Working with Compound Units Revision Notes | GCSE AQA Higher Maths

1. Why Compound Units Matter

A speed of 60 tells you nothing. 60 what? Miles per hour? Metres per second? The unit carries half the meaning. That is the whole point of compound units — they describe how two quantities relate to each other.

You already use compound units every day without thinking about it. Petrol prices in pence per litre, wages in pounds per hour, motorway speeds in miles per hour — all of these are compound units. The maths behind them is the same each time: one quantity divided by another.

The three compound units you need for GCSE are speed (distance ÷ time), rates of pay (earnings ÷ time), and unit pricing (cost ÷ quantity). Once you see that all three follow the same structure — one thing divided by another — the topic becomes much more manageable.

2. Speed as a Compound Unit

Speed measures how much distance is covered in each unit of time. Think of it as a rate — how fast distance is accumulating. The formula triangle below is a useful memory aid, but understanding the relationship is more important than memorising a triangle.

The most common trap here is mixing units. If distance is in kilometres and time is in minutes, your speed will be in km/min — which is almost certainly not what the question wants. Always check that your distance and time units match the required speed unit before substituting.

Worked example: Calculating Speed in m/s

Mohammed runs 400 m in 50 seconds. Calculate his average speed in m/s. [2 marks]

1. $$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$ — State the formula first — this earns the method mark [M1] even if arithmetic goes wrong later.
2. Distance = 400 m, Time = 50 s — both in SI base units, so speed will be in m/s. — No conversion needed here. Always confirm units match before substituting.
3. $$\text{Speed} = \frac{400}{50} = 8$$ — 400 ÷ 50 = 8. Arithmetic check: 50 × 8 = 400. [A1]
4. Speed = 8 m/s — Missing units costs a mark. Write the unit every time.

Answer: 8 m/s

Now here is where it gets interesting. Exam questions often give you time in minutes but ask for speed in m/s or km/h. That means you need to convert before you substitute — or convert your answer afterwards. Either approach works, but doing it before substituting is less error-prone.

Worked example: Speed with Unit Conversion (km/h)

Aisha cycles 15 km in 45 minutes. Calculate her average speed in km/h. [3 marks]

1. Required unit: km/h. Distance is already in km. Time must be in hours. — The target unit tells you which conversion is needed.
2. $$45 \text{ minutes} = \frac{45}{60} \text{ hours} = 0.75 \text{ hours}$$ — Divide by 60 to convert minutes to hours. [M1] for conversion.
3. $$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{15}{0.75}$$ — Both quantities now in km and hours respectively. [M1] for substitution.
4. $$\frac{15}{0.75} = 15 \div \frac{3}{4} = 15 \times \frac{4}{3} = \frac{60}{3} = 20$$ — Dividing by 0.75 is the same as multiplying by 4/3. Arithmetic check: 0.75 × 20 = 15. [A1]
5. Speed = 20 km/h — Always attach the compound unit to the final answer.

Answer: 20 km/h

3. Rates of Pay

A rate of pay works exactly like speed — it is one quantity divided by another. Instead of distance per unit time, you have money earned per unit time. The structure is identical: total amount divided by the number of units.

Worked example: Calculating and Comparing Rates of Pay

Ben earns £94.50 for working 9 hours. Priya earns £130 for working 13 hours. Who has the higher rate of pay? [3 marks]

1. $$\text{Ben's rate} = \frac{94.50}{9} = £10.50 \text{ per hour}$$ — 94.50 ÷ 9: think 9 × 10 = 90, remainder 4.50, and 4.50 ÷ 9 = 0.50, so total = £10.50. [M1]
2. $$\text{Priya's rate} = \frac{130}{13} = £10.00 \text{ per hour}$$ — 130 ÷ 13 = 10. Arithmetic check: 13 × 10 = 130. [M1]
3. £10.50 > £10.00, so Ben has the higher rate of pay. — State the comparison explicitly — the examiner needs a clear conclusion. [A1]

Answer: Ben has the higher rate of pay at £10.50/hour compared to Priya's £10.00/hour.

Notice that the comparison only works because both rates are now in the same compound unit — £ per hour. If one person's pay had been given per day and another's per hour, you would need to convert to the same unit before comparing. This is a recurring theme across all compound unit questions.

