Welcome Zoom Room Jamboard Textbook Lectures Calculator  # Math OER Mad Science: Simplifying an Expression

Our second foundational topic is Mad Science. This is our nickname for when we do things to a mathematical expression.

What is a mathematical expression?

Here are some examples:

47

20 − (2 × 5)

2x + 5

24 ÷ 8 + 102 − 10 × 15

So a mathematical expression is one or more values (numbers or variables) that are linked with valid arithmetic operations.

We should notice three important things. First, after we plug in any variables we can find the value of the entire expression. When we work the arithmetic correctly it will simplify into a single number.

Second, there are no equal signs in a mathematical expression. But the tools used to simplify an expression can be used to find y = something. This big topic will treat the expression 50 − (2 × 10) and the formula y = 50 − (2 × 10) as if there was no difference between them. Philopsophically they are distinct. But mathematically they are identical twins wearing slightly different clothes.

Third, invalid arithmetic operations do exist. You probably know at least one: the rule "never divide by zero". So an expression is like a recipe to follow, but we should not be asked to follow a gibberish recipe.

To summarize, we will explore more deeply how to use "mad science" to smoosh values together into a single, simple number. We use arithmetic: adding, subtracting, multiplying, dividing, and exponents. We will also study grouping structures and how they organize and order our arithmetic: parenthesis, fraction bars, square roots, and algorithms.

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

## Fractions

### Fraction × and ÷§2.4

#### Fraction Multiplication Without Canceling

How do we do fraction multiplication? Let's use some pictures to get a sense of what happens.

Use the picture below to show multiplying one-third by one-half. This picture starts with three vertical "cake cuts" and one-third shaded. Put the one-half in the picture by making a horizontal "cut" across the middle, and shading either above or below our horizontal cut. 1. What amount is overlap-shaded as belonging to both a verictal one-third and a horizontal one-half?

one-sixth

1. What are the denominators counting?

How many pieces is the cake cut into?

1. What are the numerators counting?

How many pieces are shaded? Or after multiplying, how many pieces are overlap-shaded?

To paraphrase, how many pieces do we bundle together into a group? In this problem there was no bundling. Each direction (horizontal or vertical) had only one shaded piece.

Use the picture below to show multiplying two-thirds by three-quarters. This time you must prepare the picture by drawing the two-thirds yourself. Use different types of shading for two of the thirds and for three of the quarters. 1. What amount is overlap-shaded as belonging to both two-thirds and three-quarters?

six-twelfths

1. What are the denominators counting?

The same thing: how many pieces is the cake cut into?

1. What are the numerators counting?

The same thing: how many pieces are shaded? Or after multiplying, how many pieces are overlap-shaded?

To paraphrase, how many pieces do we bundle together into a group? In this problem there was bundling. When we cut thirds, two of those were bundled together. When we cut fourths, three of those were bundled together.

Definition

To multiply fractions, treat the situation as two independent multiplication problems. Multiply the numerators. Multiply the denominators.

The answer will be a fraction that may or may not need to be reduced.

We call this process the fraction multiplication algorithm.

The word algorithm is merely a fancy word for "recipe" or "process". It describes the steps to follow to do a task.

Here is an example where the answer does not need to be reduced.

1. Multiply two-thirds and seven-fifths.

In the numerators, 2 × 7 = 14. In the denominators, 3 × 5 = 15. So our answer is 1415.

Here is an example where the answer does need to be reduced.

1. Multiply five-sevenths and one-tenth. (Use reducing.)

Using reducing: In the numerators, 5 × 1 = 5. In the denominators, 7 × 10 = 70. So our answer is 570, which can be reduced to 114.

#### Fraction Multiplication With Canceling

Definition

During fraction multiplication, canceling is reducing early.

(This definition is not an algorithm. It defines a word. But it does not list the steps for a recipe or process.)

Canceling avoids big numbers when multiplying. Let's review it with an example.

1. Multiply five-sevenths and one-tenth. (Use canceling.)

Using canceling: We can divide by five in the first numerator and second denominator. Then 17 × 12 = 114.

We cannot escape dividing both numerator and denominator by 5. But can chose to do it before or after multiplying.

Your turn. Here are two more examples. Attempt them before we do them together. (If you have time, try doing this problem both ways: once with reducing and once with canceling.)

1. Multiply three-fifths and one-twelfth.

Using reducing: In the numerators, 3 × 1 = 3. In the denominators, 5 × 12 = 60. So our answer is 360, which can be reduced to 120.

Using canceling: We can divide by three in the first numerator and second denominator. Then 15 × 14 = 120.

1. Multiply six-elevenths and two-thirds.

Using reducing: In the numerators, 6 × 2 = 12. In the denominators, 11 × 3 = 33. So our answer is 1233, which can be reduced to 411.

Using canceling: We can divide by three in the first numerator and second denominator. Then 211 × 21 = 411.

This example is quite important, because in shows how canceling can get a little complicated.

1. Multiply ten-fifteenths and six-eighths.

Using reducing: In the numerators, 10 × 6 = 60. In the denominators, 15 × 8 = 120. So our answer is 60120, which can be reduced to 12.

Using canceling: We can divide a numerator and denominator by 5. (Yes, it is okay when "canceling" is merely reducing one of the fractions without touching the other.) Twice we can divide a numerator and denominator by 2. (Both times involve the denominator that starts out 8.) We can divide a denominator and numerator by 3. Then 11 × 12 = 12.

How do we multiply a fraction by a mixed number?

1. Multiply 23 × 1 48

After we change 1 48 into 128 we are solving 23 × 128.

Then, in the numerators, 2 × 12 = 24. In the denominators, 3 × 8 = 24. So our answer is 2424, which can be reduced to 1. (Or you could use canceling.)

1. Multiply 1 23 × 1 48

Change 1 23 into 53, as well as changing 1 48 into 128.

Now we are solving 53 × 128.

Then, in the numerators, 5 × 12 = 60. In the denominators, 3 × 8 = 24. So our answer is 6024, which can be reduced to 52. (Or you could use canceling.)

#### Fraction Division

Remember when we explained division as making piles of a fixed size? We did fraction division problems like these:

Review of Division as Making Piles of a Fixed Size

12 ÷ 14 =

Imagine having half a piece of paper. We rip it in half again, making quarters, while setting those quarters down as "piles". We get 2 piles.

So 12 ÷ 14 = 2

4 ÷ 23 =

Imagine having 4 pieces of paper. We rip each in thirds while setting pairs of thirds down as "piles". We make 12 thirds by ripping. So we get 6 piles of two-thirds.

So 4 ÷ 23 = 6

Now what is happening to the numerators and denominators?

1. Can you see what the second numerator is doing that is "denominator-like"?

When we played with cake above, the denominators counted the number of pieces. The second numerator is also doing this.

For example, in 4 ÷ 23 = 6 the 3 meant each of the four wholes was ripped into 3 pieces.

1. Can you see what the second denominator is doing that is "numerator-like"?

When we played with cake above, the numerators counted the number of shaded parts in a group. The second denominator is also saying how many pieces made a group.

For example, in 4 ÷ 23 = 6 the 2 meant each of the piles bundled together 2 of the ripped thirds.

Definition

To divide fractions, flip the second fraction and multiply.

(Remember to reduce if needed.)

This definition is another algorithm. It describes the steps of a process. We can name it the fraction division algorithm.

Let's review fraction division with examples.

1. Simplify 12 ÷ 34

12 ÷ 34 = 12 × 43 = 23

1. Simplify three-fifths divided by one-tenth.

35 ÷ 110 = 35 × 101 = 6

1. Simplify 69 ÷ 23

69 ÷ 23 = 69 × 32 = 1

1. Divide ten-fifteenths by six-eighths.

1015 ÷ 68 = 1015 × 86 = 89

1. Simplify 1 23 ÷ 1 28

53 ÷ 108 = 53 × 810 = 43 Video example problems have three images to click on. You can see a video step-by-step answer, a written step-by-step answer, or only the answer. If you find yet more helpful videos, please let your instructor know so that this website can be updated and improved!

The Organic Chemistry Tutor

Multiplying Fractions - The Easy Way!

YouTube Problems   Simplify: 43 × 24 Rewrite as 43 × 241 and use division by 3 to cancel with the 3 and 24. Then we see 41 × 81 = 321 = 32 32

Changing these YouTube problems into a trio of links for video, steps, and only the answer is a lot of work. This is as far as I have gotten.

Are you using these? Does the textbook's videos for every example problem make this idea obsolete?

