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Introduction

Keeping Track of Books

Manuel loves to read. He found a series of mystery books that take place in the world of medieval knights, and he has been working his way through the series. There are twelve books in the series.

After the first 8 weeks of school, Manuel has finished 5 of the 12 books.

“How many have you finished?” his sister Sarina asked at breakfast.

“I have read 5 out of 12,” Manuel said. “I have seven more to go to finish the series.”

“How long is that going to take you?” Sarina asked.

Manuel had to think for a few minutes about this. Then he took out his notebook to do some figuring.

Think about this with Manuel. At this rate, how many books is he reading per week? How many weeks will it take him to read all 12?

To figure out this problem, you will need to know about rates and unit rates. In this lesson, you will learn everything that you need to know to solve this problem. By the end, you will be able to help Manuel determine how many more weeks it will take him to complete the series.

What You Will Learn

By the end of this lesson, you will be able to complete the following:

  • Identify and write equivalent rates.
  • Find unit rates.
  • Compare unit rates.
  • Model and solve real-world problems using simple equations involving rates.

Teaching Time

I. Identify and Write Equivalent Rates

What is a rate?

A rate is a special kind of ratio. It compares two different types of units, such as dollars and pounds.

Suppose you are buying turkey at a supermarket, and you pay $12 for 2 pounds of turkey. That is an example of a rate. The turkey you bought cost $6 per pound for the turkey. That is another example of a rate.

Notice the word “per”. This word signals us that we are talking about a rate.

We use rates all the time. We use them when shopping, for example, the term "per pound". We use them with gasoline, when we say "per mile". We use them with pricing when we say "$4.00 per yard of material".

Sometimes, two rates are equivalent or equal.

How can we tell if two rates are equivalent?

Two rates are equivalent if they show the same relationship between two units of measure. We can use the same strategies to find equivalent rates that we used to find equivalent ratios.

Example

Determine if these two rates are equivalent: 40 miles in 2 hours, and 80 miles in 4 hours.

You can think of this as the distance and time of two different cars. Did they both travel at an equal rate?

First, express each rate as a fraction. Be sure to keep the terms consistent. That is, if the first ratio compares miles to hours, the second ratio should also compare miles to hours.

40 \ miles \ \text{in}\ 2 \ hours &=  \frac{40 mi}{2h} = \frac{40}{2}\\80 \ miles \ \text{in} \  4 \ hours &=  \frac{80 mi}{4h} = \frac{80}{4}

In order to compare them, the denominators need to match. We are going to change the ratio, \frac{40}{2}, to a ratio with 4 as the second term, or denominator.

Since 2 \times 2 = 4, multiply both terms of the ratio \frac{40}{2} by 2.

\frac{40}{2} = \frac{40 \times 2}{2 \times 2} = \frac{80}{4}

This shows that the ratio \frac{80}{4} is equivalent to the ratio \frac{40}{2}.

This means that the rate 80 miles in 4 hours is equivalent to the rate 40 miles in 2 hours.

The two cars traveled at the same rate.

You can also cross multiply to determine if two rates are equivalent. Let’s look at an example where this strategy is applied.

Example

The machine wound 5 meters of wire in three seconds. A second machine wound 20 meters of wire in 18 seconds. Did they wind the wire at the same rate?

Determine if these two rates are equivalent: 5 meters every 3 seconds and 20 meters every 18 seconds.

First, cross multiply to determine if the rates are equivalent or not.

\frac{5m}{3 \sec} &\overset{?}{=} \frac{20 m}{18 \sec}\\\frac{5}{3} &\overset{?}{=} \frac{20}{18}\\3 \times 20 &\overset{?}{=}  5 \times 18\\60 &\overset{?}{=}90\\60 &\neq 90

Since the cross products are not equal, the rates are not equivalent.

5E. Lesson Exercises

Determine if each of the rates are equivalent to each other.

  1. 3 feet in 9 seconds and 6 feet in 18 seconds
  2. 5 miles in 30 minutes and 6 miles in 42 minutes
  3. 5 pounds for $20.00 and 8 pounds for $32.00

Take a few minutes to check your work with a peer.

II. Find Unit Rates

There is a special kind of rate which is called a unit rate. A unit rate has a denominator of 1, meaning that it is the measure for one of whatever you are talking about: 1 mile, 1 pound, 1 foot, etc. When we talk about unit rates, we are talking about singular measurement and single rates.

Example

1 pound of apples is $1.79.

