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Mad Science: Simplifying an Expression

Our second big 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)

2*x* + 5

24 ÷ 8 + 10^{2} − 10 × ^{1}⁄_{5}

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 cannot 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.

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

**1.** Use the picture below to show multiplying one-third by 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.

After re-shading half of the one-third, what amount is re-shaded?

What are the numerators counting?

What are the denominators counting?

**2.** 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.

After re-shading three-quarters of the two-thirds, what amount is re-shaded?

What are the numerators counting?

What are the denominators counting?

Fraction Multiplication

To

multiplyfractions, 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.

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

**3.** Multiply two-thirds and seven-fifths.

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

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

Canceling

During fraction multiplication,

cancelingis reducing early.

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

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

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

Your turn. Here are three 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.)

**6.** Multiply three-fifths and one-tenth.

**7.** Multiply six-elevenths and two-thirds.

**8.** Multiply ten-fifteenths and six-eighths.

How do we multiply a fraction by a whole number? Remember that we can rewrite any whole number as a fraction by writing it over a denominator of one.

**9.** Multiply ^{2}⁄_{3} × 1 ^{4}⁄_{8}

**10.** Multiply 1 ^{2}⁄_{3} × 1 ^{4}⁄_{8}

Remember when we explained division as making piles of a fixed size? We did two fraction division problems then.

Review of Two Problems

Shapeshifting #23.4 ÷^{1}⁄_{2}=Imagine having 4 pieces of paper. We rip each in half while setting those halves down as "piles". We get 8 piles.

So 4 ÷

^{1}⁄_{2}= 8

Shapeshifting #24.^{1}⁄_{2}÷^{1}⁄_{4}=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

^{1}⁄_{2}÷^{1}⁄_{4}= 2

Now what is happening to the numerators and denominators?

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

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

Fraction Division

To divide fractions, flip the second fraction and multiply.

(Remember to reduce if needed.)

Let's review fraction division with examples.

**11.** Divide one-half by three-quarters.

**12.** Divide three-fifths by one-tenth.

**13.** Divide six-ninths and two-thirds.

**14.** Divide ten-fifteenths and six-eighths.

**15.** Divide 1 ^{2}⁄_{3} ÷ 1 ^{2}⁄_{8}

Prealgebra Textbookby College of the Redwoods MathematicsEvaluating an Expression in §3.2

Multiplying and Dividing Mixed Fractions in §4.5

Prealgebraby Santa Ana CollegeExpressions vs. Equations in §2.1

Multiplying and Dividing Fractions in §4.2

Arithmeticby OpenText

Bittinger Chapter Tests, 11th Edition

Chapter 2 Test, Problem 24: Simplify:

^{4}⁄_{3}× 24Chapter 2 Test, Problem 25: Simplify: 5 ×

^{3}⁄_{10}Chapter 2 Test, Problem 26: Simplify:

^{2}⁄_{3}×^{15}⁄_{4}Chapter 2 Test, Problem 27: Simplify:

^{22}⁄_{15}×^{5}⁄_{33}Chapter 2 Test, Problem 31: Simplify:

^{1}⁄_{5}÷^{1}⁄_{8}Chapter 2 Test, Problem 32: Simplify: 12 ÷

^{2}⁄_{3}Chapter 2 Test, Problem 33: Simplify:

^{24}⁄_{5}÷^{28}⁄_{15}

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 ^{3}⁄_{4} 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.

**16.** Can we change either?

Yes.

We could change both to halves. Then we would say they ate 1 ^{1}⁄_{2} halves.

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

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

The answer "3 quarters" is a lot more friendly than the answer "1 ^{1}⁄_{2} 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.

**18.** Use 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.

We get stuck because a portion of the half is already shaded! What should we do?

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 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.

**19.** Change one-twelfth and one-sixth to have common denominators.

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.

**20.** Change one-fourteenth and one-fifteenth to have common denominators.

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.

Relatively Prime

Two or more numbers are

relatively primeif they have no common factors except for 1.

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.

**21.** Change one-twelvth and one-fifteenth to have common denominators.

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 must reduce the answer more than otherwise.

We want to use the **least common denominator** when adding fractions.

We want to use the number that is the **least common multiple** of the starting denominators.

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.

**22.** What is the least common multiple of 24, 36, and 48?

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.

