GCSE Maths Proportion & Rates of Change Revision

Overview

What you’ll learn

Calculate with speed, density and pressure and convert between compound units such as km/h and m/s.
Solve direct proportion problems with the unitary method or one multiplier.
Write and use y = kx and y = k/x equations, finding the constant of proportionality first.
Read gradients of real-life graphs as rates of change, with units.
Estimate a population from a capture–recapture sample and state the assumptions involved.

Specification checklist (AQA 8300 · Edexcel 1MA1 · OCR J560)

Statements are paraphrased from the shared DfE subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.

Foundation — every student

13.1Change freely between related standard units and compound units (speed, rates of pay, unit pricing, density and pressure) in numerical and algebraic contexts.
13.2Use direct proportion: solve problems with the unitary method or a single multiplier, and recognise direct proportion graphically as a straight line through the origin.
13.3Use inverse proportion in context: as one quantity increases the other decreases so that their product stays constant (workers and days, speed and time for a fixed journey).
13.5Interpret the gradient of a real-life graph as a rate of change, such as a cost per unit, a filling rate in litres per minute or a speed from a distance–time graph.

Higher — Higher tier only

13.4Represent direct and inverse proportion algebraically: y = kx for direct and y = k/x for inverse proportion, finding k from one known pair and using the equation to predict further values.
13.6Estimate the size of a population using the capture–recapture method and comment on the assumptions the estimate depends on.

Key points

1The three compound-measure facts: speed = distance ÷ time, density = mass ÷ volume, pressure = force ÷ area.
2km/h → m/s: divide by 3.6 (72 km/h = 20 m/s); m/s → km/h: multiply by 3.6.
3Direct proportion: both quantities change in the same ratio; the graph is a straight line through the origin.
4Unitary method: price one first — 5 kg for £6.25 → £1.25/kg → 8 kg = £10.
5Inverse proportion: one rises as the other falls so the product is constant; 8 × 15 = 120 worker-days, so 12 workers take 10 days.
6Algebraically (Higher): direct is y = kx, inverse is y = k/x — find k from one known pair.
7Exchange rates are multipliers both ways: multiply going out, divide coming home.
8Capture–recapture (Higher): estimate = first catch × second catch ÷ recaptured tagged = 40 × 60 ÷ 12 = 200.

Notes · 1 of 4

Compound measures: speed, density and pressure

One quantity per unit of another — then convert carefully.

A compound measure combines two units: speed (distance per time), density (mass per volume) and pressure (force per area). Each is a division: speed = distance ÷ time, density = mass ÷ volume, pressure = force ÷ area — and each rearranges by multiplying back up. Cover the one you want: force = pressure × area, mass = density × volume, distance = speed × time.

Conversions need both units changed. 1 km = 1000 m and 1 h = 3600 s, so km/h → m/s divides by 3.6: 72 km/h = 72 × 1000 ÷ 3600 = 20 m/s. For volumes the scale factor squares or cubes: 1 m³ = 100³ = 1 000 000 cm³. Densities convert g/cm³ → kg/m³ by × 1000, because 1 g/cm³ means a 1 cm³ cube weighing 1 g, and 1000 of those cubes (a 10 cm side) weigh 1000 g = 1 kg.

Notes · 2 of 4

Direct proportion

Same ratio both ways: find one, or find k.

Two quantities are in direct proportion when one scales, the other scales by the same factor — doubling one doubles the other. The unitary method finds the value of one first: if 5 kg of rice costs £6.25, then 1 kg costs £1.25 and 8 kg costs 8 × 1.25 = £10. Equivalently, scale by a single multiplier 8⁄5.

Graphically, direct proportion is a straight line through the origin — zero of one pairs with zero of the other, and equal steps up per equal step across. Algebraically (Higher), write y ∝ x as y = kx: the constant k is also the gradient of that line. Exchange rates behave the same way, with one trap: multiply by the rate going out (£ → €), divide by it coming home.

