Rate Problems
Rate problems use a known rate to find an amount, a time or a cost. In NSW Year 7 maths, students write the rule first, substitute, and let the units guide each step (NESA outcome MA4‑RAT‑C‑01).
You will use \(\text{amount} = \text{rate} \times \text{time}\), work backwards to find a time, apply the unitary method, calculate fuel use from L/100 km, add rates when two hoses or machines work together, and find pay at time and a half. Filling a pool, planning a long road trip or working a Sunday shift all need these steps. This subtopic is part of Year 7 Ratios, Rates and Time.
Every question with a fully worked solution.
- Rate Problems - Video - Rate problems Watch
Theory
A rate problem uses a rate to find an amount, a time or a cost. Write the rule first, then substitute, with the units matched: at \(800\) km/h, an aircraft travels \(80\) km in \(6\) minutes. This Year 7 maths page shows the rules, the unitary method, fuel, pay and two things working together.
Find an amount: amount \(=\) rate \(\times\) time. The units show it works: litres per minute \(\times\) minutes gives litres.
Work backwards: time \(=\) amount \(\div\) rate. The same idea finds hours worked from pay, or hectares from kilograms of fertiliser.
Unitary method. Given an amount for several units, find the amount for one unit first, then multiply.
Match the units. If the rate is per hour, the time must be in hours: \(20\) min \(= \dfrac{1}{3}\) h. Fuel in L/100 km means litres for every \(100\) km: fuel \(=\) (distance \(\div 100\)) \(\times\) rate.
Two things working together: their rates add. Two hoses at \(15\) L/min and \(25\) L/min fill at \(40\) L/min.
Special rates. Time and a half means \(1.5\) times the normal pay rate: at \(\$20\)/h it is \(\$30\)/h. A rate can be a percentage: losing \(4\%\) of the full amount each year, it is all gone after \(100 \div 4 = 25\) years.
Choose the rule, then check that the units give the right unit:
| You want | Rule | Units check |
|---|---|---|
| an amount | amount = rate × time | L/min × min → L |
| a time | time = amount ÷ rate | L ÷ L/min → min |
| a rate | rate = amount ÷ time | L ÷ min → L/min |
| fuel used | fuel = (distance ÷ 100) × rate | (lots of 100 km) × L/100 km → L |
| a cost or pay | cost = number of units × price per unit | h × \(\$\)/h → \(\$\) |
How to solve a rate problem
- Decide what you want: an amount, a time, a rate or a cost.
- Match the units (minutes to hours, or kilometres to lots of \(100\) km).
- Write the rule, substitute and work it out.
- Check with the units, or with the unitary method.
When two things work together
- Add their rates to get the combined rate.
- Divide the amount by the combined rate.
| \(6 \times 5\) | \(=\) | \(30\) |
9:00 am to 2:00 pm is \(5\) hours, so \(30\) mm of rain falls.
(ii) A fast train travels at \(270\) km/h. How far does it travel in \(40\) minutes?
| \(800 \div 60 \times 6\) | \(=\) | \(80\) |
| \(270 \times \dfrac{2}{3}\) | \(=\) | \(180\) |
(i) \(80\) km (\(6\) min is \(\dfrac{1}{10}\) of an hour)
(ii) \(180\) km
| \(50 \div 2\) | \(=\) | \(25\) |
| \(21 + 25\) | \(=\) | \(46\) |
Losing \(50\%\) takes \(25\) years, so Helen is \(46\).
(i) How much diesel is used?
(ii) How much does the diesel cost?
| \(350 \div 100\) | \(=\) | \(3.5\) |
| \(3.5 \times 12\) | \(=\) | \(42\) |
| \(42 \times 2.10\) | \(=\) | \(88.2\) |
(i) \(42\) L
(ii) \(\$88.20\)
(i) Both pumps run together. How many minutes does it take?
(ii) How much longer would pump A take on its own?
| \(50 + 70\) | \(=\) | \(120\) |
| \(3600 \div 120\) | \(=\) | \(30\) |
| \(3600 \div 50\) | \(=\) | \(72\) |
| \(72 - 30\) | \(=\) | \(42\) |
(i) \(30\) min
(ii) \(42\) min longer
| \(1.5 \times 16\) | \(=\) | \(24\) |
| \(5 \times 16\) | \(=\) | \(80\) |
| \(3 \times 24\) | \(=\) | \(72\) |
| \(80 + 72\) | \(=\) | \(152\) |
\(\$152\)
Common pitfalls
Frequently asked questions
How do you find an amount from a rate?
Multiply the rate by the time. \(6\) mm/h for \(5\) h is \(30\) mm.
How do you find the time from a rate?
Divide the amount by the rate. \(3600\) L at \(120\) L/min takes \(30\) min.
How do you use L/100 km?
Divide the distance by \(100\), then multiply by the rate. \(350\) km at \(12\) L/100 km uses \(3.5 \times 12 = 42\) L.
What is the unitary method?
Find the amount for one unit first, then multiply by the number of units you need.
What happens when two machines work together?
Their rates add. Find the combined rate, then divide the job by it.