**23.** Use the brute force method to add ^{1}⁄_{30} + ^{1}⁄_{42}

The second method is to make a **list of multiples** for each denominator. The smallest number on both lists will be the optimal denominator. This method is slow, but is favored by some students who prefer to work with multiples instead of factors.

**24.** Use the list of multiples method to add ^{1}⁄_{30} + ^{1}⁄_{42}

The third method is the **prime factorization method**. After using a factor tree to find the prime factorization of each denominator we find the least common multiple quickly by overlapping the two prime factorizations as much as possible.

Multiples of 30: 30, 60, 90, 120, 150, 180, **210**, 240

Multiples of 42: 42, 84, 126, 168, **210**, 252

**25.** Use the prime factorization method to add ^{1}⁄_{30} + ^{1}⁄_{42}

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

**26.** Add ^{1}⁄_{36} + ^{1}⁄_{6}

**27.** Subtract ^{3}⁄_{4} − ^{1}⁄_{12}

**28.** Add ^{1}⁄_{5} + ^{1}⁄_{4}

**29.** Add ^{1}⁄_{12} + ^{2}⁄_{15}

**30.** Subtract ^{5}⁄_{42} − ^{1}⁄_{24}

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 which factors are missing from among more than two numbers.

**31.** Add ^{1}⁄_{4} + ^{1}⁄_{6} + ^{1}⁄_{15}

Prealgebra Textbookby College of the Redwoods Mathematics

Prealgebraby Santa Ana CollegeAdd and Subtract Fractions with Common Denominators in §4.4

Add and Subtract Fractions with Different Denominators in §4.5

Arithmeticby OpenTextAdding and Subtracting Fractions and Mixed Numbers with Like Denominators in §2.5

Adding and Subtracting Fractions and Mixed Numbers with Unlike Denominators in §2.6

Bittinger Chapter Tests, 11th Edition

Chapter 3 Test, Problem 3: Simplify:

^{1}⁄_{2}+^{5}⁄_{2}Chapter 3 Test, Problem 4: Simplify:

^{7}⁄_{8}+^{2}⁄_{3}Chapter 3 Test, Problem 5: Simplify:

^{7}⁄_{10}+^{19}⁄_{100}+^{31}⁄_{1,000}Chapter 3 Test, Problem 6: Simplify:

^{5}⁄_{6}−^{3}⁄_{6}Chapter 3 Test, Problem 7: Simplify:

^{5}⁄_{6}−^{3}⁄_{4}Chapter 3 Test, Problem 8: Simplify:

^{17}⁄_{24}−^{1}⁄_{15}

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.

**32.** You cut 2 ^{1}⁄_{4} inches off a 3 ^{1}⁄_{2} inch board. How long is the remaining part? (Borrow if needed, like you are thinking about place value.)

**33.** You cut 3 ^{1}⁄_{2} inches off a 5 ^{1}⁄_{4} inch board. How long is the remaining part? (Borrow if needed, like you are thinking about place value.)

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

**34.** You cut 2 ^{1}⁄_{4} inches off a 3 ^{1}⁄_{2} inch board. How long is the remaining part? (Use improper fractions.)

**35.** You cut 3 ^{1}⁄_{2} inches off a 5 ^{1}⁄_{4} inch board. How long is the remaining part? (Use improper fractions.)

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.

Prealgebra Textbookby College of the Redwoods MathematicsAdding and Subtracting Mixed Fractions in §4.6

Prealgebraby Santa Ana CollegeAdd and Subtract Mixed Numbers in §4.6

Arithmeticby OpenTextAdding and Subtracting Fractions and Mixed Numbers with Like Denominators in §2.5

Adding and Subtracting Fractions and Mixed Numbers with Unlike Denominators in §2.6

Bittinger Chapter Tests, 11th Edition

Chapter 3 Test, Problem 19: Simplify: 10

^{1}⁄_{6}− 5^{7}⁄_{8}

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 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

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!

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!

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

termis 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.)

**36.** How many terms are in this expression?

(4 + 2) ÷ 3 + 8 × 2^{2} =

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

**37.** How many terms are in this expression?

^{1}⁄_{2} × 4 × 5 − 3 × (9 − 1) ÷ 2^{2} + 5 =

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

**38.** How many terms are in this expression?

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

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 + 2^{2}) and 3^{2} + 2^{2}. Wanting to square the sum of 3 + 2 demanded putting parenthesis around that sum.

Now we should actually simplify those three expressions.