Chapter 2 Test, Problem 25: Simplify: 5 × 310

Chapter 2 Test, Problem 26: Simplify: 23 × 154

Chapter 2 Test, Problem 27: Simplify: 2215 × 533

Chapter 2 Test, Problem 31: Simplify: 15 ÷ 18

Chapter 2 Test, Problem 32: Simplify: 12 ÷ 23

Chapter 2 Test, Problem 33: Simplify: 245 ÷ 2815 Textbook Exercises for Fraction × and ÷

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

Please read the advice on doing homework in the study skills page.

Section 2.4 (Page 128) # 1, 3, 7, 11, 13, 21, 27, 29

### Fraction + and −§3.2

You probably know that we need to have common denominators when adding fractions.

The denominator of a fraction acts like a word describing the type of thing we are considering. The fraction 34 says "I am considering fourths, and care about three of them."

In this way, "three-fourths" is very similar to the phrases "3 inches" or "3 centimeters".

Because of how denominators are conceptually like labels, trying to add fractions with unlike denominators works as badly as trying to add inches and centimeters. We need to change the fractions so their numerators are counting the same type of thing. We need to make their denominators match.

Consider the cake below. Amy gets to eat the big shaded half. Beatrice gets to eat the smaller shaded fourth. How much of the cake have they eaten? We cannot combine the shaded half and fourth into a single thing without changing one of them to match the other.

1. Can we change either?

Yes.

We could change both to halves. Then we would say they ate 1 12 halves.

We could change both to quarters. Then we would say they ate 3 quarters.

1. Why is it better to change one than the other?

The answer "3 quarters" is a lot more friendly than the answer "1 12 halves".

Tangentially, how did fraction multiplication avoid the need for common denominators? Ponder that sometime, until you have a satisfying answer. Here is a clue.

Try using the picture below to show adding one-third and one-half. This picture starts with all the "cake cuts" going in one direction, and one-third shaded. Put the one-half in the picture by making a "cut" in the other direction. Then try to also shade the top or bottom half to add its amount without double-counting any of the shaded pieces. 1. Does it work?

No. We get stuck because a portion of the half is already shaded! What should we do?

What should we do about a portion of the half already being shaded?

We need to "wrap around" as we shade in the half. This means we were forced to be aware that we are using sixths before looking for our final answer. The process itself requires sixths.

For the similar fraction multiplication problem earlier, we did not need to think about the new denominator until after the process is complete, when we are ready to write our answer. The process itself did not require sixths. Only naming our answer did.

When we ask with addition, "How much is one-half more than one-third?" then overlapping the shaded pieces is not allowed.

When we ask with multiplication, "How much is one-half of than one-third?" then the overlap is our answer.

See the difference?

#### How Common Denominators Happen

When we change the denominators of two fractions so they match, there are three situations.

Sometimes one denominator is a multiple of the other.

Twelve is a multiple of six.

1. Change one-twelfth and one-sixth to have common denominators.

We only have to change the one-sixth.

We un-reduce 16 × 22 = 212

This example is like the cake pieces Amy and Beatrice ate. We should chop up the larger-size piece so it matches the smaller-size piece.

Sometimes the denominators have no common factors.

Fourteen and fifteen have no common factors.

1. Change one-fourteenth and one-fifteenth to have common denominators.

We have to change both fractions.

We un-reduce 114 × 1515 = 15210

We un-reduce 115 × 1414 = 14210

We can always "brute force" a common denominator by multiplying the two denominator numbers.

In this example the "brute force" technique is the best we can do.

Definition

Two or more numbers are relatively prime if they have no common factors except for 1.

(Is this definition is an algorithm? Why or why not?)

For fractions with relatively prime denominators, the "brute force" technique will always be the best we can do.

Sometimes the denominators have factors in common, but neither is a multiple of the other.

Twelve and fifteen have a common factor of three.

1. Change one-twelvth and one-fifteenth to have common denominators.

We have to change both fractions.

We un-reduce 112 × 55 = 560

We un-reduce 115 × 44 = 460

In this example the "brute force" technique is not what we want. There is a common denominator less than 12 × 15 = 180.

We could use 180. But whenever we use a needlessly large common denominator we will eventually do more reducing than otherwise.

Definitions

When we look at a group of two or more fractions, the smallest number we can un-reduce every denominator to match is the least common denominator.

How do we find it? It will be the least common multiple of the starting denominators.

(Are these definitions algorithms? Why or why not?)

If we use the least common denominator we might have to reduce our answer. But it is no longer a step that will always be needed.

#### Two Methods for Finding the Least Common Multiple

Hm. We will need to find least common multiples. How do we do that?

The simplest method is to simply start making a list of multiples for each number. The smallest number the lists have in common is our answer.

1. Use a list of multiples to find the least common multiple of 10, 12, and 15.

Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, ...

Multiples of 12 are 12, 24, 48, 60, 72, ...

Multiples of 15 are 15, 30, 45, 60, 75, ...

In that example we picked 60 for our answer because it was the smallest number on every list.

The quickest method is to write the prime factorization of each number (the long way, do not use exponents to condense your answer), merge factors shared in two or more prime factorizations, and then multiply the surviving factors.

1. Use prime factorizations to find the least common multiple of 6, 10, and 15.

The prime factorization of 10 is 2 × 5. The prime factorization of 12 is 2 × 2 × 3. The prime factorization of 15 is 3 × 5

The least common multiple is 2 × 2 × 3 × 5 = 60

In that example we merged the green 2s, the red 3s, the blue 5s. Every time a factor appears in more than one prime factorization we smoosh together one instance of it from each prime factorization that has it. Notice that the second 2 in the prime factorization of 12 was not merged. Neither of the other prime factorizations also had a 2 to smoosh it with.

1. Use a list of multiples to find the least common multiple of 24, 36, and 48.

Multiples of 24 are 24, 48, 72, 96, 120, 144, 168, ...

Multiples of 36 are 36, 72, 108, 144, 180, ...

Multiples of 48 are 48, 96, 144, 192, ...

1. Use prime factorizations to find the least common multiple of 24, 36, and 48.

The prime factorization of 24 is 2 × 2 × 2 × 3. The prime factorization of 36 is 3 × 2 × 2 × 3. The prime factorization of 48 is 2 × 2 × 2 × 2 × 3

We merge the red 2s because they appear in more than one prime factorization. And we merge the green 2 × 2 × 3 because it appears in more than one prime factorization. There is also an unmerged 2 and an an unmerged 3.

The least common multiple is 2 × 3 ×2 × 2 × 2 × 3 = 144

Those examples were pretty tricky because we found the least common multiple of three numbers. Most of the time when we add fractions we are only adding two fractions. Finding the least common multiple of two numbers will be a lot easier!

#### Three Methods for Adding Fractions

Let's do the same fraction addition problem in three different ways.

The first method we'll nickname the brute force method. We find a new denominator very quickly be multiplying each old denominator by the other. This always works, but forces us to deal with large numbers.

1. Use the brute force method to add 130 + 142

We un-reduce 130 × 4242 = 421,260

We un-reduce 142 × 3030 = 301,260

Then we add 421,260 + 301,260 = 721,260 = 235

We started quickly! We could rush right into un-reducing the two fractions. But then we had to do a lot of work reducing 721,260 to get our final answer.

The second method is using the list of multiples method to find the least common denominator. This method is slow, but is favored by some students who prefer to work with multiples instead of factors.

1. Use the list of multiples method to add 130 + 142

Multiples of 30 are 30, 60, 90, 120, 150, 180, 210, 240, ... So we change 130 into 7210 (Notice that 210 is the seventh multiple in the list.)

Multiples of 42 are 42, 84, 126, 168, 210, 252, ... So we change 142 into 5210 (Notice that 210 is the fifth multiple in the list.)

Then we add 7210 + 5210 = 12210 = 235

We started slowly, finding lists of multipes. But we had a lot less reducing at the end!

The third method is using the prime factorization method to find the least common denominator. This method is quickest for students who can think quickly about factors.

1. Use the prime factorization method to add 130 + 142

The prime factorization of 30 is 2 × 3 × 5. The prime factorization of 42 is 2 × 3 × 7.

We merge the green 2 × 3. There is also an unmerged 5 and an an unmerged 7. So the least common multiple is 2 × 3 × 5 × 7 = 210.

Then we add 130 + 142 = 7210 + 5210 = 12210 = 235

Notice in that last example that we had hints about how to un-reduce the numerators of our starting gractions: the unmerged factor(s) from the other prime factorization!

Time for a few more examples. You try these before seeing the instructor's work. Use any of the three methods.

