Year 10+ Measurement

Converting Between Units

Converting Between Units

Converting Between Units

Sam is a bit confused. He's been given the speed of a fast moving tortoise as \(2 \text{ km/h}\) and needs to convert this to metres per second. He knows that there are \(1,000\) metres in one kilometre, \(60\) minutes in an hour and \(60\) seconds in a minute. So, he can work out that there are \(3,600\) seconds in an hour, but now he's not sure how to combine these together to get a speed in metres per second. He wonders whether he should calculate \(\dfrac{2 \times 1000}{3600} = 0.55555 \dots \) m/s or \(\dfrac{3600}{2000} = 1.8 \) m/s. What should Sam do?

It's understandable that Sam might be a bit confused. One way to resolve his confusion would be to think a bit and realise that he has to put the metres over the seconds because he needs to find the tortoise's speed in metres per second. So, the correct answer is \(\dfrac{2 \times 1000}{3600} = 0.55555 \dots \) m/s.

If that doesn't help, Sam can always try the technique discussed in this article. I've never seen this technique taught in an Australian school, but then I've only taught in NSW. I have been doing some online teaching for an American company, and this is the way they teach unit conversion. You might have watched an American video about problem solving, say in science, where they've spoken about "setting up a proportion" to solve the problem. They're actually talking about fractions, and this article will show you how to use them to solve unit conversion problems. You can then decide which method you like the best.

The best way to learn any new technique is to work through lots of examples, so that's what we'll do. Let's start with an easy one. I know this one isn't likely to cause you confusion, but it's just to give you an idea of how the method works before we try anything more complicated.

Example 1

Suppose you are asked to convert \(5\) m to centimetres (cm). We all know this is easy: there are \(100\) cm in one metre, so the answer is \(5 \times 100 = 500 \) cm, but let's see how the other method works.

There are three steps to the new method:

  1. Set up the "proportion": write the conversion as a fraction and set it equal to \(1\).
  2. Multiply the quantity to be converted by this fraction (this is OK - you're just multiplying by \(1\))
  3. Cancel any units that occur in both the numerator and denominator (top and bottom) of the answer

Step 1
We know there are \(100\) cm in \(1\) m, so we write

\(\dfrac{100 \text{ cm}}{1 \text{ m}} = 1\)
We've come up with a new form of \(1\)!

Step 2
Now multiply the quantity we need to convert by our new form of \(1\):

\(5 \text{ m} \times \dfrac{100 \text{ cm}}{1 \text{ m}} = \dfrac{5 \text{ m} \times 100 \text{ cm}}{1 \text {m}} = \dfrac{500 \text{ m} \times \text{ cm}}{1 \text{ m}}\)
The right hand side looks a bit funny now, but we'll fix that up in the next step.

Step 3
Cancel the units that are both on the top and the bottom (those are the metres)

\(\dfrac{500 \text{ m} \times \text{ cm}}{1 \text{ m}} = 500 \text{ cm}\)
It looks like this method gives us the same answer as our original method. I'm sure you're asking why anyone would do this, but it'll become a bit more apparent when we try a harder example.

If you still get confused and write the conversion the wrong way around, you'll end up not being able to cancel out. You'll know you're wrong and will be able to go back and fix it up. For example, if we wrote our conversion fraction as

\(\dfrac{1 \text{ m}}{100 \text{ cm}} = 1\),
we'd end up with
\(\dfrac{500 \text{ m} \times \text{ m}}{100 \text{ cm}}\)
and wouldn't be able to cancel the units. Then we could say, "whoops" and turn \(\dfrac{1 \text{ m}}{100 \text{ cm}} = 1\) up the other way to give \(\dfrac{100 \text{ cm}}{1 \text{ m}} = 1\), and multiply by this new fraction to find the correct answer.

Example 2

Let's use this technique to solve Sam's tortoise problem from the beginning of the article. To avoid further confusion for Sam, we'll do it in two steps, but when Sam gets good at this method (should he decide to use it), he can do it all in one line. I'll show you how to do that at the end.

