Welcome Zoom Room ToDoodles Calculator  Application Topics    # Math OER Justice: Solving an Equation Our third foundational topic is Justice. This is our nickname for when we do the same thing to both sides of an equation.

Recall that we called an equation with the format y = something an algorithm. These worked like a recipe. To follow the recipe we plug in numbers, simplify, and get an answer.

A more complicated equation is a puzzle, not a recipe.

y − 3 = 47

length × width = 36

age + 3 = age × 2

Our math toolbox is nearly full.

We know how to "shapeshift" a number by rounding or by changing formats (fraction, decimal, percent, and measurement units).

We know how to use "mad science" to simplify an expression using arithmetic (adding, subtracting, multiplying, dividing, and exponents) and grouping structures (parenthesis, fraction bars, square roots, and algorithms).

Now we get one last tool in our toolbox. For the three above example equations the key is to do the same thing to both sides of the equation. But an equation is like a tangle of yarn. Yes, we can "pull at it" by doing the same thing to both sides of the equation. But some types of pulling will make it more tangled instead of less.

So we must study different situations to see which action, or series of actions, will unravel the equation and reveal the answer we want.

To summarize, we will explore more deeply how to use "justice" to be not only fair (doing the same action to both sides of the equation) but wise (picking the action that unravels the tangle). This exploration will take us through the first three subsections (Fractions, Ratios, and Percentages).

Then we will shift our focus twice. In the Measurement subsection we wil study situations where we "build" an equation, even though the equation we build is not solved by doing the same series of actions to both sides. In the Patterns subsection we will think about where most of the equations used in real life come from.

As we study this topic, work on making helpful and organized notes, so you have handy the comments, formulas, and example problems you need.

## Fractions

### One Step Equations§11.2, §11.3

The first equations we study are one-step equations.

We only need to do one thing to both sides of the equation to solve the puzzle.

#### Multiply and Divide

First consider when on one side of the equals sign a letter is either multiplied or divided by a number. On the other side of the equals sign is a number.

To solve these, "undo" what is attached to the letter by doing the opposite.

1. Solve y × 5 = 35

We undo a "multiply by 5" by doing a "divide by 5". We want to be fair and treat both sides of the equation the same.

On the left side y × 5 ÷ 5 = y and then on the right side 35 ÷ 5 = 7

The result is y = 7

When solving this type of problem it often helps to remember that when multiplying two items the order does not matter. We can swap the order to make the formula look better.

1. Solve 5 × y = 35

We can swap the order of multiplying 5 and y.

So we can rewrite this equation to match the previous one. It is another y × 5 = 35

1. Solve 4 × b = 20

We undo a "multiply by 4" by doing a "divide by 4". We want to be fair and treat both sides of the equation the same.

On the left side b × 4 ÷ 4 = b and then on the right side 20 ÷ 4 = 5

The result is b = 5 How does this picture help us solve for b? Where is division for both sides hiding?

In the picture, we can divide the large rectangle on the right that weighs 20 into four pieces that each weigh 5. Then we can see the correspondence: each "b" rectangle on the left matches with a "5" rectangle on the right.

So far we have only started with multiplication, and used division to undo it.

What about the other way around?

1. Solve p ÷ 6 = 8

We undo a "divide by 6" by doing a "multiply by 6". We want to be fair and treat both sides of the equation the same.

On the left side p ÷ 6 × 6 = p and then on the right side 8 × 6 = 48

The result is p = 48 How does this picture help us solve for p? Where is multiplication for both sides hiding?

In the picture, we can make a complete circle called "p" on the left side by making six copies of the shaded slice. We can also make a complete circle on the right side with six copies of 8. Then we can see the correspondence: the circle named "p" will equal the circle of size 48.

When we solve an equation we should write each step on its own line.

Use the vertical Format

3 × y = 210

÷ 3       ÷ 3

y = 70

Writing each step on its own line makes clear what you were thinking in each step. This helps you check your work, contribute in a study group, earn partial credit on tests, and most importantly use your work later in the term to refresh your memory about how to solve that problem.

Later, in future math classes, writing each step on its own line will also helps avoid careless errors in more complicated problems.

Students who try to cram everthing into one line run into trouble.

Do Not Use the Horizontal Format

Solve 3 × y ÷ 3 = 210 ÷ 3 = 70

This only looks okay because we are using different colors and write very neatly.

If we did not add those cosmetic details...

Solve 3 × y ÷ 3 = 210 ÷ 3 = 70

Now we have trouble even identifying what the original problem was!

Watch how I write this next problem on the board.

1. Solve y × 9 = 189

I used one color for the equations, and a second color for the intermediate "what we are doing to both sides" descriptive lines. In our class you need not use colors, but you should at least write each step on its own line.

If you take an algebra class one of your goals will be to eventually wean yourself from always writing the "what we are doing to both sides" descriptive lines. The instructor will write fewer of these steps on the board. You will train your eye to "see" those steps even if they are not actually written.

But that is for a later algebra class. In this class we will always include the "what we are doing to both sides" descriptive lines.

There are two other reasons to use the vertical format.

First, it promotes doing homework in two or three columns per page. This often saves paper. By their nature, homework problems are seldom as wide as a page.

Second, putting work in that shape makes it easier to do scratch work off on the side. Watch how that helps me stay organized when solving for y when fraction arithmetic happens.

1. Solve y × 8 = 23

Notice that there are many possible ways to write the step of dividing both sides by 8. The clearest is to use the vertical format and write either ÷ 8 or /8 on its own line, as we just did.

Do not use parenthesis to incorrectly mean "do this to the entire equation".

3 × y = 210

(3 × y = 210) ÷ 3

Our process involves doing the same thing separately to each side of the equation. Putting the entire equation in parenthesis might make logical sense, but it is bad grammar because it implies we are not modifying each side of the equation separately.

Do not use parenthesis on each side of the equation improperly.

3 × y = 210

÷ 3 (3 × y) = (210) ÷ 3

The right hand side is legitimate. But the left hand side begins confusingly with the ÷ symbol.

In a future math class studying algebra you will encounter other incorrect ways, for more complicated equations.

Here we show that writing ÷ 3 to the right of each side of the equation can be incorrect.

3 × y + 3 = 210

3 × y + 3 ÷ 3 = 210 ÷ 3

This violates the distributive property, which you will learn about in an algebra class.

One step equations that involve addition and subtraction are very similar.

First we solve an equation that has addition.

1. Solve u + 9 = 200

We undo an "add 9" by doing a "subtract 9". We want to be fair and treat both sides of the equation the same.

On the left side u + 9 − 9 = u and then on the right side 200 − 9 = 191

The result is u = 191

Can you draw a balance scale picture that describes solving the equation u + 9 = 200?

Next we solve an equation that has subtraction.

1. Solve x − 12 = 75

We undo a "subtract 12" by doing an "add 12". We want to be fair and treat both sides of the equation the same.

On the left side x − 12 + 12 = x and then on the right side 75 + 12 = 87

The result is x = 87

Can you draw a balance scale picture that describes solving the equation x − 12 = 75?

Let's do a few more example problems.

1. Solve 50 = v + 4.5

We undo an "add 4.5" by doing a "subtract 4.5". We want to be fair and treat both sides of the equation the same.

On the right side v + 4.5 − 4.5 = v and then on the left side 50 − 4.5 = 45.5

The result is 45.5 = v which if we want we can rewrite as v = 45.5

1. Solve 45 = y − 15

We undo a "subtract 15" by doing an "add 15". We want to be fair and treat both sides of the equation the same.

On the right side y − 15 + 15 = y and then on the left side 45 + 14 = 60

The result is 60 = y which if we want we can rewrite as y = 60

1. Solve w ÷ ½ = 8

We undo a "divide by ½" by doing a "multiply by ½". We want to be fair and treat both sides of the equation the same.

On the left side w ÷ ½ × ½ = w and then on the right side 8 × ½ = 4

The result is w = 4

#### Other Arithmetic

Besides the four fundamental arithmetic operations (addition, subtraction, multiplication, and division) there are other arithmetic operations that have opposites. We can also create one step equations using those. But those are not part of our class.

Nevertheless, here is one as a token example.

1. Solve 9 = x2

This problem asks, "What number, when multiplied by itself, equals 9?" The answer is 3. Bittinger Chapter Tests, 11th Edition

Chapter 1 Test, Problem 28: Solve: 28 + x = 74

Chapter 1 Test, Problem 29: Solve: 169 ÷ 13 = n

Chapter 1 Test, Problem 30: Solve: 38 × y = 532

Chapter 1 Test, Problem 31: Solve: 381 = 0 + a

Chapter 2 Test, Problem 34: Solve: 78 × x = 56

Chapter 2 Test, Problem 35: Solve: t × 25 = 710

Chapter 3 Test, Problem 9: Solve: 14 + y = 4

Chapter 3 Test, Problem 10: Solve: x + 23 = 1112

Chapter 4 Test, Problem 32: Solve: 4.8 × y = 404.448

Chapter 4 Test, Problem 33: Solve: x + 0.018 = 9 Textbook Exercises for One Step Equations

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 1.5 (Page 60) # 99, 101, 103

Section 5.4 (page 278) # 11, 13, 15, 17, 19

Section 11.2 (Page 585) # 11, 13, 23, 47, 49

Section 11.3 (Page 593) # 1, 9, 11, 51, 53

Section 11.5 (Page 616) # 29, 31 (officially too hard for our class, but you can do these!)