1. Add 136 + 16

136 + 16 = 136 + 66 = 736

1. Subtract 34112

34112 = 912112 = 812 = 23

1. Add 15 + 14

15 + 14 = 420 + 520 = 920

1. Add 112 + 215

112 + 215 = 560 + 860 = 1360

1. Subtract 542124

542124 = 201687168 = 13168

Note that the same techniques work when we have more than two fractions. The problem just takes longer because it is more work to find the prime factorization of more than two numbers.

1. Add 14 + 16115

14 + 16115 = 1560 + 1060460 = 2160 = 720 Bittinger Chapter Tests, 11th Edition

Chapter 3 Test, Problem 3: Simplify: 12 + 52

Chapter 3 Test, Problem 4: Simplify: 78 + 23

Chapter 3 Test, Problem 5: Simplify: 710 + 19100 + 311,000

Chapter 3 Test, Problem 6: Simplify: 5636

Chapter 3 Test, Problem 7: Simplify: 5634

Chapter 3 Test, Problem 8: Simplify: 1724115 Textbook Exercises for Fraction + and −

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

Please read the advice on doing homework in the study skills page.

Section 3.2 (Page 159) # 1, 5, 9, 13, 21, 23, 27, 29, 33, 35, 39, 47, 51, 55

### Mixed Number Subtraction§3.3

There are two ways to think about subtracting with mixed numbers.

We could treat the fractions as a place value column and borrow from the 1's column if we need to do so.

1. You cut 2 14 inches off a 3 12 inch board. How long is the remaining part? (Borrow if needed, like you are thinking about place value.)

Subract the whole numbers and fractions separately, then squish the answers together.

3 12 − 2 14 = 3 − 2 and then 1214 = 1 and then 14 = 1 14

1. You cut 3 12 inches off a 5 14 inch board. How long is the remaining part? (Borrow if needed, like you are thinking about place value.)

Subract the whole numbers and fractions separately, then squish the answers together.
5 14 − 3 12 = 5 14 − 3 24 = 5 − 3 and then 1424. We get stuck with not enough fourths to start that subtraction!

We borrow from the whole numbers by changing 5 into both 4 and 44.
Now we have plenty of fourths: the one we started with and four more (so five total)

So 4 − 3 = 1 and then 5424 = 34, for a final answer of 1 34

We could change both mixed numbers to improper fractions and then subtract.

1. You cut 2 14 inches off a 3 12 inch board. How long is the remaining part? (Use improper fractions.)

3 12 − 2 14 = 7294 = 14494 = 54

1. You cut 3 12 inches off a 5 14 inch board. How long is the remaining part? (Use improper fractions.)

5 14 − 3 12 = 21472 = 214142 = 74

It is helpful to be fluent with both these methods of subtracting mixed numbers. For some problems the first method will be easier. For other problems the second method will be easier.

Notice that treating the fractions as a place value column naturally resulted in a mixed number answer, whereas changing both mixed numbers to improper fractions naturally resulted in an improper fraction answer. Bittinger Chapter Tests, 11th Edition

Chapter 3 Test, Problem 19: Simplify: 10 16 − 5 78 Textbook Exercises for Mixed Number Subtraction

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

Please read the advice on doing homework in the study skills page.

Section 3.3 (Page 170) # 1, 9, 13, 21, 29, 49, 51, 53, 55

### Order and Terms§1.7, §2.4, §3.4

#### Potential Problems with PEMDAS

Many math students are taught to memorize the acronym PEMDAS to help solve resolve order of operations.

PEMDAS

P = Parenthesis
E = Exponents
M = Multiplication
D = Division
A = Addition
S = Subtraction

This acronym can be very helpful if it was taught well.

Unfortunately, PEMDAS can cause problems if it was taught poorly.

Which answer for 8 ÷ 4 × 2 is Correct?

Do the 4 × 2 first because M comes before D

8 ÷ 4 × 2 = 8 ÷ 4 × 2 = 8 ÷ 8 = 1

or

Do the 8 ÷ 4 first because we go left to right

8 ÷ 4 × 2 = 8 ÷ 4 × 2 = 2 × 2 = 4

1. In the above box, which answer for 8 ÷ 4 × 2 is correct?

The second is correct. Multiplication and division have equal priority, and should be done left-to-right.

To solve an order of operations problem we cannot simply do P-E-M-D-A-S in order! This is because some operations have equal priority. These must be done left-to-right when only these are left to do.

Multiplication and division have equal priority.

Addition and subtraction have equal priority.

Stay alert!

#### Fixing PEMDAS?

Also note that PEMDAS does not tells us what to do with fractions. Fraction bars are actually another grouping symbol, just like parenthesis. We could rewrite fractions problems as parenthesis problems. We could try to "fix" PEMDAS by rewriting it.

(PF)E(MD)(AS)

PF = Parenthesis and Fraction Bars
E = Exponents
MD = Multiplication and Division
AS = Addition and Subtraction

Parenthesis and fraction bars have first priority. Then exponents. Then multiplication and division. Then addition and subtraction. Work from left to right with items of equal priority.

Unfortunately, this is stupidly awkward. There must be something better!

#### Terms

The best way to deal with order of operations is to not replace the (AS) step of PEMDAS with a new kind of thinking.

Terms

A term is an portion of an expression separated from the equal sign only by addition or subtraction.

How many terms are in these expressions? (Hint: Try underlining them.)

1. How many terms are in this expression?

(4 + 2) ÷ 3 + 8 × 22 =

Two. They are separated by the only + symbol outside of parenthesis.

1. How many terms are in this expression?

12 × 4 × 5 − 3 × (9 − 1) ÷ 22 + 5 =

Three. The first stops at the − symbol outside of parenthesis. The final 5 is its own term.

1. How many terms are in this expression?

11 + 8 × 2 − 14 ÷ 7 + (19 − 4) ÷ 3 − 90 ÷ 32 =

Five.

Why are terms the easiest way to think about order of operations?

First, each term is totally independent until as the very last step we combine them. It does not matter which term we simplify first.

Second, within a term we only have two priorities: first grouping structures (parenthesis, fraction bars, and exponents) and then going left-to-right with multiplication and division.

As a tangential comment, exponents are not actually grouping structures, but they demand using them. When we see an exponent we must know clearly how much it covers. So mathematicians include exponents when talking about grouping structures.

Consider the expression (3 + 2)2. It is different from the expressions (3 + 22) and 32 + 22. Wanting to square the sum of 3 + 2 demanded putting parenthesis around that sum.

Now we should actually simplify those three expressions.

1. Simplify this expression.

(4 + 2) ÷ 3 + 8 × 22 =

2 + 32 = 34

1. Simplify this expression.

12 × 4 × 5 − 3 × (9 − 1) ÷ 22 + 5 =

10 − 6 + 5 = 9

1. Simplify this expression.

11 + 8 × 2 − 14 ÷ 7 + (19 − 4) ÷ 3 − 90 ÷ 32 =

11 + 16 − 2 + 5 − 10 = 20

As one last comment about terms, remember that we can cancel factors but we cannot cancel terms.  Bittinger Chapter Tests, 11th Edition

Chapter 1 Test, Problem 41: Simplify: 35 − 1 × 28 ÷ 4 + 3

Chapter 1 Test, Problem 42: Simplify: 102 − 22 ÷ 2

Chapter 1 Test, Problem 43: Simplify: (25 − 15) ÷ 5

Chapter 1 Test, Problem 44: Simplify: 24 + 24 ÷ 12

Chapter 4 Test, Problem 50: Simplify: 256 ÷ 3.2 ÷ 2 − 1.56 + 78.325 × 0.02

Chapter 4 Test, Problem 51: Simplify: (1 − 0.08)2 + 6 × [5 × (12.1 − 8.7) + 10 × (14.3 − 9.6)] Textbook Exercises for Order and Terms

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

Please read the advice on doing homework in the study skills page.

Section 1.7 (Page 84) as much of # 27-73 odd as helps you, 77, 79, 81

Section 2.4 (Page 129) # 31, 43, 45, 47, 49, 51

Section 3.4 (Page 178) # 1, 3, 9, 13, 39, 41, 43, 45, 51

### Goeswith Chains(none)

Understanding order of operations and terms allows us to smush together numbers using the basic arithmetic operations: +, −, ×, and ÷.

There is a very useful but less well-known fifth arithmetic operation that looks like . It also smushes numbers together. But it is used with rates instead of with plain numbers.

We will call it goeswith because it was invented by the famous yet ficticious mathematician and mad scientist, Lucille Hyperious Goeswith. Other math books and websites might call it the "proportional operator".

