Compound Measures: Speed, Density and Pressure

A compound measure combines two different units into a single rate — like miles per hour or grams per cubic centimetre. They are a core part of the Rates of Change strand in OCR GCSE Mathematics (J560) and appear on every paper. This lesson covers the three most-examined compound measures — speed (distance and time), density (mass and volume) and pressure (force and area) — together with general unit rates such as cost per litre. The same triangular relationship governs all three.

This lesson builds AO1 fluency in rearranging and substituting into rate formulae, AO2 reasoning when you choose and convert units sensibly, and AO3 problem-solving in journeys with several stages.

Key Vocabulary

Term	Meaning
Compound measure	A measure made from two units, e.g. km/h, g/cm³, N/m²
Speed	Distance travelled per unit time
Density	Mass per unit volume
Pressure	Force per unit area
Unit rate	"How much per one", e.g. cost per litre
Average speed	Total distance $\div$÷ total time over a whole journey

The Three Key Formulae

Each compound measure links three quantities, and you can rearrange the formula to find any one of them. The three measures share the same structure — a "per" quantity equals one thing divided by another — which means that if you understand one of them, you understand all three. The word "per" in the unit is a strong clue: "miles per hour" literally means miles divided by hours, and "grams per cubic centimetre" means grams divided by volume.

$\text{speed} = \dfrac{\text{distance}}{\text{time}}, \qquad \text{density} = \dfrac{\text{mass}}{\text{volume}}, \qquad \text{pressure} = \dfrac{\text{force}}{\text{area}}.$speed=timedistance​,density=volumemass​,pressure=areaforce​.

Rearranged: distance $=$= speed $\times$× time; mass $=$= density $\times$× volume; force $=$= pressure $\times$× area. A helpful way to remember the three rearrangements of each formula is the "formula triangle": write the top quantity (distance, mass or force) above a line, and the other two side by side beneath it. Cover the quantity you want, and what remains shows you whether to multiply or divide. Covering distance leaves "speed $\times$× time"; covering time leaves "distance over speed". The same triangle works for all three measures.

Quantity	Formula	Common units
Speed	$\dfrac{\text{distance}}{\text{time}}$timedistance​	m/s, km/h, mph
Density	$\dfrac{\text{mass}}{\text{volume}}$volumemass​	g/cm³, kg/m³
Pressure	$\dfrac{\text{force}}{\text{area}}$areaforce​	N/cm², N/m² (pascals)

Speed, Distance and Time

Speed is the most familiar compound measure, and the relationship between speed, distance and time is examined on every paper. The most important practical skill is handling time: a journey time given in hours and minutes must be converted to a decimal number of hours (or to minutes) before it goes into the formula, and the conversion trips up many candidates. Half an hour is $0.5$0.5 hours, a quarter of an hour is $0.25$0.25 hours, and $20$20 minutes is $\dfrac{20}{60} = \dfrac{1}{3}$6020​=31​ of an hour. Get the time right and the rest of the calculation is straightforward.

Worked Example 1

A cyclist travels $90$90 km in $4$4 hours. Work out the average speed.

$\text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{90}{4} = 22.5 \text{ km/h}.$speed=timedistance​=490​=22.5 km/h.

Answer: $22.5$22.5 km/h. This is an average speed: the cyclist may have gone faster downhill and slower uphill, but $22.5$22.5 km/h is the steady speed that would cover the same distance in the same time.

Worked Example 2

A train travels at $120$120 km/h for $2.5$2.5 hours. Work out the distance travelled.

$\text{distance} = \text{speed} \times \text{time} = 120 \times 2.5 = 300 \text{ km}.$distance=speed×time=120×2.5=300 km.

Answer: $300$300 km.

Worked Example 2b

A plane flies at $540$540 km/h. Work out how far it travels in $40$40 minutes.

The time must be in hours to match the speed: $40$40 minutes $= \dfrac{40}{60} = \dfrac{2}{3}$=6040​=32​ of an hour.

$\text{distance} = 540 \times \dfrac{2}{3} = 360 \text{ km}.$distance=540×32​=360 km.

Answer: $360$360 km.

Common error: putting $40$40 straight into the formula as if it were hours. The units of speed (km per hour) decide the units of time you must use.

Worked Example 3

A car covers $150$150 miles at an average speed of $60$60 mph. Work out the time taken.

$\text{time} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{150}{60} = 2.5 \text{ hours} = 2 \text{ hours } 30 \text{ minutes}.$time=speeddistance​=60150​=2.5 hours=2 hours 30 minutes.

Answer: $2$2 hours $30$30 minutes.

Common error: writing $2.5$2.5 hours as "$2$2 hours $50$50 minutes". The decimal $0.5$0.5 of an hour is $30$30 minutes, not $50$50; multiply the decimal part by $60$60.

