## 2022 AMC 8

Time limit: 40 minutes

Typeset by: LIVE, by Po-Shen Loh

https://live.poshenloh.com/past-contests/amc8/2022

Copyright: Mathematical Association of America. Reproduced with permission.

1.

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?

$$10$$

$$12$$

$$13$$

$$14$$

$$15$$

###### Solution(s):

Let's first consider the following:

Using the Pythagorean theorem, we know how to solve for $$x$$ relatively easily: $a^2+b^2=c^2 \iff 1^2+1^2=x^2$$\iff x=\sqrt{2}$ With that in mind, let's try and find the area of the purple square: $A=x^2=(\sqrt{2})^2=2$ We now know that the area of this slanted, small square is equal to 2. With that in mind, let's look at the full picture:

We can clearly see that there are 5 of these same small purple squares that make up the logo. We already know that the area of one of the small purple squares is equal to 2, so the total area $$A_t$$ is: $A_t=5(A)=5(2)=10$ Thus, the answer is A.

2.

Consider these two operations:

\begin{align*} a\,\blacklozenge\,b &= a^2 - b^2 \\ a \star b &= (a - b)^2 \end{align*}

Compute the value:

$(5\,\blacklozenge\,3) \star 6$

$$-20$$

$$4$$

$$16$$

$$100$$

$$220$$

###### Solution(s):

Given the definitions of $$\blacklozenge$$ and $$\star,$$ we can directly substitute these operations for said definitions to get: \begin{align} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align}

3.

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?

$$0$$

$$1$$

$$2$$

$$3$$

$$4$$

###### Solution(s):

We are given that $$a < b < c, abc = 100.$$

As such, we know that $$a^3 Considering \(\sqrt[3]{100},$$ we know that $$4^3=64<100$$ and $$5^3=125>100,$$ and therefore, we can conclude that $$a\ge 4.$$ As $$a$$ is a positive integer, we have four possible values of $$a:1,2,3,4.$$ As $$a$$ is a factor of $$100,$$ we can eliminate the possibility of $$a$$ being $$3,$$ and are now left with three cases: $$a=1,2,4.$$

If $$a=1,$$ then $$bc =100 .$$ Since $$b < c,$$ we know $$b^2 < bc=100$$ so $$1 < b < 10 .$$ Since $$b$$ is a factor of $$100$$ that satisfies this constraint, $$b$$ must be either $$2,4,5.$$ This creates the solutions of $$(1,2,50),(1,4,25),$$ or $$(1,5,20.)$$

If $$a=2,$$ then $$bc =50 .$$ Since $$b < c,$$ we know $$b^2 < bc=50$$ so $$2 < b < 8 .$$ Since $$b$$ is a factor of $$50$$ that satisfies this constraint, $$b$$ must be $$5.$$ This creates the solution of $$(2,5,10.)$$

If $$a=4,$$ then $$bc =25 .$$ Since $$b < c,$$ we know $$b^2 < bc=25$$ so $$4 < b < 5 .$$ This creates no integer solutions for $$b.$$

We therefore have 4 total solutions.

4.

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?

###### Solution(s):

First reflect

M

over the line $$q,$$ from which we obtain the following:

Then reflect

M

over the line $$p,$$ from which we obtain the following:

5.

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?

$$1$$

$$2$$

$$3$$

$$4$$

$$5$$

###### Solution(s):

If Bella was $$6$$ five years ago, then she is $$11$$ right now.

If the kitten was a newborn five years ago, then it is $$5$$ right now.

Since the sum of all $$3$$ ages is $$30,$$ Anna's age is $30-5-11 = 14$

Since Anna is $$14$$ and Bella is $$11,$$ she is $$3$$ years older than Bella.

6.

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?

$$4$$

$$5$$

$$6$$

$$7$$

$$8$$

###### Solution(s):

Since one of the outer number is $$4$$ times the other, we can make one of the numbers $$x$$ and the other be $$4x.$$

Since the numbers are equally spaced, the middle number is the average of the outer two.