To use ∝ we follow a three step algorithm.

• First we list out a buch of rates
• Second we write our goal, to identify where to start
• Third, we link the rates together in a chain (probably un-reducing or reducing some rates to fit the pieces together nicely)

Let's look at an example.

1. Lucy drove from Eugene to Los Angeles. On average, she drove 7 hours per day, at 60 miles per hour. Gasoline cost 3 dollars per gallon. Her old car gets 20 miles per gallon of gasoline. How much did she spend each day on gasoline?

First we list the rates. It helps to write them as fractions. Second we write our goal. For this problem 1 day = _____ dollars. This tells us to start our chain by writing 1 day and keep going until we get to dollars.

Third we make our goeswith chain. The only rate that involves days is 7 hours per day.

1 day ∝ 7 hours ∝ ?

The other rate that involves hours is her average speed. We un-reduce it by multiplying the top and bottom by 7 to get 420 miles per 7 hours. Then we can continue building our goeswith chain.

1 day ∝ 7 hours ∝ 420 miles ∝ ?

The other rate that involves miles is her old car's disappointing mileage. We un-reduce it by multiplying the top and bottom by 21 to get 420 miles per 21 gallons. Then we can continue building our goeswith chain.

1 day ∝ 7 hours ∝ 420 miles ∝ 21 gallons ∝ ?

The other rate that involves gallons is the price of gasoline. We un-reduce it by multiplying the top and bottom by 21 to get 63 dollars per 21 gallons. Then we can continue building our goeswith chain.

1 day ∝ 7 hours ∝ 420 miles ∝ 21 gallons ∝ 63 dollars

The goeswith chain is complete. We now have five items that go with 1 day, including the dollar amount we were looking for.

Let's look at more examples.

1. Six friends want to order pizza. Each person will eat four slices. A pizza has 12 slices and costs \$14. How much will they spend? 6 people ∝ 24 slices ∝ 2 pizzas ∝ \$28

1. A pest control worker needs to add permethrin to 1 quart of water. He knows to use 8 teaspoons of permethrin per gallon of water. How much permethrin should he use? 1 quart water ∝ 0.25 gallon water ∝ 2 teaspoons permethrin

1. Your car has a 12 gallon gasoline tank and has a mileage of 30 miles per gallon. When driving cross-country you are at a quarter tank of gasoline remaining and see a sign that says "Next gas 120 miles". Will you make it? 120 miles ∝ 4 gallons gas ∝ 13 tank (which is unfortunately more gas than you have left in the tank)

1. Luke smokes half a pack of cigarettes each day. Each pack costs \$6. How much will he spend per year on cigarettes? 1 year ∝ 365 days ∝ 182.5 packs ∝ \$1,095

Goeswith chains can be really useful! The process is pretty foolproof. It tells us when to multiply and when to divide. Yes, we have to write a lot. But if our intuition about when to multiply or divide is not accurate, it is worth writing more to avoid making a mistake.

We recently studied one step measurement conversions. Goeswith chains allow us to solve problems with more than one step.

Unfortunately, goeswith chains are terribly impractical. All of that fraction un-reducing and reducing takes a lot of time. There must be a more efficient way to do this type of problem!

Fortunately, there is a better algorithm! We will study it soon. It is called unit analysis. However, unit analysis does nothing to help build an understanding for why this type of math problem smushes rates together two at a time to eventually form a chain that ends with the desired answer. So for the sake of building up a deeper understanding and intuition we look at goeswith chains now, as a temporary step between one step measurement conversions and unit analysis. None yet Textbook Exercises for Goeswith Chains

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

Please read the advice on doing homework in the study skills page.

(Our textbook has no exercises for this topic.)

### Ten Exercises

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

1. A cookie recipe calls for 23 cup of flour. If you are making a double batch, how much flour will you use?

2. What is four-fifths times twenty?

3. What is four-fifths times two-thirds?

4. What is two-thirds divided by three-halves?

5. 2215 ÷ 335 =

6. What is the least common multiple of 12 and 18?

7. What is five-sixths plus three-fourths?

8. What is five-sixths minus three-fourths?

9. A piece of fabric that is 1 34 yards long is cut into seven equal pieces. How long is each piece?

10. Grandma Jorgensen left 23 of her 78 pound silver bullion bar to her son Lloyd. Lloyd gave each of his four children a 14 share. How much silver did each of Lloyd's children receive?

### Random Exercises

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

## Percentages

### Percent Of(none)

#### Copies of Something

The word of in a word problem tells us to multiply.

This makes sense. Making copies is what multiplication does. When visiting a candy store, when you tell the clerk "I would like ten of those" then you are saying "I would like ten copies of one of those".

1. Tickets to an amusement park cost \$25. A family buys 5 of them. What was the total cost?

\$25 × 5 = \$125

The phrase "5 of them" turned into × 5.

1. Grading a certain math test takes the instructor 7 minutes. A class of 30 students takes the test. How long will the instructor be grading?

7 minutes × 30 = 210 minutes

The phrase "a class of 30" turned into × 30.

#### A Fraction of Something

The way the English language works also makes it very natural to put the word "of" into a word problem with a fractional amount.

1. In a sports program with 350 children, three-sevenths of the children are boys. How many boys are in the sports program?

350 total × 37 are boys = 150 boys

The phrase "three-sevenths of the children" turned into × 37.

1. Reanna eats five-eights of a pizza. Then her daughter, Mitzi, eats one-half of what is left. How much of the whole pizza did Mitzi eat?

38 remaining × 12 of that = 316 of a pizza

The phrase "one-half of what is left" turned into × 12.

#### A Percentage of Something

The third occasion when the English language naturally uses "of" to mean multiplication is with the phrase "percent of..."

Use RIP LOP to change the percentage into a decimal. Then multiply.

1. What is 40% of 200?

200 × 0.4 = 80

The phrase "40% of..." turned into × 0.4.

The "Percent Of..." Algorithm

When we are asked to find the percentage of an amount, we can use RIP LOP on the percentage and then multiply.

The amount could be a normal number.

1. What is 60 percent of 200?

200 × 0.6 = 120

The amount could be a labeled amount.

1. What is 44% of 50 hours?

50 hours × 0.44= 22 hours

The amount could even be another percentage. Treat the other percertage symbol as a label.

1. What is 10% of 50%?

50 percent × 0.1 = 5 percent = 5%

When comparing "percent of..." answers be wary of when they have the same or different initial amounts.

1. Two politicians are discussing health policy. Consider these two claims, which might not actually be true. (a) 16.3% of Americans are without health insurance. (b) Only 55.9% of adults receive employer provided health insurance. Can both those claims be true? Are the two claims equivalent (they say the same thing), in conflict (they cannot both be true), or not compatible (they are talking about different things)?

The initial amount in the first claim is "all Americans" which includes children. And the percentage measures how many have any health insurance.

The initial amount in the first claim is "adult Americans" which is a smaller group. And the percentage measures how many have one kind of health insurance.

The two claims are not compatible.

#### The Switcheroo Trick

Here is a trick that allows you to solve more "percent of..." problems without needing to use a calculator. Impress your friends!

Remember that when multiplying two number the order of the numbers does not matter. Also remember that "percent" merely means to divide by 100 at the end. Let's put these two facts together.

Definition

The Switcheroo Trick says that when multiplying a number in percent format by a normal number we can switch which number is the percentage.

As an example, we can solve "What is 4% of 50?" by doing 4% × 50. However, finding 4% of a number might be tricky for you without a calculator. But finding 50% is easy—that is half of the number. So we switch which number is the percentage. 4 × 50% = 2.

1. What is 24% of 25?

We can switcheroo to ask, "What is 25% of 24?" Then with mental arithmetic we can more easily see that one-quarter of 24 is 6.

1. A computer that normally costs \$900 is on sale for 8% off. How much is the discount?

We can switcheroo to ask, "What is 900% of 8?" Then with mental arithmetic we can more easily see that 9 times 8 is a \$72 discount.

#### Tipping

A common application for "percent of..." word problems is tipping at a restaurant. You can learn a few shortcuts to impress your friends.

• To find 10% of a value, move the decimal point one place to the left.
• To find 20% of a value, first find 10% of the value and then double that.
• To find 5% of a value, first find 10% of the value and then halve that.
• To find 15% of a value, first find 10% of the value, then find 5% of the value, and then add those together.

Before continuing, make sure you are not confused. We know that being in percent format is equivalent to two decimal point scoots. Why does the shortcut for finding 10% only involve only one decial point scoot?

Have you figured that out?