This is a unit rate. It tells us that for 1 pound of apples, we will pay $1.79. Based on this number, we can figure out what we will pay for 2 pounds, 3 pounds, 5 pounds, etc.

This problem told us what the unit rate was, but sometimes you will need to figure out the unit rate. How do we figure out the unit rate?

To figure out the unit rate, we have to write the ratio as a fraction with a denominator of one.

Example

Rochelle ran 15 miles in 2 hours. Express her speed as a unit rate.

To express this as a unit rate, we need to figure out Rochelle’s speed for one mile. Begin by setting up the rate as a fraction.

15 \ miles \ \text{in}\ 2 \ hours = \frac{15mi}{2h} = \frac{15}{2}

The second number in a unit rate is 1. Since 2 \div 2 = 1, you can divide both terms by 2 to write this as a unit rate.

\frac{15mi}{2h} = \frac{15}{2} = \frac{15 \div 2}{2 \div 2}

Use long division to divide 15 by 2.

& \overset{ \ \ 7.5}{2 \overline{ ) {\ 15.0 \;}}}\\& \quad \underline{-14}\\& \quad \ \ 10\\& \ \ \ \underline{-10}\\& \qquad \ 0 So, \frac{15 mi}{2h} = \frac{15 \div 2}{2 \div 2} = \frac{7.5}{1} = \frac{7.5mi}{1h} or 7.5 miles per hour.

The unit rate is 7.5 miles per hour.

A quicker strategy for finding a unit rate is to simply divide the first term by the second term.

To find the unit rate for Example 3, you could have simply divided 15 miles by 2 hours to find the miles per hour. Actually, that that is exactly what we did. The method that we used showed why we can simply divide the numerator by the denominator.

Try this out on this example.

Example

Kyle ran 4 miles in 28 minutes. How fast did he run per mile?

“Per mile” let’s us know that we are looking for his rate for 1 mile. That is the unit rate.

We write a fraction ratio and divide.

& \frac{min}{miles} = \frac{28}{4}\\& 28 \div 4 = 7\\&\frac{7}{1}

Kevin ran 1 mile in 7 minutes. That is the unit rate.

5F. Lesson Exercises

Find each unit rate.

  1. 5 pounds of apples for $2.00
  2. 45 miles for 3 gallons of gasoline
  3. 18 inches in 2 minutes

Take a few minutes to check your work with a neighbor.

III. Compare Unit Rates

Sometimes, we may need to compare unit rates. For example, you may want to compare two unit prices to determine which one is the better buy.

Example

Felicia needs to buy sugar. She could buy a 16-ounce box of sugar for $1.12, or she could buy a 24-ounce box of sugar for $1.44. Which is the better buy? How many cents per ounce cheaper is that buy?

Find the unit price, per ounce, for the 16-ounce box. Remember, you can find the unit price by dividing the first term by the second term.

\$1.12 \ \text{for}\ 16 \ oz = \frac{\$1.12}{16 oz} && & \overset{ \ \ \$0.07}{16 \overline{ ) {\$1.12 \;}}}\\&&& \quad \underline{-1.12}\\&&& \qquad \quad 0

Find the unit price, per ounce, for the 24-ounce box.

\$1.44 \ \text{for} \ 24 \ oz = \frac{\$1.44}{24oz} && & \overset{ \ \ \$0.06}{24 \overline{ ) {\$1.44 \;}}}\\&&& \quad \underline{-1.44}\\&&& \qquad \quad 0

Since $0.06 < $0.07, the 24-ounce box is cheaper, at $0.06 per ounce.

Subtract to find how much cheaper per ounce the better buy is $0.07 - $0.06 = $0.01

The 24-ounce box of sugar is the better buy. Felicia will pay $0.01 less per ounce if she buys the 24-ounce box.

Once you can figure out the unit rate, you can easily compare which unit rate is greater and which is less. Let’s continue looking at how to apply unit rates to real-life examples in the next section.

IV. Model and Solve Real-World Problems Using Simple Equations Involving Rates

We can use this information to solve real-world problems. Remember that when you are working with a problem, we read the problem first and then write an equation that we can solve.

Example

Maria ran 5 miles in 60 minutes. Her brother Mario ran the same distance in half the time. What is Mario’s rate?

To figure this out, first notice that we are looking for Mario’s unit rate. To find this, we need to start with figuring out Maria’s. We know that Mario ran the same distance in half the time, so we can say that Mario’s unit rate is one-half of Maria’s. Mario runs 1 mile in one-half the time it takes Maria to run one mile.

\frac{1}{2} (Maria’s unit rate) = Mario’s unit rate

First, we need to figure out Maria’s unit rate.