This means the average of $$x$$ and $$4x$$ is 15, so $$\dfrac{x+4x}{2} =15 .$$ This means $$\ 5x=30 ,$$ so $$x=6.$$ Our outer numbers therefore are $$6,24,$$ so the smaller one is 6.

7.

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.)

$$0.6$$

$$10$$

$$1800$$

$$7200$$

$$36000$$

###### Solution(s):

Given our definition of a megabyte as $$8000$$ kilobits, we know $$4.2$$ megabytes is $$8000\cdot 4.2$$ kilobits. This can be rewritten as $$(8000 \cdot \dfrac{3}{40})\cdot(4.2 \cdot \dfrac{40}{3})$$ kilobits which leads to us having $$600\cdot56$$ kilobits. Since we know there are $$56$$ kilobits per second, we have $$600$$ seconds. This leads us to the answer of $$10$$ minutes.

8.

What is the value of:

$\dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}$

$$\displaystyle \dfrac{1}{462}$$

$$\displaystyle \dfrac{1}{231}$$

$$\displaystyle \dfrac{1}{132}$$

$$\displaystyle \dfrac{2}{213}$$

$$\displaystyle \dfrac{1}{22}$$

###### Solution(s):

Since every integer from $$3$$ to $$20$$ occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only $$1$$ and $$2$$ left as numerators and $$21$$ and $$22$$ left as denominators.

The remaining fraction is $$\dfrac{1 \cdot 2}{21 \cdot 22} .$$ This simplifies to $$\dfrac{2}{462} = \dfrac{1}{231}$$

9.

A cup of boiling water ($$212^\circ$$ F) is placed to cool in a room whose temperature remains constant at $$68^\circ$$ 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?

$$77$$

$$86$$

$$92$$

$$98$$

$$104$$

###### Solution(s):

The current difference is $$212-68 = 144 .$$

Since we have $$15$$ minutes while halving every $$5$$ minutes, we halve the difference $$3$$ times.

This means the difference is multiplied by $$(\dfrac{1}{2})^3 = \dfrac{1}{8} ,$$ so our new difference is $$\dfrac{1}{8}\cdot144=18.$$ This makes our final temperature $$68+18=86.$$

10.

One sunny day, Ling decided to take a hike in the mountains. She left her house at $$8$$ a.m., drove at a constant speed of $$45$$ miles per hour, and arrived at the hiking trail at $$10$$ a.m. 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?

###### Solution(s):

She drives $$45$$ miles per hour for the first $$2$$ hours, so she travels $$90$$ miles in the first two hours. This leaves choices $$A,E.$$

Since she hikes for $$3$$ hours, she starts going back at $$1$$pm.

She drives $$90$$ miles at $$60$$ mph, so she gets back in $$\dfrac{90}{60} = 1.5$$ hours, so she is back at $$2$$:$$30.$$

11.

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?

$$34$$

$$38$$

$$41$$

$$44$$

$$47$$

###### Solution(s):

Since there are $$10$$ pieces, there were $$9$$ locations where a bite was made. Since we have $$9$$ bites and $$3$$ inches are removed per bite, a total of $$27$$ inches were removed.

With $$27$$ inches removed and $$17$$ inches remaining, we know we started with $$27+17 = 44$$ inches.

12.

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

$$\displaystyle \dfrac{1}{16}$$

$$\displaystyle \dfrac{1}{8}$$

$$\displaystyle \dfrac{1}{4}$$

$$\displaystyle \dfrac{3}{8}$$

$$\displaystyle \dfrac{1}{2}$$

###### Solution(s):

This is the same problem as if we made the spinner A make the tens digit and spinner B make the ones digit.

If the tens digit is 5,6,7, or 8, and the ones digit is 1,2,3, or 4, the only possible ways to get a perfect square is 6,4 or 8,1.