When we find 10% we end up multipying by × 0.1

We know from our examination of decimal point scoots that × 0.1 is the same as one scoot to the left.

Let's practice using those shortcuts.

1. A restaurant meal costs \$25. You want to leave a 20% tip. What is 20% of \$25?

10% of \$25 becomes \$25 × 0.1 = \$2.50

But we want 20% instead of that 10%, so to get twice as much we double the 10% result and get a \$5 tip.

1. A restaurant meal costs \$20. You want to leave a 15% tip. What is 15% of \$20?

10% of \$20 becomes \$20 × 0.1 = \$2

Next we see that 5% of \$20 is half that, which is \$1

But we want 15% so we combine the 10% result and the 5% result and get a \$3 tip.

1. A restaurant meal costs \$30. How much is a 15% tip?

10% of \$30 becomes \$30 × 0.1 = \$3.00

Next we see that 5% of \$30 is half that, which is \$1.50

But we want 15% so we combine the 10% result and the 5% result and get a \$4.50 tip.

Notice that the word "of" did not appear in the previous problem. But we could rephrase the problems to ask "What is 15% of \$30?"

In other words, merely having the option to rephrase a situation using the phrase "percent of..." is still an instruction to multiply.

Tipping is one instance when estimation is very useful. If my restaurant bill was \$21.87, the tip will not change much if I estimate based on \$22.00.

1. A restaurant meal costs \$39.80. Estimate a 15% tip.

Estimate by using \$40 for the meal cost.

10% of \$40 becomes \$40 × 0.1 = \$4

But we want 15% so we use one-and-a-half that result and get a \$6 tip.

1. A restaurant meal costs \$23.75. Estimate a 20% tip.

Estimate by using \$24 for the meal cost.

10% of \$24 becomes \$24 × 0.1 = \$2.40

But we want 20% so we double that result and get a \$4.80 tip.

#### Retail Sale Percentages

Another common application for "percent of..." word problems is a retail item that goes on sale with a percentage price reduction.

1. A book normally costs \$12. It is on sale for 20% off. What is 20% of \$12?

20% of \$12 becomes \$12 × 0.2 = \$2.40 discount

1. A dress normally costs \$60. It is on sale for 35% off. How much is the reduction?

35% of \$60 becomes \$60 × 0.35 = \$21 reduction

Price reductions are another situation where estimation is very useful.

1. A jacket normally costs \$79.99. It is on sale for 60% off. Estimate the reduction in price.

Estimate by using \$80 for the jacket cost.

60% of \$80 becomes \$80 × 0.6 = \$48 reduction

Notice that the word "of" did not appear in the previous two problems. But we could rephrase the problems to ask "What is 35% of \$60?" and "Estimate 60% of \$79.99."

Once again, merely having the option to rephrase a situation using the phrase "percent of..." is still an instruction to multiply.

#### The Guppy Problem

All of these "percent of..." problems told us a percentage. We were multiplying that percentage by an amount.

It is possible to write an English sentence using the words "percent of..." that asks for the percentage instead. This is a different type of problem. We will learn about it later.

Guppies—Asking for the Percentage

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

Notice that this problem did not tell us a percentage. The "percent of..." algorithm does not work. There is no percentage to RIP LOP as the first step of that algorithm. The Organic Chemistry Tutor

How To Find The Percent of a Number Fast!

Percentages Made Easy! Textbook Exercises for Percent Of

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

Please read the advice on doing homework in the study skills page.

(Our textbook has no exercises for this topic.)

### Percent Change§6.5

Many retail sale situations ask you to find the percentage.

Compared to the problems we just did, we are working backwards. So we do the opposite of multiplying, which is dividing. And instead of starting with RIP LOP we finish with RIP LOP.

1. A candy bar that normally costs 75¢ is on sale for 15¢ off. What is the percent of the decrease?

Think of 15¢ and 75¢ as being a part and whole.

Perhaps imagine the fraction 15¢75¢

Then 15¢ ÷ 75¢ = 0.2 = 20% decrease

Often the monetary amount of the decrease is hidden. We must subtract to find it.

1. A candy bar that normally costs 80¢ is on sale for 60¢. What is the percent of the decrease?

The decrease is 80¢ − 60¢ = 20¢

Perhaps imagine the fraction 20¢80¢

Then 20¢ ÷ 80¢ = 0.25 = 25% decrease

To avoid confusion, it is helpful to memorize a formula.

Percent Change Formula

percent change = change ÷ original

Then use RIP LOP to change the decimal answer into percent format.

This definition is an algorithm dressed up as a formula. The steps of the algorithm are "change ÷ original". We can write those steps as a formula by putting in front "percent change =". Notice that any formula of similar structure, which says "your answer is what happens when you correctly simplify this expression" is an algorithm dressed up as a formula.

This formula reminds us to find the change if the word problem is worded to only give us the old and new values. It also reminds us to divide by the original instead of dividing incorrectly by the new value.

The percent change formula works for increases as well as decreases.

1. The house purchased for \$120,000 increases in value by \$15,000. What is the percent of the increase?

We do change ÷ original = \$15,000 ÷ \$120,000 = 0.125 = 12.5% increase

So far so good. Unfortunately, the word problems get trickier.

We already saw one issue to keep us on our toes: problems can show or hide the numeric amount of the change.

1. The value of a house increases from \$150,000 to \$170,000. What is the percent of the increase?

The increase is \$170,000 − \$150,000 = \$20,000.

Then change ÷ original = \$20,000 ÷ \$150,000 ≈ 0.13 = 13% increase

Also, some problems ask for the change, but others ask for the change combined with the original amount.

1. A painting originally worth \$200 increases in value by 3%. By how much did its value change? (Find the change, then stop.)

3% of \$200 becomes \$200 × 0.03 = \$6 increase

1. A painting originally worth \$200 increases in value by 3%. What is its new value? (Find the change, then add it to the original.)

3% of \$200 becomes \$200 × 0.03 = \$6 increase

Then add \$200 + \$6 = \$206 new value

Let's do a variety of problems to make sure you understand how the wording can ask for different things.

1. If the population of the city of Eugene was 163,000 in 2015, and it grows by 5% over the next ten years, what will the population be in 2025?

5% of 163,000 becomes 163,000 × 0.05 = 8,150 increase

Then add 163,000 + 8,150 = 171,150 people for the new value

1. Eugene's population was 145,000 in 2005. (And from the last problem 163,000 in 2015.) By what percentage did Eugene's population increase from from 2005 to 2015?

The increase as a number was 163,000 − 145,000 = 18,000 people

So the percent change is change ÷ original = 18,000 ÷ 145,000 ≈ 0.12 = 12% increase

1. A 40-ounce bag of tea that normally costs \$24 is on sale for 20% off. What is the current cost per pound?

First we change the ounces to pounds. 40 ounces ÷ 16 ounces per pound = 2.5 pounds

Then we find the current price, which is 80% of \$24. \$24 × 0.8 = \$19.20

Then we find the unit rate by dividing. \$19.20 ÷ 2.5 = \$7.68 per pound

1. A firearm originally worth \$100 appreciates (goes up in value) by 12% before it is sold. What is the selling price?

12% of \$100 becomes \$100 × 0.12 = \$12 increase

Then add \$100 + \$12 = \$112 new value

Here is a diagram to help us remember both issues when doing "percent change..." problems.  Plettski Productions Math

Percent of Increase and Decrease

Mathispower4u

Percent of Change

Determine a Percent of Change (Decrease)

Determine a Percent of Change (Increase)

Bittinger Chapter Tests, 11th Edition

Chapter 6 Test, Problem 9: The number of foreign children adopted by Americans declined from 20,679 in 2006 to 19,292 in 2007. Find the percent of the decrease.

Chapter 6 Test, Problem 13: The marked price of a DVD player is \$200 and the item is on sale for 20% off. What are the discount (in dollars) and sale price?

Chapter 6 Test, Problem 19: A television that normally costs \$349 is on sale for \$299. What is the discount in dollars? What is the discount rate? Textbook Exercises for Percent Change

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

Please read the advice on doing homework in the study skills page.

Section 6.5 (Page 347) # 1, 3, 9, 11, 13, 15, 17

### The One Plus Trick(none)

Notice that in the previous problem, when we calculated \$100 original value × 0.12 rate = \$12 change, the initial amount of \$100 was used but then disappeared. We had to add it back as a second step.

What if it did not disappear?

We know that multiplying any number by 1 does not change it. Let's use that.

The One Plus Trick

When we multiply by a percentage increase, we can first add 1 to that percentage to keep the original amount around.

