Rates, Unit Rates & Conversions
Master rates, unit rates and conversions for NSW Year 12 Mathematics Standard 2. A rate compares two quantities with different units, like \(\$/\text{kg}\) or \(\text{km/h}\), and a unit rate gives the amount per one unit — the key to comparing and scaling rates.
You will find unit rates, convert speeds between km/h and m/s using \(3.6\), convert flow rates such as L/h and mL/min, and decide the best buy by comparing the price per common unit — core skills for Standard 2 ratios and rates.
Theory
Rates and conversions are part of the ratios and rates topic in Year 12 Standard 2 (NSW). A rate compares quantities with different units; a unit rate gives the amount per one unit. This guide shows how to find unit rates, convert between rates such as km/h and m/s, and compare best buys using the price per common unit.
A rate compares two quantities measured in different units, so it always carries a unit — for example \(\$/\text{kg}\), \(\text{km/h}\) or \(\text{L/min}\). This is what separates a rate from a ratio, which compares quantities of the same kind and has no unit.
A unit rate gives the amount for one of the second quantity: you find it by dividing, such as \(\$16.50\) per 1 hour or \(16\) km per 1 litre. Unit rates make rates easy to compare and to scale up or down.
Converting a rate changes its units without changing its value. In this Year 12 Standard 2 (NSW) topic you convert speeds between \(\text{km/h}\) and \(\text{m/s}\), convert flow rates such as \(\text{L/h}\) and \(\text{mL/min}\), and use unit rates to decide the best buy.
For quantities \(A\) and \(B\) measured in different units, the unit rate is the amount per one unit of \(B\):
Convert between the two common speed units by \(3.6\):
For a best buy, write every option as a price per common unit and pick the smallest:
How to solve a rate or conversion problem
- Identify the two quantities and their units — this tells you whether it is a rate (different units) and what unit the answer needs.
- Find the unit rate by dividing, so you have the amount per one unit.
- Convert if needed: km/h \(\div 3.6\) for m/s (\(\times 3.6\) back), and change other rates one unit at a time (e.g. L/h \(\times 1000 \div 60\) for mL/min).
- Compare or scale: for a best buy choose the smallest price per common unit; to scale, multiply the unit rate by the new amount.
Best-buy example — orange juice sold in three sizes:
| Size | Price | Price per litre |
|---|---|---|
| \(1~\text{L}\) | \(\$3.00\) | \(\$3.00\) |
| \(2~\text{L}\) | \(\$5.20\) | \(\$2.60\) |
| \(3~\text{L}\) | \(\$8.40\) | \(\$2.80\) |
The \(2~\text{L}\) bottle is the best buy at \(\$2.60\) per litre — the biggest pack is not the cheapest here.
Divide the total pay by the hours for the unit rate, then multiply by the new hours.
| \(\text{pay per hour}\) | \(=\) | \(\dfrac{148.50}{9} = \$16.50\) |
| \(\text{pay for }5\text{ h}\) | \(=\) | \(5 \times \$16.50 = \$82.50\) |
The rate is \(\$16.50\) per hour, so \(5\) hours pays \(\$82.50\).
Speed \(=\) distance \(\div\) time; convert to km/h by multiplying by \(3.6\).
| \(\text{speed}\) | \(=\) | \(\dfrac{200}{25} = 8~\text{m/s}\) |
| \(\) | \(=\) | \(8 \times 3.6 = 28.8~\text{km/h}\) |
The sprinter runs at \(8~\text{m/s}\), i.e. \(28.8~\text{km/h}\).
Find the price per kilogram of each, then compare.
| \(2\text{ kg}\) | \(\to\) | \(\dfrac{7.60}{2} = \$3.80\text{/kg}\) |
| \(5\text{ kg}\) | \(\to\) | \(\dfrac{18.00}{5} = \$3.60\text{/kg}\) |
| \(\$3.60\) | \(<\) | \(\$3.80\) |
The \(5~\text{kg}\) bag is better value at \(\$3.60\) per kg.
Convert to a common unit (litres per hour) to compare, and use time \(=\) volume \(\div\) rate.
| \(\text{(i)}\) | \(=\) | \(18 \times 60 = 1080~\text{L/h}\) |
| \(\text{(ii) time}\) | \(=\) | \(\dfrac{5400}{18} = 300~\text{min} = 5~\text{h}\) |
| \(\text{(iii)}\) | \(\) | \(1200 > 1080~\text{L/h}\) |
(i) \(1080~\text{L/h}\), (ii) \(5\) hours, (iii) the second pump (\(1200~\text{L/h}\)) is faster.
Common pitfalls
Frequently asked questions
What is the difference between a ratio and a rate?
A ratio compares two quantities of the same kind and has no unit, such as 3 to 2. A rate compares quantities of different kinds and always carries a unit, such as 8 litres per 100 km or 60 kilometres per hour.
What is a unit rate?
A unit rate is the amount for exactly one of the second quantity. You find it by dividing, for example $16.50 per one hour or 16 kilometres per one litre. Unit rates make it easy to compare rates and to scale them up or down.
How do you convert km/h to m/s?
Divide the speed in kilometres per hour by 3.6 to get metres per second. For example, 36 km/h divided by 3.6 is 10 m/s. To go the other way, multiply metres per second by 3.6 to get kilometres per hour.
Why do you divide by 3.6 to change km/h into m/s?
Because 1 km/h equals 1000 metres over 3600 seconds, which simplifies to 1 divided by 3.6 metres per second. So dividing km/h by 3.6 converts to m/s, and multiplying by 3.6 converts back.
How do you work out the best buy?
Write each option as a price per common unit, such as dollars per litre or dollars per kilogram, by dividing the price by the quantity. The option with the smallest unit price is the best value, even if it is not the cheapest pack overall.
How do you convert L/h to mL/min?
Convert one unit at a time. Multiply by 1000 to change litres to millilitres, then divide by 60 to change per hour into per minute. For example, 1.2 L/h is 1200 mL/h, which is 20 mL/min.