There are 4x4=16 possible combinations that are equally likely, and 2 of them satisfy the property. Therfore, our answer is $$\dfrac{2}{16} = \dfrac{1}{8}$$

13.

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 $$28$$”

$$6$$

$$7$$

$$8$$

$$9$$

$$10$$

###### Solution(s):

Let the first number be $$x.$$ Then the second number is $$2x+c$$ where both $$x$$ and $$c$$ are positive integers.

Since their sum is $$28,$$ we know $$3x+c=28.$$
This also means $$c = 28-3x$$

With $$x > 0,$$ the smallest possible value of $$x$$ is $$1.$$
With $$c > 0 ,$$ we know $$28 -3x > 0$$ so $$\dfrac{28}{3} > x.$$

This means the maximum value of $$x$$ is $$9.$$

Since $$x$$ can be any number from $$1$$ to $$9,$$ we have $$9$$ solutions.

14.

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

$$1$$

$$4$$

$$12$$

$$24$$

$$120$$

###### Solution(s):

First, I claim that every E must be in an odd position.
This is because bringing any two Es together would bring them right next to each other.

This means the B,K,P, and R must each be in one of the even positions.

There are $$4$$ choices for a letter in the second position, $$3$$ choices for position for B, $$3$$ remaining choices for a position in K, $$2$$ remaining choices for a position in P, and $$1$$ remaining choice for a position in R.

Therefore, there are $$4 \cdot 3\cdot2\cdot1=24$$ possible choices.

15.

László 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?

$$1$$

$$2$$

$$3$$

$$4$$

$$5$$

###### Solution(s):

For each weight, we find the lowest price.

At $$1$$ ounce, we have a price slightly over $$1$$ dollar, so the price per ounce is greater than $$1.$$

At $$2$$ ounces, we have a price of $$2$$ dollars, so the price per ounce is $$1.$$

At $$3$$ ounces, we have a price of about $$2.5$$ dollars, so the price per ounce close to $$\dfrac{2.5}{3}$$ which is close to $$0.83.$$

At $$4$$ ounces, we have a price close to $$3.9$$ dollars, so the price per ounce close to $$\dfrac{3.9}{4}$$ which is close to $$0.97.$$

At $$5$$ ounces, we have a price close to $$4.5$$ dollars, so the price per ounce close to $$\dfrac{4.5}{5}$$ which is close to $$0.90.$$

The price per ounce at $$3$$ ounces is the lowest.

16.

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?

$$24$$

$$25$$

$$26$$

$$27$$

$$28$$

###### Solution(s):

Let the numbers be $$a,b,c,d$$ in order.

Since the average of $$a$$ and $$b$$ is $$21,$$ we know their sum is $$21\cdot2 = 42.$$
Since the average of $$c$$ and $$d$$ is $$30,$$ we know their sum is $$30\cdot2 = 60.$$
Since $$a+b = 42$$ and $$c+d = 60 ,$$ we know \begin{align*}a+b+c+d &= 42 +60 \\&= 112\end{align*}

Now, with the average of $$b$$ and $$c$$ being $$26,$$ we know their sum is $$26\cdot2=52.$$ This means $$b+c = 52.$$

Subtracting this result from the sum of all the terms yields $$a+d =50.$$
Since $$\dfrac{a+d}{2}=25,$$ our answer is $$25.$$

17.

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:

$8!! = 2 \times 4 \times 6 \times 8$

What is the units digit of the following sum?

$2!! + 4!! + \cdots + 2022!!$

$$0$$

$$2$$

$$4$$

$$6$$

$$8$$

###### Solution(s):

If we take $$n!!$$ for an even $$n$$ that is greater or equal to $$10,$$ then $$10$$ is one of the numbers we multiply by. Since $$10$$ is a factor of $$n!!,$$ we know that the units digit of $$10$$ is $$0,$$ which means that it doesn't affect our result.

