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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.
The first equations we study are onestep equations.
We only need to do one thing to both sides of the equation to solve the puzzle.
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.
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.
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
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?
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...
Hard to Read Horizontal Format
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.
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.
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.
Please avoid bad math grammar.
Bad Math Grammar #1
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.
Bad Math Grammar #2
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.
Bad Math Grammar #3
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.
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.
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.
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
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
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
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.
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: ^{7}⁄_{8} × x = 56
Chapter 2 Test, Problem 35: Solve: t × ^{2}⁄_{5} = ^{7}⁄_{10}
Chapter 3 Test, Problem 9: Solve: ^{1}⁄_{4} + y = 4
Chapter 3 Test, Problem 10: Solve: x + ^{2}⁄_{3} = ^{11}⁄_{12}
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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
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!)
We will only consider two kinds of two step equations, which "fix" problems with one step equations.
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.
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
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:
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.
We could begin by dividing both sides by 30. But this creates 1 ÷ y = ^{3}⁄_{30} which is increasing the complexity.
It is better to begin by multiplying both sides by y.
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 onestep 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 wellunderstood 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.
To get u by itself we want to remove a ×20. So we do ÷20 to both sides. 0.4 ÷ 20 = 0.02
To get v by itself we want to remove a ÷20. So we do ×20 to both sides. 0.4 × 20 = 8
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
To get x by itself we want to remove a ×4. So we do ÷4 to both sides. ^{1}⁄_{8} ÷ 4 = ^{1}⁄_{32}
To get y by itself we want to remove a ÷4. So we do ×4 to both sides. ^{1}⁄_{8} × 4 = ^{1}⁄_{2}
To get 4 by itself we want to remove a ÷z. So we do ×z to both sides.
The equation becomes 4 = ^{1}⁄_{8} × z, which is now a one step equation.
To get z by itself we want to remove a × ^{1}⁄_{8}. So we do ÷ ^{1}⁄_{8} to both sides. 4 ÷ ^{1}⁄_{8} = 32
Similar shenanigans can happen with subtraction.
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.
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?
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 wellunderstood 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.
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
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
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
To get x by itself we want to remove a + ^{1}⁄_{8}. So we do − ^{1}⁄_{8} to both sides. ^{1}⁄_{2} − ^{1}⁄_{8} = ^{3}⁄_{8}
To get y by itself we want to remove a − ^{1}⁄_{8}. So we do + ^{1}⁄_{8} to both sides. ^{1}⁄_{2} + ^{1}⁄_{8} = ^{5}⁄_{8}
To get ^{1}⁄_{3} by itself we want to remove a −z. So we do +z to both sides.
The equation becomes ^{1}⁄_{3} = ^{1}⁄_{9} + z, which is now a one step equation.
To get z by itself we want to remove a + ^{1}⁄_{9}. So we do − ^{1}⁄_{9} to both sides. ^{1}⁄_{3} − ^{1}⁄_{9} = ^{2}⁄_{9}
None yet
Textbook Exercises for Two Step Equations
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
(Our textbook has no exercises for this topic.)
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 = ^{16}⁄_{10}
2. Solve 2.6 + y = ^{16}⁄_{10}
3. Solve y − ^{1}⁄_{20} = ^{1}⁄_{5}
4. Solve ^{1}⁄_{20} − y = ^{1}⁄_{5}
5. Solve y × 12 = 0.6
6. Solve 12 × y = 0.6
7. Solve y ÷ ^{1}⁄_{10} = 0.5
8. Solve ^{1}⁄_{10} ÷ y = 0.5
9. Solve y − 2.8 = 3.2 + 8.7
10. Solve 2.8 − y = 3.2 + 8.7
I need to make the random exercises for Justice Fractions.
This problem is more difficult than our one or twostep equations! Can you solve it anyway?
Definition
A proportion is an equation of the format "ratio equals ratio" (or "rate equals rate").