Notes · 3 of 4

Inverse proportion

One up, the other down — the product stays fixed.

In inverse proportion one quantity rises as the other falls in the same ratio, so the product stays constant: more workers, fewer days; faster speed, shorter time. Eight workers take 15 days, so the job is 8 × 15 = 120 worker-days; twelve workers take 120 ÷ 12 = 10 days. Check the direction: if your answer grows when the input grew, you used direct proportion by mistake.

Fixed-distance journeys behave identically: at 40 km/h a 120 km trip takes 3 hours; at 60 km/h it takes 120 ÷ 60 = 2 hours — speed × time is always 120. Algebraically (Higher), inverse proportion is y = k/x: find k as the product of one known pair, then use the equation forwards and backwards.

Notes · 4 of 4

Gradients as rates of change, and capture–recapture (Higher)

Steepness is a speed of change; tagged samples size a population.

On real-life graphs the gradient is a rate of change with units: a cost graph rising £30 over 4 items has gradient £7.50 per item; a tank graph rising 24 litres in 8 minutes fills at 3 litres per minute. Steeper means faster. On a distance–time graph the gradient is a speed — the same compound measure from a picture.

Capture–recapture estimates a whole population from one marked sample. Catch and tag 40 fish, release, then later catch 60 and find 12 are tagged. The tagged fraction of the second catch, 12⁄60 = ⅕, estimates the tagged fraction of the pond, so the population is 40 × 60 ÷ 12 = 200. The estimate is only honest if the population is closed (no deaths, births or migration) and the tagged animals mix evenly — exam answers earn a mark for saying which assumption matters.

Revise

Key terms

The words examiners expect you to use precisely.

compound unit: A unit combining two measures, such as km/h, g/cm³ or N/m².speed = distance ÷ time; density = mass ÷ volume; pressure = force ÷ area.
unitary method: Solving a proportion problem by finding the value of one unit first, then scaling.£6.25 for 5 kg → £1.25 per kg → £10 for 8 kg.
direct proportion: Two quantities where one scales by the same factor as the other, so their ratio is constant.The graph is a straight line through the origin; y = kx.
inverse proportion: Two quantities where one increases as the other decreases, so their product is constant.More workers, fewer days; y = k/x.
constant of proportionality: The fixed number k linking two proportional quantities.k is the gradient for y = kx and the product xy for y = k/x.
capture–recapture: Estimating a population size from the proportion of tagged animals recaptured in a second sample.estimate = first catch × second catch ÷ recaptured tagged.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Dividing by 3.6 when converting m/s → km/h.
That is the km/h → m/s direction; the number must grow going back to km/h.
Instead: 25 m/s = 25 × 3.6 = 90 km/h; ÷ 3.6 one way, × 3.6 the other.
Mistake: Treating 1 m³ as 1000 cm³.
All three dimensions convert, not just one.
Instead: 1 m = 100 cm, so 1 m³ = 100³ = 1 000 000 cm³.

