5,10,20,25,50,100.

We apply casework to a:

If a=1, then there are 3 cases:

- b=2,c=50

- b=4,c=25

- b=5,c=20

If a=2, then there is only 1 case:

- b=5,c=10

In total, there are 3+1=\boxed{\textbf{(E) } 4} ways to choose distinct positive integer values of a,b,c.

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Problem 4 Easy

The letter M in the figure below is first reflected over the line q and then reflected over the line p. What is the resulting image?

  • A.

  • B.

  • C.

  • D.

  • E.

Answer:E

When M is first reflected over the line q, we obtain the following diagram:

When M is then reflected over the line p, we obtain the following diagram:

Therefore, the answer is \boxed{\textbf{(E)}}.

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Problem 5 Easy

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 30 years. How many years older than Bella is Anna?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:C

Five years ago, Bella was 6 years old, and the kitten was 0 years old.

Today, Bella is 11 years old, and the kitten is 5 years old. It follows that Anna is 30-11-5=14 years old.

Therefore, Anna is 14-11=\boxed{\textbf{(C) } 3} years older than Bella.

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Problem 6 Easy

Three positive integers are equally spaced on a number line. The middle number is 15, and the largest number is 4 times the smallest number. What is the smallest of these three numbers?

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    7

  • E.

    8

Answer:C

Solution 1

Let the smallest number be x. It follows that the largest number is 4x.

Since x,15, and 4x are equally spaced on a number line, we have

\begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*}

 

Solution 2

Let the common difference of the arithmetic sequence be d. Consequently, the smallest number is 15-d and the largest number is 15+d. As the largest number is 4 times the smallest number, 15+d=60-4d\implies d=9. Finally, we find that the smallest number is 15-9=\boxed{\textbf{(C) } 6}.

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Problem 7 Easy

When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about 56 kilobits per second. Approximately how many minutes would the download of a 4.2-megabyte song have taken at that speed? (Note that there are 8000 kilobits in a megabyte.)

  • A.

    0.6

  • B.

    10

  • C.

    1800

  • D.

    7200

  • E.

    36000

Answer:B

Solution 1

Notice that the number of kilobits in this song is 4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.

We must divide this by 56 in order to find out how many seconds this song would take to download: \frac{8 \times 7 \times 6 \times 100}{56} = \frac{56 \times 6 \times 100}{56} = 6 \times 100 = 600.

Finally, we divide this number by 60 because this is the number of seconds to get the answer \frac{600}{60}=\boxed{\textbf{(B) } 10}.

 

Solution 2

We seek a value of x that makes the following equation true, since every other quantity equals 1.

\frac{x\ \min }{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \min } = 1.

Solving yields x=\boxed{\textbf{(B) } 10}.

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Problem 8 Easy

What is the value of

\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?

  • A.

    \frac{1}{462}

  • B.

    \frac{1}{231}

  • C.

    \frac{1}{132}

  • D.

    \frac{2}{213}

  • E.

    \frac{1}{22}

Answer:B

Solution 1

Note that common factors (from 3 to 20, inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes

\frac{1}{{3}}\cdot\frac{2}{{4}}\cdot\frac{{3}}{{5}}\cdots\frac{{18}}{{20}}\cdot\frac{{19}}{21}\cdot\frac{{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.

 

Solution 2

The original expression becomes

\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.

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Problem 9 Easy

A cup of boiling water (212^{\circ}\text{F}) is placed to cool in a room whose temperature remains constant at 68^{\circ}\text{F}. Suppose the difference between the water temperature and the room temperature is halved every 5 minutes. What is the water temperature, in degrees Fahrenheit, after 15 minutes?

  • A.

    77

  • B.

    86

  • C.

    92

  • D.

    98

  • E.

    104

Answer:B

Initially, the difference between the water temperature and the room temperature is 212-68=144 degrees Fahrenheit.

After 5 minutes, the difference between the temperatures is 144\div2=72 degrees Fahrenheit.