Here is an example of a proportion: ^{7 miles}⁄_{2 hours} = ^{35 miles}⁄_{10 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 twofourths is equal to fivetenths 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 onefourths, and cut each into 10 parts, I would have 20 pieces that were onefortieths. Similarly, if I had 5 pieces that were onetenths, and cut each into 4 parts, I would also have 20 pieces that were onefortieths. Both cutting processes give me 20 pieces that are onefortieths.
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.
This leads to another definition.
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.
Does 3 × 35 equal 7 × 15? Yes. The ratios are proportional. The proportion is true.
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.
Does 7 × 7.2 equal 9 × 5.4? No. The ratios are not proportional. The proportion is false.
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.)
Does ⅓ × 2 equal 3 × ½? In other words, does ^{2}⁄_{3} equal ^{3}⁄_{2} ?
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 rewrite the previous problem in a way that might be easier to read when typed:
Does ⅓ × 2 equal ½ × 3? In other words, does ^{2}⁄_{3} equal ^{3}⁄_{2} ?
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.
Does ^{1}⁄_{3} × ^{3}⁄_{5} equal ^{2}⁄_{5} × ^{2}⁄_{3}? In other words, does ^{3}⁄_{15} equal ^{4}⁄_{15} ?
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 ^{3}⁄_{15} when multiplying. We could have written it as ^{1}⁄_{5} but that would only have made the problem harder!
Does ^{2}⁄_{3} × ^{6}⁄_{4} equal ^{3}⁄_{4} × ^{4}⁄_{3}? In other words, does ^{12}⁄_{12} equal ^{12}⁄_{12} ?
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 ^{7}⁄_{8} is proportional to ^{63}⁄_{72}
Chapter 5 Test, Problem 11: Check if ^{1.3}⁄_{3.4} is proportional to ^{5.6}⁄_{15.2}
Textbook Exercises for Proportions and Cross Multiplying
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
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
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!
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
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...
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.
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
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
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 ^{8}⁄_{12} = ^{y}⁄_{9} 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 ^{8}⁄_{3} = ^{(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.
Chapter 5 Test, Problem 12: Solve: ^{9}⁄_{4} = ^{27}⁄_{x}
Chapter 5 Test, Problem 13: Solve: ^{150}⁄_{2.5} = ^{x}⁄_{6}
Chapter 5 Test, Problem 14: Solve: ^{x}⁄_{100} = ^{27}⁄_{64}
Chapter 5 Test, Problem 15: Solve: ^{68}⁄_{y} = ^{17}⁄_{25}
Textbook Exercises for Proportions with Variables
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 5.4 (Page 284) # 9, 11, 13, 15, 21, 23, 25, 31, 33, 35, 51, 53
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 crossmultiplying 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.
The proportion is ^{120 cal}⁄_{1 serv} = ^{y cal}⁄_{3.5 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!)
The proportion is ^{5 ml}⁄_{25 lbs} = ^{y ml}⁄_{80 lbs}
After cross multiplying, the equation without fractions is y × 25 = 5 × 80
The division step tells us y = 5 × 80 ÷ 25 = 16 ml
The proportion is ^{12 pages}⁄_{16 min} = ^{y pages}⁄_{300 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
The proportion is ^{3 problems}⁄_{11 min} = ^{20 problems}⁄_{y min}
After cross multiplying, the equation without fractions is y × 3 = 11 × 20
The division step tells us y = 11 × 20 ÷ 3 ≈ 73 minutes
The proportion is ^{8 feet}⁄_{$14.49} = ^{158 feet}⁄_{y 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
The proportion is ^{4 cups}⁄_{5 days} = ^{y cups}⁄_{365 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.
The proportion is ^{24 tagged in entire pond}⁄_{y total in entire pond} = ^{3 tagged in second catch}⁄_{19 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
The proportion is ^{260 tagged in entire lake}⁄_{y total in entire lake} = ^{20 tagged in second catch}⁄_{144 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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 5.5 (Page 291) # 1, 3, 5, 7, 9, 13, 19, 21
Now you know three methods to solve a proportional rate.
