Math OER Zoom Room Jamboard Lectures Textbook 


Our third big Math 20 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).
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.
1. Solve 5 × y = 35
We "undo" a multiplyby5 with a divideby5. To be fair, we treat both sides of the equation the same.
2. Solve 4 × b = 20
How does this picture help us solve for b? Where is division for both sides hiding?
We "undo" a multiplyby4 with a divideby4. To be fair, we treat both sides of the equation the same.
In thie 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.
3. Solve p ÷ 6 = 8
How does this picture help us solve for b? Where is multiplication for both sides hiding?
We "undo" a divideby6 with a multiplyby6. To be fair, we treat both sides of the equation the same.
In thie 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.
4. Solve y × 9 = 189
I used both black and blue writing. In our class you need not use colors, but you should write a lot, like I did.
If you take an algebra class one of your goals will be to eventually wean yourself from always using the steps I did in blue. The instructor will write fewer of these steps on the board. You will train your eye to "see" the steps I write in blue even if they are not actually written.
There are two other reasons to use the vertical format.
First, it promotes doing homework in two 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.
5. Solve y × 8 = ^{2}⁄_{3}
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.
6. Solve u + 9 = 200
To get u by itself we want to remove a +9. So we subtract 9 from both sides. 200 − 9 = 191
7. Solve 50 = v + 5
To get v by itself we want to remove a +5. So we subtract 5 from both sides. 50 − 5 = 45
8. Solve 50 = 5 + w
To get w by itself we want to remove a +5. So we subtract 5 from both sides. 50 − 5 = 45
9. Solve x − 9 = 79
To get x by itself we want to remove a −9. So we add 9 to both sides. 79 + 9 = 88
10. Solve 45 = y − 15
To get y by itself we want to remove a −15. So we add 15 to both sides. 45 + 15 = 60
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.
11. Solve 9 = x^{2}
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.
12. Solve y × 3 = 9
To get y by itself we want to remove a ×3. So we do ÷3 to both sides. 9 ÷ 3 = 3
13. Solve 3 × y = 9
This is the same problem as before! The order of multiplication does not matter. y × 3 is the same thing as 3 × y
To get y by itself we want to remove a ×3. So we do ÷3 to both sides. 9 ÷ 3 = 3
But for division we can get stuck if the equation starts with a ÷ y
14. Solve y ÷ 3 = 30
To get y by itself we want to remove a ÷3. So we do ×3 to both sides. 30 × 3 = 90
15a. 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 = ^{3}⁄_{30} which is increasing the complexity.
It is better to begin by multiplying both sides by y.
15b. Solve 30 ÷ y = 3
To get 30 by itself we want to remove a ÷y. So we do ×y to both sides.
The equation becomes 30 = 3 × y, which is now a one step equation.
To get y by itself we want to remove a ×3. So we do ÷3 to both sides. 30 ÷ 3 = 10
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.
16. 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
17. 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
18. 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
19. Solve x × 4 = ^{1}⁄_{8}
To get x by itself we want to remove a ×4. So we do ÷4 to both sides. ^{1}⁄_{8} ÷ 4 = ^{1}⁄_{32}
20. Solve y ÷ 4 = ^{1}⁄_{8}
To get y by itself we want to remove a ÷4. So we do ×4 to both sides. ^{1}⁄_{8} × 4 = ^{1}⁄_{2}
21. Solve 4 ÷ z = ^{1}⁄_{8}
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.
22. 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.
23a. 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?
23b. 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 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.
24. 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
25. 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
26. 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
27. Solve x + ^{1}⁄_{8} = ^{1}⁄_{2}
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}
28. Solve y − ^{1}⁄_{8} = ^{1}⁄_{2}
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}
29. Solve ^{1}⁄_{3} − z = ^{1}⁄_{9}
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.)
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.
30. ^{3}⁄_{7} ≟ ^{15}⁄_{35}
Does 3 × 35 equal 7 × 15? Yes. The ratios are proportional. The proportion is true.
31. ^{4}⁄_{9} ≟ ^{12}⁄_{28}
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.