**39.** Simplify this expression.

(4 + 2) ÷ 3 + 8 × 2^{2} =

2 + 32 = **34**

**40.** Simplify this expression.

^{1}⁄_{2} × 4 × 5 − 3 × (9 − 1) ÷ 2^{2} + 5 =

10 − 6 + 5 = **9**

**41.** Simplify this expression.

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

11 + 16 − 2 + 5 − 10 = **20**

As one last comment about terms, remember that we can cancel *factors* but we cannot cancel *terms*.

Prealgebra Textbookby College of the Redwoods MathematicsAddition and Subtraction in §1.2

Multiplication and Division in §1.3

Order of Operations with Fractions in §4.7

Prealgebraby Santa Ana CollegePEMDAS explained carefully in §2.1

Arithmeticby OpenTextYou might have noticed that this third OER likes to use parenthesis to show multiplication, as well as a grouping structure. It does point this out on page 1.83 and warn you to be alert.

Bittinger Chapter Tests, 11th Edition

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

Chapter 1 Test, Problem 42: Simplify: 10

^{2}− 2^{2}÷ 2Chapter 1 Test, Problem 43: Simplify: (25 − 15) ÷ 5

Chapter 1 Test, Problem 44: Simplify: 2

^{4}+ 24 ÷ 12Chapter 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)]

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".

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

**43.** 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?

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

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

350 total × ^{3}⁄_{7} are boys = 150 boys

**45.** 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?

^{3}⁄_{8} remaining × ^{1}⁄_{2} of that = ^{3}⁄_{16} of a pizza

The other 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.

**46.** What is 40% of 200?

0.4 × 200 = 80

**47.** What is 60 percent of 200?

0.6 × 200 = 120

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?

Let's practice using those shortcuts.

**48.** A restaurant meal costs $25. How much is a 20% tip?

**49.** A restaurant meal costs $20. How much is a 15% tip?

**50.** A restaurant meal costs $30. How much is a 15% tip?

Notice that the word "of" did not appear in the previous three problems. But we could rephrase the problems to ask "What is 20% of $25?" and "What is 15% of $20?" and "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.

**51.** A restaurant meal costs $39.80. Estimate a 15% tip?

**52.** A restaurant meal costs $23.75. Estimate a 20% tip?

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

**53.** A book normally costs $12. It is on sale for 20% off. How much is the reduction?

**54.** A dress normally costs $60. It is on sale for 30% off. How much is the reduction?

Notice that the word "of" did not appear in the previous two problems. But we could rephrase the problems to ask "What is 20% of $12?" and "What is 30% of $60?"

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

Prealgebra Textbookby College of the Redwoods MathematicsFind a Given Percent of a Given Number in §7.2

Prealgebraby Santa Ana CollegeSolve General Applications of Percent in §6.2

Arithmeticby OpenText

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.

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

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

**56.** A candy bar that normally costs 80¢ is on sale for 60¢. What is the percent of the 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 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.

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

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

We already saw one issue to keep us on our toes: when the problem asks for the change as a percentage, does it show or hide the change as a number?

We also saw (earlier) problems that asked for the change as a number. The matching issue is whether the problem wants merely that numeric change or if we must combine it with the original amount.

**58.** You put $200 in a money market account for a year and earn 3% annual interest. How much interest is earned during the year? (Find the interest, then stop.)

**59.** You put $200 in a money market account for a year and earn 3% annual interest. How much is the account worth at the end of the year? (Find the interest, then add it to the original.)

**60.** A $100 loan lasts one year. It has a 12% annual interest rate. What is the total amount paid at the end of this loan?

$100 loan × 0.12 rate = $12 interest. Then we add together the two amounts to pay. $100 loan + $12 interest = **$112 total paid**

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

Prealgebra Textbookby College of the Redwoods MathematicsFind a Percent Given Two Numbers in §7.2

Prealgebraby Santa Ana CollegeSolve General Applications of Percent in §6.2

Arithmeticby OpenText

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?

Notice that in the previous problem, when we calculated $100 prinicipal × 0.12 rate = $12 interest, 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 prinicipal × **1**.12 rate = $112 total paid

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 loan amount around, and doing × 0.12 to find the interest.

**61.** A $5,000 loan lasts one year. It has an 8% annual interest rate. What is the total amount paid at the end of this loan?

$5,000 loan × 1.08 rate = **$5,400 total paid**

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.