After 10 minutes, the difference between the temperatures is 72\div2=36 degrees Fahrenheit.

After 15 minutes, the difference between the temperatures is 36\div2=18 degrees Fahrenheit. At this point, the water temperature is 68+18=\boxed{\textbf{(B) } 86} degrees Fahrenheit.

 

Remark:

Alternatively, we can condense the solution above into the following equation:68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.

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Problem 10 Easy

One sunny day, Ling decided to take a hike in the mountains. She left her house at 8 \, \text{AM}, drove at a constant speed of 45 miles per hour, and arrived at the hiking trail at 10 \, \text{AM}. After hiking for 3 hours, Ling drove home at a constant speed of 60 miles per hour. Which of the following graphs best illustrates the distance between Ling’s car and her house over the course of her trip?

  • A.

  • B.

  • C.

  • D.

  • E.

Answer:E

Solution 1 (Analysis) Note that:

- At {8 \, \text{AM},} Ling's car was {0} miles from her house.

- From 8 \, \text{AM} to 10 \, \text{AM}, Ling drove to the hiking trail at a constant speed of 45 miles per hour.

~~~It follows that at {10 \, \text{AM},} Ling's car was {45\cdot2=90} miles from her house.

- From 10 \, \text{AM} to 1 \, \text{PM}, Ling did not move her car.

~~~It follows that at {1 \, \text{PM},} Ling's car was still {90} miles from her house.