Show 10 more mistakes

Mistake: Converting g/cm³ → kg/m³ with the wrong factor.
The cubic centimetre shrinks a thousand times faster than the gram grows.
Instead: Multiply by 1000: 8 g/cm³ = 8000 kg/m³.
Mistake: Using direct proportion for a workers-and-days problem.
More workers means fewer days — the quantities move oppositely.
Instead: Inverse proportion: the product (worker-days) is fixed.
Mistake: Writing y = kx for inverse-proportion data.
Inverse pairs have constant products, not constant ratios.
Instead: The equation is y = k/x, with k = xy from one known pair.
Mistake: Calling any straight-line graph direct proportion.
A line with a non-zero intercept pairs zero of x with a non-zero cost.
Instead: Direct proportion needs a straight line through the origin.
Mistake: Rounding the unit price halfway through.
Mid-rounding errors multiply in later steps.
Instead: Keep exact fractions until the final answer, then round money to 2 d.p.
Mistake: Multiplying by the exchange rate in both directions.
The rate is a per-pound price, not a universal multiplier.
Instead: £ → € multiplies; € → £ divides by the same rate.
Mistake: Computing speed as time ÷ distance.
Speed is distance covered per unit of time, not the reverse.
Instead: 120 km in 1.5 h is 120 ÷ 1.5 = 80 km/h.
Mistake: Using cm² areas in a pressure calculation with newtons.
Pascals are newtons per square metre.
Instead: 250 cm² = 250 ÷ 10 000 = 0.025 m² before dividing.
Mistake: Combining capture–recapture numbers by adding.
The method rests on equal fractions, not on counting heads.
Instead: Estimate = first catch × second catch ÷ recaptured tagged: 40 × 60 ÷ 12.
Mistake: Dropping units from a compound-measure answer.
A number without a unit does not answer a “find the speed/pressure” question.
Instead: Write km/h, g/cm³ or Pa in the answer line — the unit often carries its own mark.

Check yourself

Quick quiz

30 questions with instant explanations. About 5 minutes.

Question 1 of 3

Question 1: current question
Question 2: not answered
Question 3: not answered

Calculate speed from distance and time

A car travels 120 km in 1.5 hours. Its average speed is

Practise

Topical questions

18 exam-style questions with model answers and mark schemes.

A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier questions are tagged.

Practise topical questions

Exam practice

Practice paper

A 30-mark paper in the style of a GCSE paper, with a separate mark scheme. Give yourself 40 min.

Question paper

Chapter 13 Practice Paper

30 marks · 40 min · 5 pages

10 multiple-choice questions and structured questions with answer lines. Chapter 13: speed, density and pressure, converting compound units, direct and inverse proportion with the unitary method and y = kx / y = k/x equations, rates of change from graphs, and capture–recapture estimation.

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Mark scheme

PRO

Chapter 13 Practice Paper — mark scheme

Answers · Marking points · 3 pages

Model answers with a mark for each creditworthy point — for marking at home or in class.

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How is this chapter examined?

Paper 1

non-calculator · 1 h 30

Exact fractions, surds and multiples of π expected

Papers 2 and 3

calculator · 1 h 30 each

Non-exact answers to 3 s.f., angles to 1 d.p.

Tiers

Foundation (grades 1–5) · Higher (grades 4–9)

Higher assumes all Foundation content

Want a full paper? Try the Paper 1-style (Foundation), Papers 2 & 3 style (Foundation), Paper 1-style (Higher), Papers 2 & 3 style (Higher) sets.

Questions

Frequently asked questions

What is the difference between direct and inverse proportion?

Direct proportion: both quantities rise together in the same ratio, so their division is constant (cost per kilogram). Inverse proportion: one rises as the other falls, so their product is constant (workers × days). Graphically, direct proportion is a straight line through the origin; inverse proportion is a curve that never touches either axis.

How do I decide whether to multiply or divide in a proportion problem?

Use the unitary method: find the value of one unit first, then ask whether you need more or fewer of them. If 250 g makes 6 cups, one gram makes 6⁄250 cups — multiplying by grams when you have more, dividing when you have less. Writing the sentence with units keeps the direction honest.

Do I need the algebra (y = kx and y = k/x) at Foundation?

No — Foundation questions can be answered with the unitary method or a multiplier. The equations are Higher-tier equipment: write the ∝ statement as an equation with k, find k from one known pair, and then the question becomes substitution.

Why do speed, density and pressure questions lose so many marks?

Almost always on units. Convert before calculating (42 minutes is 0.7 hours; 250 cm² is 0.025 m²) and write the unit on the final answer — many mark schemes award a mark for the unit alone. Remember the two big conversions: ÷ 3.6 for km/h to m/s, and × 1000 for g/cm³ to kg/m³.

GCSE Maths Proportion & rates revision