32. ^{7}⁄_{9} ≟ ^{5.4}⁄_{7.2}
Does 7 × 7.2 equal 9 × 5.4? No. The ratios are not proportional. The proportion is false.
33. ^{1.2}⁄_{1.8} ≟ ^{4.99}⁄_{7.56}
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.)
33. ^{⅓}⁄_{3} ≟ ^{½}⁄_{2}
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:
34. (33 again) Is the ratio ^{1}⁄_{3} to 3 proportional to the ratio ^{1}⁄_{2} to 2?
35. Is the ratio ^{1}⁄_{3} to ^{1}⁄_{2} proportional to the ratio 3 to 2?
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.
36. Is the ratio ^{1}⁄_{3} to ^{2}⁄_{5} proportional to the ratio ^{2}⁄_{3} to ^{3}⁄_{5}?
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!
37. Is the ratio ^{2}⁄_{3} to ^{3}⁄_{4} proportional to the ratio ^{4}⁄_{3} to ^{6}⁄_{4}?
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!
38. ^{y}⁄_{32} = ^{3}⁄_{8}
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
39. ^{1.3}⁄_{1.2} = ^{y}⁄_{96}
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.
40. ^{144}⁄_{y} = ^{9}⁄_{4}
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
41. ^{y}⁄_{39} = ^{8}⁄_{13}
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
42. ^{52}⁄_{91} = ^{4}⁄_{y}
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.
43. 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 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!)
44. 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 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
45. Suzie can read 12 pages in sixteen minutes. How many pages can she read in five hours?
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
46. 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 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
47. Maria drinks 4 cups of coffee every 5 days. How many cups of coffee is this per year?
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.
48. 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 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
49. 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 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
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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
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.
50. 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"
51. 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
52. Onehalf 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.
53. 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.
Kahn Academy
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}
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 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
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. A certain dog medicine contains 3/7 of a gram of active ingredient in every 120 gram dose. For a small dog, how much active ingredient is in a 7 gram dose?
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.
54. 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.
55. 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%.
56. 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 second method is to write the percent sentence into a proportion. The steps are always the same.
Let's redo the same examples.
57. Use the proportion method to solve: What is 35% of 60?
The percent sentence becomes the proportion ^{35}⁄_{100} = ^{y}⁄_{60}
Solve for y. The answer is 21.
58. Use the proportion method to solve: 12 is what percent of 5?
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.)
59. Use the proportion method to solve: 5 is 2% of what?
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.
60. A young couple earns money by improving a "fixerupper" 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?
61. 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 ^{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.
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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.)
62. 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?
 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
63. 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 ^{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
64. 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 miles ≈ 59 miles
The proportion method makes it ^{15}⁄_{100} = ^{y}⁄_{396} and of course the answer is the same: about 59 miles.
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 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%.
66. 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?
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%.
67. 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 ^{18}⁄_{100} = ^{y}⁄_{$41.50} and of course the answer is the same: about $7.47.
68. 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 ^{29}⁄_{100} = ^{7}⁄_{y} and of course the answer is the same: about 24 grams.
69. 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 ^{y}⁄_{100} = ^{60}⁄_{230} and of course the answer is the same: about 26%.
70. 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 ^{y}⁄_{100} = ^{$2.25}⁄_{$1.50} and of course the answer is the same: about 150%.
71. 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 ^{y}⁄_{100} = ^{$0.35}⁄_{$1.65} and of course the answer is the same: about 21%.
72. 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 ^{y}⁄_{100} = ^{$0.07}⁄_{$3.71} and of course the answer is the same: about 2%.
73. 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 ^{5}⁄_{100} = ^{y}⁄_{$249} and of course the answer is the same: $12.45.
74. 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 ^{y}⁄_{100} = ^{$35.50}⁄_{$499} and of course the answer is the same: about 7%.
75. 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 ^{7}⁄_{100} = ^{y}⁄_{$3,900} and of course the answer is the same: $273.
76. 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 ^{y}⁄_{100} = ^{$80}⁄_{$1,500} and of course the answer is the same: about 5.3%.
77. 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 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. What percent of 160 is 150?