### Two Step Equations(none)

We will only consider two kinds of two step equations, which "fix" problems with one step equations.

#### Starting with ÷ y

When we have a multiplication one step equation, it does not matter whether the side of the equation with the multiplication has the y or the number written first.

1. Solve y × 3 = 9

We undo a "multiply by 3" by doing a "divide by 3". We want to be fair and treat both sides of the equation the same.

On the left side y × 3 ÷ 3 = y and then on the right side 9 ÷ 3 = 3

The result is y = 3

1. Solve 3 × y = 9

This is the same problem as before! The order of multiplication does not matter. We can rewrite 3 × y so it becomes y × 3.

On the left side y × 3 ÷ 3 = y and then on the right side 9 ÷ 3 = 3

The result is y = 3

But for division we can get stuck if the equation starts with a ÷ y

This equation can be solved in one step:

1. Solve y ÷ 3 = 30

We undo a "divide by 3" by doing a "multiply by 3". We want to be fair and treat both sides of the equation the same.

On the left side y ÷ 3 × 3 = y and then on the right side 30 × 3 = 90

The result is y = 90

But the next equation does not work like the ones above. It cannot be solved it with one step.

1. Look at 30 ÷ y = 3 but do not solve it yet.

We could begin by dividing both sides by 30. But this creates 1 ÷ y = 330 which is increasing the complexity.

It is better to begin by multiplying both sides by y.

1. Solve 30 ÷ y = 3

We undo a "divide by y" by doing a "multiply by y". We want to be fair and treat both sides of the equation the same.

On the left side 30 ÷ y × y = 30 and then on the right side we have 3 × y. The result is 30 = 3 × y

This is now a familiar one-step equation. We divide both sides by 3 and find the answer is 10 = y.

Notice that we created 30 = 3 × y which looked nicely familiar. We changed a problematic division equation into a well-understood multiplication equation.

This trick always works. Let's write it in a box.

How to Fix Starting with ÷ y

To solve an equation that looks like a one step equation but it starts with ÷ y, begin by multiplying both sides by y.

Here are a few more example problems.

1. Solve u × 20 = 0.4

To get u by itself we want to remove a ×20. So we do ÷20 to both sides. 0.4 ÷ 20 = 0.02

1. Solve v ÷ 20 = 0.4

To get v by itself we want to remove a ÷20. So we do ×20 to both sides. 0.4 × 20 = 8

1. Solve 20 ÷ w = 0.4

To get 20 by itself we want to remove a ÷w. So we do ×w to both sides.

The equation becomes 20 = 0.4 × w, which is now a one step equation.

To get w by itself we want to remove a ×0.4. So we do ÷0.4 to both sides. 20 ÷ 0.4 = 50

1. Solve x × 4 = 18

To get x by itself we want to remove a ×4. So we do ÷4 to both sides. 18 ÷ 4 = 132

1. Solve y ÷ 4 = 18

To get y by itself we want to remove a ÷4. So we do ×4 to both sides. 18 × 4 = 12

1. Solve 4 ÷ z = 18

To get 4 by itself we want to remove a ÷z. So we do ×z to both sides.

The equation becomes 4 = 18 × z, which is now a one step equation.

To get z by itself we want to remove a × 18. So we do ÷ 18 to both sides. 4 ÷ 18 = 32

#### Starting with − y

Similar shenanigans can happen with subtraction.

1. Solve y − 7 = 10

To get y by itself we want to remove a −7. So we do +7 to both sides. 10 + 7 = 17

That problem worked great. Adding 7 to both sides solved the puzzle.

1. Look at 10 − y = 7 but do not solve it yet.

Subtracting 10 from both sides is not the best way to begin. It creates y = 7 − 10 which is increasing the complexity.

What do you think is the right way to begin?

1. Solve 10 − y = 7

To get 10 by itself we want to remove a −y. So we do +y to both sides.

The equation becomes 10 = 7 + y, which is now a one step equation.

To get y by itself we want to remove a +7. So we do −7 to both sides. 10 − 7 = 3

Notice that we created 10 = 7 + y which looked nicely familiar. We changed a problematic subtraction equation into a well-understood addition equation.

This trick always works. Let's write in in a box.

How to Fix Starting with − y

To solve an equation that looks like a one step equation but it starts with − y, begin by adding y to both sides.

Here are a few more example problems.

1. Solve u + 0.8 = 1.4

To get u by itself we want to remove a +0.8. So we do −0.8 to both sides. 1.4 − 0.8 = 0.6

1. Solve v − 0.8 = 1.4

To get v by itself we want to remove a −0.8. So we do +0.8 to both sides. 1.4 + 0.8 = 2.2

1. Solve 2 − w = 1.4

To get 2 by itself we want to remove a −w. So we do +w to both sides.

The equation becomes 2 = 1.4 + w, which is now a one step equation.

To get w by itself we want to remove a +1.4. So we do −1.4 to both sides. 2 − 1.4 = 0.6

1. Solve x + 18 = 12

To get x by itself we want to remove a + 18. So we do − 18 to both sides. 1218 = 38

1. Solve y18 = 12

To get y by itself we want to remove a − 18. So we do + 18 to both sides. 12 + 18 = 58

1. Solve 13z = 19

To get 13 by itself we want to remove a −z. So we do +z to both sides.

The equation becomes 13 = 19 + z, which is now a one step equation.

To get z by itself we want to remove a + 19. So we do − 19 to both sides. 1319 = 29 None yet Textbook Exercises for Two Step Equations

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

(Our textbook has no exercises for this topic.)

### Ten Exercises

Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?

1.Solve    y + 2.6 = 1610

2. Solve    2.6 + y = 1610

3. Solve    y120 = 15

4. Solve    120y = 15

5. Solve    y × 12 = 0.6

6. Solve    12 × y = 0.6

7. Solve    y ÷ 110 = 0.5

8. Solve    110 ÷ y = 0.5

9. Solve    y − 2.8 = 3.2 + 8.7

10. Solve    2.8 − y = 3.2 + 8.7

### Random Exercises

I need to make the random exercises for Justice Fractions.

#### Bonus Problem

This problem is more difficult than our one- or two-step equations! Can you solve it anyway?

## Ratios

### Proportions and Cross Multiplying§5.4

Definition

A proportion is an equation of the format "ratio equals ratio" (or "rate equals rate").

Here is an example of a proportion: 7 miles2 hours = 35 miles10 hours

Note that proptions are much easier to read if the ratios are not written as "slanted fractions" the way HTML forces web page typing to do. During lecture we will rewrite these problems on the board as the vertical fractions that are easier to work with.

The most common thing to do with a proportion is to play a game involving "multiply in an x shape". Many students have seen this already and are good at doing it. But we should still discuss the technique.

Consider these two pictures. In each, two ratios claim to be equal. But only the top problem's equality is true. The pictures claim you can check if ratios are equal by multiplying in an x shape.  Why does this trick work? In a group, think about common denominators until you develop an explanation for the top picture that involves putting together the 2 and 10, and the 5 and 4.

What is happening when we check if two-fourths is equal to five-tenths by multiplying in an x shape? Why are we putting the 2 and 10 together? Or the 5 and 4 together?

If I had 2 pieces that were one-fourths, and cut each into 10 parts, I would have 20 pieces that were one-fortieths. Similarly, if I had 5 pieces that were one-tenths, and cut each into 4 parts, I would also have 20 pieces that were one-fortieths. Both cutting processes give me 20 pieces that are one-fortieths.

In other words, the "multiply in an x shape" trick is simply telling you to find common denominators using the brute force method.

So a proportion can be false, like the second picture.

Definition

Two ratios are proportional if they are equal.

The word "proportional" is just a fancy new term for the old concepts of "equal" or "equivalent fractions".

Let's do some problems about checking if a proportion is true.

Remember to be better than this webpage, and write your fractions vertically instead of diagonally!

First, a couple problems in which all of the numbers are whole numbers.

1. 37  ≟  1535

Does 3 × 35 equal 7 × 15? Yes. The ratios are proportional. The proportion is true.

1. 49  ≟  1228

Does 4 × 28 equal 9 × 12? No. The ratios are not proportional. The proportion is false.

Second, in which some or all of the numbers are decimals.

1. 79  ≟  5.47.2

Does 7 × 7.2 equal 9 × 5.4? No. The ratios are not proportional. The proportion is false.

1. 1.21.8  ≟  4.997.56

Does 1.2 × 7.56 equal 1.8 × 4.99? No. The ratios are not proportional. The proportion is false.

Strangely, we need to do the "multiply in an x shape" trick more than once if both diagonals of the "x shape" include fraction arithmetic. (Unless we can simply see the answer, which might happen with the next example.)

1. 3  ≟  ½2

Does ⅓ × 2 equal 3 × ½? In other words, does 23 equal 32 ?

You can probably see the answer. But if you could not, perhaps because the numbers were trickier with decimals or something, then use the "multiply in an x shape" trick again.

Does 2 × 2 equal 3 × 3? No. The ratios are not proportional. The proportion is false.

