Edexcel · GCSE Maths · 1MA1 · Foundation and Higher
M24 · Direct and inverse proportion
Revision notes, worked examples and methods for direct and inverse proportion.
Revision notes ready · Quizzes and videos coming soon.
Foundation shows shared notes. Higher includes shared notes and the labelled Higher extensions. Changing tier starts the notes again.
Revise the key ideas
Direct and inverse relationships
Direct proportion means y changes by the same factor as x: y = kx for constant k. The graph is a straight line through the origin. A straight line with a non-zero intercept is not direct proportion. Direct and inverse proportion
Worked example: Five notebooks cost £12.50 at a constant unit price. One costs £2.50, so eight cost £20. Here cost = 2.5 × number of notebooks.
For direct proportion, yx is constant. Doubling x doubles y; tripling x triples y. Test ratios rather than differences.
Inverse proportion means y = kx, so xy is constant. For positive quantities, doubling x halves y. Its graph is a reciprocal curve, not a straight line.
Worked example: Four equally productive workers take 9 hours for a fixed job. Six workers take (4 × 9) ÷ 6 = 6 hours, assuming work is shared perfectly and each worker's rate is unchanged.
For a fixed distance, time is inversely proportional to speed. This needs a constant journey length; a general time-versus-speed situation may not be inverse proportion.
Using equations
To use a proportional equation, find the constant from known values, then substitute the new input. Keep the complete relation, not just the value of k.
Worked example: y = 24x. When x = 3, y = 8; when x = 8, y = 3. x = 0 is not allowed.
Read the context carefully: a fixed starting fee, changing productivity or a changing total can invalidate a proportional model.
Higher — powers and constructing models
If y is directly proportional to x2, write y = kx2. If y is inversely proportional to x2, write y = kx2. State which power the question specifies.
Worked example: y ∝ x2, and y = 18 when x = 3. Then 18 = 9k, so k = 2 and y = 2x2. At x = 5, y = 50.
Worked example: y ∝ 1x2, and y = 12 when x = 2. Then k = 48, so y = 48x2. At x = 4, y = 3.
If y ∝ √x and y = 15 at x = 9, then k = 5 and y = 5√x. To find x when y = 20, √x = 4, so x = 16.
Plotting y against x2 gives a straight line through the origin for y = kx2. Its gradient is k. Plotting against 1x does the same for inverse proportion.
Test yourself
Quiz coming soon
Practice questions with explained answers will be added here.
For now, cover the worked answers, try the calculations yourself, then compare each step. Include units and reasons where needed.
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