- From 1 \, \text{PM}, Ling drove home at a constant speed of 60 miles per hour. So, she arrived home 90\div60=1.5 hour later.

~~~It follows that at {2:30 \, \text{PM},} Ling's car was {0} miles from her house.

Therefore, the answer is \boxed{\textbf{(E)}}.

 

Solution 2 (Elimination)

Ling's trip took 2 hours, thus she traveled for 2 \times 45=90 miles. Choices \textbf{(B)}, \textbf{(C)}, and \textbf{(D)} are eliminated. Ling drove 45 miles per hour (mph) to the mountains, and 60 mph back to her house, so the rightmost slope must be steeper than the leftmost one. Choice \textbf {(A)} is eliminated. This leaves us with \boxed{\textbf{(E)}}.

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Problem 11 Medium

Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating 3 inches of pasta from the middle of one piece. In the end, he has 10 pieces of pasta whose total length is 17 inches. How long, in inches, was the piece of pasta he started with?

  • A.

    34

  • B.

    38

  • C.

    41

  • D.

    44

  • E.

    47

Answer:D

Solution 1

Let's say that the first strand of pasta he had is x. The first time he takes a bite of this strand will make it 2 pieces. This is x - 3. The second time he takes the bites of EACH strand will become (x - 3) * 3(2). The 2 is for the two bites he took, 1 for each strand. This equation is showing that for every bite that he is taking, he is taking off 3 inches. This is being showed in the 3(2). He took 2 bites, each taking off 3 inches. Now, the problem is stating that their are 10 pieces of pasta and the TOTAL length is 17 inches. We have to plug in the total number of bites to the equation (x - 3) * 3( ). To find this out, we can see that for the number of strands there are, he took 1 less bite than that number. For example, if there are 2 pieces, he took 1 bite to get those. Since there are 10 pieces, he took 9 bites. The equation then becomes, (x - 3) * 3(9) = 17. To solve this, x = 44. Therefore, the answer is \boxed{\textbf{(D) } 44}.

 

Solution 2

If there are 10 pieces of pasta, Henry took 10-1=9 bites. Each of these 9 bites took 3 inches of pasta out, so all his bites in total took away 9\cdot 3 = 27 inches of pasta. Thus, the original piece of pasta was 27+17=\boxed{\textbf{(D) } 44} inches long.

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Problem 12 Medium

The arrows on the two spinners shown below are spun. Let the number N equal 10 times the number on Spinner \text{A}, added to the number on Spinner \text{B}. What is the probability that N is a perfect square number?

  • A.

    \dfrac{1}{16}

  • B.

    \dfrac{1}{8}

  • C.

    \dfrac{1}{4}

  • D.

    \dfrac{3}{8}

  • E.

    \dfrac{1}{2}

Answer:B

Solution 1

First, we calculate that there are a total of 4\cdot4=16 possibilities. Now, we list all of two-digit perfect squares. 64 and 81 are the only ones that can be made using the spinner. Consequently, there is a \frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}} probability that the number formed by the two spinners is a perfect square.

 

Solution 2

There are 4 \cdot 4 = 16 total possibilities of N. We know N=10A+B, which A is a number from spinner A, and B is a number from spinner B. Also, notice that there are no perfect squares in the 50s or 70s, so only 4-2=2 values of N work, namely 64 and 81. Hence, \frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}.

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Problem 13 Medium

How many positive integers can fill the blank in the sentence below?

"One positive integer is            more than twice another, and the sum of the two numbers is 28."

  • A.

    6

  • B.

    7

  • C.

    8

  • D.

    9

  • E.

    10

Answer:D

Let m and n be positive integers such that m>n and m+n=28. It follows that m=2n+d for some positive integer d. We wish to find the number of possible values for d.

By substitution, we have (2n+d)+n=28, from which d=28-3n. Note that n=1,2,3,\ldots,9 each generate a positive integer for d, so there are \boxed{\textbf{(D) } 9} possible values for d.

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Problem 14 Medium

In how many ways can the letters in \textbf{BEEKEEPER} be rearranged so that two or more \textbf{E}s do not appear together?

  • A.

    1

  • B.

    4

  • C.

    12

  • D.

    24

  • E.

    120

Answer:D

All valid arrangements of the letters must be of the form

\textbf{E}\underline{\hspace{1.5mm}}\textbf{E}\underline{\hspace{1.5mm}}\textbf{E}\underline{\hspace{1.5mm}}\textbf{E}\underline{\hspace{1.5mm}}\textbf{E}

The problem is equivalent to counting the arrangements of \textbf{B},\textbf{K},\textbf{P}, and \textbf{R} into the four blanks, in which there are 4!=\boxed{\textbf{(D) } 24} ways.

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Problem 15 Medium

Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:C

Solution 1 

We are looking for a black point, that when connected to the origin, yields the lowest slope. The slope represents the price per ounce. We can visually find that the point with the lowest slope is the blue point. Furthermore, it is the only one with a price per ounce significantly less than 1. Finally, we see that the blue point is in the category with a weight of \boxed{\textbf{(C) } 3} ounces.

 

Solution 2 (Elimination)

By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram:

We then proceed in the same way that we had done in Solution 1. Following the steps, we figure out the blue dot that yields the lowest slope, along with passing the origin. We then can look at the x-axis(in this situation, the weight) and figure out it has \boxed{\textbf{(C) } 3} ounces.

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Problem 16 Hard