We could re-write the previous problem in a way that might be easier to read when typed:

1. Is the ratio 13 to 3 proportional to the ratio 12 to 2?
1. Is the ratio 13 to 12 proportional to the ratio 3 to 2?

Does ⅓ × 2 equal ½ × 3? In other words, does 23 equal 32 ?

You can probably see the answer. But if you could not, perhaps because the numbers were trickier with decimals or something, then use the "multiply in an x shape" trick again.

Does 2 × 2 equal 3 × 3? No. The ratios are not proportional. The proportion is false.

1. Is the ratio 13 to 25 proportional to the ratio 23 to 35?

Does 13 × 35 equal 25 × 23? In other words, does 315 equal 415 ?

Remember that we can think of "fifteenths" as a label, like apples or miles. No, 3 of them is not the same as 4 of them. The ratios are not proportional. The proportion is false.

Notice that we did not reduce 315 when multiplying. We could have written it as 15 but that would only have made the problem harder!

1. Is the ratio 23 to 34 proportional to the ratio 43 to 64?

Does 23 × 64 equal 34 × 43? In other words, does 1212 equal 1212 ?

Yes. The ratios are proportional. The proportion is true.

Notice that we did not reduce fractions, either before multiplying or after multiplying. That would only have made the problem longer! Parvin Taraz

Proportions and Cross Multiplying

Bittinger Chapter Tests

Chapter 5 Test, Problem 10: Check if 78 is proportional to 6372

Chapter 5 Test, Problem 11: Check if 1.33.4 is proportional to 5.615.2 Textbook Exercises for Proportions and Cross Multiplying

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 5.4 (Page 284) # 1, 3, 5, 7 (do not worry about this book's jargon words "means" and "extremes")

#### Proportions with Variables

Section 5.4 (Page 284) # 9, 11, 13, 15, 21, 23, 25, 31, 33, 35, 51, 53

### Proportions with Variables§5.4

Checking if the ratios in a potential proportion are really equal is only slightly interesting. Much more interesting is when we are told three of the four values in a proportion and must solve for the missing value. We still use the "multiply in an x shape" game. The process does not change if the ratios include decimals or mixed numbers.

Remember to be better than this webpage, and write your fractions vertically instead of diagonally!

1. y32  =  38

The proportion turns into the one step equation y × 8 = 96

To get y by itself we want to remove a ×8. So we do ÷8 to both sides. 96 ÷ 8 = 12

1. 1.31.2  =  y96

The proportion turns into the one step equation 124.8 = y × 1.2

To get y by itself we want to remove a ×1.2. So we do ÷1..2 to both sides. 124.8 ÷ 1.2 = 104

Notice that we have the power to make two choices...

• It does not matter which diagonal is on which side of the equality. So we can pick which diagonal to write to the left of the equal sign.
• Within each diagonal, it does not matter which number is first or second when multiplied. So we can pick which to write before the multiplication symbol.

We can put those two choices together and agree to always write the diagonal with the variable to the left of the equal sign, and to always put the variable before the multiplication symbol.

1. 144y  =  94

The proportion turns into the one step equation y × 9 = 576

To get y by itself we want to remove a ×9. So we do ÷9 to both sides. 576 ÷ 9 = 64

1. y39  =  813

The proportion turns into the one step equation y × 13 = 312

To get y by itself we want to remove a ×13. So we do ÷13 to both sides. 312 ÷ 13 = 24

1. 5291  =  4y

The proportion turns into the one step equation y × 52 = 364

To get y by itself we want to remove a ×52. So we do ÷52 to both sides. 364 ÷ 52 = 7

There is an important warning about the "multiply in an x shape" game. The following warning is only for students who have been taught a certain "shortcut", who have been taught to include division with the multiplying. If you have not been taught this "shortcut" then the warning will not make sense. Please ignore it! It is not for you.

Some students know a supposed shortcut that allows solving for x in one step: multiply diagonally and then divide by the other number. It may seem faster to do this than to always write out the "multiply in an x shape" step.

Let us solve 812 = y9 both ways to compare the differences.

You are advised to not use this shortcut! If the problem was even slightly harder the shortcut would hide options about how multiple ways to solve the problem. Don't build bad habits that will cause trouble in later classes.

Consider 83 = (y + ⅓)2.

When we multiply in an x shape we get 8 × 2 = 3 × (y + ⅓)

We could change that into either 16 = 3 × (y + ⅓) or 16 = 3y + 1.

The options lead to different natural next steps. The "shortcut" always picks the first option. So the habit of always solving proportions using the shortcut will later on force your to follow one path (which might be the hard one) instead of noticing both options.

This is why in this class we clearly define:

Definition

Cross multiplying is the "multiply in an x shape" step for dealing with a proportion.

Cross multiplying is usually followed by a step involving division. This is always true in our class. It is not always true in later math classes.

Note that some textbooks or websites call the combined process "cross multiplying", all the way from starting the problem until getting the answer. But wrapping the division step into what you name "cross multiplying" makes it harder to talk about the actual cross multiplying step while analyzing a problem written on the board.

(Our textbook avoids this isssue by using jargon involving "means" and "extremes". You can ignore that jargon.)

For now, work on good habits. Approach proportions by writing three steps.

• the cross multiplying step
• the division step (if solving for a letter) Chapter 5 Test, Problem 12: Solve: 94 = 27x

Chapter 5 Test, Problem 13: Solve: 1502.5 = x6

Chapter 5 Test, Problem 14: Solve: x100 = 2764

Chapter 5 Test, Problem 15: Solve: 68y = 1725 Textbook Exercises for Proportions with Variables

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 5.4 (Page 284) # 9, 11, 13, 15, 21, 23, 25, 31, 33, 35, 51, 53

### Proportion Word Problems§5.5

Just as the previous sections involving checking the correctness of proportions before we tried to solve for the missing value with proportions, now we will check if word problems are correctly put into proportions before we try to solve any word problems.

We begin by examining the patterns that differentiate correct and incorrect proportions. Below are four situations, each involving a pair of events. For each situation four possible proportions are listed. In groups, use cross-multiplying to check which proportions are correct. When most groups are done we will discuss what patterns people found. The pattern your group should have found was that the two events needs to be "kept together" symmetrically, either vertically or horizontally. If the two events are spread out upside down compared to each other then the proportion will not be correct.

If the two events are spread out diagonally then the proportion will not be correct. The pattern your group should have found while checking if a setup was correct is that the two events needs to be "kept together" symmetrically.

Most students remember this with the rule The labels on the right must match the labels on the left—do not flip them!

Here are more proportion word problems. As we solve them look for the symmetry we just discussed.

1. One brand of microwave popcorn has 120 calories per serving. A whole bag of this popcorn has 3.5 servings. How many calories are in a whole bag of this microwave popcorn?

The proportion is 120 cal1 serv = y cal3.5 serv

After cross multiplying, the equation without fractions is y = 120 × 3.5 = 420 calories

(For this example you could probably see to multiply even without setting up a proportion. An easy example is still an appropriate way to get used to a new process!)

1. When pediatricians prescribe acetaminophen to children, they prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of the child’s weight. If Zoe weighs 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?

The proportion is 5 ml25 lbs = y ml80 lbs

After cross multiplying, the equation without fractions is y × 25 = 5 × 80

The division step tells us y = 5 × 80 ÷ 25 = 16 ml

1. Suzie can read 12 pages in sixteen minutes. How many pages can she read in five hours?

The proportion is 12 pages16 min = y pages300 min (notice how we had to change 5 hours into 300 minutes, so the words on top would match)

After cross multiplying, the equation without fractions is y × 16 = 12 × 300

The division step tells us y = 12 × 300 ÷ 16 = 225 pages

1. Scott can do three test problems in eleven minutes. How long would it take him to finish a test with twenty problems?

The proportion is 3 problems11 min = 20 problemsy min

After cross multiplying, the equation without fractions is y × 3 = 11 × 20

The division step tells us y = 11 × 20 ÷ 3 ≈ 73 minutes

1. Eight feet of pressure-treated 2×10 board costs \$14.49. How much will it cost to buy 158 feet of the board?

The proportion is 8 feet\$14.49 = 158 feety dollars

After cross multiplying, the equation without fractions is y × 8 = 158 × \$14.49

The division step tells us y = 158 × \$14.49 ÷ 8 ≈ \$286.18

1. Maria drinks 4 cups of coffee every 5 days. How many cups of coffee is this per year?

The proportion is 4 cups5 days = y cups365 days (notice how we had to change 1 year into 365 days, so the words on top would match)

After cross multiplying, the equation without fractions is y × 5 = 4 × 365

The division step tells us y = 4 × 365 ÷ 5 = 292 cups

Some proportion problems are really tricky. These are the "catch and release" problems. Everyone's natural intuition about labels for rates is of absolutely no help in creating "symmetrical" labels for the two rates in these proportions. So don't feel bad that these are hard. They are tricky for everyone.

Let's look at two examples of "catch and release" problems.

1. At the beginning of a study 24 fish are caught, tagged, and released back into a large pond containing fish. A few days later, 19 fish are caught from the pond and three of them have tags. Using this ratio, how many total fish would you expect to be living in the pond?

