## AMC 8 2022

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 1 point for each correct answer. There is no penalty for wrong answers.
- No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
- Figures are not necessarily drawn to scale.
- You will have
**40 minutes**working time to complete the test.

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?

$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

$\textbf{A}$

The symbol covers 4 1-inch squares and 12 half squares. So the area of the symbol is $4\times1+12\times0.5=10$ square inches.

Consider these two operations:\begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*}What is the value of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$

$\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220$

$\textbf{D}$

\begin{align*} (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\ &= 16 \, \bigstar \, 6 \\ &= (16-6)^2 \\ &=100. \end{align*}

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?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$

$\textbf{E}$

If $a=1$, then $(b,c)$ could be $(2,50)$, $(4,25)$, or $(5,20)$;$\newline$

If $a=2$, then $(b,c)$ could be $(5,10)$;$\newline$

If $a=4$ or above, we can not find any suitable $(b,c)$ pairs.$\newline$

So the answer is $3+1=4$.

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

$\textbf{E}$

When $\textbf{M}$ is first reflected over the line $q,$ we obtain the following diagram:

When $\textbf{M}$ is then reflected over the line $p,$ we obtain the following diagram:

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?

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

$\textbf{C}$

Five years ago, Bella was $6$ years old, and the kitten was $0$ years old. Today, Bella is $6+5=11$ years old, and the kitten is $5$ years old. So Anna is $30-11-5=14$ years old. Therefore, Anna is $14-11=3$ years older than Bella.

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?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

$\textbf{C}$

Let the difference between these integers be $x$. Then the smallest value is $15-x$, while the largest value to be $15+x$. Hence, we have $$15+x=4(15-x)\Rightarrow x=9$$ So the smallest value is $15-9=6$.

When the World Wide Web first became popular in the $1990$s, 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.)

$\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000$

$\textbf{B}$

Since 4.2-megabyte is equal to $4.2\times8000=33600$ kilobits, the time we need to download a 4.2-megabyte song is $33600/56=600$ seconds, or $600/60=10$ minutes.

What is the value of\[\frac{1}{3}\times\frac{2}{4}\times\frac{3}{5}\times\frac{18}{20}\times\frac{19}{21}\times\frac{20}{22}?\]

$\textbf{B}$

$$\frac{1}{3}\times\frac{2}{4}\times\frac{3}{5}\times\frac{18}{20}\times\frac{19}{21}\times\frac{20}{22}=\frac{20!}{22!/2!} = \frac{20! \times 2!}{22!} = \frac{20! \times 2}{20! \times 21 \times 22} = \frac{2}{21 \times 22} = \frac{1}{21 \times 11} =\frac{1}{231}$$

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?

$\textbf{(A)} 77\qquad\textbf{(B)} 86\qquad\textbf{(C)} 92\qquad\textbf{(D)} 98\qquad\textbf{(E)} 104$

$\textbf{B}$

The temperature difference in the beginning is $212-68=144^{\circ}\text{F}$. After 15 minutes, the temperature is halved for 3 times, which is $144\div2^3=18^{\circ}\text{F}$. Hence, the final temperature of the water is $68+18=86^{\circ}\text{F}$.

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?

$\textbf{E}$

From $8 \, \text{am}$ to $10 \, \text{am}$, Ling drove at a constant speed of 45 miles per hour, which means she traveled for $45\times2=90$ miles. Three hours later ($1 \, \text{pm}$), she drove back at a speed of 60 miles per hour. It takes $90/60=1.5$ hours to come back home. So Ling arrived home at $2:30 \, \text{pm}$.

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?

$\textbf{(A)} ~34\qquad\textbf{(B)} ~38\qquad\textbf{(C)} ~41\qquad\textbf{(D)} ~44\qquad\textbf{(E)} ~47$

$\textbf{D}$

For every time he bites, the number of pieces will go up by 1. Since he started with one piece of pasta and finally got 10 pieces, he had bit for $10-1=9$ times. In every bite he eats 3 inches of pasta. So the original length of the pasta is $17+9\times3=44$ inches.

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?

$\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}$

$\textbf{B}$

Among the $4\times4=16$ possible results, we notice than $N$ lies between $51$ and $84$. The square numbers in such range $64$ and $81$, which are both achievable by Spinner A and B. So the possibility is $\dfrac{2}{16}=\dfrac18$.

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

“One positive integer is $\underline{\ }\underline{\ }\underline{\ }\underline{\ }$ more than twice another, and the sum of the two numbers is $28$.”

$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

$\textbf{D}$

Let $x$ and $28-x$ be the two positive integers. Then we have $$28-x>2x\Rightarrow x<9\dfrac13$$ So the positive integer $x$ could be $1,2,3,4,5,6,7,8,9$.

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

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$

$\textbf{D}$

Since 5 $\textbf{E}$s should be separated by 4 distinct letters $\textbf{B, K, P, R}$, all valid arrangements of the letters must be of the form$$\textbf{E}\underline{\ }\underline{\ }\textbf{E}\underline{\ }\underline{\ }\textbf{E}\underline{\ }\underline{\ }\textbf{E}\underline{\ }\underline{\ }\textbf{E}$$ Hence, we have $4!=24$ ways.

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?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

$\textbf{C}$

The slope of the line represents the price per ounce. Hence, we need to find the line with the smallest slope, which is the red line in the diagram. The blue spot on the red line has a weight of 3 ounces.