Four numbers are written in a row. The average of the first two is 21, the average of the middle two is 26, and the average of the last two is 30. What is the average of the first and last of the numbers?

  • A.

    24

  • B.

    25

  • C.

    26

  • D.

    27

  • E.

    28

Answer:B

Solution 1 (Arithmetic)

Note that the sum of the first two numbers is 21\cdot2=42, the sum of the middle two numbers is 26\cdot2=52, and the sum of the last two numbers is 30\cdot2=60.

It follows that the sum of the four numbers is 42+60=102, so the sum of the first and last numbers is 102-52=50. Therefore, the average of the first and last numbers is 50\div2=\boxed{\textbf{(B) } 25}.

 

Solution 2 (Algebra)

Let a,b,c, and d be the four numbers in that order. We are given that

\begin{align*} \frac{a+b}{2} &= 21, &(1) \\ \frac{b+c}{2} &= 26, &(2) \\ \frac{c+d}{2} &= 30, &(3) \end{align*}

and we wish to find \frac{a+d}{2}.

We add (1) and (3), then subtract (2) from the result:

\frac{a+d}{2}=21+30-26=\boxed{\textbf{(B) } 25}.

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Problem 17 Hard

If n is an even positive integer, the double factorial notation n!! represents the product of all the even integers from 2 to n. For example, 8!! = 2 \cdot 4 \cdot 6 \cdot 8. What is the units digit of the following sum?2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!

  • A.

    0

  • B.

    2

  • C.

    4

  • D.

    6

  • E.

    8

Answer:B

Solution 1

Notice that once n>8, the units digit of n!! will be 0 because there will be a factor of 10. Thus, we only need to calculate the units digit of

2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.

We only care about units digits, so we have 2+8+8+8\cdot8, which has the same units digit as 2+8+8+4. The answer is \boxed{\textbf{(B) } 2}.

 

Solution 2 (Solution 1 worded differently)

We can see that after 8!! in the sequence, the units digit is always 0 (every value after 8!! is a multiple of 10). Therefore, our answer is the sum of the units digits of 2!!, 4!!, 6!!, and 8!! respectively. This sum is equal to 2 + 8 + 8 + 4, or \boxed{\textbf{(B) } 2}.

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Problem 18 Hard

The midpoints of the four sides of a rectangle are (-3,0), (2,0), (5,4), and (0,4). What is the area of the rectangle?

  • A.

    20

  • B.

    25

  • C.

    40

  • D.

    50

  • E.

    80

Answer:C

Solution 1

The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.

Let A=(-3,0), B=(2,0), C=(5,4), and D=(0,4). Note that A,B,C, and D are the vertices of a rhombus whose diagonals have lengths AC=4\sqrt{5} and BD=2\sqrt{5}. It follows that the dimensions of the rectangle are 4\sqrt{5} and 2\sqrt{5}, so the area of the rectangle is 4\sqrt{5}\cdot2\sqrt{5}=\boxed{\textbf{(C) } 40}.

 

Solution 2

If a rectangle has area K, then the area of the quadrilateral formed by its midpoints is \frac{K}{2}.

Define points A,B,C, and D as Solution 1 does. Since A,B,C, and D are the midpoints of the rectangle, the rectangle's area is 2[ABCD]. Now, note that ABCD is a parallelogram since AB=CD and \overline{AB}\parallel\overline{CD}. As the parallelogram's height from D to \overline{AB} is 4 and AB=5, its area is 4\cdot5=20. Therefore, the area of the rectangle is 20\cdot2=\boxed{\textbf{(C) } 40}.

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Problem 19 Hard

Mr. Ramos gave a test to his class of 20 students. The dot plot below shows the distribution of test scores.

Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students 5 extra points, which increased the median test score to 85. What is the minimum number of students who received extra points?

(Note that the median test score equals the average of the 2 scores in the middle if the 20 test scores are arranged in increasing order.)

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    5

  • E.

    6

Answer:C

We notice that 13 students have scores under 85 currently and only 5 have scores over 85. We find the median of these two numbers, getting:

13-5=8

\frac{8}{2}=4

13-4=9

Thus, we realize that 4 students must have their score increased by 5.

So, the correct answer is \boxed{\textbf{(C)}4}.

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Problem 20 Hard

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number x in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of x?

  • A.

    -1

  • B.

    5

  • C.

    6

  • D.

    8

  • E.

    9

Answer:D

Solution 1

The sum of the numbers in each row is 12. Consider the second row. In order for the sum of the numbers in this row to equal 12, the two shaded numbers must add up to 13:

If two numbers add up to 13, one of them must be at least 7: If both shaded numbers are no more than 6, their sum can be at most 12. Therefore, for x to be larger than the three missing numbers, x must be at least 8. We can construct a working scenario where x=8:

So, our answer is \boxed{\textbf{(D) } 8}.

 

Solution 2

The sum of the numbers in each row is -2+9+5=12, and the sum of the numbers in each column is 5+(-1)+8=12.

Let y be the number in the lower middle. It follows that x+y+8=12, or x+y=4.

We express the other two missing numbers in terms of x and y, as shown below:

We have x>x-1, x>y+10, and x>y. Note that the first inequality is true for all values of x. We only need to solve the second inequality so that the third inequality is true for all values of x. By substitution, we get x>(4-x)+10, from which x>7.

Therefore, the smallest possible value of x is \boxed{\textbf{(D) } 8}.

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Problem 21 Hard

Steph scored 15 baskets out of 20 attempts in the first half of a game, and 10 baskets out of 10 attempts in the second half. Candace took 12 attempts in the first half and 18 attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?