Let's do the same problem in using three methods.
First we find the unit rate using both pieces of information from the wellknown 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.
The proportion is ^{36 feet}⁄_{1.8 minutes} = ^{y feet}⁄_{12 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
Notice that the wellknown situation becomes the way we modify the second situation that has only one number.
^{12 minutes}⁄_{1} = ^{12 minutes}⁄_{1} × ^{36 feet}⁄_{1.8 minutes} = ^{432 feet}⁄_{1.8} = 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).
Many of the proportion problems we solved above involve reducing or unreducing.
When this happens the number we are unreducing 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.
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"
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
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.
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.
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 ^{7}⁄_{8} is proportional to ^{63}⁄_{72}
Chapter 5 Test, Problem 11: Check if ^{1.3}⁄_{3.4} is proportional to ^{5.6}⁄_{15.2}
Chapter 5 Test, Problem 12: Solve: ^{9}⁄_{4} = ^{27}⁄_{x}
Chapter 5 Test, Problem 13: Solve: ^{150}⁄_{2.5} = ^{x}⁄_{6}
Chapter 5 Test, Problem 14: Solve: ^{x}⁄_{100} = ^{27}⁄_{64}
Chapter 5 Test, Problem 15: Solve: ^{68}⁄_{y} = ^{17}⁄_{25}
Textbook Exercises for Scale Factors
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 5.6 (Page 299) # 1, 3, 5, 7, 9, 17, 19, 23, 25, 27, 31
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 1}⁄_{Initial Amount 2} = ^{Other Amount 1}⁄_{Other 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.
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.
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.
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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
(Our textbook has no exercises for this topic.)
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 lefthanded. A certain class has nine lefthanded students. How large would you estimate the class is if its proportion of lefthanded students is average?
7. The United States debttoGDP 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.
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.
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 first method is to translate the percent sentence into an equation. As usual in math:
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.
The percent sentence is translated into y = 0.35 × 60
Solve for y. The answer is 21.
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%.
The percent sentence is translated into 5 = 0.02 × y
Solve for y. The answer is 250.
The second method is to write the percent sentence into a proportion. The steps are always the same.
Let's redo the same examples.
The percent sentence becomes the proportion ^{35}⁄_{100} = ^{y}⁄_{60}
Solve for y. The answer is 21.
The percent sentence becomes the proportion ^{y}⁄_{100} = ^{12}⁄_{5}
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.)
The percent sentence becomes the proportion ^{2}⁄_{100} = ^{5}⁄_{y}
Solve for y. The answer is 250.
Mathispower4u
Solving Percent Problems Using The Percent Equation
YouTube Problems
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?
Notice that percent sentences appear three different patterns:
(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".
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.
$105,000 is what percent of $65,000?
The translate method makes it $105,000 = y × $65,000 (and needs RIP LOP as a final step)
The proportion method makes it ^{y}⁄_{100} = ^{$105,000}⁄_{$65,000} (and does not use RIP LOP)
Either way, the answer is about 162%.
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 oldfashioned 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.
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.
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.
^{8}⁄_{10} = 0.8 = 80%
None yet
Textbook Exercises for Percent Sentences
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 6.2 (Page 327) # 1, 3, 5, 9, 11, 13, 17, 19, 25, 27, 29, 31, 3, 35, 37, 45
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.)
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?
 58% of what is 300?
 300 is 58% of what?
 What is 58% of 300?
 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
We ask, "What is 58% of 300?"
The translate method makes it y = 0.58 × 300 = 174 students
The proportion method makes it ^{58}⁄_{100} = ^{y}⁄_{300} and of course the answer is the same: 174 students.
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.
Your turn to create a word problem solvable with a percent sentence.
Create Your Own
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.
 Pick one of these patterns
 Make up numbers for Y and Z
 Invent a word problem for those numbers
 Trade problems and race to solve them
We ask, "What is 15% of 396 miles?"