This means it suffices to compute the units digit of $$2!!+4!!+6!!+8!!,$$ which is equivalent to: \begin{align*}&2+ 2(4)+2(4)(6)\\&+2(4)(6)(8) \\&= 2+8+48+384 \\&= 442\end{align*} The units digit therefore is $$2$$

18.

The midpoints of the four sides of a rectangle are:

$(-3, 0),$

$(2, 0),$

$(5, 4),$

$(0, 4).$

What is the area of the rectangle?

$$20$$

$$25$$

$$40$$

$$50$$

$$80$$

###### Solution(s):

Allow: $A=(-3,0)$$B=(2,0)$$C=(5,4)$$D=(0,4)$ This gives us the following rhombus:

Now, moving away from the rhombus in question, let's consider more generally the relationship between a quadrilateral and the figure formed by its midpoints. Observe the arbitrary quadrilateral $$PQRS,$$ its diagonals (dashed red), and the figure formed by its midpoints:

Note that the actual coordinates of $$P,Q,R,S$$ don't matter, rather, as long as $$PQRS$$ is a convex quadrilateral.

Let's first consider the triangle $$\triangle QSP.$$ Notice that $$\overline{PW} = \dfrac12 \overline{PQ}$$ as $$W$$ is the midpoint of $$\overline{PQ},$$ and similarly, $$\overline{PZ}=\dfrac12 \overline{PS}.$$ Also notice that the angle $$\angle SPQ$$ is contained within both $$\triangle QSP$$ and $$\triangle WZP.$$ Therefore, we know that the two triangles are similar, and consequently, the side $$\overline{WZ}$$ and the diagonal $$\overline{SQ}$$ are parallel, and $$\overline{WZ} = \dfrac12 \overline{SQ}.$$ We can use this same reasoning to show that the side $$\overline{XY}$$ and the diagonal $$\overline{SQ}$$ are parallel, and $$\overline{XY} = \dfrac12 \overline{SQ}.$$ Furthermore, if we consider the triangles formed by the other diagonal $$PR,$$ we can reapply this reasoning to show that the sides $$\overline{WX},\overline{YZ}$$ and the diagonal $$\overline{PR}$$ are parallel, and $$\overline{WX} = \overline{YZ}=\dfrac12 \overline{PR}.$$

Therefore, $$WXYZ$$ is a parallelogram where each of the unequal sides' lengths are equal to half the length of the corresponding parallel diagonal. With that in mind, notice that if we were to create a triangle out of $$W,X,$$ and the midpoint of $$\overline{PR},$$ then the area of the right part of the parallelogram would be half the area of the new triangle formed, which would be $$\dfrac14$$ the area of $$\triangle PQR.$$ Similarly, it is clear that the area of the bottom left part of the parallelogram is half the area of $$\triangle PRS.$$ Adding those two parts up shows that the area of $$WXYZ$$ is half that of $$PQRS.$$

With that in mind, let's go back to the original problem. Calculating the area of $$ABCD$$: \begin{align}\text{Area} &= \overline{AB}\cdot \overline{MD}\\ &= 5\cdot 4\\ &= 20\end{align} And as the area of $$ABCD$$ is half that of the quadrilateral whose midpoints created it, we can conclude that the parent quadrilateral has an area of $$2\cdot 20=40.$$

19.

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.)

$$2$$

$$3$$

$$4$$

$$5$$

$$6$$

###### Solution(s):

To make the median equal to $$85$$ the average between the $$10$$th and $$11$$th highest scores must be $$85.$$ Therefore, we either have the $$10$$th best and $$11$$th best scores being $$85$$ or having the $$10$$th best score be above $$85$$ and the bottom and the $$11$$th best score be under $$85.$$ The second option involves not having any scores of $$85.$$

If we move all the scores out of $$85$$ we have only $$7$$ scores greater than $$85.$$ We can't move any other score to be greater than $$85.$$ Therefore, this scenario can't happen.