  • A.

    7

  • B.

    8

  • C.

    9

  • D.

    10

  • E.

    11

Answer:C

Solution 1 (Inequalities)

Let x be the number of shots that Candace made in the first half, and let y be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have x+y=10+15=25. In addition, we have the following inequalities:

\frac{x}{12}<\frac{15}{20} \implies x<9,

and

\frac{y}{18}<\frac{10}{10} \implies y<18.

Pairing this up with x+y=25 we see the only possible solution is (x,y)=(8,17), for an answer of 17-8 = \boxed{\textbf{(C) } 9}.

 

Solution 2 (Answer Choices)

Clearly, Steph made 15 + 10 = 25 shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is \boxed{\textbf{(C) } 9}.

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Problem 22 Hard

A bus takes 2 minutes to drive from one stop to the next, and waits 1 minute at each stop to let passengers board. Zia takes 5 minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus 3 stops behind. After how many minutes will Zia board the bus?

  • A.

    17

  • B.

    19

  • C.

    20

  • D.

    21

  • E.

    23

Answer:A

Solution 1

Initially, suppose that the bus is at Stop 0 (starting point) and Zia is at Stop 3.

We construct the following table of 5-minute intervals:

\renewcommand{\arraystretch}{1.5} \begin{array}{|c||c|c|} \hline \textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ \hline\hline 5\ \textbf{Minutes} & \text{Stop}\ 2\ \text{(Waiting)} & \text{Stop}\ 4 \\ \hline 10\ \textbf{Minutes} & \text{Midpoint between Stops}\ 3\ \text{and}\ 4 & \text{Stop}\ 5 \\ \hline 15\ \textbf{Minutes} & \text{Stop}\ 5\ \text{(Leaving)} & \text{Stop}\ 6 \\ \hline \end{array}

Note that Zia will wait for the bus after 15 minutes, and the bus will arrive 2 minutes later.

Therefore, the answer is 15+2=\boxed{\textbf{(A) } 17}.

 

Solution 2

Since Zia will wait for the bus if the bus is at the previous stop, we can create an equation to solve for when the bus is at the previous stop. The bus travels \frac{1}{3} of a stop per minute, and Zia travels \frac{1}{5} of a stop per minute. Now we create the equation, \frac{1}{3}m = \frac{1}{5}m + 3 - 1 (the -1 accounts for us wanting to find when the bus reaches the stop before Zia's). Solving, we find that m=15. Now Zia has to wait 2 minutes for the bus to reach her, so our answer is 15+2=\boxed{\textbf{(A) } 17}.

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Problem 23 Hard

A \triangle or \bigcirc is placed in each of the nine squares in a 3-by-3 grid. Shown below is a sample configuration with three \triangles in a line.

How many configurations will have three \triangles in a line and three \bigcircs in a line?

  • A.

    39

  • B.

    42

  • C.

    78

  • D.

    84

  • E.

    96

Answer:D

Solution 1 (Casework)

Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.

We take casework:

Case 1: 3 lines: In this case, the lines would need to be 2 of one shape and 1 of another, so there are \frac{3!}{2} = 3 ways to arrange the lines and 2 ways to pick which shape has only one line. In total, this is 3\cdot 2 = 6.

Case 2: 2 lines: In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are 3! = 6 ways to arrange the lines and 2^3-2 = 6 ways to choose the last line. (We subtract 2 from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is 6\cdot 6 = 36.

Finally, we add and multiply: 2(36+6)=2(42)=\boxed{\textbf{(D) }84}.

 

Solution 2 We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are 3 ways to choose a column with all \bigcirc's and 2 ways to choose a column with all \triangle's. The third column can be filled in 2^3=8 ways. Therefore, we have a total of 3\cdot2\cdot8=48 cases. However, we overcounted the cases with 2 complete columns of with one symbol and 1 complete column with another symbol. This happens in 2\cdot3=6 cases. 48-6=42. However, we have to remember to double our answer, giving us \boxed{\textbf{(D) }84} ways to complete the grid.

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Problem 24 Hard

The figure below shows a polygon ABCDEFGH, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that AH = EF = 8 and GH = 14. What is the volume of the prism?

  • A.

    112

  • B.

    128

  • C.

    192

  • D.

    240

  • E.

    288

Answer:C

While imagining the folding, \overline{AB} goes on \overline{BC}, \overline{AH} goes on \overline{CI}, and \overline{EF} goes on \overline{FG}. So, BJ=CI=8 and FG=BC=8. Also, \overline{HJ} becomes an edge parallel to \overline{FG}, so that means HJ=8.

Since GH=14, then JG=14-8=6. So, the area of \triangle BJG is \frac{8\cdot6}{2}=24. If we let \triangle BJG be the base, then the height is FG=8. So, the volume is 24\cdot8=\boxed{\textbf{(C)} ~192}.

 

Remark

After folding polygon ABCDEFGH on the dotted lines, we obtain the following triangular prism:

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Problem 25 Hard

A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops what is the probability that the cricket has returned to the leaf where it started?

  • A.

    \frac{2}{9}

  • B.

    \frac{19}{80}

  • C.

    \frac{20}{81}

  • D.

    \frac{1}{4}

  • E.

    \frac{7}{27}

Answer:E

Solution 1 (Casework)

Let A denote the leaf where the cricket starts and B denote one of the other 3 leaves. Note that:

- If the cricket is at A, then the probability that it hops to B next is 1.

- If the cricket is at B, then the probability that it hops to A next is \frac13.

- If the cricket is at B, then the probability that it hops to B next is \frac23.

We apply casework to the possible paths of the cricket:

1. A \rightarrow B \rightarrow A \rightarrow B \rightarrow A

