Converting Rates
Converting rates means changing the units of a rate without changing the rate itself, for example from kilometres per hour to metres per second. This NSW Year 9 Core maths subtopic revises and extends Stage 4 work under NESA outcome MA4‑RAT‑C‑01.
You will change the top unit, the bottom unit or both, one at a time, as in \(120\) L/min to mL/s and \(50\) g/m\(^2\) to kg/ha. You will use the factor \(3.6\) between m/s and km/h and see where it comes from, quote fuel use in litres per \(100\) km, convert a price per gram to dollars per kilogram, and use rates to find totals and times. Problems include a driver's reaction distance, filling a tank and deciding which walker is faster. It is part of Year 9 Core Working with Numbers.
Every question with a fully worked solution.
- Converting Rates - Video - Converting Rates Watch
Theory
A rate has two units, such as km/h or \(\$\) per kg. This Year 9 maths page revises how to convert a rate by changing one unit at a time, how to change between km/h and m/s, and how to use a converted rate to find a distance, a filling time or a price.
A rate compares two quantities in different units, such as \(60\) km/h or \(\$4\) per kg. It has a top unit (kilometres, dollars) and a bottom unit (per hour, per kilogram).
To convert a rate, change one unit at a time. Changing the top unit changes the rate the same way: kilograms to grams multiplies the top by \(1000\), so the rate is multiplied by \(1000\).
Changing the bottom unit works the opposite way. Per hour to per minute divides by \(60\), because one minute gets only a sixtieth of what one hour gets.
Two rates can only be compared once they are in the same units.
The speed conversion, built from \(1\) km \(= 1000\) m and \(1\) h \(= 3600\) s:
Using a rate:
| Change | Effect on the rate | Example |
|---|---|---|
| top: kg to g | \(\times 1000\) | \(2\) kg/h \(= 2000\) g/h |
| top: mL to L | \(\div 1000\) | \(500\) mL/s \(= 0.5\) L/s |
| bottom: per h to per min | \(\div 60\) | \(120\) km/h \(= 2\) km/min |
| bottom: per min to per day | \(\times 1440\) | \(1\) L/min \(= 1440\) L/day |
How to convert a rate
- Write the rate as a fraction: top unit over bottom unit.
- Change the top unit. The rate moves the same way as the number on top.
- Change the bottom unit. The rate moves the opposite way.
- Check the size makes sense, then use the rate: multiply by a time for a total, divide a total by the rate for a time.
(i) \(250\) mL/min to L/h and
(ii) \(2.5\) kg/m to g/cm.
(i) Bottom unit first: an hour is \(60\) minutes, so \(60\) times as much flows. Then the top unit: millilitres to litres divides by \(1000\).
| \(250 \times 60\) | \(=\) | \(15\,000 \text{ mL/h}\) |
| \(15\,000 \div 1000\) | \(=\) | \(\mathbf{15} \text{ L/h}\) |
(ii) Write the rate as a fraction and change each unit inside it:
| \(\dfrac{2.5 \text{ kg}}{1 \text{ m}}\) | \(=\) | \(\dfrac{2500 \text{ g}}{100 \text{ cm}}\) |
| \(\) | \(=\) | \(\mathbf{25} \text{ g/cm}\) |
Ask each time: should the number get bigger or smaller?
(i) Convert \(54\) km/h to m/s.
(ii) A runner moves at \(8\) m/s. What is this in km/h?
(i) Kilometres to metres multiplies by \(1000\); per hour to per second divides by \(3600\):
| \(54 \times 1000 \div 3600\) | \(=\) | \(\mathbf{15} \text{ m/s}\) |
The one-step version is \(54 \div 3.6 = 15\).
(ii) Going from m/s to km/h, multiply by \(3.6\):
| \(8 \times 3.6\) | \(=\) | \(\mathbf{28.8} \text{ km/h}\) |
Check the size: a km/h figure is always bigger than the same speed in m/s.
(i) A driver's reaction time is \(0.8\) s and the car is doing \(60\) km/h. How far does the car travel before the brakes go on, to one decimal place?
(ii) A tap runs at \(600\) mL/s. How many minutes does it take to fill a \(270\) L tank?
(i) The time is in seconds, so change the speed to m/s first:
| \(60 \div 3.6\) | \(=\) | \(16.666\ldots \text{ m/s}\) |
| \(\text{distance}\) | \(=\) | \(16.666\ldots \times 0.8\) |
| \(\) | \(\approx\) | \(\mathbf{13.3} \text{ m}\) |
(ii) Put the tank in millilitres, divide by the rate to get seconds, then change to minutes:
| \(270 \text{ L}\) | \(=\) | \(270\,000 \text{ mL}\) |
| \(\text{time}\) | \(=\) | \(270\,000 \div 600 = 450 \text{ s}\) |
| \(\) | \(=\) | \(450 \div 60 = \mathbf{7.5} \text{ min}\) |
Keep the unrounded speed until the last line.
(i) Find the price per kilogram.
(ii) Find the cost of \(250\) g.
(iii) Salami costs \(2.5\) cents per gram. Which is cheaper per kilogram?
(i) A kilogram is ten lots of \(100\) g:
| \(1.60 \times 10\) | \(=\) | \(\mathbf{\$16.00} \text{ per kg}\) |
(ii) \(250\) g is a quarter of a kilogram:
| \(16.00 \times 0.25\) | \(=\) | \(\mathbf{\$4.00}\) |
(iii) Grams to kilograms multiplies by \(1000\), then cents to dollars divides by \(100\):
| \(2.5 \times 1000\) | \(=\) | \(2500 \text{ c/kg}\) |
| \(2500 \div 100\) | \(=\) | \(\$25.00 \text{ per kg}\) |
In the same units, \(\$16\) per kg is less than \(\$25\) per kg, so the ham is cheaper.
Common pitfalls
Frequently asked questions
How do you convert km/h to m/s?
Divide by \(3.6\). This is the same as multiplying by \(1000\) to change kilometres to metres and dividing by \(3600\) to change per hour to per second.
How do you convert m/s to km/h?
Multiply by \(3.6\). For example \(10\) m/s is \(36\) km/h.
How do you convert a rate with two units?
Change one unit at a time. Changing the top unit changes the rate the same way. Changing the bottom unit changes it the opposite way.
Why is the factor 3.6?
One metre per second is \(3600\) metres in an hour, and \(3600\) m is \(3.6\) km.
How do you work out how long something takes from a rate?
Get the total and the rate into matching units, then divide the total by the rate.