Therefore, we must have the $$10$$th and $$11$$th best scores be $$85.$$ There are currently $$7$$ scores greater than or equal to $$85,$$ so we must add $$4$$ more scores so the $$11$$th best score is $$85.$$

20.

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 $$x?$$

$$-1$$

$$5$$

$$6$$

$$8$$

$$9$$

###### Solution(s):

First, by adding the numbers on the top row, we know the sum of the rows and columns are $$12.$$

Since the sum of the numbers in the first column is $$12$$ and we know that one of the numbers is $$-2,$$ we know the sum of the other two is $$14.$$ This means that the number above $$x$$ is $$14-x.$$

Since the sum of the numbers in the bottom row is $$12$$ and we know that one of the numbers is $$8,$$ we know the sum of the other two is $$4.$$ This means that the number to the right of $$x$$ is $$4-x.$$

Since the sum of the numbers in the middle row is $$12$$ and we know that two of them are $$14-x$$ and $$-1,$$ we know the remaining number is $$x-1.$$

We know that $$x > 14-x$$ and $$x > 4-x$$ as $$x$$ is the greatest number. This leads us to know $$x > 7 .$$ Since $$x > x-1$$ is always true, our only restriction is $$x > 7 .$$ This means $$x=8$$ is our minimum solution.

21.

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?

$$7$$

$$8$$

$$9$$

$$10$$

$$11$$

###### Solution(s):

They both shot 30 baskets total, so if they have the same make percentage, then they made the same total amount. This means Candace also made $$25$$ total shots.

Now, let $$f$$ be the number of shots Candace made in the first half and let $$s$$ be the number of shots Candace made in the second half.
We know she made $$25$$ shots, so $$f+s=25.$$

By the condition that she had a lower percentage in each half, we know $$\dfrac{f}{12} < \dfrac{15}{20}$$ and $$\dfrac{s}{18} < \dfrac{10}{10}.$$ With cross multiplication, we know $$f < 9$$ and $$s < 18.$$ Since $$f,s$$ are whole numbers, we have the restriction that $$f \leq 8, s \leq 17.$$

With our knowledge that $$f+s=25$$ and the condition that $$f \leq 8, s \leq 17,$$ the only possible solution is $$f=8$$ and $$s=17.$$
This makes our answer $$s-f = 9.$$

22.

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?

$$17$$

$$19$$

$$20$$

$$21$$

$$23$$

###### Solution(s):

Since a bus stops takes $$2$$ minutes to drive and $$1$$ minute at every stop, the bus takes $$3$$ minutes at every stop.

Zia can only stop at every $$5$$ minute interval as that is when she arrives at a bus stop and chooses whether to stop or continue.
Therefore, we can consider the time Zia stops by looking at each of their locations at these intervals.

After $$0$$ minutes, Zia is $$3$$ stops from where the bus started.
After $$5$$ minutes, Zia is $$4$$ stops from where the bus started. Meanwhile, the bus is $$2$$ stops from where it started, waiting. Zia therefore doesn't stop here.
After $$10$$ minutes, Zia is $$5$$ stops from where the bus started. Meanwhile, the bus is between the $$3$$rd and $$4$$th stops from where it started. Zia therefore doesn't stop here.
After $$15$$ minutes, Zia is $$6$$ stops from where the bus started. Meanwhile, the bus is $$5$$ stops from where it started, about to leave. Since the bus is at the previous stop, she will wait here.

It will then take $$2$$ more minutes to get to Zia, so the total time was $$17$$ minutes.

23.

A $$\bigtriangleup$$ or $$\bigcirc$$ is placed in each of the nine squares in a $$3 \times 3$$ grid. Shown below is a sample configuration with three $$\bigtriangleup$$'s in a line.

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

$$39$$

$$42$$

$$78$$

$$84$$

$$96$$

###### Solution(s):

First, there can't be a horizontal line of one shape and a vertical line of another shape.
This can't happen as it would require a position to take both shapes.