~~~The probability for this case is 1\cdot\frac13\cdot1\cdot\frac13=\frac19.

2. A \rightarrow B \rightarrow B \rightarrow B \rightarrow A

~~~The probability for this case is 1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.

Together, the probability that the cricket returns to A after 4 hops is \frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.

 

Solution 2 (Casework)

We can label the leaves as shown below:

Carefully counting cases, we see that there are 7 ways for the cricket to return to leaf A after four hops if its first hop was to leaf B:

1. A \rightarrow B \rightarrow A \rightarrow B \rightarrow A

2. A \rightarrow B \rightarrow A \rightarrow C \rightarrow A

3. A \rightarrow B \rightarrow A \rightarrow D \rightarrow A

4. A \rightarrow B \rightarrow C \rightarrow B \rightarrow A

5. A \rightarrow B \rightarrow C \rightarrow D \rightarrow A

6. A \rightarrow B \rightarrow D \rightarrow B \rightarrow A

7. A \rightarrow B \rightarrow D \rightarrow C \rightarrow A

By symmetry, we know that there are 7 ways if the cricket's first hop was to leaf C, and there are 7 ways if the cricket's first hop was to leaf D. So, there are 21 ways in total for the cricket to return to leaf A after four hops.

Since there are 3^4 = 81 possible ways altogether for the cricket to hop to any other leaf four times, the answer is \frac{21}{81} = \boxed{\textbf{(E) }\frac{7}{27}}.

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