The proportion is 24 tagged in entire pondy total in entire pond = 3 tagged in second catch19 total in second catch

After cross multiplying, the equation without fractions is y × 3 = 24 × 19

The division step tells us y = 24 × 19 ÷ 3 ≈ 152 fish

1. To determine the number of fish in a lake, a park ranger catches 260 fish, tags them, and returns them to the lake. Later, 144 fish are caught, and it is found that 20 of them are tagged. Estimate the number of fish in the lake.

The proportion is 260 tagged in entire lakey total in entire lake = 20 tagged in second catch144 total in second catch

After cross multiplying, the equation without fractions is y × 20 = 260 × 144

The division step tells us y = 260 × 144 ÷ 20 ≈ 1,872 fish Chapter 5 Test, Problem 16: An ocean liner traveled 432 kilometers in 12 hours. At this rate, how far would it travel in 42 hours?

Chapter 5 Test, Problem 17: A watch loses 2 minutes in 10 hours. At this rate, how much will it lose in 24 hours?

Chapter 5 Test, Problem 18: On a map, 3 inches represents 225 miles. If two cities are 7 inches apart on the map, how far are they apart in reality?

Chapter 5 Test, Problem 21: A grocery store special sells ingredients for a traditional turkey dinner for eight people for \$33.81. How much should it cost if that deal applied to a dinner for fourteen people? Textbook Exercises for Proportion Word Problems

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 5.5 (Page 291) # 1, 3, 5, 7, 9, 13, 19, 21

### Three Methods to Solve a Proportional Rate

Now you know three methods to solve a proportional rate.

Let's do the same problem in using three methods.

#### Find an Intermediate Unit Rate

1. A bug crawls 36 feet in 1.8 minutes. How many feet would it crawl in 12 minutes?

First we find the unit rate using both pieces of information from the well-known situation. 36 feet ÷ 1.8 minutes = 20 feet per minute.

Then we use the number from the second situation. 20 feet per minute × 12 minutes = 240 feet.

#### Proportion

1. A bug crawls 36 feet in 1.8 minutes. How many feet would it crawl in 12 minutes?

The proportion is 36 feet1.8 minutes = y feet12 minutes

After cross multiplying, the equation without fractions is y × 1.8 = 36 × 12

The division step tells us y = 36 × 12 ÷ 1.8 ≈ 240 feet

#### Unit Analysis

1. A bug crawls 36 feet in 1.8 minutes. How many feet would it crawl in 12 minutes?

Notice that the well-known situation becomes the way we modify the second situation that has only one number.

12 minutes1 = 12 minutes1 × 36 feet1.8 minutes = 432 feet1.8 = 240 feet

So...

Which method do you like best? Why?

You do not need to be fluent with all three methods. But you should be able to understand all three, for when you are watching a lecture (the instructor might use any of the three methods) or doing group work (your group members might use any of the three methods).

### Scale Factors§5.6 Many of the proportion problems we solved above involve reducing or un-reducing.

When this happens the number we are un-reducing with is called the scale factor.

In the first example to the right, we can say we scaled up the speed of 40 miles per hour by 2.

In the second example to the right, we can say we scaled up the parking meter cost of 25 cents per 15 minutes by 3.

We do not need to notice the scale factor to solve a proportion problem with a missing value. We can always cross multiply.

However, when we are asked to do an entire set of similar problems it can be efficient to notice and use the scale factors.

We can find scale factors by doing division. Identify which amounts are "new" and which are "original". Then do new ÷ original.

1. An elementary school student is doing a measurement project at the playground when she notices something surprising. She is 4' tall, but her shadow is 12' tall. Her three project partners are 4' 2" tall, 4' 4" tall, and 4' 6" tall. How long are their three shadows?

The scale factor is new ÷ original = 12 feet ÷ 4 feet = 3

The three heights, when changed to inches, are 50", 52", and 54". (Why is this step needed?)

So the three shadows will be 50" × 3 = 150" = 12' 6", 52" × 3 = 156" = 13', and 54" × 3 = 162" ≈ 13' 6"

1. A web page designer needs to include images that are 600 pixels wide. This page will have four images. The original images all have a width of 750 pixels, but four different heights: 600 pixels, 720 pixels, 800 pixels, and 1,200 pixels. What will the four new heights be?

The scale factor is new ÷ original = 600 pixels ÷ 750 pixels = 0.8

So the four new heights will be 600 × 0.8 = 480 pixels, 720 × 0.8 = 576 pixels, 800 × 0.8 = 640 pixels, and 1,200 × 0.8 = 960 pixels

1. One-half inch on my map represents 3 miles in real life. Three hiking trails shown on the map are 1.5 inches long, 1.75 inches long, and 2.1 inches long. How long are these hiking trails in real life?

The scale factor is new ÷ original = 3 ÷ 0.5 = 6

So the three hike lengths will be 1.5 × 6 = 9 miles, 1.75 × 6 = 10.5 miles, and 2.1 × 6 = 12.6 miles

Some problems are best to solve with scaling instead of cross multiplyying simply because the problem gives us the scale factor.

1. An elementary school student is doing a measurement project at the playground when she notices something surprising. At that time, every shadow is five times the height of its object. If she is 4' 2" tall, and her teacher is 5' 6" tall, how long are their shadows?

The scale factor is 5.

The two heights, when changed to inches, are 50" and 66". (Why is this step needed?)

So the two shadows will be 50" × 5 = 250" = 20' 10" and 66" × 5 = 330" ≈ 27' 6"

Scaling can happen with two objects (a girl and her shadow), when an object changes size (the digital images were shrunk), or when something is measured differently (moving from a daily to annual amount).

We can think of simple interest problems as using two scale factors. We scale the principal by both the interest rate and the years of time.

Many problems that use scaling involve pictures. The picture below is from the website MathIsFun. Click on that link or the picture below to go to a page where you can drag a picture of a butterfly to resize it and see the appropriate scale factor.

In general, these pictures are called scale diagrams. When the pictures are geometric shapes, they are also called similar figures.

When using scale factors be wary of how often they are "triggered" over time.

1. Two science fiction fans are discussing space elevators. Consider these two claims, which might not actually be true. (a) The maximum strength of cable has doubled from 1950 to today. (b) Every year since 1950 the maximum strength of cable has doubled. Are those claims two ways of saying the same thing?

The first claim has only one instance of doubling. Whatever the strength of cable was in 1950, we are now at that × 2.

The second claim has only an instance of doubling every year. Whatever the strength of cable was in 1950, we are now at that × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2...

The second claim results in a very much bigger answer!

If you already know how to graph lines, you can see that the slope of a line is another example of a situation that involves scaling.

You may have memorized the formula "rise over run" to find slope. This is simply using "top ÷ bottom" to find a unit rate. In the example below, the rise is 7 and the run is 5, so the slope is 7 ÷ 5 = 1.4 Every point on the line is a scaled version of the measured points.

We could move along the line "one slope" to move (+1, +1.4) from any known point.

We could move along the line "two slopes" to move (+2, +2.8) from any known point.

We could move along the line "three slopes" to move (+3, +4.2) from any known point.

We could move along the line "five slopes" to move (+5, +7) from any known point. The graph above shows this!

Does it help to think of the slope as unit rate that we can multiply by a scale factor? Bittinger Chapter Tests, 11th Edition

Chapter 5 Test, Problem 1: Write the ratio "85 to 97" in fraction notation. Do not simplify.

Chapter 5 Test, Problem 2: Write the ratio "0.34 to 124" in fraction notation. Do not simplify.

Chapter 5 Test, Problem 6: A twelve pound shankless ham contains sixteen servings. What is the rate in servings per pound?

Chapter 5 Test, Problem 7: A car will travel 464 miles on 14.5 gallons of gasoline in highway driving. What is the rate in miles per gallon?

Chapter 5 Test, Problem 8: A sixteen ounce bag of salad greens costs \$2.39. Find the unit price in cents per ounce.

Chapter 5 Test, Problem 10: Check if 78 is proportional to 6372

Chapter 5 Test, Problem 11: Check if 1.33.4 is proportional to 5.615.2

Chapter 5 Test, Problem 12: Solve: 94 = 27x

Chapter 5 Test, Problem 13: Solve: 1502.5 = x6

Chapter 5 Test, Problem 14: Solve: x100 = 2764

Chapter 5 Test, Problem 15: Solve: 68y = 1725 Textbook Exercises for Scale Factors

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 5.6 (Page 299) # 1, 3, 5, 7, 9, 17, 19, 23, 25, 27, 31

### Variation

Another name for proportional situations is direct variation. What is "direct" is that an increase or decrease affects both numbers.

For example, 80 miles every 2 hours is proportional to both 120 miles every 3 hours (scaling up using ×1.5 to both numbers) and also 40 miles every 1 hour (scaling down using ÷2 to both numbers).

We spent a lot of time with proportional situations because the happen so often in real life. You can think of more situations that involve direct variation.

At a constant rate, the longer you drive the farther you go.

At a constant pace and elevation, the more time you spend jogging the more calories you burn.

At a constant price, the more candy bars you buy the more money you spend.