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?

$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$

$\textbf{B}$

Let $a,b,c,$ and $d$ be the four numbers in that order. We are given that\begin{align*} \dfrac{a+b}{2} &= 21 &(1) \\ \dfrac{b+c}{2} &= 26 &(2) \\ \dfrac{c+d}{2} &= 30 &(3) \end{align*}and we wish to find $\dfrac{a+d}{2}.$

We add $(1)$ and $(3),$ then subtract $(2)$ from the result:\[\frac{a+d}{2}=21+30-26=25\]

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

$\textbf{(A)} ~0\qquad\textbf{(B)} ~2\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~8$

$\textbf{B}$

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\[2!!+4!!+6!!+8!! = 2+8+48+384\]We only care about units digits, which is $2+8+8+4=22.$ The answer is $2.$

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?

$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$

$\textbf{C}$

By drawing the graph, we get the area of the figure formed by connecting these midpoints, which is 20. Since the area is half the area of the rectangle, the area of the rectangle is 40.

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?$\newline$

(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.)

$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$

$\textbf{C}$

Before Mr. Ramos added scores, the median was $\dfrac{80+80}{2}=80$. There are two cases now:

Case $1$: The middle two scores are $80$ and $90$. To do this, we firstly suppose that the two students who got $85$ are awarded the extra $5$ points to empty the 85 column. We then realize that it is impossible to move students from below-85 side to above-85 side with only extra 5 points. Therefore, we reject this case.

Case $2$: The middle two scores are both $85$. To do this, we simply need to suppose that some of the students who got $80$ are awarded the extra $5$ points. Note that there are $8$ students who got $75$ or less. Therefore there must be only $1$ student who got $80$ so that the middle two scores are both $85$. Therefore our answer is $4$.

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 $x$?

$\textbf{(A) } {-}1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

$\textbf{D}$

The sum of each row and column is 12. Hence, the the number to the right of $x$ is $4-x$, the number above $x$ is $14-x$, and the number in the center of the grid is $x-1$. Since $x$ is the greatest in the 4 numbers, we notice that $14-x$ is greater than $4-x$, and $x$ is greater than $x-1$. So we only need $x>14-x\Rightarrow x>7$. Hence, the smallest value for $x$ is 8.

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?

$\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

$\textbf{C}$

Let $x$ be the number of shots that Candace made in the first half, and $y$ 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 $x+y=10+15=25.$ Given that Steph scored a higher percentage of baskets than Candace in each half, 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 $17-8 =9.$

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?

$\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$

$\textbf{A}$

Let the initial position of the bus be Stop 1, and the initial position of Zia be Stop 4. 5 minutes later, when Zia just arrived Stop 5, the bus just arrived Stop 3. Another 5 minutes passed, when Zia just arrived Stop 6, the bus was at the midpoint of Stop 4 and 5. After another 5 minutes, when Zia just arrived Stop 7, the bus was ready to leave Stop 6. Hence Zia would wait for 2 minutes until the bus arrived Stop 7. The total time is $5+5+5+2=17$ minutes.

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 $\triangle$s in a line.

How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

$\textbf{D}$

When the $\triangle$ line is in a row, the $\bigcirc$ line can not be in a column or in a diagonal. Hence, both the $\triangle$ line and $\bigcirc$ line must be in rows or in columns. By symmetry, we consider only the situation in rows, and the counts for the situation in columns must be the same.

We have $\dbinom{3}{1}=3$ ways to choose a row for the $\triangle$ line, and then $\dbinom{2}{1}=2$ ways to choose a row for the $\bigcirc$ line. The rest line have $2^3=8$ ways to be filled with $\triangle$ and $\bigcirc$. This makes $3\times2\times8=48$ ways. However, we double count the cases when there are two rows of $\triangle$ and one row of $\bigcirc$, or two rows of $\bigcirc$ and one row of $\triangle$. We have $\dbinom{3}{1}=3$ ways to choose the row which is unique that only appears once, and then decide whether the $\triangle$ or $\bigcirc$ be filled in the unique row. The rest two rows are now determined. Hence, the number of repetitive cases is $3\times2=6$. By subtracting it, we get the real cases for the situation in rows as $48-6=42$.

The final result in both the situations of rows and columns is $42\times2=84$.

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?

$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$

$\textbf{C}$

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

Hence, the volume is $\dfrac12\times6\times8\times8=192$.

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?

$\textbf{(A) }\dfrac{2}{9}\qquad\textbf{(B) }\dfrac{19}{80}\qquad\textbf{(C) }\dfrac{20}{81}\qquad\textbf{(D) }\dfrac{1}{4}\qquad\textbf{(E) }\dfrac{7}{27}$

$\textbf{E}$

Let $P_n$ be the probability that the cricket would return back to the starting point after $n$ hops. Hence, $1-P_n$ means the probability that the cricket is not on the original leaf after $n$ hops. Then, we get the recursive formula\[P_n = \frac13(1-P_{n-1})\]because if the cricket is not on the original leaf, then there is a $\dfrac13$ probability that it will make it back. With this formula and the fact that $P_1=0$ (After one hop, the cricket can never be back to the original leaf.), we have\[P_2 = \frac13, P_3 = \frac29, P_4 = \frac7{27},\]so our answer is $\dfrac{7}{27}$.