Sam has a bit of trouble with unit conversions, so I think he should break it up as follows:

  1. Convert from km/h to m/h
  2. Convert from m/h to m/s
Let's see how this works.

1. Convert km/h to m/h
\( \begin{align*} \dfrac{2 \text{ km}}{\text{h}} \times \dfrac{1000 \text{ m}}{1 \text{ km}} = \dfrac{2000 \text{ km} \times \text{ m}}{1 \text{ h} \times \text {km}} \end{align*} \)
Now cancel out the common units from the top and bottom (km this time) to give
\(\dfrac{2,000 \text{ m}}{\text{ h}}\).

1. Convert m/h to m/s
We know we want the \(\text{h}\) on top of our conversion fraction because we have an \(\text{h}\) downstairs in our m/h fraction, and we want to be able to cancel them later. This tells us which way up to put the fraction that will convert from hours to seconds. So now we multiply:
\(\dfrac{2,000 \text{ m}}{\text{ h}} \times \dfrac{1 \text{ h}}{3,600 \text{ s}} = \dfrac{2,000 \text{ m} \times \text{ h}}{3,600 \text{ s} \times \text{ h}}\)
Now cancel out the common units from the top and bottom (h this time) to give
\(\dfrac{2,000 \text{ m}}{3,600 \text{ s}} = 0.55555 \dots \text{ m/s}\).
So, that tortoise is rocketing along at \(0.555\dots\) m/s. That really is fast for a tortoise.

Note: If you got the hours to seconds conversion fraction wrong and put it up the other way, you'd end up with

\(\dfrac{2,000 \text{ m}}{\text{ h}} \times \dfrac{3,600 \text{ s}}{1 \text{ h}} = \dfrac{2,000 \times 3,600 \times \text{ m} \times \text{ s}}{1 \text{ h} \times \text{ h}}\)
and wouldn't be able to cancel. So, you'd realise you'd made an error straight away and would be able to go back and fix it.

Getting Clever and Doing it All in One Hit
When you've done lots of these conversion examples doing this technique, you can start to take short cuts like doing the conversion in one step. Make sure you can cancel the units you don't want in your final answer from the top and the bottom. Sam's conversion becomes:
\(\dfrac{2 \text{ km}}{\text{h}} \times \dfrac{1000 \text{ m}}{1 \text{ km}}\times \dfrac{1 \text{ h}}{3,600 \text{ s}} = \dfrac{2,000 \text{ km} \times \text{ m} \times \text{ h}}{3,600 \text{ s} \times \text{ km} \times \text{ h}},\)
and the units we don't want any more (the km and the h) magically cancel each other out to leave us with
\( \dfrac{2,000 \text{ m} }{3,600 \text{ s} } = 0.555 \dots \text{ m/s}.\)

Example 3

There's time for one last example. This time, we'll use the method to convert an imperial unit into a metric one. Gertie the snail is doing some low flying along the garden path at a speed of \(0.25\) inches per hour. What is her speed in centimetres per second?

We know that \(1 \text{ inch}\) is equal to \(2.54 \) centimetres, so we can set up the following product of fractions for the conversion, making sure that the unwanted units will eventually cancel each other out:

\(\dfrac{0.25 \text{ in}}{\text{h}} \times \dfrac{2.54 \text{ cm}}{1 \text{ in}} \times \dfrac{1 \text{ h}}{3,600 \text{ s}} = \dfrac{0.25 \times 2.54 \times \text{ in} \times \text{ cm} \times \text{ h}}{3,600 \times \text{h} \times \text{in}\times \text{s}} = 0.00021166\dots \text{ cm/s}.\)


If you decide to use this method, make sure you write down all your working. The steps are:

  1. Set up the "proportion": write the conversion as a fraction and set it equal to \(1\).
  2. Multiply the quantity to be converted by this fraction (this is OK - you're just multiplying by \(1\))
  3. Cancel any units that occur in both the numerator and denominator (top and bottom) of the answer
Remember that you want the unwanted units to cancel each other out from the top and bottom of your final product of fractions. This should help you to decide which way up the fractions go. If you get it wrong, you'll end up with units that won't cancel. That's OK. You can just flip them over and go back to fix the calculation up.


In these chapters, you will learn more about measurement topics such as 

  • Metric numbers
  • Speed
  • Conversions

and several other topics


Year 10 or higher, some chapters are suitable for Year 8+ students

Learning Objectives

Learn about measurement and related topics

Author: Subject Coach
Added on: 28th Sep 2018

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