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Australia
Year 7

5.04 Rates

Lesson

Rates

A rate is a measure of how quickly one measurement changes with respect to another. Some commonly used rates in our everyday lives are speed, which measures distance per time, and the price of food, which is often measured in dollars per kilogram.

When we combine two different units into a single compound unit we call this a unit rate. We can write these compound units using a slash ( / ) between the different units, so "metres per second" becomes "\text{m/s}" and "dollars per kilogram" becomes "\text{\$/kg}". This compound unit represents the division of one measurement by another to get a rate.

Not all compound units are written using a slash and instead use the letter "p" to represent "per". For example, "beats per minute" uses the compound unit bpm and "frames per second" uses fps.

Rates have two components, the numeric value and the compound unit. The compound unit tells us which units are being measured and the numeric value tells us how quickly the numerator unit changes with respect to the denominator unit. When constructing a rate we usually start with just a fraction of measurements. Whenever we can, simplify the fraction to get a whole number value for the rate.

Examples

Example 1

Write the following as a unit rate:

91 people per 7 buses

Worked Solution
Create a strategy

Find the number of people per 1 bus.

Apply the idea

If we divide both numbers in the rate by 7, the rate remains in the same proportion.

\displaystyle \text{Unit rate}\displaystyle =\displaystyle \dfrac{91}{7}Divide 91 by 7
\displaystyle =\displaystyle 13 \text{ people/bus}Simplify
Idea summary

A rate is a measure of how quickly one measurement changes with respect to another.

When we combine two different units into a single compound unit we call this a unit rate.

Apply rates

Now that we know how to make our rates, it's time to use them. Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another.

A rate of 10 metres per second (10 \text{ m/s}) is not the same as a rate of 10 seconds per metre (10\text{ s/m}).

In fact, 10 \text{ m/s}=\dfrac{1}{10} \text{ s/m}. When we flip the compound unit we also need to take the reciprocal of the numeric value.

Examples

Example 2

On a road trip, Tracy drives with an average speed of 90 \text{ km/hr}. How far does she travel in 8 hours?

Worked Solution
Create a strategy

Multiply the rate by 8.

Apply the idea

Tracy travels at 90 \text{ km/h}. This means that the car travels 90 kilometres per 1 hour. So we need to multiply the rate by 8 to find the how far the car travels in 8 hours.

\displaystyle \text{Distance}\displaystyle =\displaystyle 90 \times 8Multiply 90 by 8
\displaystyle =\displaystyle 720 \text{ km}
Idea summary

Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another.

Convert rates

When applying rates it's important to make sure that we are applying the right one.

Examples

Example 3

Consider the following rate:

192 metres per 240 seconds

a

Express this is as a unit rate in terms of metres and seconds.

Worked Solution
Create a strategy

Divide both numbers in the rate by 240.

Apply the idea
\displaystyle \text{Unit rate}\displaystyle =\displaystyle \dfrac{192\text{ m}}{240\text{ sec}}Divide 192 by 240
\displaystyle =\displaystyle \dfrac{4}{5} \text{ m/sec}Simplify
b

Express this is as a unit rate in terms of metres and minutes.

Worked Solution
Create a strategy

Convert the seconds to minutes.

Apply the idea

240 seconds equals 4 minutes, so we can replace the seconds in the original rate.

\displaystyle \text{Unit rate}\displaystyle =\displaystyle \dfrac{192\text{ m}}{240 \text{ sec}}Write the original rate
\displaystyle =\displaystyle \dfrac{192\text{ m}}{4\text{ min}}Write the denominator in minutes
\displaystyle =\displaystyle 48 \text{ m/min}Simplify
Idea summary

Before we apply rates, we should convert the units to the indicated unit rate.

Outcomes

ACMNA173

Recognise and solve problems involving simple ratios

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