4. Unit Pricing

Unit pricing answers the question: which deal gives you more for your money? To compare two products fairly, you need to find the cost per single unit — per gram, per litre, per item. Once both prices are expressed per the same unit, the comparison is straightforward.

There are two equivalent approaches: find the price per unit (cost ÷ quantity), or find the quantity per pound (quantity ÷ cost). Both work — but you must use the same approach for both products being compared, otherwise the comparison is meaningless.

Worked example: Comparing Unit Prices

A 400 g jar of coffee costs £3.20. A 600 g jar of the same coffee costs £4.50. Which jar offers better value? [3 marks]

1. Find the price per gram (pence per gram) for each jar. — Converting to pence avoids small decimals. Either pence/g or £/g works — be consistent.
2. $$\frac{320 \text{ p}}{400 \text{ g}} = 0.80 \text{ p/g}$$ — £3.20 = 320 p. Then 320 ÷ 400 = 0.8 p per gram. [M1]
3. $$\frac{450 \text{ p}}{600 \text{ g}} = 0.75 \text{ p/g}$$ — £4.50 = 450 p. Then 450 ÷ 600 = 0.75 p per gram. [M1]
4. 0.75 p/g < 0.80 p/g, so the 600 g jar is cheaper per gram. — Lower price per gram = better value. State the conclusion clearly. [A1]

Answer: The 600 g jar offers better value at 0.75 p/g compared to 0.80 p/g for the 400 g jar.

5. Converting Between Compound Units

Sometimes a question gives you a speed in m/s and asks for km/h, or gives a price in £/kg and asks for pence/g. Converting compound units looks intimidating, but there is a reliable method: convert the numerator and denominator separately, then simplify.

Think of the compound unit as a fraction. If you change the top (numerator unit) or the bottom (denominator unit), the numerical value changes accordingly. The key conversions to know are: 1 km = 1000 m, 1 hour = 3600 seconds, 1 kg = 1000 g, and £1 = 100 pence.

Worked example: Converting m/s to km/h

A sprinter runs at 10 m/s. Convert this speed to km/h. [2 marks]

1. $$10 \text{ m/s} = \frac{10}{1000} \text{ km/s} = 0.01 \text{ km/s}$$ — 1 km = 1000 m, so divide the distance part by 1000. [M1]
2. $$0.01 \text{ km/s} = 0.01 \times 3600 \text{ km/h} = 36 \text{ km/h}$$ — 1 hour = 3600 seconds. There are 3600 seconds in each hour, so multiply by 3600 to get km per hour. Arithmetic check: 0.01 × 3600 = 36.
3. Speed = 36 km/h — Sanity check: 36 km/h is a reasonable sprinting speed (roughly 22 mph). [A1]

Answer: 36 km/h

There is a shortcut worth knowing: to convert m/s to km/h, multiply by 3.6. To convert km/h to m/s, divide by 3.6. This comes directly from the two-step process above (÷1000 then ×3600 = ×3.6). On a calculator-allowed exam question, this shortcut saves time — but In questions without a calculator, show the full conversion steps.

6. Formula Reference Bank

All three compound unit types follow the same mathematical structure. Here is a compact reference to use during revision — notice how the pattern 'total ÷ number of units' runs through all of them.

7. Deliberate Practice

These questions cover all three compound unit types. Work through each one before checking the answer — the method mark is available even if your arithmetic slips, so always show your working.

8. Exam Technique and Common Mistakes

Compound unit questions are reliable mark-earners — the method is always the same. The marks that get dropped are almost always due to unit errors, not conceptual misunderstanding. Here is where to focus your attention in the exam.

One more thing worth flagging: dividing by a number less than 1 gives a result that is larger than the original, not smaller. For example, if time is 0.5 hours, then Distance ÷ 0.5 doubles the distance value. This surprises students who expect division to always reduce a number — but dividing by a fraction is the same as multiplying by its reciprocal.

9. Exam-Style Multi-Part Question

Exam questions on compound units often combine two or three skills in one problem. This multi-part question brings together speed calculation, unit conversion, and rate of pay — the full set of skills from this lesson.

Part (c) is a good example of the error-carried-forward (ECF) principle. If you used an incorrect time in part (a) but then applied it consistently in part (c), you would still earn the method mark in part (c). Always show your working — method marks are available even when the final answer is wrong.