But other real life situations work in the opposite way. We call these either indirect variation or inverse variation. When one number goes up with multiplication, the other number goes down with division. Consider frozen orange juice concentrate. For the sake of efficient packaging and shipping, it is packaged with quadruple concentration. Mixing it with three cans of water brings the concentration down to the normal concentration for orange juice. The volume changes by ×4 (getting bigger) while the concentration changes by ÷ 4 (getting smaller).

Using other words, in proportional situations with direct variation each situation's two numbers divide to the same amount (the unit rate).

Direct Variation (Proportional, Scaling Up or Down)

In a situation with direct variation,

Initial Amount 1Initial Amount 2 = Other Amount 1Other Amount 2

But in situations with indirect/inverse variation each situation's two numbers always multiply to the same amount (which usually has no name).

Indirect/Inverse Variation

In a situation with indirect/inverse variation,

Initial Amount 1 × Initial Amount 2 = Other Amount 1 × Other Amount 2

Try to think of more situations that involve indirect/inverse variation.

The more people equally share a cake, the smaller is each person's slice.

The more people equally pay for a dinner, the less each person pays.

The more people equally help to fold a thousand paper cranes, the fewer each person must fold.

In a chemistry laboratory, many chemicals are stored as a stock solution that has a high concentration. Similar to frozen orange juice concentrate, these stock solutions must be dilluted before use.

1. A chemical is stored in a 25% concentration stock solution. Your experiment needs 500 mL of a 10% concentration solution. How much stock solution is used? (Note that the rest of the 500 mL will be the water you add to dillute appropriately.)

We need a dillution with Amount 1 × Amount 2 = 500 mL × 0.1 concentration = 50

For each situation the two numbers must multiply to the same amount. So for the stock solution we need y mL × 0.25 concentration = 50

Solving for y tells us to use 200 mL of stock solution (and the other 300 mL will be the water we add).

Another example from a science laboratory is Boyle's Law, which states that the pressure and volume of an ideal gas vary with indirect/inverse variation. Notice that we do not need to understand the unit of pressure with the label mmHg as long as that unit is used in both situations. But that's a short and interesting read, about the history of measuring air pressure with barometers.

1. A flexible containter of 200 mL of an ideal gas is stored at a pressure of 750 mmHg. If the pressure is changed to 250 mmHg, what is the resulting volume?

We have Amount 1 × Amount 2 = 200 mL × 750 mmHg = 150,000

For each situation the two numbers must multiply to the same amount. So we need y mL × 250 mmHg = 150,000

Solving for y tells us the new volume will be 600 mL.

1. A respiratory therapist is administering oxygen to a patient. Each tank has 250 mL of oxygen at a pressure of 1,000 mmHg. Her equipment reduces the pressure the patient recieves to 200 mmHg. What volume will be provided by each tank?

We have a tank with Amount 1 × Amount 2 = 250 mL × 1,000 mmHg = 250,000

For each situation the two numbers must multiply to the same amount. So the patient will receive y mL × 200 mmHg = 250,000

Solving for y tells us each tank will provide 1,250 mL of oxygen before it is empty and a new tank must be swapped into the equipment. None yet Textbook Exercises for Two Step Equations

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

(Our textbook has no exercises for this topic.)

### Ten Exercises

Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?

1. Is the ratio "2.4 to 3.6" proportional to the ratio "1.8 to 2.7"?

2. Solve the proportion: "7 to 1/4" is proportional to "28 to what?"

3. Fifteen hours of studying before a test lets you score 75 points. At that rate, how many points would you expect from studying 18 hours?

4. Julia's car can drive 120 miles on 4.5 gallons of gas. While driving across the country the tank gets down to 0.9 gallons. How many miles are left, for her to find a gas station?

5. Typically 5 people produce 13 kilograms of garbage each day. How many kilograms of garbage are produced each day by the 346,560 people in Lane County?

6. In a class of 40 students, on average six will be left-handed. A certain class has nine left-handed students. How large would you estimate the class is if its proportion of left-handed students is average?

7. The United States debt-to-GDP ratio is currently 255 to 200. If the U.S. debt is currently 28.8 trillion dollars, what is the gross domestic product?

8. A 25 pound turkey serves 18 people. How many pounds does each serving weigh?

9. A 25 pound turkey serves 18 people. What is the unit rate of servings per pound?

10. To determine the number of deer in a game preserve, a forest ranger catches, tags, and releases 318 deer. Later he catches 168 deer and sees that 56 of them are tagged. Use a proportion to estimate the number of deer in the game preserve.

### Random Exercises

Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.

## Percentages

### Percent Sentences§6.2

Percent sentences are the simplest word problems that involve percents.

We will soon see that the key to doing harder percent word problems is to first translate them into percent sentences before trying to write an equation.

So percent sentences are both a kind of problem and a tool to solve other problems.

Percent Sentence

A Percent Sentence is a short word problem that includes the words is, of, what, and %.

Those four words can appear in any order.

Here are sample percent sentences. (We will solve them later.)

What is 35% of 60?

12 is what percent of 5?

5 is 2% of what?

There are two different methods for solving percent sentences. You only need to master one method. On the homework and on tests you can always do which method you choose.

#### The Translation Method

The first method is to translate the percent sentence into an equation. As usual in math:

• of means multiply
• is means equals
• what means y

Do not translate the word percent into something. Instead, use RIP LOP to move between decimal format and percent format mentally, with two decimal point scoots.

1. Use the translation method to solve: What is 35% of 60?

The percent sentence is translated into y = 0.35 × 60

Solve for y. The answer is 21.

1. Use the translation method to solve: 12 is what percent of 5?

The percent sentence is translated into 12 = y × 5

We solve for y and we get 2.4.

But the problem asked for an answer in percent format. So use RIP LOP to see the answer is 240%.

1. Use the translation method to solve: 5 is 2% of what?

The percent sentence is translated into 5 = 0.02 × y

Solve for y. The answer is 250.

#### The Proportion Method

The second method is to write the percent sentence into a proportion. The steps are always the same.

• Write the "skeleton" of the proportion, fraction bar equals fraction bar
• Make the left ratio represent the percent by writing a value over 100
• The bottom right is the value that follows the word "of"
• The last value goes on the upper right

Let's redo the same examples.

1. Use the proportion method to solve: What is 35% of 60?

The percent sentence becomes the proportion 35100 = y60

Solve for y. The answer is 21.

1. Use the proportion method to solve: 12 is what percent of 5?

The percent sentence becomes the proportion y100 = 125

Solve for y. The answer is 240. Notice we do not use RIP LOP. We write 240%.

(The proportion includes writing the percentage over 100. This is already one of our four replacements for the percent symbol. So we do not need RIP LOP.)

1. Use the proportion method to solve: 5 is 2% of what?

The percent sentence becomes the proportion 2100 = 5y

Solve for y. The answer is 250. Mathispower4u

Solving Percent Problems Using The Percent Equation

Chapter 6 Test, Problem 5: What is 40% of 55?

Chapter 6 Test, Problem 6: What percent of 80 is 65?

Chapter 6 Test, Problem 21: 0.75% of what number is 300?

#### Three Patterns

Notice that percent sentences appear three different patterns:

• "what" first: What is Y percent of Z?
• "what" second: Y is what percent of Z?
• "what" third: Y is Z percent of what?

(In these pattenrs Y and Z are two numbers.)

We could try to memorize rules for what arithmetic steps happen in each pattern. But this is too much work! It is much easier to simply learn either the translation method or the proportion method since those two methods can always be used.

However, we should notice that in every patten the word "is" appears before the word "of". This is important! We like that!

Not every percent sentence is friendly enough to have "is" appear before "of". All three patterns have an alternate form in which the "of" apperas before the "is".

• "what" first: What is Y percent of Z? — the same as "Y percent of Z is what?"
• "what" second: Y is what percent of Z? — the same as "What percent of Z is Y?"
• "what" third: Y is Z percent of what? — the same as "Z percent of what is Y?"

It is not important to memorize how the three patterns have alternate forms. Both the translation method and the proportion method work in all situations. We are fully prepared!

Yet when we write our own percent sentences we should be polite and always have "is" appear before "of". For most people this looks and reads more natural. Be careful! This nice picture falsely implies that the part/change/new amount is always smaller than the whole/original/baseline amount. But that is not true! Real life is not so simple. Prices go up, as well as going on sale. People gain weight, as well as losing weight. Investments appreciate, as well as depreciate.

1. A young couple earns money by improving a "fixer-upper" home. They buy it the home for \$65,000. After months of repairs and improvements they sell the home for \$105,000. A friend asks them, "What is the new worth as a percentage of the value you paid for it?" Rewrite this situation as a percent sentence.

\$105,000 is what percent of \$65,000?

1. Continuing the previous problem, solve your percent sentence using your preferred method.

The translate method makes it \$105,000 = y × \$65,000 (and needs RIP LOP as a final step)

The proportion method makes it y100 = \$105,000\$65,000 (and does not use RIP LOP)

#### How Many Vowels Worksheet Now that we can do "X is what percent of Y?" type problems, we can make pie charts.

Let's use a worksheet named How Many Vowels?.

Today people can make a pie chart using a computer. But doing the old-fashioned process is still a useful project to help cement our understanding of percentages.

We will make a bar chart first, and then use scissors and tape to turn the bar chart into a pie chart.

