Unit Rate – Examples and Its Uses

Key Concepts

• Solving constant speed problems.

• Solving unit price problems.

• Using an equation to represent unit rate problems.

5.7 Solve unit rate problems 

Unit rate: 

Unit rate can be defined as the amount of one quantity for every 1 unit of another quantity. 

Rates can be converted to unit rates. Below is an example depicting how to convert rates to unit rates. 

Note: A unit can also be defined as a rate having 1 in the denominator. 

5.7.1 Solve constant speed problems 

Constant speed: 

Over a period of time travelled, if the speed remains the same, then it is defined as constant speed. 

Example 1: 

A speed boat can go 123 miles in 3 hours, how far can it go in 8 hours? 

Method: 1 

Solution:  

Step 1: Form a table to write rates that are equivalent to 

125  miles 3 hours125  miles 3 hours

Step 2: Divide the rate by 3 to bring it to a unit rate and write equivalent rates. 

Step 3: On analyzing the table, we observe that boat covers 328 miles in 8 hours. 

Method: 2 

Step 1: Write the rate as a fraction:

12331233

Step 2: Divide both terms of the rate by the same non-zero number to find a unit rate. 

123 miles 3 hours=     123 ÷ 3 3 ÷3123 miles 3 hours=     123 ÷ 3 3 ÷3

=

    41 miles 1 hour    41 miles 1 hour

Step 2: Multiply both terms of the rate by 8 to find how far the boat can go in 8 hours. 

41×8 1×8=     328 miles   8 hours41×8 1×8=     328 miles   8 hours

Therefore, we conclude that boat can cover 328 miles in 8 hours. 

Example 2:  

Sam can walk 8 miles in 2 hours. How many miles can he walk in 6 hours? 

Solution:  

Step 1: Form a table to write rates that are equivalent to 

8  miles 2 hours8  miles 2 hours

Step 2: Divide the rate by 2 to bring it to a unit rate and write equivalent rates. 

Step 3: On analyzing the table, we observe that Sam covers 24 miles in 6 hours. 

Method: 2 

Step 1: Write the rate as a fraction:

    8   2    8   2

Step 2: Divide both terms of the rate by the same non-zero number to find a unit rate. 

8 miles 2 hours=      8 ÷ 2    2 ÷ 2

=

    4 miles 1 hour

Step 2: Multiply both terms of the rate by 6 to find how far Sam can go in 6 hours. 

4×6 1×6=     24  miles   6 hours

Therefore, we conclude that Sam can cover 24 miles in 6 hours. 

5.7.2 Solve unit price problems 

Example 1: 

If a family of 8 people paid $44 for the tickets. What would be the cost of tickets for 5 members? 

Method: 1 

Solution:  

Step 1: Form a table to write rates that are equivalent to 

$44  8 persons$44  8 persons

Step 2: Divide the rate by 8 to bring it to a unit rate and write equivalent rates. 

Step 3: On analyzing the table, we observe that $27.5 are charged for 5 persons. 

Method: 2 

Step 1: Write the rate as a fraction:

   $44    8   $44    8

Step 2: Divide both terms of the rate by the same non-zero number to find a unit rate. 

$44 8 persons=      44 ÷ 8    8 ÷ 8$44 8 persons=      44 ÷ 8    8 ÷ 8

=

    $5.5 1 person    $5.5 1 person

Step 2: Multiply both terms of the rate by 5, to find cost of tickets for 5 persons. 

5.5×5 1×5=     $27.55 persons5.5×5 1×5=     $27.55 persons

Therefore, we conclude that $27.5 are charged for 5 persons. 

Example 2:  

 6 books cost $4, find the cost of 30 books. 

Method: 1 

Solution:  

Step 1: Form a table to write rates that are equivalent to 

6 books $46 books $4

Step 2: Divide the rate by 4 to bring it to a unit rate and write equivalent rates. 

Step 3: On analyzing the table, we observe that $30 are charged for 20 books. 

Method: 2 

Step 1: Write the rate as a fraction:

   $6    4   $6    4

Step 2: Divide both terms of the rate by the same non-zero number to find a unit rate. 

$6 4 books=      6 ÷ 4    4 ÷ 4$6 4 books=      6 ÷ 4    4 ÷ 4

=

    $1.5 1 book    $1.5 1 book

Step 2: Multiply both terms of the rate by 20, to find the number of books we can purchase for $30 

1.5×20 1×20=     $3020 books1.5×20 1×20=     $3020 books

Therefore, we conclude that $30 are charged for 20 books. 

5.7.3 Use an equation to represent unit rate problems 

Example 1: 

A car travels at a constant speed of 80 miles in 4 hours. How much distance can the car cover in 6 hours? 

Solution: 

Step 1: Find the unit rate.  

20 miles per hour is the unit rate.  

Step 2: Find the distance.  

Distance = rate × time  

D = 20 × 6 

Distance = 120 miles 

Therefore, distance travelled by the car in 6 hours would be 120 miles

Exercise:

1. Steve paid $14 for 7 crates of oranges. At the same rate how much would 15 crates of oranges cost?

2. A sub-marine travels 20 miles in 2; hour. Write an equation to find how long would it take to travel 57 miles.

 3. A jet plane can fly 1500 miles in 50 minutes. Find the distance it can cover at the same rate in 3 hours.

4. A peacock can run 4 miles in 12 minutes. How far can an ostrich run in 30 minutes?

5. Berries sell for $0.64 per pound. Establish an equation to show the relationship between the total cost, and the number of pounds of berries.

6. A basketball player can run 60 yards in 30 seconds. If he maintains the same rate of speed, how far can he run in 50 seconds?

7. The price of an 8-minute phone call is $2.40. Establish an equation to find the cost of 20-minute phone call.

8. A photographer charges $8 for 36 pictures. At the same rate, how much will he charge for 180 pictures?

9. A motorist rides at the rate 45 miles per hour. How long will the motorist take to ride 225 miles?

10. Cheetah can run 120 miles per hour. How much time will cheetah require to run 30 miles?

What have we learned?

• Solving constant speed problems.

• Solving unit price problems. Using an equation to represent unit rate problems.