\frac{60 \ min}{5 \ miles} = \frac{12 \ min}{1 \ mile}

Maria runs 1 mile in 12 minutes.

Mario runs the same distance in half the time, so he runs one mile in 6 minutes.

Now let’s apply what we have learned about rates to our introduction problem.

Real-Life Example Completed

Keeping Track of Books

Here is the original problem once again. Reread it and underline any important information.

Manuel loves to read. He found a series of mystery books that take place in the world of medieval knights, and he has been working his way through the series. There are twelve books in the series.

After the first 8 weeks of school, Manuel has finished 5 of the 12 books.

“How many have you finished?” his sister Sarina asked at breakfast.

“I have read 5 out of 12,” Manuel said. “I have seven more to go to finish the series.”

“How long is that going to take you?” Sarina asked.

Manuel had to think for a few minutes about this. Then he took out his notebook to do some figuring.

Think about this with Manuel. At this rate, how many books is he reading per week? How many weeks will it take him to read all 12?

There are two questions here to figure out. Let’s tackle the first one. We need to figure out how many books Manuel is reading per week. The word “per” lets us know that we want to figure out the number of books for one week. Let’s write a ratio to compare books read to weeks.

\frac{5 \ books}{8 \ weeks}

Next, we need to figure out how many books he read per week. To do this, we divide 5 by 8.

\overset{ \ \ .625}{8 \overline{ ) {5.0 \;}}}

Manuel reads .625 books per week. We could also say that he reads a little more than \frac{1}{2} book per week.

Given this rate, how long will it take him to read 12 books? We can set up a pair of ratios to figure this out.

\frac{.625 \ books}{1 \ week}=\frac{12 \ books}{x \ weeks}

If we divide 12 by 1.6, we will have the number of weeks.

\overset{ \ \qquad 19.2}{.625 \overline{ ) {12.0 \;}}}

Manuel will finish the series in 19.2 weeks. Since he has already been reading for 8 weeks, he can subtract this number from 19.2. 19.2 - 8 = 11.2. Manuel will finish the series in 11.2 more weeks.

Vocabulary

Rate
A special kind of ratio that compares two different quantities.
Per
a word that signals you that a rate is being used.
Unit Rate
a rate that is compared to 1. A rate can be for 1 pound, 1 mile, 1 second, 1 of any unit.

Technology Integration

James Sousa, Rates and Unit Rates

James Sousa, Determine Unit Rate

James Sousa, Determine a Unit Rate in Gallons per Minute

James Sousa, Determine the Best Buy Using Unit Rate

Time to Practice

Directions: Write an equivalent rate for each rate.

1. 2 for $10.00

2. 3 for $15.00

3. 5 gallons for $12.50

4. 16 pounds for $40.00

5. 18 inches for $2.00

6. 5 pounds of blueberries for $20.00

7. 40 miles in 80 minutes

8. 20 miles in 4 hours

9. 10 feet in 2 minutes

10. 12 pounds in 6 weeks

Directions: Now take each of the rates in numbers 1–10 and find the unit rate for each.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Directions: Determine whether each is equivalent or not.

21. 60 units in 3 minutes and 80 units in 4 minutes

22. $16 for 8 pounds and $22 for 10 pounds

23. 50 kilometers in 2 hours and 75 kilometers in 3 hours

Directions: Solve each problem.

24. Max paid $45 for 15 gallons of gasoline. What was the cost per gallon of gasoline?

25. Mr. Brown paid $8.28 for 12 cans of green beans. Express this cost as a unit price.

26. A train travels 480 kilometers in 3 hours. Express this speed as a unit rate.

27. Mrs. Jenkins paid $50 for 40 square feet of carpeting. What was the cost per square foot of carpeting?

28. A copy machine can produce 310 copies in 5 minutes. How many copies can the machine produce per minute?

29. Nadia needs to buy some cheddar cheese. An 8-ounce package of cheddar cheese costs $2.40. A 12-ounce package of cheddar cheese costs $3.36.

a. Find the unit price for the 8-ounce package.

b. Find the unit price for the 12-ounce package.

c. Which is the better buy? How many cents per ounce cheaper is the better buy?

30. Joe drove 141 miles in 3 hours. His cousin Amy drove 102 miles in 2 hours. Assume both cousins were driving at constant speeds.

a. How fast was Joe driving, in miles per hour?

b. How fast was Amy driving, in miles per hour?

c. Who was driving at a faster rate of speed? How much faster?  

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Feb 22, 2012

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