How to Convert Metric Units and Solve Compound Measures
This page explains how to convert metric units and solve compound measure questions in GCSE and IGCSE Mathematics. You will learn how to handle length, area and volume conversions, and apply them to speed, density and pressure using clear exam-style methods. Worked examples are included throughout, followed by practice questions.
kilo = 1 000 (×1 000)
centi = \(\tfrac{1}{100}\) (÷100)
milli = \(\tfrac{1}{1000}\) (÷1 000)
When the length factor is \(k\), the area factor is \(k^2\) and the volume factor is \(k^3\). Square or cube the linear conversion factor.
Speed \(=\dfrac{\text{distance}}{\text{time}}\), Density \(=\dfrac{\text{mass}}{\text{volume}}\), Pressure \(=\dfrac{\text{force}}{\text{area}}\).
Always convert units before substituting.
Converting metric units: the method
- Identify the direction of the conversion: smaller unit to larger, or larger to smaller.
- Multiply when moving to a smaller unit (e.g. m to cm). Divide when moving to a larger unit (e.g. cm to m).
- For area, square the linear factor (\(100^2 = 10\,000\) for m to cm). For volume, cube it (\(100^3 = 1\,000\,000\)).
Length conversion factors
Why area and volume use different factors
Scaling one direction by a factor of \(k\) scales length by \(k\), area by \(k^2\), and volume by \(k^3\) because each dimension is scaled independently.
1D: length
Scale by \(10\): length becomes \(10\times\) bigger.
2D: area
Scale each side by \(10\): area scales by \(10^2 = 100\).
3D: volume
Scale each edge by \(10\): volume scales by \(10^3 = 1000\).
Worked examples: unit conversions
💡 Example 1: metres and centimetres
1a. Convert \(7\text{ m}\) to cm.
\[\begin{aligned}7\text{ m} &= 7\times100\text{ cm}\\ &= 700\text{ cm}\end{aligned}\]1b. Convert \(450\text{ cm}\) to m.
\[\begin{aligned}450\text{ cm} &= \dfrac{450}{100}\text{ m}\\ &= 4.5\text{ m}\end{aligned}\]What's happening?
- Moving from m to cm is moving to a smaller unit, so the number gets bigger. Multiply by 100.
- Moving from cm to m is moving to a larger unit, so the number gets smaller. Divide by 100.
- The factor 100 comes from \(1\text{ m} = 100\text{ cm}\).
💡 Example 2: kilometres and metres
2a. Convert \(3\text{ km}\) to m.
\[\begin{aligned}3\text{ km} &= 3\times1000\text{ m}\\ &= 3000\text{ m}\end{aligned}\]2b. Convert \(2500\text{ m}\) to km.
\[\begin{aligned}2500\text{ m} &= \dfrac{2500}{1000}\text{ km}\\ &= 2.5\text{ km}\end{aligned}\]What's happening?
- km to m: multiply by 1000 (km is bigger, so the number goes up).
- m to km: divide by 1000 (km is bigger, so the number goes down).
- The \(\div 1000\) rule follows from the prefix: kilo = 1000.
💡 Example 3: area conversions (\(\text{m}^2\) and \(\text{cm}^2\))
Rule. \(1\text{ m} = 100\text{ cm}\), so \(1\text{ m}^2 = 100^2\text{ cm}^2 = 10{,}000\text{ cm}^2\).
3a. Convert \(5\text{ m}^2\) to \(\text{cm}^2\).
\[\begin{aligned}5\text{ m}^2 &= 5\times10{,}000\text{ cm}^2\\ &= 50{,}000\text{ cm}^2\end{aligned}\]3b. Convert \(4500\text{ cm}^2\) to \(\text{m}^2\).
\[\begin{aligned}4500\text{ cm}^2 &= \dfrac{4500}{10{,}000}\text{ m}^2\\ &= 0.45\text{ m}^2\end{aligned}\]What's happening?
- The linear factor between m and cm is 100. For area, you square it: \(100^2 = 10{,}000\).
- Think of a square that is \(1\text{ m}\times1\text{ m}\). That is \(100\text{ cm}\times100\text{ cm} = 10{,}000\text{ cm}^2\).
- Always square the linear factor for area conversions.
💡 Example 4: volume conversions (\(\text{m}^3\) and \(\text{cm}^3\))
Rule. \(1\text{ m} = 100\text{ cm}\), so \(1\text{ m}^3 = 100^3\text{ cm}^3 = 1{,}000{,}000\text{ cm}^3\).
4a. Convert \(2\text{ m}^3\) to \(\text{cm}^3\).
\[\begin{aligned}2\text{ m}^3 &= 2\times1{,}000{,}000\text{ cm}^3\\ &= 2{,}000{,}000\text{ cm}^3\end{aligned}\]4b. Convert \(6000\text{ cm}^3\) to \(\text{m}^3\).
\[\begin{aligned}6000\text{ cm}^3 &= \dfrac{6000}{1{,}000{,}000}\text{ m}^3\\ &= 0.006\text{ m}^3\end{aligned}\]What's happening?
- The linear factor is 100. For volume, cube it: \(100^3 = 1{,}000{,}000\).
- A cube that is \(1\text{ m}\times1\text{ m}\times1\text{ m}\) contains \(100\times100\times100 = 1{,}000{,}000\) smaller cubes of \(1\text{ cm}^3\).