So we can redo the previous problem as \$100 original value × 1.12 rate = \$112 new value

That is great! Sticking a 1 in front of the percentage is trivial. That is much nicer than our first method of solving the problem that involved two steps.

Remember, × 1.12 is a combination of doing × 1.0 to keep the initial amount around, and doing × 0.12 to find the change.

1. A wedding ring worth \$5,000 increases in value by 8%. What is the new value?

\$5,000 original value × 1.08 rate = \$5,400 new value

1. A \$200 necklace increases in value by 4%. What is the new value of the necklace?

\$200 original value × 1.04 rate = \$208 new value

1. Last year's model of Zapitall microwave cost \$140. This year's model has more features and costs 18% more. What is the cost of the newer model?

\$140 original value × 1.18 rate = \$165.20 new value

1. An apartment's monthly rent was \$750 before it was raised by 5%. What is the new monthly rent?

\$750 original value × 1.05 rate = \$787.50 new value

#### The One Minus Trick

For a percent decrease problem we can use the same trick but it looks slightly different. We still add one. But the change we are combining with 1 is negative. So we need to subtract the percentage from 1.

The One Minus Trick

When we multiply by a percentage decrease, we can first subtract the percentage from 1 to keep the original amount around.

1. A bracelet worth \$200 decreases in value by 15%. What is the new value?

\$200 × (10.15) = \$200 × 0.85 = \$170

1. A brand new car cost \$22,500. After one month its value goes down 10%. What is that month-old value?

\$22,500 × (10.1) = \$22,500 × 0.9 = \$20,250

Try to explain in your own words the process for increasing a quantity by 10%.

Try to explain in your own words the process for increasing a quantity by 50%.

Is it possible to increase a quantity by 100%? Explain what that would mean. None yet Textbook Exercises for the One Plus Trick

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

Please read the advice on doing homework in the study skills page.

(Our textbook has no exercises for this topic.)

### Ten Exercises

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

1. What is 25% of 228?

2. What is 40% of 750?

3. The price tag (before sales tax) on an item says \$100. The sales tax rate is 5%. What is the total price (including tax)?

4. During a sale, a dress decreased in price from \$90 to \$72. What was the percent of decrease?

5. An investment increases in value from \$200 to \$216. What is the percent increase?

6. A pinball machine that normally sells for \$3,999 is on sale for \$3,150. What is the rate of discount?

7. An investment of \$3,000 increases in value by 4%. What is the increase?

8. An investment of \$4,000 increases in value by 8%. What is the new value?

9. An investment of \$5,000 decreases in value by 2.4%. What is the new value?

10. An investment of \$6,000 decreases in value by 2%. Then it decreases by another 5%. What is the final value?

### Random Exercises

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

## Measurement

### Unit Analysis§7.1, §7.2, §7.3

Remember that one-step conversions were the wrong tool for some measurement unit conversion problems. We needed a new technique if we did not know a single rate that translated from the old unit to the new unit.

Our new technique is a five-step process called Unit Analysis (nicknamed "Canceling Words"). It is a simple and powerful tool. Here are the steps without context, but only for later reference. To first meet the new technique it is best to see it in action in an example.

Unit Analysis

The process of Unit Analysis uses five steps to convert measurement units.

• write the given measurement as a fraction
• write some "empty" rates without numbers
• fill in the numbers for the rates
• multiply
• simplify the fraction answer

Unit Analysis is an algorithm whose process is too complicated to be written as a formula.

Here is a simple example of Unit Analysis. We could do this problem much more efficiently as a one-step conversion. But we will use it now merely as a model of the five step process.

Detailed Example of Needless Unit Analysis

How many inches is 3.7 feet?

Our first step is to write the given measurement as a fraction. If it is not already a fraction we put the value over 1. Our second step is to write some "empty" rates without numbers, multiplying. We do this until each unit we want to get rid of appears in a numerator-and-denominator pair so it will cancel. Our third step is to fill in the numbers for the rates. To do this we need to memorize or look up the unit conversion rates. Our fourth step is to multiply. This works just like any other fraction multiplication. Notice how units cancel. We do not normally reduce before multiplying. Our fifth step is to simplify the fraction answer . Often the denominator is not 1, so we simplify by doing "numerator ÷ denominator" like old unit rate problems. That was very methodical: if we follow the five steps we have almost no chance of making a mistake.

Now for a problem when we actually need to use Unit Analysis because the unit conversion has more than one step.

Detailed Example of Needing Unit Analysis

How many miles per hour is 200 inches per minute?

Our first step is to write the given measurement as a fraction. If it is not already a fraction we put the value over 1. Our second step is to write some "empty" rates without numbers, multiplying. We do this until each unit we want to get rid of appears in a numerator-and-denominator pair so it will cancel.

(Be alert! We want distance on top and time on bottom. So we must start that way! Starting with the fraction upside down is our biggest potential danger—besides a careless calculator button error it is the only way to mess up.) Our third step is to fill in the numbers for the rates. To do this we need to memorize or look up the unit conversion rates. Our fourth step is to multiply. This works just like any other fraction multiplication. Notice how units cancel. We do not normally reduce before multiplying. Our fifth step is to simplify the fraction answer . Often the denominator is not 1, so we simplify by doing "numerator ÷ denominator" like old unit rate problems. To summrize, Unit Analysis is an important process because it works no matter how many rates are needed to translate from the old units to the new units. The process keeps track of when to multiply and when to divide, so we do not have to keep track in our heads. All we need to do is look up (or memorize) the unit conversion rates and line them up.

Now for more example problems from the textbook. We will always use Unit Analysis, even for the problems that are one-step unit conversions.

1. How many feet is 9.6 yards?

9.6 yards1 = 9.6 yards1 × feetyard = 9.6 yards1 × 3 feet1 yard = 28.8 feet1 = 28.8 feet

1. How many inches is 3 yards?

3 yards1 = 3 yards1 × feetyard × inchesfeet = 3 yards1 × 3 feet1 yard × 12 inches1 foot = 108 inches1 = 108 inches

1. How many feet is 1 inch?

1 inch1 = 1 inch1 × feetinches = 1 inch1 × 1 foot12 inches = 112 foot (or if you prefer a decimal ≈ 0.83 foot)

1. How many inches is 6 13 yard?

First change 6 13 yard into 193 yard.

193 yard = 19 yard3 = 19 yard3 × feetyard × inchesfeet = 19 yards3 × 3 feet1 yard × 12 inches1 foot = 228 inches1 = 228 inches

1. How many pints is 11 gallons?

11 gallons1 = 11 gallons1 × quartsgallons × pintsquarts = 11 yards1 × 4 quarts1 gallon × 2 pints1 quart = 88 pints1 = 88 pints

#### A Timely Warning

Except for inches to centimeters, the unit conversion rates that move between Standard and SI are rounded. Because they are rounding that happens before the problem is finished they introduce some error. Let's do the same problem in two different ways to see this happen.

1. 3.171 quarts is how many liters? (Use 1 liter = 1.057 quarts.)

3.171 quarts1 = 3.171 quarts1 × litersquarts = 3.171 quarts1 × 1 liters1.057 quarts = 3 liters1 = 3 liters

1. 3.171 quarts is how many liters? (Use 1 quart = 0.946 liters.)

3.171 quarts1 = 3.171 quarts1 × litersquarts = 3.171 quarts1 × 0.946 liters1 quarts = 2.999766 liters13 liters

Is the answer slightly less than three liters or not? It is actually slightly more! But because our original number had quite a few decimal places the unit conversion rates we used did not have enough decimal places to produce answers with enough accuracy.

If your original numbers have many decimal places (thus asking your answers to also have many decimal places) you will need unit conversion rates with lots of decimal places. Mathispower4u

American Unit Conversion

Find the Number of Meters Traveled in 3 seconds Given Kilometers Per Hour

KetzBook

Unit Conversion the Easy Way (Dimensional Analysis)

Bittinger Chapter Tests, 11th Edition

Chapter 8 Test, Problem 1: How many inches is 4 feet?

Chapter 8 Test, Problem 2: How many feet is 4 inches?

Chapter 8 Test, Problem 5: How many meters is 200 yards?

Chapter 8 Test, Problem 6: How many miles is 2,400 kilometers?

Chapter 8 Test, Problem 11: How many ounces is 4 pounds?

Chapter 8 Test, Problem 12: How many pounds is 4.11 tons?

Chapter 8 Test, Problem 16: How many minutes is 5 hours?

Chapter 8 Test, Problem 17: How many hours is 15 days?