2. $39 is what percent of $50?
3. 56.32 is 64% of what?
4. 80% of what is 16?
5. 42 is 30% of what number?
6. What number is 150% of ^{30}⁄_{45}?
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 weights 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.
78. 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
79. 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 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!
80. 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 puzzlelike area problem.
81. 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:
82. What is the area of this square?
length × width = 16
Here is the same problem with extra numbers.
83. 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 puzzlelike area problems.
84. 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
85. 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.
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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
Negative numbers are less than than zero.
−5 is less than −2 even though 5 is more than 2.
A number line can help keep track both whole numbers and numbers with decimal digits.
Here are 3.5 and −2.8 on a number line.
To add and subtract with both positive and negative numbers it is helpful to think of money.
One technique is to try making up a story to go along with each addition and subtraction problem.
Example of a Money Story Adding a Negative
Solve: 30 + (−12) =
Here is one possible story:
"I have $30 in my hand and am going shopping. I know that I owe my friend $12. A debt has been added. So in my mind I plan my shopping as if I had only $18, because that debt means $12 of the $30 is not mine to spend. So 30 + (−12) = 18."
Example of a Money Story Subtracting a Negative
Solve: 18 − (−12) = ?
Here is one possibility:
"I am about to go shopping to spend $18. I owe my friend $12 and have that money with me too. But when I finally see my friend he says, 'Never mind the debt. It's been seven years anyway. Just keep the $12.' A debt has been removed. So I change how I think about the money. I don't have only $18 to spend, but all $30. So 18 − (−12) = 30."
Instead of a story, you can also use a number line to keep track of adding and subtracting with both positive and negative numbers.
86. Use a number line to find the value of the expression 4 − 6 + (−3) − (−10) − 2
Start at 4. Go left 6, left 3 more, then right 10, then left 2. The answer is 3.
Remember from the equation 18 − (−12) = 30 that subtracting a negative worked just like adding. I get more money, whether I am paid or have a debt subtracted away.
There is a saying that goes "a negative of a negative is a positive". Why is this true?
Imagine that the front wall of our room is a number line. Zero is in the middle. Now we need a student to volunteer to show us by walking the values of +3 and −3.
Draw a number line. Have a volunteer stand at "zero", facing positive, before continuing.
How would our volunteer show us +3?
He or she would walk forward 3 steps. Note that this location along the front wall represents +3 on our number line.
How would our volunteer show us −3?
He or she would walk backwards 3 steps. Note that this location along the front wall represents −3 on our number line.
How else could our volunteer show us −3, without any backward steps?
He or she could pivot 180 degrees, and then walk forwards 3 steps. Note that results in the same location along the front wall representing −3 on our number line.
What happens if our volunteer does both kinds of negative? That is, if he or she represents −(−3) by both turning around 180 degrees and then taking three backwards steps?
Our volunteer winds up at the +3 location on our number line. This is another way to see that − (−3) = 3.
Using arithmetic, a number is made negative by × (−1)
Consider the expression 10 − x.
If we plug in x = 20 then we get 10 − 20 = −10.
If we plug in x = −20 then we get 10 − (−20) = 30.
So, is the expression 10 − x positive or negative?
It depends upon what we plug in to x!
In general, it does not make sense to say that x is positive or negative. It depends upon what we plug in to x.
This is why in college math we write −5 instead of ⁻5. Once we use variables instead of normal numbers, then every time we see a − sign we are subtracting, but it might not be negative!
That is a very weird statement!
The Very Weird Statement
Once we use variables instead of normal numbers, then every time we see a − sign we are subtracting, but it might not be negative!
Compare the mathematical expression 5 − x and the mathematical expression 7 − (−x). Both involve subtraction. Neither necessarily involves a negative amount.
None yet
Textbook Exercises for ariables and Negatives
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 10.1 (Page 529) # 3, 5, 23, 27, 81, 83, 93, 95, 97
Section 10.2 (Page 537) # 3, 5, 11, 13, 49, 57, 59
Section 10.3 (Page 544) # 5, 7, 9, 11, 47, 49, 51, 55, 65, 67, 69, 75
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.