- The pattern: length ×100, area ×10 000, volume ×1 000 000.
Compound measures: the method
- Write the formula for the quantity required (Speed, Density, or Pressure).
- Identify the target units for the final answer (e.g. m/s, kg/m\(^3\), Pa).
- Convert all given values into the target units before substituting into the formula.
- Substitute and calculate. Show one equals per line.
Useful shortcuts: km/h to m/s: multiply by \(\tfrac{5}{18}\). g/cm\(^3\) to kg/m\(^3\): multiply by 1000. When working in Pa (\(\text{N/m}^2\)), always convert area to m\(^2\) first.
Worked examples: compound measures
💡 Example 5: speed
Given. A car travels \(18\text{ km}\) in \(45\text{ min}\). Find the speed in \(\text{m s}^{-1}\).
Step 1. Convert distance and time to the target units (m and s).
\[\begin{aligned}18\text{ km} &= 18\times1000 = 18{,}000\text{ m}\\45\text{ min} &= 45\times60 = 2700\text{ s}\end{aligned}\]Step 2. Substitute into the formula.
\[\begin{aligned}\text{speed} &= \dfrac{\text{distance}}{\text{time}}\\[4pt] &= \dfrac{18{,}000}{2700}\\[4pt] &= 6.\overline{6}\text{ m s}^{-1}\end{aligned}\]Answer. \(6.67\text{ m s}^{-1}\) (2 d.p.)
What's happening?
- The target unit is m s\(^{-1}\), so distance must be in metres and time in seconds.
- Both conversions happen before substitution, not during.
- Shortcut check: \(18\text{ km/h} = 5\text{ m/s}\) and \(18\text{ km}\) in \(45\text{ min}\) is equivalent to \(24\text{ km/h}\). Then \(24\times\tfrac{5}{18} = 6.\overline{6}\text{ m/s}\). Same answer.
💡 Example 6: density
Given. Mass \(= 750\text{ g}\), volume \(= 300\text{ cm}^3\). Find density in \(\text{kg/m}^3\).
Step 1. Convert to kg and m\(^3\).
\[\begin{aligned}750\text{ g} &= \dfrac{750}{1000} = 0.75\text{ kg}\\300\text{ cm}^3 &= \dfrac{300}{1{,}000{,}000} = 0.0003\text{ m}^3\end{aligned}\]Step 2. Substitute into the formula.
\[\begin{aligned}\text{density} &= \dfrac{\text{mass}}{\text{volume}}\\[4pt] &= \dfrac{0.75}{0.0003}\\[4pt] &= 2500\text{ kg/m}^3\end{aligned}\]What's happening?
- The question asks for kg/m\(^3\), so both mass and volume must be in SI units first.
- The volume conversion uses the cube factor: \(1\text{ m}^3 = 1{,}000{,}000\text{ cm}^3\).
- Shortcut: if density is in g/cm\(^3\), multiply by 1000 to get kg/m\(^3\). Here \(750/300 = 2.5\text{ g/cm}^3\), and \(2.5\times1000 = 2500\text{ kg/m}^3\).
💡 Example 7: pressure
Given. Force \(= 180\text{ N}\), area \(= 12\text{ cm}^2\). Find pressure in Pa.
Step 1. Convert area to m\(^2\) (force is already in N, which is correct for Pa).
\[12\text{ cm}^2 = \dfrac{12}{10{,}000} = 0.0012\text{ m}^2\]Step 2. Substitute into the formula.
\[\begin{aligned}P &= \dfrac{F}{A}\\[4pt] &= \dfrac{180}{0.0012}\\[4pt] &= 150{,}000\text{ Pa}\\[4pt] &= 1.5\times10^5\text{ Pa}\end{aligned}\]What's happening?
- 1 Pa = 1 N/m\(^2\), so area must be in m\(^2\) before using the formula.
- The area conversion uses the square factor: \(1\text{ m}^2 = 10{,}000\text{ cm}^2\), so divide by 10 000.
- The large answer (150 000 Pa) is realistic for pressure. Writing it in standard form is cleaner: \(1.5\times10^5\text{ Pa}\).
🔑 Key points
- Prefixes: kilo = 1000, centi = \(\tfrac{1}{100}\), milli = \(\tfrac{1}{1000}\).
- Length: km → m ×1000; m → cm ×100; cm → mm ×10. Reverse to divide.
- Area: square the length factor. Volume: cube it.
- Compound measures: convert all units into the target system before substituting.
- Speed km/h to m/s: ×\(\tfrac{5}{18}\). Density g/cm\(^3\) to kg/m\(^3\): ×1000.
- Pressure is in Pa = N/m\(^2\). Always convert area to m\(^2\) first.
⚠️ Common pitfalls
- Using the linear factor (×100) for area instead of squaring it (×10 000). The factor changes with dimension.
- Substituting values in mixed units (e.g. mass in g but volume in m\(^3\)). Convert everything first.
- Dividing when you should multiply, or vice versa. Ask: am I going to a smaller unit (multiply) or a larger one (divide)?
- Forgetting that 1 km/h is not the same as 1 m/s. Never cancel the units without converting.
- Leaving pressure in cm\(^2\) when the formula requires m\(^2\) for a Pa answer.