#### The Complicated Shortcuts

Most students like using a reliable process that always works. We have just learned two: the translation method and the proportion method. If either of those makes you happy, great! You have a routine you like. Skip this next thing.

A few students like juggling a bunch of specific shortcuts. Shortcuts can feel clever and powerful. For these students, it seems worth the extra effort to keep track of many rules, and to pay attention to when to use each rule.

If you are that kind of student, here are the shortcuts for percent sentences. You would develop this intuition anyway after doing a bunch of problems using the translation method or the proportion method.

Percent Sentence Shortcuts

Any percent sentence involves three values: a number that is the part/change/new amount, a percentage, and a number that is the whole/original/baseline amount.

• If you are missing the part/change/new amount, multiply the percentage and the whole/original/baseline amount.
• If you are missing the percentage amount, divide the part/change/new amount by the whole/original/baseline amount.
• If you are missing the whole/original/baseline amount, divide the part/change/new amount by the percentage.

If you like shortcuts, please be wary. In other books or websites you might encounter different percent sentence shortcuts that only work when the part/change/new amount is smaller.

#### The Fraction Trick

Some word problems include with the words "What percent of...?"

These can be solved the ways we learned above, changing them into a percent sentence and then using either the Translation Method or the Proportion Method.

Guppies with Percent Sentences

In a tank of 10 fish, 8 are guppies. What percent of the fish are guppies?

First make into a percent sentence with is before of.

Ask, "8 is what percent of 10?"

Then solve. Let's use the Translation Method for the sake of brevity.

8 = y × 10

0.8 = y

80% = y

But we can also think of these problems as asking for a fraction. Look for a part divided by a whole. As before, the whole always follows the word of.

Guppies with Fraction Trick

In a tank of 10 fish, 8 are guppies. What percent of the fish are guppies?

Consider the part and whole.

realize that 8 is the part, 10 is the whole

Write this as a fraction, then change it into a percent.

810 = 0.8 = 80%

None yet Textbook Exercises for Percent Sentences

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 6.2 (Page 327) # 1, 3, 5, 9, 11, 13, 17, 19, 25, 27, 29, 31, 3, 35, 37, 45

### Percent Word Problems§6.3, §6.4

Recall the definition of a Percent Sentence.

Definition

A Percent Sentence is a short word problem that includes the words is, of, what, and %.

Those four words can appear in any order.

Here is a long word problem to translate into a percent sentence. (Only translate the word problem. Do not solve it.)

1. In a recent survey, 58% of Edgewood students said they prefer tablets to laptops. If there are 300 Edgewood students, then how many students prefer tablets?

There is more than one correct translation!

Check if a Setup is Correct

Which percent sentences correctly translate the problem? How do we know they are right?

1. 58% of what is 300?
2. 300 is 58% of what?
3. What is 58% of 300?
4. 58% of 300 is what?

The last two options are valid translations.

The first two options falsely imagine a huge group bigger than 300. They ask, "A bit more than half of what huge group is 300?"

But the situation has no group bigger than 300. We are looking for a group smaller than 300 because the students who prefer tablets are only a part of the total students.

Now let's solve the problem

1. In a recent survey, 58% of Edgewood students said they prefer tablets to laptops. If there are 300 Edgewood students, then how many students prefer tablets?

We ask, "What is 58% of 300?"

The translate method makes it y = 0.58 × 300 = 174 students

The proportion method makes it 58100 = y300 and of course the answer is the same: 174 students.

#### Shortcut for the Proportion Method

When solving percent word problems the four steps of the proportion method turn into this diagram: If you prefer the proportion method you can memorize that diagram and skip writing the percent sentence.

Your job is still to read the word problem identify the "part as amount", "part as percent", and "whole". Two of these will be numbers you know. The last will be something to solve for.

#### Work Backwards to Create Your Own

Your turn to create a word problem solvable with a percent sentence.

Recall that percent sentences appear three different patterns:

• what first: What is Y percent of Z?
• what second: Y is what percent of Z?
• what third: Y is Z percent of what?

As a group, follow these four steps.

1. Pick one of these patterns
2. Make up numbers for Y and Z
3. Invent a word problem for those numbers
4. Trade problems and race to solve them

#### More Example Problems

1. William France's car can drive 396 miles on a full tank of gas. His gas tank is currently 15% full. How many more miles can he drive before he runs out of gas?

We ask, "What is 15% of 396 miles?"

The translate method makes it y = 0.15 × 396 miles59 miles

The proportion method makes it 15100 = y396 and of course the answer is the same: about 59 miles.

1. 65. During the school year Eugene has a population of 180,700. The University of Oregon has 20,600 undergraduate students. What percentage of Eugene's school year population are undergraduate University of Oregon students?

We ask, "20,600 is what percent of 180,700?"

The translate method makes it 20,600 = y × 180,700. We divide both sides by 180,700 and then use RIP LOP to get 11.4% (notice we keep three non-rounded digits, to match the problem's original numbers)

The proportion method makes it y100 = 20,600180,700 and of course the answer is the same: about 11.4%.

1. The University of Oregon has 20,600 undergraduate students and 3,800 graduate students. What percentage of the students at the University of Oregon are undergraduate students?

The translate method makes it 20,600 = y × 24,400. We divide both sides by 180,700 and then use RIP LOP to get 84.4% (again we keep three non-rounded digits, to match the problem's original numbers)

The proportion method makes it y100 = 20,60024,400 and of course the answer is the same: about 84.4%.

1. Cierra and her sister enjoyed a special dinner in a restaurant, and the bill was \$41.50. If she wants to leave 18% of the total bill as her tip, how much should she leave?

We ask, "What is 18% of \$41.50?"

The translate method makes it y = 0.18 × \$41.50\$7.47

The proportion method makes it 18100 = y\$41.50 and of course the answer is the same: about \$7.47.

1. One serving of Chocolate Frosted Sugar Bomb cereal has 7 grams of fiber, which is 29% of the recommended daily amount. What is the total recommended daily amount of fiber?

We ask, "7 is 29% of what?"

The translate method makes it 7 = 0.29 × y. We divide both sides by 0.29 to get about 24 grams

The proportion method makes it 29100 = 7y and of course the answer is the same: about 24 grams.

1. A muffin package says each muffin has 230 calories, of which 60 calories are from fat. What percent of the total calories are from fat?

We ask, "60 is what percent of 230?"

The translate method makes it 60 = y × 230. We divide both sides by 230 and then use RIP LOP to get 26%

The proportion method makes it y100 = 60230 and of course the answer is the same: about 26%.

1. In 1995, the standard bus fare in Chicago was \$1.50. In 2008, the standard bus fare was \$2.25. Find the percent increase.

We ask, "\$2.25 is what percent of \$1.50?"

The translate method makes it \$2.25 = y × \$1.50. We divide both sides by \$1.50 and then use RIP LOP to get 150%

The proportion method makes it y100 = \$2.25\$1.50 and of course the answer is the same: about 150%.

1. At the campus coffee cart, a medium coffee costs \$1.65. Leslie brings \$2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?

We ask, "\$0.35 is what percent of \$1.65?" (Why are we using \$0.35 instead of \$2.00?)

The translate method makes it \$0.35 = y × \$1.65. We divide both sides by \$1.65 and then use RIP LOP to get about 21%

The proportion method makes it y100 = \$0.35\$1.65 and of course the answer is the same: about 21%.

1. The average price of a gallon of gas in one city in June 2014 was \$3.71. The average price in that city in July was \$3.64. Find the percent decrease.

We ask, "\$0.07 is what percent of \$3.71?" (Why are we using \$0.07 instead of \$3.64?)

The translate method makes it \$0.07 = y × \$3.71. We divide both sides by \$3.71 and then use RIP LOP to get about 2%

The proportion method makes it y100 = \$0.07\$3.71 and of course the answer is the same: about 2%.

1. Alison was late paying her credit card bill of \$249. She was charged a 5% late fee. What was the amount of the late fee?

We ask, "What is 5% of \$249?"

The translate method makes it y = 0.05 × \$249. We multiply to get \$12.45

The proportion method makes it 5100 = y\$249 and of course the answer is the same: \$12.45.

1. John bought a new tablet for \$499 plus tax. He was surprised when the tax amount was \$35.50. What was the sales tax rate?

We ask, "\$35.50 is what percent of \$499?"

The translate method makes it \$35.50 = y × \$499. We divide both sides by \$499 and then use RIP LOP to get about 7%

The proportion method makes it y100 = \$35.50\$499 and of course the answer is the same: about 7%.

1. Louie is a travel agent. He receives 7% commission when he books a cruise for a customer. How much commission will he receive for booking a \$3,900 cruise?

We ask, "What is 7% of \$3,900?"

The translate method makes it y = 0.07 × \$3,900. We multiply to get \$273

The proportion method makes it 7100 = y\$3,900 and of course the answer is the same: \$273.

1. Melinda earned an \$80 commission when she sold a \$1,500 stove. What is her commission rate?

We ask, "\$80 is what percent of \$1,500?"

The translate method makes it \$80 = y × \$1,500. We divide both sides by \$1,500 and then use RIP LOP to get about 5.3% (commission rates are often measured to the tenth of a percent)

The proportion method makes it y100 = \$80\$1,500 and of course the answer is the same: about 5.3%.