**62.** An investment of $200 decreases by 15%. What is the new value?

$200 × (1 − 0.15) = $200 × 0.85 = $170

Prealgebra Textbookby College of the Redwoods MathematicsBalance Formula Using Simple Interest in §7.5

Prealgebraby Santa Ana Collegenone

Arithmeticby OpenTextnone

**Interest** is a fee the lender charges the borrower.

For your savings account or a government bond, you are loaning money to the bank or government, and you collect the interest. For your debts, a bank or credit card is loaning money to you, and they collect the interest.

The initial amount of a loan is called the **principal**.

The loan with the least math would borrow money for one year, and at the end of that year the borrower pays back both the loan and interest.

**63.** A $500 loan lasts one year. It has a 2% annual interest rate. What is the total amount paid at the end of this annual loan?

63.$500 loan × 1.02 rate =$110 total paid

So far our "story" about interest has been the simplest case:

- A certain amount was loaned (the principal).
- An annual fee is charged (the interest).
- The loan lasts one year, to match the time period that naturally goes with the fee.

Let's modify that third issue. What if the loan lasts multiple years?

For this type of loan it remains true that the fee is only paid once, at the very end of the loan. Then borrower finally pays back both the loan and several years' copies of than annual interest fee.

**64.** A $500 loan lasts 5 years. It has a 2% annual interest rate. What is the total amount paid at the end of this loan?

64.More than one year! We must do this with two steps.First we find the overall interest

5 years × 0.02 interest per year = 0.1 overall interest rate

Then we can use the One Plus Trick.

$500 prinicipal × 1.1 overall rate =

$550 total paid

For that loan the interest payments all stayed the same. That type of loan is called **simple interest**.

Many government bonds do work like that. So do most payday loans. Also, banks sometimes offer "certificates of deposit" that have simple interest. In all these situations, no interest payments are made until the end of the loan. Then the principal is returned along with a copy of the interest fee for each year of the loan.

Let's modify that third issue again. What if the loan lasts less than a year?

What should happen if the borrower desired a loan that lasted only six months?

There are two sensible responses.

Perhaps the person or bank loaning the money would say, "Too bad! The fee is for a one-year loan. Sure, you can pay off the loan early. But you still owe me the fee." That makes perfect sense from the point of view of contract law.

But that would not make a happy customer, and would be bad for business. So the other sensible response is actually what happens in real life with simple interest. If the loan only lasts for half a year, then the borrower only pays half the annual fee.

**65.** A $500 loan lasts six months. It has a 2% annual interest rate. How much interest is owed?

65.First we find the annual interest, as we did before.$500 prinicipal × 0.02 annual rate = $10 annual interest

Then we can divide by 2 since the loan only lasts half a year.

$10 annual interest ÷ 2 =

$5 interest for half a year

Actually, any time span can be converted into years.

**66.** A $2,000 loan lasts for 200 days. It has a 20% annual interest rate. How much interest will be due?

66.Notice that we can write 200 days as^{200}⁄_{365}of a year.$2,000 prinicipal × 0.2 rate = $400 interest per year

$400 interest per year × (200 ÷ 365) ≈

$219.18 interestNotice that the parenthesis in the previous line are not needed for order of operations. They only help our eyes identify their contents as something that used to be written as a fraction.

The previous problem only asked for the interest, *not* for the total payment. When we are only asked to find the interest we can use a formula.

Simple Interest Formula

Simple Interest = Principal × Annual Interest Rate × Years

You can compare the previous problem's work to the formula. Do you see how they match up? The formula is just describing what we were already doing.

Unfortunately, this formula is traditionally written in a more confusing manner.

Traditional Simple Interest Formula

I=P×r×t

Pleae be careful with this version of the formula!

Remember that the interest it gives is

simple interest.Remember that the rate is an

annual interest ratethat needs RIP LOP.Remember that the time is always

measured in yearsand might need unit conversion if initially provided in days, weeks, or months.

Whichever version you use, rember that the formula *only solves for interest*. If the word problem asks for the total, you must add on the principal as a final step.

**67.** A $3,500 loan lasts for 9 months. It has a 4.5% annual interest rate. How much interest will be due? Use the simple interest formula.

67.Notice that we can write 9 months as 0.75 of a year.

I=P×r×t= $3,500 prinicipal × 0.045 rate × 0.75 years =$118.13

**68.** A $100 loan lasts 18 months and has 3% annual simple interest. What is the total amount of interest?

68.First we change the time span into years. 18 ÷ 12 = 1.5 yearsThen we use the formula. We multiply the principal, annual interest rate, and number of years.