Chapter 8 Test, Problem 18: How many quarts is 64 pints?

Chapter 8 Test, Problem 19: How many ounces is 10 gallons?

Chapter 8 Test, Problem 20: How many ounces is 5 cups? Textbook Exercises for Unit Analysis

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

Please read the advice on doing homework in the study skills page.

Section 7.1 (Page 371) # 13, 41, 43, 45, 47, 49, 51

Section 7.2 (Page 382) # 55, 59, 65

Section 7.3 (Page 388) # 35, 39, 41

### Polygon Perimeter and Area§8.1, § 8.2

#### Three Definitions

There are two common ways to measure the size of a shape: perimeter and area.

Perimeter

The perimeter of a shape is the distance around its edges.

Area

The area of a shape is how much room the shape's surface takes up.

We'll start by talking about perimeter.

Let's explore the perimeter of polygons.

Polygon

Polygons are closed shapes whose edges are straight lines.

Here are a bunch of polygons: squares, rectangles, triangles, parallelograms, and trapezoids. 1. Draw a polygon with seven edges.
1. Draw a shape with four edges that is not a polygon.

#### Adding Up Side Lengths

Perimeter is the distance around a shape. There is nothing more complicated to say, since all polygon perimeter problems are simply adding up the lengths of sides. Finding a perimeter is easy. Just mark a corner (so you don't carelessly forget an edge) and begin adding.

1. Find the perimeter of these two polygons.

For the first 3 inches + 4 inches + 5 inches = 14 inches

For the second it might help to label the sides that start without labels. 3 + 6.1 + 3 + 6.1 = 18.2 inches There are only two possible ways to be tricky.

The first way to be tricky with perimeter is demonstrated with this problem:

1. A rectangle has sides 2.5 feet long and 20 inches wide. What is its perimeter in inches?

We need to first change the sides labeled with 2.5 feet so they are labeled with 30 inches.

Then 30 inches + 20 inches + 30 inches + 20 inches = 100 inches

The trickiness comes from hiding a measurement unit conversion problem inside the perimeter problem.

The first way to be tricky is to ask a complicated question while being obtuse about providing all the required information, like this problem:

1. When my grandfather turned 70 he started to walk 3 miles each morning. Now he is 75 and we have no idea where he is.

Oops. That is not a math problem. Let's try again.

1. When my grandfather turned 70 he started to walk around the neighborhood each morning. Below is a picture of his route. How many miles does he walk each week? This problem is a great excuse to introduce a six-step problem solving process.

• Determine what you are looking for
• Draw pictures
• Name things
• Make equations
• Solve the equations
• Check your answer

Step one is Determine what you are looking for. Read the problem two or three times so you understand it and notice all the details. Write down (or pretend to write) in an English sentence what you are looking for. Don't wander off track and forget what you are looking for.

Consider our example problem. We are looking for a number of miles. We need to add a bunch of numbers.

Step two is Draw pictures. Draw a picture or diagram of the situation. Then label things in the picture or diagram! Your visualization will not be much help without labels.

Consider our example problem. The diagram is already provided. Hooray!

But two sides are not labeled. We can use subtraction to find the missing lengths. For the top we added ½ mile + 1 mile = 1 ½ mile

For the far left we subtracted ¾ mile − ½ mile = ¼ mile

Step three is Name Things. Write (or pretend to write) English sentences to give a one letter name to each quantity. Be aware of when two quantities are related to each other and can be expressed using the same letter. Check that each piece of information you are given is really relevant to the problem.

Consider our example problem. We are not told where on the route he starts (and stops). But that does not matter. We should just pick a corner as our "start" and label it. Let's use the top left corner.

Step four is Make equations. Express each relationship you know as an equation. Write (or pretend to write) an English sentence explaining each equation you write.

Consider our example problem. The sum is 1 ½ mile + ¾ mile + 1 mile + ½ mile + ½ mile + ¼ mile

Step five is Solve the equations.

Consider our example problem. The two ½'s add to 1. The ¼ and ¾ also add to 1. The total is 1 mile + 1 mile + 1 mile + 1 ½ miles = 4 ½ miles

Step six is Check your answer. Check that your answer is a reasonable amount. Make sure that your final answer is in units that make sense.

Consider our example problem. An answer of four ½ miles seems reasonable.

What can you do if you get stuck? Here are five ways to try to get unstuck.

• Imagine that you are in the problem's setting, to help use your intuition. (Maybe picture yourself tiny and in the diagram.)
• Explain to a friend in very clear English what you have done and where you are stuck.
• Look for intermediate goals: quantities to find that will help you find others.
• If the problem involves large numbers, try solving the problem with smaller numbers instead. Once you see the steps to use, go back and do those steps with the original numbers.
• If the problem is complicated, try simplifying it and solving the simpler problem, to get ideas on how to approach the original problem.

For most people, the hardest part of a word problem is drawing the picture, naming the amounts, and especially setting up the equation.

Also remember which English words and phrases correspond with which arithmetic operations. Addition is usually "sum", "total", "increased by", or "more than". Subtraction is usually "difference", "how much more", "decreased by", or "less than". Multiplication is usually "of" or "times". Division usually lacks a phrase but is about finding equal portions.

#### Area for Rectangles and Squares

Finding the area of a rectangle is easy. The formula Area = length × width is well known to most students. Squares are just the same.

Take three paper rectangles to use while we figure out how other area formulas relate to the area for rectangles.

#### Area for Triangles

With a partner, take one of your paper rectangles and invent a way to use it to explain how to measure the area this triangle. Imagine that all you know is how to find the area of a rectangle.

1. How can you make this triangle out of a rectangle? How much of the rectangle does the triangle use? One fold will make this kind of triangle out of a rectangle. We can see the triangle uses up half of the rectangle. So its area is half of the rectangle's area.

The previous triangle had one side straight up and down. How about this other triangle with two slanted sides?

1. How can you make this triangle out of a rectangle? How much of the rectangle does the triangle use? Two folds will make this kind of triangle out of a rectangle. Which still is half of the rectangle, as we can see by moving the small piece over. So for any triangles—whether they have a vertical side or not—we can see the triangle uses up half of the rectangle. Any triangle's area is half of the area of the rectangle that snugly contains it.

For historical reasons we normally do not write Area = 12 × length × width for the area of a triangle. Instead of "length" and "width" we call those measuremets "base" and "height".

Why does Area = 12 × base × height make people happier? No idea.

Students who prefer decimals to fractions will use the variation Area = base × height ÷ 2.

Notice that finding the area of a triangle requires its height, not diagonal edge lengths.

#### Area for Parallelograms

With a partner, take another of your paper rectangles and invent a way to use it to explain how to measure the area this parallelogram. As before, imagine that all you know is how to find the area of a rectangle.

1. How can you make this parallelogram out of a rectangle? How much of the rectangle does the parallelogram use?  We can make a rectangle out of a parallelogram by cutting off a triangular shape and sliding it over. We could reverse this to make a paralellogram from a rectangle.

The area of a paralellogram is thus the same as the area of the rectangle.

Notice that finding the area of a parallelogram requires its height, not diagonal edge length.

#### Area for Trapezoids

One last time, with a partner, take another of your paper rectangles and invent a way to use it to explain how to measure the area this trapezoid. As before, imagine that all you know is how to find the area of a rectangle.

1. How can you this trapezoid to a single previously studied shape (one rectangle, triangle, or parallelogram)?

Sure, we could cut it apart into two triangles and a rectangle but that is too much work.  The area of a trapezoid is thus half of the area of a bigger parallelogram. This bigger parallelogram has a base whose length is equal to top + bottom for the original trapezoid.

We can write Area = (top + bottom) × height ÷ 2.

Notice that finding the area of a trapezoid requires both horizontal edges and its height, but not either diagonal edge length.

The formula for the area of a trapezoid can instead be written using an average. It looks like "average the horizontal sides, then multiply by the height" if we rewrite it as Area = (top + bottom) ÷ 2 × height.

As a special challenge, try proving that the area of a trapezoid equals its perimeter times one-quarter its height. (If you need a hint, click here.)

#### Summary Page

We just invented a bunch of algorithms! These polygon area formulas should make sense to you now.

If you want to double-check your understanding before the final exam, you can use this summary page. What is the area of each shape? How could you prove it to a first-grader using scissors and perhaps a crayon?  Mathispower4u

Perimeter and Area Formulas

Bittinger Chapter Tests, 11th Edition

Chapter 9 Test, Problem 1: A rectangle has length 9.4 cm and width 7.01 cm. Find its perimeter and area.

