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Year 12 Maths Standard 1 (2027) Ratios and rates

Introduction to Rates & Unit Rates

20 practice questions 0 video lessons Theory + worked examples

Get started with rates for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-05). A ratio compares amounts in the same unit, while a rate compares amounts in different units — such as kilometres per hour, litres per minute or dollars per kilogram.

You will see how a rate differs from a ratio, turn information into a unit rate (the amount for one unit), find a unit price, and use a unit rate to work out how much, how many or how long — everyday skills for pay rates, shopping, typing speed and travel.

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Theory

A rate compares two quantities measured in different units, such as kilometres and hours. A unit rate gives the amount for one unit, like \(90\text{ km/h}\) or \(\$2.50\) per kilogram. This Year 12 Standard 1 (NSW) guide shows how to find a unit rate and use it to work out amounts and times.

A ratio compares quantities in the same unit, such as \(3\) cups of flour to \(1\) cup of sugar. A rate compares quantities in different units, so it always has a unit: \(\text{km/h}\), \(\text{L/min}\), \(\$\)/kg or words/min.

A unit rate is the amount for one unit. Find it by dividing the total amount by the number of units, for example \(180\text{ km}\) in \(2\text{ h}\) is \(90\text{ km/h}\). Then use it: how much? multiply the unit rate by the number of units; how many or how long? divide the amount by the unit rate.

A unit rate for a cost, such as \(\$4.00\) per kg, is called a unit price.

On a graph of cost against mass, each line rises by its unit rate for every \(1\text{ kg}\), so the steeper line has the greater cost per kilogram. This is a core Year 12 Standard 1 (NSW) Ratios and rates skill.

Cost against mass for fruit A and fruit BTwo straight lines from the origin. Fruit B rises to 20 dollars at 5 kg (4 dollars per kg); fruit A rises to 12.50 dollars at 5 kg (2.50 dollars per kg). At 2 kg, B costs 8 dollars and A costs 5 dollars. 1 2 3 4 5 5 10 15 20 m (kg) C ($) B: $4/kg A: $2.50/kg
At \(2\text{ kg}\), B costs \(\$8\) (\(\$4\) per kg) and A costs \(\$5\) (\(\$2.50\) per kg). The steeper line has the greater unit rate.
Double number line: 180 km in 2 hours is 90 km in 1 hourTwo number lines one above the other. Distance 0, 90, 180 km lines up with time 0, 1, 2 hours, showing a unit rate of 90 km per hour. 0 90 180 km 0 1 2 h unit rate: 90 km per 1 h
Halve both lines: \(180\text{ km}\) in \(2\text{ h}\) is the same rate as \(90\text{ km}\) in \(1\text{ h}\).

Find a unit rate by dividing the total by the number of units:

\[\text{unit rate} = \dfrac{\text{total amount}}{\text{number of units}}\]
unit rate=total amountnumber of units

Use the unit rate to find an amount, for example distance \(=\) speed \(\times\) time:

\[\text{amount} = \text{unit rate} \times \text{number of units}\]
amount=unit rate×number of units

Or find the number of units (how many, how long):

\[\text{number of units} = \dfrac{\text{amount}}{\text{unit rate}}\]
number of units=amountunit rate
The “per” unit goes on the bottom. Price per kg \(=\) dollars \(\div\) kilograms; speed in km/h \(=\) kilometres \(\div\) hours.

How to solve a rate problem

  1. Identify the two quantities and their units, and decide which unit comes after “per”.
  2. Convert units if needed (for example grams to kilograms) so they match the rate.
  3. Divide to find the unit rate, then multiply for an amount or divide for a number of units.
  4. Write the unit with your answer and check it makes sense.
Example 1 — Writing a unit rate
A bus travels \(156\text{ km}\) in \(2\) hours at a steady speed. Write its speed as a unit rate in \(\text{km/h}\).
Solution

Divide the distance by the number of hours.

\(\text{speed}\)\(=\)\(\dfrac{156\text{ km}}{2\text{ h}}\)
\(\therefore\ \text{speed}\)\(=\)\(78\text{ km/h}\)
1562=78 km/h

The speed is \(78\text{ km/h}\).

Example 2 — Using a rate to find an amount
A garden sprinkler uses water at \(12\text{ L/min}\). How much water does it use in \(25\) minutes?
Solution

Amount \(=\) unit rate \(\times\) number of minutes.

\(\text{water}\)\(=\)\(12 \times 25\)
\(\therefore\ \text{water}\)\(=\)\(300\text{ L}\)
12×25=300 L

The sprinkler uses \(300\text{ L}\).

Example 3 — How long?
Priya is paid \(\$24.50\) per hour. How many hours must she work to earn \(\$294\)?
Solution

Number of hours \(=\) total pay \(\div\) pay per hour.

\(\text{hours}\)\(=\)\(\dfrac{294}{24.50}\)
\(\therefore\ \text{hours}\)\(=\)\(12\)
\(\text{check: } 12 \times \$24.50\)\(=\)\(\$294\)
29424.50=12

Priya must work \(12\) hours.

Example 4 — Price per kilogram from grams
A \(600\text{ g}\) bag of mixed nuts costs \(\$9.00\). (i) Find the price per kilogram. (ii) Find the cost of \(2.5\text{ kg}\) of the same nuts.
Solution

Change grams to kilograms, divide, then multiply.

\(600\text{ g}\)\(=\)\(0.6\text{ kg}\)
\(\text{(i) price per kg}\)\(=\)\(\dfrac{9.00}{0.6} = \$15.00\)
\(\text{(ii) cost}\)\(=\)\(2.5 \times \$15.00\)
\(\therefore\ \text{cost}\)\(=\)\(\$37.50\)
$15.00,$37.50

(i) \(\$15.00\) per kg; (ii) \(\$37.50\).

Common pitfalls

Divide the right way round. The unit you want “per” goes on the bottom: price per kg \(=\) dollars \(\div\) kilograms, not kilograms \(\div\) dollars.
Match the units first. Change grams to kilograms (or mL to L) when the rate is per kg or per L: \(750\text{ g}=0.75\text{ kg}\).
Always write the unit. A rate needs its unit: \(50\) words/min, not just \(50\).

Frequently asked questions

What is a unit rate?

A unit rate is the amount for one unit of something, such as 90 km per hour, $24.50 per hour or $15 per kilogram. You find it by dividing the total amount by the number of units.

What is the difference between a rate and a ratio?

A ratio compares quantities measured in the same unit, such as 3 cups of flour to 1 cup of sugar, so it has no unit. A rate compares quantities in different units, such as kilometres and hours, so it always has a unit like km/h.

How do you find a unit rate?

Divide the total amount by the number of units, keeping the unit you want 'per' on the bottom. For example 156 km in 2 hours is 156 divided by 2, which is 78 km/h.

How do you use a rate to find an amount?

Multiply the unit rate by the number of units. A sprinkler using 12 litres per minute for 25 minutes uses 12 times 25, which is 300 litres.

How do you find the price per kilogram from grams?

Change the grams to kilograms first, then divide the price by the number of kilograms. A 600 g bag costing $9 is 0.6 kg, and 9 divided by 0.6 is $15 per kilogram.