Simple Interest =

P×r×t= $100 principal × 0.03 annual rate × 1.5 years =$4.50 total interest

**69.** An $750 bond pays a 2% annual simple interest rate, and matures in five years. How much will it be worth when it matures?

69.Be careful! This problem did not ask for the interest. It asked for the total!So we need to find the principal with the formula, and then add back the principal.

Simple Interest =

P×r×t= $750 principal × 0.02 annual rate × 5 years = $75Then we add $750 principal + $75 interest =

$825 total due

**70.** An $400 loan has a 5% annual simple interest rate and lasts 200 days. How much will the borrower need to repay at the end of that time?

70.Be careful again! This problem also did not ask for the interest. It asked for the total, so we need to find the principal with the formula, and then add back the principal.Simple Interest = $400 principal × 0.05 annual rate × (200 ÷ 365) years ≈ $10.96

Then we add $400 principal + $10.96 interest =

$410.96 total due

Prealgebra Textbookby College of the Redwoods Mathematics

Prealgebraby Santa Ana CollegeSolve Simple Interest Applications in §6.4

Arithmeticby OpenText

Bittinger Chapter Tests, 11th Edition

Chapter 6 Test, Problem 14: What is the simple interest on a principal of $120 at the annual interest rate of 7.1% for one year?

Chapter 6 Test, Problem 15: A city orchestra invests $5,200 at 6% annual simple interest. How much is in the account after half a year?

InspireMath Tutorials

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 Analysisuses 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

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.

**71.** How many feet is 9.6 yards?

**72.** How many inches is 3 yards?

**73.** How many feet is 1 inch?

**74.** How many inches is 6 ^{1}⁄_{3} yard?

**75.** How many pints is 11 gallons?

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.

**76.** 3.171 quarts is how many liters? (Use 1 liter = 1.057 quarts.)

**77.** 3.171 quarts is how many liters? (Use 1 quart = 0.946 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.

Prealgebra Textbookby College of the Redwoods MathematicsUnit Conversion: American System in §6.3

Prealgebraby Santa Ana CollegeSystems of Measurement in §7.5

Arithmeticby OpenText

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?

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

Perimeter

The

perimeterof a shape is the distance around its edges.

Area

The

areaof 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

Polygonsare closed shapes whose edges are straight lines.

Here are a bunch of polygons: squares, rectangles, triangles, parallelograms, and trapezoids.

**78.** Draw a polygon with seven edges.

**79.** Draw a shape with four edges that is not a polygon.

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.

**80.** Find the perimeter of these two polygons.

There are only two possible ways to be tricky.

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

**81.** A rectangle has sides 2.5 feet long and 20 inches wide. What is its perimeter in 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:

**82.** 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.

**82.** 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.

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.

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.

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.

**83.** 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?

**84.** 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* = ^{1}⁄_{2} × *length* × *width* for the area of a triangle. Instead of "length" and "width" we call those measuremets "base" and "height".

Why does *Area* = ^{1}⁄_{2} × *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.

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.

**85.** How can you make this parallelogram out of a rectangle? How much of the rectangle does the parallelogram use?

One cut and slide will make a rectangle out of a parallelogram.

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.

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.

**86.** 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*.

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?

Prealgebra Textbookby College of the Redwoods Mathematics

Prealgebraby Santa Ana College

Arithmeticby OpenText

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

^{7}⁄_{8}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.

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:

Consider this small square, of area *A* = *r*^{2}.

If we make four of them, the area is then *A* = 4 × *r*^{2}.

Now we ask a second question about circles: how large a circle can we fit into the big
square of area 4 × *r*^{2}?

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 × *r*^{2} and *A* = 4 × *r*^{2}.

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 × radius^{2}

*A* = π × *r*^{2}

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.

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 ^{22}⁄_{7}). 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.

**87.** Find the perimeter and area of this circle.

**88.** Find the perimeter and area of this circle.

**89.** Find the perimeter and area of this circle.

**90.** Find the perimeter and area of this shape.

Prealgebra Textbookby College of the Redwoods MathematicsThe Circle in §5.3 (hidden in the middle of the section)

Prealgebraby Santa Ana CollegeFind the Circumference and Area of Circles in §5.3 (again hidden in the middle of the section)

Arithmeticby OpenTextCircumference and Area of Circles in §3.3.1 (also hidden in the middle of the section, what's up with not giving circles their own section?)

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?