This means we can consider only when the lines are horizontal and only when the lines are vertical.
Since rotating the shapes will take the lines from horizontal to vertical, we only need to check how many vertical lines there are as there are equally as many horizontal lines.

Now, when finding configurations, we can have $$3$$ separate vertical lines or only $$2$$ vertical lines. We can split this into cases.

Case 1: $$3$$ lines: There are $$2$$ choices for the first line (which could be all triangles and all circles), $$2$$ choices for the second line, and $$2$$ choices for the third line. This would make $$2\cdot2\cdot2 = 8$$ choices. However, there are two cases to ignore in which each line is the same. These cases would ensure that only one shape has a line. Therefore, with $$3$$ lines, we have $$8-2=6$$ total choices.

Case 2: $$2$$ lines: Since there are two lines and we need at least one line of each shape, each shape can only have one line. There are $$3$$ ways to choose a spot for the line of triangles a $$2$$ ways to choose a spot for the line of circles. Now, with the remaining 3 spots, we need to ensure that a line isn't creates or it would fall into the other case. This would mean we have $$2\cdot2\cdot2-2 =6$$ choices for the remaining line. Totally, we have $$3\cdot2\cdot6=36$$ choices with $$2$$ lines.

This means we have $$6+36=42$$ configurations with vertical lines. As shown before, we have the same number of horizontal configurations, so we have $$42$$ horizontal configurations. This makes $$84$$ total cases.

24.

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?

$$112$$

$$128$$

$$192$$

$$240$$

$$288$$

###### Solution(s):

Since $$EF$$ maps to $$GF$$ on the map, we know $GF=EF=8.$ Since $$GFCB$$ is a rectangle with $$GF$$ and $$BC$$ being opposite, we know $BC=GF=8.$ Since $$AB$$ maps to $$BC$$ on the map, we know $AB=BC=8.$ Since $$HJBA$$ is a rectangle with $$HJ$$ and $$AB$$ being opposite, we know $HJ=AB=8.$ Since $$HJBA$$ is a rectangle with $$BJ$$ and $$AJ$$ being opposite, we know $BJ=AH=8.$ Since $$GJ = GH - JH,$$ we know $GJ = 14-8=6.$ The area of $$BJG$$ is $$\dfrac{6\cdot8}{2} = 24.$$ Now, we can make this prism complete with bases of $$BJG$$ and $$CIF.$$ The volume of a prism is the area of the base times the altitude to the base. We can use $$BJG$$ as our base and $$GF$$ as our altitude. This would make our volume equal to $[BJG]\cdot GF = 24\cdot8=192.$

25.

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?

$$\displaystyle \dfrac{2}{9}$$

$$\displaystyle \dfrac{19}{80}$$

$$\displaystyle \dfrac{20}{81}$$

$$\displaystyle \dfrac{1}{4}$$

$$\displaystyle \dfrac{7}{27}$$

We begin by defining the action of the cricket jumping to the starting leaf as $$A$$ and the action of it jumping to any of the other three leaves as $$B.$$ With this in mind, note that when the cricket is on the first leaf, the probability of it jumping to the first leaf, $$P(A),$$ is zero, and the probability that it jumps to any of the other leaves, $$P(B),$$ is $$1.$$ Similarly, notice that when the cricket is not on the first leaf, $$P(A)=\dfrac13$$ and $$P(B)=\dfrac23.$$
We want to find the total probability of the cricket landing on the first leaf, and as such, we want to find the total probability that the last node in the diagram is $$A_4.$$ This means that we want to find $$P(\sum A_4)$$: \begin{align*} &= P_{(1)}+P_{(4)} \\ & = \left(1\cdot \dfrac13 \cdot 1 \cdot \dfrac 13\right)\\ &+ \left( 1\cdot \dfrac23 \cdot \dfrac23 \cdot \dfrac13 \right)\\ &= \dfrac19 +\dfrac{4}{27}\\ &=\dfrac{7}{27} \end{align*}