Density, Mass and Volume

Density measures how much mass is packed into a given volume — it is what makes a small lead weight feel heavy while a large piece of polystyrene feels light. The formula has exactly the same shape as speed, so the same triangle and the same multiply-or-divide reasoning apply. The usual units are grams per cubic centimetre (g/cm³) or kilograms per cubic metre (kg/m³).

Worked Example 4

A block of metal has mass $480$480 g and volume $60$60 cm³. Work out its density.

$\text{density} = \dfrac{\text{mass}}{\text{volume}} = \dfrac{480}{60} = 8 \text{ g/cm}^3.$density=volumemass​=60480​=8 g/cm3.

Answer: $8$8 g/cm³.

Worked Example 5

Gold has a density of $19.3$19.3 g/cm³. Work out the mass of a gold bar with volume $50$50 cm³.

$\text{mass} = \text{density} \times \text{volume} = 19.3 \times 50 = 965 \text{ g}.$mass=density×volume=19.3×50=965 g.

Answer: $965$965 g.

Worked Example 5b

A wooden plank has mass $4{,}200$4,200 g and density $0.6$0.6 g/cm³. Work out its volume.

Rearrange the formula to make volume the subject:

$\text{volume} = \dfrac{\text{mass}}{\text{density}} = \dfrac{4{,}200}{0.6} = 7{,}000 \text{ cm}^3.$volume=densitymass​=0.64,200​=7,000 cm3.

Answer: $7{,}000$7,000 cm³. Notice the answer is larger than the mass figure, which makes sense: dividing by a density below $1$1 increases the result, and light wood does take up a lot of space for its mass.

Pressure, Force and Area

Pressure tells you how concentrated a force is over an area. The same force spread over a large area gives a low pressure (think of snowshoes or a tractor's wide tyres), while the same force on a tiny area gives a high pressure (think of a knife edge or a stiletto heel). Once again the formula has the familiar "one thing over another" shape, so the triangle and the multiply-or-divide reasoning carry straight over. Pressure is measured in newtons per square metre (N/m², also called pascals) or newtons per square centimetre (N/cm²).

Worked Example 6

A force of $200$200 N acts on an area of $8$8 cm². Work out the pressure.

$\text{pressure} = \dfrac{\text{force}}{\text{area}} = \dfrac{200}{8} = 25 \text{ N/cm}^2.$pressure=areaforce​=8200​=25 N/cm2.

Answer: $25$25 N/cm².

Common error: multiplying force by area. Pressure is force divided by area — a small area concentrates the force, giving a higher pressure (which is why a drawing pin works).

Worked Example 6b

A box weighs $480$480 N and stands on a square base of side $0.4$0.4 m. Work out the pressure on the floor in N/m².

First find the area of the base: $0.4 \times 0.4 = 0.16$0.4×0.4=0.16 m².

$\text{pressure} = \dfrac{\text{force}}{\text{area}} = \dfrac{480}{0.16} = 3{,}000 \text{ N/m}^2.$pressure=areaforce​=0.16480​=3,000 N/m2.

Answer: $3{,}000$3,000 N/m².

Common error: forgetting to work out the area first. Pressure questions in two dimensions usually need an area calculation before the pressure formula.

Unit Rates

Many "best value", rate and "how much per one" questions just need a unit rate — the amount per one of something. The unit rate is the bridge between compound measures and the proportion work earlier in this course: speed is "distance per one hour", density is "mass per one cubic centimetre", and a best-buy comparison is "cost per one gram". Once you can find a unit rate, you can compare options fairly and scale to any quantity.

Worked Example 7

A printer prints $480$480 pages in $6$6 minutes. Work out its printing rate in pages per minute.

$\dfrac{480}{6} = 80 \text{ pages per minute}.$6480​=80 pages per minute.

Answer: $80$80 pages per minute. A unit rate is simply direct proportion expressed as "amount per one", and once you know it you can scale to any number of minutes.

Worked Example 7b

Two taps fill containers at different rates. Tap A delivers $36$36 litres in $4$4 minutes; tap B delivers $50$50 litres in $5$5 minutes. Work out which tap is faster.

Find each rate in litres per minute:

Tap A: $\dfrac{36}{4} = 9$436​=9 litres per minute.

Tap B: $\dfrac{50}{5} = 10$550​=10 litres per minute.

Since $10 > 9$10>9, tap B is faster.

Answer: tap B, at $10$10 litres per minute.

Extended Worked Examples

Worked Example 8: Average speed over a whole journey

A delivery van drives $60$60 km in $1$1 hour, then $90$90 km in $2$2 hours. Work out the average speed for the whole journey.

Average speed uses total distance over total time, not the average of the two speeds.