1. Jen can buy a bag of dog food for \$35 at two different stores. One store offers 5% cash back on the purchase plus \$5 off her next purchase. At the other store she can use a 20% off coupon. Which deal is better for her?

The first store gives her \$1.75 back plus the \$5 coupon for a total of \$6.75.

The second store gives her \$7.00 back, which is greater. Bittinger Chapter Tests, 11th Edition

Chapter 6 Test, Problem 8: Garrett Atkins, third baseman for the Colorado Rockies, got 175 hits during the 2008 baseball season. This was about 28.64% of his at-bats. How many at-bats did he have?

Chapter 6 Test, Problem 10: There are about 6,603,000,000 people living in the world toay, and approximately 4,002,000,000 live in Asia. What percent of people live in Asia?

Chapter 6 Test, Problem 11: The sales tax rate in Oklahoma is 4.5%. How much tax is charged on a pruchase of \$560? What is the total price?

Chapter 6 Test, Problem 12: Noah's commission rate is 15%. What is the commission from the sale of \$4,200 worth of merchandise? Textbook Exercises for Percent Word Problems

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 6.3 (Page 334) # 1, 3, 5, 7, 9, 19, 21

Section 6.4 (Page 340) # 3, 5, 7, 9, 11, 15, 17

### Ten Exercises

Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?

1. 56.32 is 64% of what number?

2. 42 is 30% of what number?

3. What number is 150% of 3045?

4. Out of 245 racers who started the Junction City Marathon, 203 completed the race, 38 gave up, and 4 were disqualified. What percentage of racers did not complete the marathon?

5. Patrick left a tip of \$8 on a restaurant bill of \$50. What percent tip is that?

6. A project on Kickstarter.com was aiming to raise \$15,000 for a precision coffee press. They ended up with 714 supporters, and raised 567% of their goal. How much did they raise?

7. A student got 35 problems correct on a test with 45 problems. What is his percentage score?

8. A salesman earns a 40% commission. One week he earns \$552 in commission. How much did he sell?

9. In my city 85% of the people who take a driver's licence test pass the first time. In January 289 people passed the test. How many people took the test?

10. At the zoo an elephant is put on a diet until it weighs only 91% of its original 9,671 pounds. What weight was the diet's goal?

### Random Exercises

Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.

## Measurement

### Temperature§7.4 Because of a strange bit of history we cannot use proportions or Unit Analysis for temperature conversion between Celsius and Fahrenheit. Instead we need to use (but not memorize) formulas.

First we need a formula to switch from Celsius to Fahrenheit. Here are three equivalent and equally workable options. Pick your favorite and ignore the other two.

• F = 1.8 × C + 32
• F = 95 × C + 32
• F = (C + 40) × 95 − 40

We also need a formula to switch from Fahrenheit to Celsius. Again, here are three equivalent and equally workable options. Pick your favorite and ignore the other two.

• C = (F − 32) ÷ 1.8
• C = (F − 32) × 59
• C = (F + 40) × 59 − 40

The last formula of each group was created in 2005 by Robert Warren. He thinks they are easier to remember. They are based on the coincidence that -40 °C is also -40 °F.

1. A hot tub is 39 °C. What is this temperature in degrees Fahrenheit?

F = 1.8 × C + 32 = 1.8 × 39 + 32 = 102.2 °F

1. The temperature on a June afternoon was 73 °F. What is this temperature in degrees Celsius?

C = (F − 32) ÷ 1.8 = (73 − 32) ÷ 1.8 = 22.8 °C Chapter 8 Test, Problem 22: Convert 95°F to Celsius.

Chapter 8 Test, Problem 23: Convert 59°C to Fahrenheit. Textbook Exercises for Temperature

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 7.4 (Page 397) # 23, 25, 49

### Area Puzzles§8.2

Many problems make finding area puzzle-like. Sometimes we can "stick together" small pieces to find a big area. Sometimes we can "remove" a small piece from a big area to get the shape in question. And sometimes either method will work!

1. Find the perimeter and area of this shape. When finding the area, which plan did you use?

This "subtract pieces" plan? (big rectangle) − (small rectangle) = (6 × 2.5) − (2 × 1.5) = 15 − 3 = 12 square inches

This "glue pieces together" plan? (left rectangle) + (right rectangle) = (1 × 2) − (2.5 × 4) = 2 + 10 = 12 square inches

Or this other "glue pieces together" plan? (top rectangle) + (bottom rectangle) = (1 × 6) − (1.5 × 4) = 6 + 6 = 12 square inches

All of those work! Which plan seems most natural varies from person to person. Our brains are not all built the same!

Let's do another example of a puzzle-like area problem.

1. Find the area of this shape. (rectangle) + (triangle) = (length × width) + (½ × base × height) = (10 × 12) + (½ × 10 × 6) = 120 + 30 = 150 square inches

Here is a "heads up" warning. When solving geometry problems do not get confused if the diagram provides too many numbers!

Consider this problem:

1. What is the area of this square? length × width = 16

Here is the same problem with extra numbers.

1. What is the area of this square? length × width = 16

The extra numbers do nothing! The area does not change. The problem does not magically change from an area problem into a perimeter problem merely because all the sides were labeled.

Be wary! Keep the formulas in mind. Ignore extra numbers.

Let's do two more examples of puzzle-like area problems.

1. Find the area of this shape. (big triangle) + 6 × (little triangle) = (½ × base × height) + 6 × (½ × base × height) = (½ × 6 × 6) + 6 × (½ × 1 × 1) = 18 + 3 = 21 square centimeters

We could also photocopy the shape, rotate the copy, and fit them together to make a rectangle with sides 6 cm and 7cm. That big rectangle thus has an area of 42 sq. cm., so half of it is our original shape with an area of 21 square centimeters

1. Find the area of the sidewalk, which is only on two sides of the building. The picture can be confusing! Try drawing the footprint of the building instead.

The problem is easy once you draw flat rectangles. (big rectangle) − (small rectangle) = (113.4 × 75.4) − (110 × 72) = 8,550.36 − 7,920 ≡ 630 square feet

The queen of area puzzles is Catriona Shearer. You can read an interview with her on the website Math With Bad Drawings. She has a book too.

The king of area puzzles that only involve rectangles is Naoki Inaba. More of his easier puzzles are here. You can also buy a book of them. None yet Textbook Exercises for Area Puzzles

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 8.2 (Page 431) # 9, 11, 15, 23, 29, 35

### Prisms§8.3, §8.4

Prisms

A prism is a three-dimensional shape formed by holding one copy of a polygon above another and then connecting matching corners.

According to this definition the polygon copies start as the top and bottom of the prism. We often rotate a prism so the two polygons look like the front and back of the shape.

(Be aware that according to this definition a cylinder is not a prism because its top and bottom are copies of a circle instead of a polygon. But many people use the phrase "cylindrical prism". We will not study cylinders.)

#### Prism Surface Area

We just finished looking at flat area puzzles. The surface area of prisms is a different kind of area puzzle. You can image the prism was created by folding flat surfaces to make a three-dimensional shape.

Working backwards, we can "unfold" a prism to make its surfaces lie flat again.

A nice slideshow by Shyanne Delaney has some examples. Time for some example problems.

First, imagine a rectangular box whose sides have lengths 2 meters, 2.5 meters, and 4 meters.

1. How much cardboard is needed for all of the box's surfaces?

??? Next, imagine an old-fashioned "soldier tent" with fabric on all four sides and its floor. The triangular front and back of the tent is 4 feet high and 6 feet across the base. The two rectangular sides make the tent 8 feet deep, and are 5 feet tall (this is taller than the height of the tent because they slant). The floor of the tent is another fabric rectangle.

1. How much fabric is needed for all of the tent's surfaces?

??? The puzzle lovers Peter and Serhiy Grabarchuk tried to make this unfolding fun on their old (now discontinued) website.

1. Which of the unfolded shapes can fold to make the triangular prism?

??? #### Prism Volume

While we are looking at prisms, we might as well think about the volume of prisms.

We can think of a rectangular prism as a skyscraper whose footprint is a certain size rectangle, and whose height count how many stories tall it is.

Each story tall is another copy of the ground floor. So after we find the area of the ground floor (using a formula for the area of a flat shape), we can multiply by the number of stories to find the overall size of the skyscraper.

1. How many small cubes would be needed to create the large wooden cube below, whose sides all equal 10 small cubes?

??? One famous skyscraper is 432 Park Avenue in Manhattan, which was the third-tallest building in the United States when constructed. Its ground floor has area 412,637 square feet. Its height is 1,396 feet.

1. What is the volume of 432 Park Avenue?

??? Your turn for one more volume example.

1. What is the volume of the rectangular box above?

???

#### Prism Volume Puzzles

We can add up a stack of rectangular blocks to find the total volume in a block tower.

This works even if the tower is "sloppy" because the blocks are not all the same size, or are not lined up well. We combine three-dimensional pieces in a manner quite similarly to how we did earleir solved flat area puzzles.

Maybe think of the towers made by the toddler blocks. Or maybe think of the towers made by the Tower Blocks Game by Stever Gardener. 1. Let's invent an example.

???

