PDF Lesson 4 Problem Solving: Solving Word Problems Using Unit Rates

4 Lesson

Problem Solving:

Solving Word Problems Using Unit Rates

Problem Solving: Solving Word Problems Using Unit Rates

How much for just one?

Often we may want to know the value of just one of something. The value of one of something is called the unit rate.

We use this concept often without actually thinking about unit rate. For instance, in the grocery store, it is easier if we compare the cost of just one unit, such as an ounce or a pound.

If we see a sign advertising four boxes of cereal for $10, we might ask ourselves how much it costs for just one box.

Let's look at a problem involving unit rate.

Example 1 Find the unit rate. Problem:

It costs $6 to buy 3 cartons of milk. What is the cost of 1 carton?

Set up a proportion with a variable.

Cost Carton

$6 3

=

x 1

x is the cost for 1 carton.

Complete the proportion by finding the value of x.

Cost Carton

$6 3

=

$2 1

One carton of milk costs $2. The unit rate is $2 per carton.

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Another way we talk about unit rate is when we use the term miles per hour. This term means the number of miles we travel in one hour. Miles per hour is a unit rate. It's the value of just one of something every one hour. Example 2 shows this situation.

Example 2

Find the unit rate.

Problem: Quentin's dad drove him to a soccer tournament. They drove 240 miles in 4 hours. About how far did they drive in just 1 hour? What was the rate in miles per hour?

Set up a proportion with a variable.

Miles Hour

240 4

=

m 1

m is the miles traveled in 1 hour.

Complete the proportion by finding a value for m.

Miles Hour

240 4

=

60 1

We see now that m = 60. This means Quentin and his dad drove 60 miles in 1 hour. Another way of saying this is "60 miles per hour."

The unit rate is 60 miles per hour.

Notice that each proportion starts with the units written in words.

Writing the words out is a good habit to practice with this type of problem. We want to remind ourselves what the numbers stand for. That way, when we solve for a variable, we know exactly what that variable represents. In the example about Quentin and his dad, the variable m stands for miles. In the previous example, the variable x stands for the cost in dollars.

Example 3 shows another problem involving unit rate.

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Example 3

Find the unit rate.

Problem: We are at the state fair. We use tickets to pay for the rides. Each ride requires the same number of tickets. We can take 4 rides on 20 tickets. How many tickets does it take for 1 ride?

Set up a proportion with a variable.

Tickets Ride

20 4

=

t 1

t stands for the number of tickets.

Complete the proportion by finding the value for t.

Tickets Ride

20 4

=

5 1

We need 5 tickets to take 1 ride.

We can see that t = 5.

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How do we solve word problems using unit rates?

Sometimes we have to compute the unit rate for something to find the

better deal. We usually expect that items marked "3 for $

"

or "5 for $

" are better deals than buying just one item.

However, this is not always the case.

To compare these different pricing methods, we find the unit rate. Example 1 shows how we analyze this type of situation to determine the better deal.

Example 1

Find the unit rate.

Problem: Monica needs soup. The store has a special--5 cans for $10. If she buys just 1 can of soup, the cost is $2.20. Which is the better deal?

We compare the cost of 1 can of soup for $2.20 to the special deal of 5 cans for $10. We set up a proportion to find the unit rate.

Cost Number of Cans

$10 5

=

x 1

? First we ask ourselves, "1 ? ? = 5?" The answer is 5.

? Then we need to find the value for x in the statement, "x ? 5 = $10."

It's $2. We complete the proportion by filling in the value for x.

Cost Number of Cans

$10 5

=

$2 1

The unit price is $2.

We now compare the unit rate of $2 to the cost for 1 can--$2.20. Buying soup at 5 for $10 is the better deal because it's $2 per can, not $2.20 per can.

Grocery stores often use this pricing method. We expect items marked

"3 for $

" to be the best deals. However, that is not always

the case.

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Example 2 shows a different situation. Sometimes the special pricing methods are not the best deal.

Example 2

Find the unit rate. Problem:

Marcus needs to wear ties for his new job. At the department store, ties are $19 each or 3 for $60.

We need to compare the cost of just 1 tie to the special deal by finding

the unit rate. We set up the proportion like this:

Cost Number of Ties

$60 3

=

x 1

? First we ask ourselves, "1 ? ? = 3?" The answer is 3.

? Then we need to find the value for x in this statement, "x ? 3 = $60."

The cost is $20. We complete the proportion by filling in the value for x.

Cost Number of Ties

$60 3

=

$20 1

When we compare the two pricing methods, we see that the "special deal" is not the better deal.

The unit rate is $20 per tie.

If we buy the ties individually, they are $19 per tie.

Problem-Solving Activity

Turn to Interactive Text,

page 108.

Reinforce Understanding

Use the mBook Study Guide to review lesson concepts.

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