Skip to content
VirtusAcademy

Direct Proportion

FoundationHigherAQAEdexcelOCR

Practise direct proportion with this free Foundation and Higher GCSE Maths worksheet from Virtus Academy. You'll work through solving direct proportion problems, building confidence for your exam, and every question comes with worked solutions in the mark scheme. Suitable for AQA, Edexcel and OCR. In direct proportion, doubling one quantity doubles the other.

Free downloads

These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

Foundation

WorksheetFoundation · PDFMark SchemeFoundation · PDF

Higher

WorksheetHigher · PDFMark SchemeHigher · PDF

Topic overview

Two quantities are in direct proportion when doubling one doubles the other. Their graph is a straight line through the origin, and the ratio between them stays constant.

The relationship is written \(y = kx\), where \(k\) is the constant of proportionality. Finding \(k\) from a given pair of values is usually the first step, after which any other pair can be calculated.

The phrase directly proportional signals this structure. Cost against quantity, distance against time at constant speed, and mass against volume for a given material are all everyday examples.

Revision notes

The relationship

If \(y\) is directly proportional to \(x\), then \(y = kx\) for some constant \(k\).

The symbol \(\propto\) means is proportional to, so \(y \propto x\) becomes \(y = kx\) once the constant is introduced.

Finding the constant

Substitute a known pair of values and solve for \(k\).

If \(y = 12\) when \(x = 3\), then \(12 = 3k\), so \(k = 4\) and the formula is \(y = 4x\).

Using the formula

With \(k\) known, substitute any value to find its partner.

Using \(y = 4x\): when \(x = 7\), \(y = 28\); and when \(y = 40\), \(x = 10\). The graph is a straight line through the origin with gradient \(k\).

Key points

  • Direct proportion means \(y = kx\).
  • The graph is a straight line through the origin.
  • \(k\) is the constant of proportionality.
  • Find \(k\) by substituting a known pair.
  • Doubling one quantity doubles the other.
  • The ratio between the quantities stays constant.

Worked examples

Example 1

\(y\) is directly proportional to \(x\), and \(y = 20\) when \(x = 4\). Find \(k\).

Working

\[y = kx\]write the relationship
\[20 = 4k\]substitute the known values
\[k = 5\]solve for k

Example 2

Using \(y = 5x\), find \(y\) when \(x = 9\).

Working

\[y = 5 \times 9\]substitute into the formula
\[= 45\]state the answer

Example 3

Using \(y = 5x\), find \(x\) when \(y = 65\).

Working

\[65 = 5x\]substitute the known value
\[x = 13\]divide both sides by 5

Common mistakes

  • Forgetting to find the constant first.

    You cannot use the formula until k is known from a given pair.

  • Assuming a straight line always means direct proportion.

    The line must pass through the origin, so y = 2x + 3 is not direct proportion.

  • Adding instead of multiplying.

    Direct proportion multiplies by a constant, it does not add one.

  • Dividing the wrong way when finding k.

    k = y ÷ x, so substitute carefully and solve.

Exam tips

  • Write y = kx before substituting anything.
  • Find k first, then answer the question asked.
  • Check the graph passes through the origin.
  • Sense-check: if x increases, y should increase too.

Key terms

Direct proportion
A relationship where \(y = kx\).
Constant of proportionality
The fixed multiplier \(k\).
Origin
The point \((0,0)\).
Proportional
Changing at a constant ratio.

Related topics

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.