The translate method makes it y = 0.15 × 396 miles ≈ 59 miles
The proportion method makes it ^{15}⁄_{100} = ^{y}⁄_{396} and of course the answer is the same: about 59 miles.
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 nonrounded digits, to match the problem's original numbers)
The proportion method makes it ^{y}⁄_{100} = ^{20,600}⁄_{180,700} and of course the answer is the same: about 11.4%.
We ask, "20,600 is what percent of 24,400?" (Do you see why we added the undergraduate and graduate students together?)
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 nonrounded digits, to match the problem's original numbers)
The proportion method makes it ^{y}⁄_{100} = ^{20,600}⁄_{24,400} and of course the answer is the same: about 84.4%.
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 ^{18}⁄_{100} = ^{y}⁄_{$41.50} and of course the answer is the same: about $7.47.
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 ^{29}⁄_{100} = ^{7}⁄_{y} and of course the answer is the same: about 24 grams.
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 ^{y}⁄_{100} = ^{60}⁄_{230} and of course the answer is the same: about 26%.
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 ^{y}⁄_{100} = ^{$2.25}⁄_{$1.50} and of course the answer is the same: about 150%.
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 ^{y}⁄_{100} = ^{$0.35}⁄_{$1.65} and of course the answer is the same: about 21%.
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 ^{y}⁄_{100} = ^{$0.07}⁄_{$3.71} and of course the answer is the same: about 2%.
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 ^{5}⁄_{100} = ^{y}⁄_{$249} and of course the answer is the same: $12.45.
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 ^{y}⁄_{100} = ^{$35.50}⁄_{$499} and of course the answer is the same: about 7%.
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 ^{7}⁄_{100} = ^{y}⁄_{$3,900} and of course the answer is the same: $273.
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 ^{y}⁄_{100} = ^{$80}⁄_{$1,500} and of course the answer is the same: about 5.3%.
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 atbats. How many atbats 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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 6.3 (Page 334) # 1, 3, 5, 7, 9, 19, 21
Section 6.4 (Page 340) # 3, 5, 7, 9, 11, 15, 17
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 ^{30}⁄_{45}?
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?
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.
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.
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.
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.
F = 1.8 × C + 32 = 1.8 × 39 + 32 = 102.2 °F
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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 7.4 (Page 397) # 23, 25, 49
Many problems make finding area puzzlelike. 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!
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 puzzlelike area problem.
(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:
length × width = 16
Here is the same problem with extra numbers.
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 puzzlelike area problems.
(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
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 oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 8.2 (Page 431) # 9, 11, 15, 23, 29, 35
Prisms
A prism is a threedimensional 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.)
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 threedimensional 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.
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Next, imagine an oldfashioned "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.
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The puzzle lovers Peter and Serhiy Grabarchuk tried to make this unfolding fun on their old (now discontinued) website.
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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.
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One famous skyscraper is 432 Park Avenue in Manhattan, which was the thirdtallest building in the United States when constructed. Its ground floor has area 412,637 square feet. Its height is 1,396 feet.
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Your turn for one more volume example.
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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 threedimensional 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.
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Textbook Exercises for One Step Equations
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 8.3 (Page 442) # 3, 23
Section 8.4 (page 452) # 3, 5, 27
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 "PacMan" shape?
5. A halfcircle 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 donutshaped 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?
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.
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.
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 numbervalue for that place. We can then look at the table to analyze the pattern and create a formula.
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.
Each row adds three toothpicks in a C shape, and there is always one extra toothpick at the far right edge to close the rightmost 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.
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 rightmost 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.
Khan Academy
Time for some area patterns.
The pattern is y = n × n, which can also be written y = n^{2}.
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.
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.
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.
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.
Tagentially, watch this.
Mark Willis
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 reallife applications. Here are three pattern problems that seem tricky until you realize how the answer is made by tweaking the Triangle Formula.
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.
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.
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.
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 extrawide 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."
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.