Lesson 4. Solving Word Problems Using Unit Rates. How much for just one? Problem Solving: Problem Solving: Solving Word Problems Using Unit Rates

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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.

CartonCost ^$6₃ = ₁^x

Complete the proportion by finding the value of x.

CartonCost ^$6₃ = ^$2₁

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

x is the cost for 1 carton.

Problem Solving: Solving Word Problems Using Unit Rates

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256

Unit 3 • Lesson 4

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.

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?

Miles

Hour ²⁴⁰₄ = ^m₁

Complete the proportion by finding a value for m.

Miles

Hour ²⁴⁰₄ = ⁶⁰₁

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.

m is the miles traveled in 1 hour.

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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?

Tickets

Ride ²⁰₄ = ₁^t

Complete the proportion by finding the value for t.

Tickets

Ride ²⁰₄ = ⁵₁

We need 5 tickets to take 1 ride.

t stands for the number of tickets.

We can see that t = 5.

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258 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.

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.

Number of CansCost ^$10₅ = ₁^x

• 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.

Number of CansCost ^$10₅ = ^$2₁

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.

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:

Number of Ties^Cost ^$60₃ = ₁^x

• 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.

Number of TiesCost ^$60₃ = ^$20₁

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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260

Activity 1

Simplify the ratios.

1. ₁₄⁷

1

2

^2. ¹²³

1

4

^3. ²⁰⁴

1 5

4. ₁₅⁵

1

3

^5. ⁴⁸⁸

1 6

Activity 2

Set up the unit rate problems as proportions. Tell what the variable represents. Show the units in words. Find the unit rate.

1. Nguyen paid $40 for 4 CDs. If all 4 cost the same amount, what was the cost of just 1 CD?

2. Sheldon paid $180 for 6 pairs of contact lenses. What was the cost of just 1 pair of lenses?

3. Britt can do 45 sit-ups in 3 minutes. How many sit-ups can she do in just 1 minute?

Activity 3

Tell the better deal in each case by finding the unit rate.

1. What’s the better deal, 1 apple for $.50, or 3 for $1?

2. What’s the better deal, 1 pair of jeans for $60 or 3 for $200?

3. What’s the better deal, 1 T-shirt for $19 or 2 for $40?

4. What’s the better deal, 1 CD for $15 or 2 for $25?

5. What’s the better deal, 3 juices for $1 or $.75 for 1 juice in the vending machine?

Model ₁₂⁶ Answer: ₁₂⁶ = ¹₂

Model It costs $8 for 4 sandwiches. How much does it cost for just 1 sandwich?

Answer: _Sandwich^Cost ^$8₄ = ₁^x X stands for the cost of 1 sandwich.

SandwichCost $8

4 = ^$2₁ The cost of one sandwich is $2.

Cost CD

$40

4 $10 1

Cost Pair

$180

6 $30 1

Sit

ups

Minutes 45 3 15

1

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Activity 4 • Distributed Practice Solve.

1. ⁴₅ + ²₅ = x

1

5

^2. 35 − w = 19

3. 4.75 + 2.98 = z 4. a + 385 = 410 5. ¹₂ · ³₅ = b

3

10

^6. 139.7 − 48.19 = c

7. ⁴₅ ÷ ¹₅ = d 8. e ÷ 7 = 50