Chapter 9 Test, Problem 2: A square has sides of length 4 78 inches. Find its perimeter and area.

Chapter 9 Test, Problem 3: A parallelogram has base 10 cm and height 2.5 cm. Find its area.

Chapter 9 Test, Problem 4: A triangle has base 8 meters and height 3 meters. Find its area.

Chapter 9 Test, Problem 5: A trapezoid has a bottom 8 feet long, top 4 feet long, and height of 3 feet. Find its area.

Chapter 9 Test, Problem 36: A rectangle has length 8 feet and width 3 inches. Find its area in square feet.

Chapter 9 Test, Problem 37: A triangle has base 5 yards and height 3 inches. Find its area in square feet. Textbook Exercises for Polygon Perimeter and Area

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

Please read the advice on doing homework in the study skills page.

Section 1.2 (Page 23) # 85, 89, 91

Section 1.5 (Page 60) # 105, 107, 109, 113

Section 2.4 (Page 129) # 57, 59, 61

Section 8.1 (Page 422) # 1, 3, 5, 7, 9, 11, 13, 35, 37, 39

Section 8.2 (Page 430) # 1, 3, 5, 7, 17, 19, 21

### Metaphorical Rectangles

For many students, finding the area of a rectangle as length × width makes a lot of sense when they see a drawing. This makes a link in our brain. Multiplying two things can be pictured in our minds as a rectangle's area.

We can use drawings of a rectangle's area to help explain why different examples of multiplication make sense.

#### The Traditional Multiplication Algorithm

Let's start with the traditional multiplication algorithm.

First we use the traditional multiplication algorithm to find 31 × 25 without a calculator. In our answer each of the four multiplication steps was given its own color. You can see why while looking at the rectangle version of this multiplication. That rectangle diagram does not include the final answer.

We still need to add up the numbers in its four parts. 600 + 20 + 150 + 5 = 775.

Notice how the in the traditional multiplication algorithm we added those same four numbers but in a different way.

Now it can be your turn with another multiplication problem.

1. Use the traditional multiplication algorithm to find 33 × 54 without a calculator. (Notice where carrying happens.) 1. Draw the rectangle version of 33 × 54. (Notice where carrying happens.) Play around with this more, on your own.

What happens if one number has three digits?

How special is the ghostly gray zero?

We saw earlier how to use place value to think about mixed number subtraction. Can you draw a rectangle diagram for 4 ¼ × 8 ½ = 36 ⅛?

#### Word Problems

When using a rectangle to help picture 31 × 25 we needed to realize that 31 = 30 + 1 to label one side as two lengths, and 25 = 20 + 5 to label the other side as another two lengths.

Word problems that similarly involve both addition and multiplication can sometimes be written as rectangle diagrams.

1. An elementary school class has 14 boys and 15 girls. The class visits a farm, and each student is given 2 pears and 4 apples. How many total pieces of fruit were given to the students? #### Algebra

Explaining how rectangle diagrams work with algebra would involve more algebra than you might have already learned.

But just in case you have already seen enough algebra to understand, here is one final rectangle diagram. Maybe that makes you happy?

If that type of algebra is new to you, but makes sense so far, you can search online for a video about the "distributive property".

### Circle Circumference and Area§8.1, §8.2

#### π Hiding in a Circumference

Time to talk about about circles.

Consider your arm. How many "arm heights" does it take to go around your arm?  How about your head? How many "head widths" does it take to go around your head?  Our answers about measuring our body parts will vary, because arms and heads are not very circular.

If we were measuring actual circles, the answer to the question "How many circle widths does it take to go around the outside?" would be a number bigger than 3 but less than 4.

We need some more careful language.

Circle Width

The width of a circle, going through the center, is its diameter.

Circle Perimeter

The perimeter of a circle is called its circumference.

(Don't ask me why. "Perimeter" was already a perfectly usable word for this.)

So we can reword what we said. If we were measuring actual circles, the answer to the question "How many diameters does it take to go around the circumference?" would be a number bigger than 3 but less than 4.

If we were to write this as formulas, the answer is between C = 3 × d and C = 4 × d.

What number between 3 and 4 is the correct coefficient?

This answer for actual circles is roughly 3.14159 but its decimal digits keep on going forever without any repeating pattern. We call this number π (also written as "pi").

Thus we can look really fancy and professional and write a formula, using either words or just letters.

Circumference = π × diameter

C = π × d

But all this formula really says is "the distance around a circle is a bit more than three times its width".

This animation from Wikipedia shows what is happening nicely: #### π Hiding in an Area

Consider this small square, of area A = r2. If we make four of them, the area is then A = 4 × r2. Now we ask a second question about circles: how large a circle can we fit into the big square of area 4 × r2? That little line going from the circle's center to its edge (half a diameter) is really useful. Let's give it a name.

Radius

Half the diameter is the radius.

It looks like the circle covers more than 3 of the little squares, but not all 4.

So the area of the circle is between A = 3 × r2 and A = 4 × r2.

What number between 3 and 4 is the correct coefficient?

The answer is the same number as the answer to the circumference question: π!

Area = π × radius2

A = π × r2

Again, this looks fancy and professional but really only says "a circle covers a bit more than three squares with side length equal to the circle's radius".

Wonderful! If π was not the answer to both the circumference question and the area question then we would need two buttons on our calculator instead of merely one.

Are the circle formulas algorithms? They can be. Math problems can give you a radius or diameter and ask you to treat the formulas as a process to follow to calculate the circumference or area. But keep in mind that these formulas are also plain descriptions of how circles behave. A bit more than three diameters are required to cover the circumference. A circle covers a bit more than three squares whose side length is the radius.

Naturally, rounding π before the end of a problem will cause incorrect answers. To develop optimal math habits, you should always use the calculator key for π so that you do not round in the middle of the problem. However, some textbooks or websites use a rounded version of π (either 3.14 or 227). This means your answers will not quite match because yours are more accurate.

A final note of warning: some students learn the circumference formula as C = π × 2 × r. This makes it look a lot like the area formula: both have a π, an r, and a 2. Don't get mixed up! The easiest way to avoid a careless mistake is to use the circumference formula with a diameter.

1. Find the perimeter and area of this circle.

Circumference = π × diameter = π × 4 inches ≈ 12.6 inches

Area = π × radius2 = π × 22 square inches ≈ 12.6 square inches 1. Find the perimeter and area of this circle.

Circumference = π × diameter = π × 2 inches ≈ 6.3 inches

Area = π × radius2 = π × 12 square inches ≈ 3.1 square inches 1. Find the perimeter and area of this circle.

Circumference = π × diameter = π × ½ inches ≈ 1.57 inches

Area = π × radius2 = π × ½2 square inches ≈ 0.78 square inches 1. Find the perimeter and area of this shape.

First we break apart the shape into a half-circle and a rectangle.

For the perimeter, add halfway around the circle with three sides of the rectangle.

For the area, add half of the circle's area with all of the rectangle.

The perimeter is about 14.6 inches.

The answer is about 16.9 inches.  Bittinger Chapter Tests, 11th Edition

Chapter 9 Test, Problem 6: A circle has a radius of one-eighth of an inch. What is its diameter?

Chapter 9 Test, Problem 7: A circle has a diameter of 18 centimeters. What is its radius? Textbook Exercises for Circle Circumference and Area

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

Please read the advice on doing homework in the study skills page.

Section 8.1 (Page 422) # 15, 17, 21, 23

Section 8.2 (Page 529) # 13, 15, 17, 23, 31, 37

### Ten Exercises

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

1. Two years is how many minutes?

2. A football field is 4,844 cm wide. What is that width in yards? (1 meter ≈ 1.09361 yards)

3. A plane flying 6,500 feet per minute is going about how many miles per hour?

4. Including the end zones, a football field is 360 feet long and 160 feet wide. If you run three laps around a football field, about what percentage of a mile do you run?

5. You want to buy Sculpey III modeling clay online. At amazon.com it costs \$7.29 for 8 ounces. At amazon.co.uk it costs £2.20 for 57 grams. (1 ounce ≈ 28.35 grams. If £1 ≈ \$1.31), which is a better buy?

6. A triangle has base 4 m and height 3.5 m. What is the area?

7. Most billiard tables are twice as long as they are wide. What is the perimeter of a billiard table that measures 4.5 feet by 9 feet?

8. A parallelogram has base 2.3 cm and height 3.5 cm. What is the area?

9. A trapezoid has base 25 cm, top 16 cm, and height 35 cm. What is the area?

10. The standard backyard trampoline has a diameter of 14 feet. What is its area?

### Random Exercises

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