1. Let's invent another example.

??? Textbook Exercises for One Step Equations

This list of recommended odd-numbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more odd-numbered problems. If your math foundation is strong, do fewer.

Section 8.3 (Page 442) # 3, 23

Section 8.4 (page 452) # 3, 5, 27

### Ten Exercises

Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?

1. 86 degrees Celsius is what temperature in degrees Fahrenheit?

2. 44 degrees Fahrenheit is what temperature in degrees Celsius?

3. A square with sides 10 feet long has the top quarter removed (like a triangular "bite" taken out of the top). What is the remaining area?

4. A circle of radius 6 cm has its southwest quarter removed. What is the perimeter of that "Pac-Man" shape?

5. A half-circle window has diameter of 8 feet. What is its perimeter?

6. You want to install a two foot wide sidewalk around a circular swimming pool. The diameter of the pool is 30 feet. What is the area of the donut-shaped sidewalk, rounded to the nearest square foot?

7. Clarabelle's Confusing Pizza Parlor sells a 20 inch diameter pizza for \$18.99, and a 40 cm diameter pizza for \$14.99. Which is the better buy? (1 inch = 2.54 centimeters.)

8. How large a circle (how big an area?) can fit inside a rectangle of base of 12 feet and height of 5 feet?

9. A square is cut in half. The perimeter of the resulting rectangle is 30 feet. What was the area of the original square?

10. The circumference of a quarter is 7.85 cm. What is its area?

### Random Exercises

Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.

## Patterns

We have looked at lots of formulas. But before concluding our foundational math topics, we should take a step back to look at what a formula really is.

### Linear Patterns

Where do formulas come from? How are they invented? Why do they work? How do they make sense?

All those questions have the same answer: patterns.

The easiest way to watch patterns become formulas is to make a table. In each row put which place in the pattern we are at, what the pattern looks like, and the number-value for that place. We can then look at the table to analyze the pattern and create a formula.

1. This pattern is an increasing number of X shapes, each made by four toothpicks. Can you explain the formula that describes how many toothpicks it takes to make a row of X shapes? Each row adds four toothpicks. The pattern is y = n × 4.

By the way, we use n as the formula's input letter instead of x because of tradition. Using n shows that only "normal" counting numbers are inputs—never fractions, decimals, or negative numbers. A formula with x would allow those.

Here are more patterns to analyze and turn into formulas. To help you meet your classmates, work on each pattern with a different partner, and introduce yourselves as you work quietly together.

1. Can you explain the formula that describes how many toothpicks it takes to make a row of boxes? Each row adds three toothpicks in a C shape, and there is always one extra toothpick at the far right edge to close the right-most box. The pattern is y = (3 × n) + 1.

Notice that we do not need the parenthesis. The order of operations for arithmetic already has us multiply before adding. But the parenthesis do help communicate where the pattern came from.

1. This pattern looks like a row of houses that gets longer and longer. Each row adds five toothpicks in the shape of a house with no right wall, and there is always one extra toothpick at the far right edge to close the right-most house. The pattern is y = (5 × n) + 1.

As before, we do not need the parenthesis because the order of operations already has us multiply before adding. But the parenthesis do help communicate where the pattern came from.

#### In-Class Activity

1. Create a new pattern in which each row is bigger than the previous row by the same amount. Write it as a table or a story problem. Challenge your classmates to find your pattern. Patterns and Mathematical Reasoning - Tables

Patterns and Mathematical Reasoning - Townhouses

### Area Patterns

Time for some area patterns.

1. How does the number of tiles in a square increase? Do we know this formula's name? The pattern is y = n × n, which can also be written y = n2.

This pattern is "tautological" because the formula does what its name says. The reason we call an exponent of two squaring a number because it makes a square whose side length is the number.

1. The area of a rectangle is equal to its length times its width. We cannot rediscover that whole formula with only one table. But we can find a specific version of it. What is the pattern for these rectangular areas? In this pattern the rectangles have sides of length n and (n + 1). The area formula multiplies these sides.

Our answer is y = n × (n + 1).

Remember that the formula A = l × w is also tautological. In our illustration is the second rectangle three rows of two tiles, or two columns of three tiles? Either way, making copies of an amount is simply what multiplcation does by definition.

1. How about a triangle of tiles? This pattern seems harder than the previous two! But there is a trick that makes it easy. You can find the trick by comparing this pattern to the rectangle pattern. Notice that each triangle in this pattern is half the size of the corresponding rectangle in the previous pattern. Since the previous pattern was y = n × (n + 1), we want half of that. We need to divide by two at the end.

Our answer is y = n × (n + 1) ÷ 2

The formula we just found is called the Triangle Formula. Outside of a math classroom it is not as famous as the Square Formula or the Rectangle Area Formula. But it does deserve its own name because it is very useful.

#### In-Class Activity

We have all seen how dust floats in the air. Each dust particle has weight and is pulled down by gravity—but the upward force of air resistance can be equally strong. Just like water strider bugs can walk on water because they do not weigh enough to sink through the surface tension of water they stand on, a dust particle can "stand" on the air below it.

1. Get into groups of three or four students. Use the table below to explain why rain falls but clouds float. Tagentially, watch this. Mark Willis

Finding the nth term of the triangular numbers

### The Triangle Formula

We just found the Triangle Formula.

The Triangle Formula

The triangle pattern goes 1, 3, 6, 10,... with each step increasing additively by one more than the previous step.

Its formula is y = n × (n + 1) ÷ 2

The Triangle Formula appears surprisingly often in real-life applications. Here are three pattern problems that seem tricky until you realize how the answer is made by tweaking the Triangle Formula.

1. This pattern involves how many toothpicks are in a triangle that grows downward. Each step in the pattern adds another row to the bottom of the previous triangle. Each step in the pattern is three times as big as the Triangle Pattern. So we need to multiply by three at the end.

The pattern is y = n × (n + 1) ÷ 2 × 3.

1. Now toothpicks make squares of increasing size. What is the pattern for how many toothpicks are in each square? (We are not looking at the area of the squares.) Each step in the pattern is four times as big as the Triangle Pattern. So we need to multiply by four at the end.

If we divide by two and then multiply by four, the overall result is simply multiplying by two. Instead of ÷ 2 × 4 we can simply do × 2.

The pattern is y = n × (n + 1) × 2.

1. What is the most number of pieces you can make with straight cuts on a pizza? You will have to cut messy and not have every cut go through the center! One cut must make 2 pieces. Two cuts cannot make more than four pieces. The picture below shows a way three cuts can make seven pieces. Four cuts can make eleven pieces! And so on. Each step in the pattern is one more than the Triangle Pattern. So we need to add one at the end.

The pattern is y = n × (n + 1) ÷ 2 + 1.

### Ten Exercises

Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?

1. Triangular tables are placed in a row to seat more people. One table has 3 seats. Two tables have 4 seats. Create a formula where we put in the number of tables (as n) and get out the number of seats (as y). 2. Now we switch to square tables. We still make a row of tables to seat more people. One table has 4 seats. Two tables have 6 seats. Create a formula where we put in the number of tables (as n) and get out the number of seats (as y). 3. This shape is sort of like a V or W that gets wider with more wiggles. What is the pattern for how many squares are in each row? (As an optional, extra challenge you can also find the pattern for how many toothpicks are in each row!) 4. How about this extra-wide plus shape? What is the pattern for how many squares are in each row? 5. How about this hollow diamond shape? What is the pattern for how many squares are in each row? 6. This shape looks somewhat like the stand that holds up a road construction sign. With each step it gets longer in each direction. What is the pattern for how many cubes are in each row? 7. There are many versions of an old story about the inventor of the game of chess. One version appears below. On which day will the total grains of rice exceed 3 million?

The Chessboard Story King Radha of India was bored of backgammon, and desired a new game. Sessa, his minister invented chess. King Radha was pleased and asked Sessa what he desired in payment.

Sessa asked that a single grain of rice be placed on the first square of the chessboard, two grains on the second square, four grains on the third, and so on, doubling each time.

King Radha saw that this would require far more rice than his kingdom would ever produce, and had Sessa executed for impudence.

8. Two trains are approaching each other on parallel tracks. Both are traveling at 30 miles per hour. What is the overall speed at which they approach?

9. Continuing the previous problem, the two trains start 9 miles apart. How many minutes does it take for them to pass each other?

10. Continuing the previous problem, a fly zooms back and forth from the headlight of one train to headlight of the other. It starts when the trains are 9 miles apart. By the time it arrives at the other train, the two trains have gotten closer. It instantly reverses direction and heads back to the first train. And so on. The fly moves at 20 miles per hour. How far does it travel before the trains pass?

The Trains and Fly Story

One day at Los Alamos, Richard Feynman noticed something interesting. When he asked a physicist to solve the Trains and Fly problem they all used the shortcut (as above) and got the answer immediately. When he asked a mathematician, they all calculated the fly's trip bit by bit and finding the answer took several minutes.

Eventually Feynman brought the Trains and Fly problem to the most astounding mathematician of the group, John von Neumann, who immediately answered.

"That's not right!" protested Feynman. "You're a mathematician. You're supposed to sum the series, not use the shortcut!"

"What shortcut?" asked von Neumann. "I did sum the series."

### Random Exercises

Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.