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Inverse Proportion

FoundationHigherAQAEdexcelOCR

Inverse Proportion is a key ratio topic at GCSE Maths. This Foundation and Higher worksheet gives you exam-style questions on solving inverse proportion problems, with a full mark scheme so you can check every method mark. Suitable for AQA, Edexcel and OCR. Download the free PDF and answers below. In inverse proportion, doubling one quantity halves the other.

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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 inverse proportion when increasing one decreases the other in the same ratio. Doubling one halves the other, and their product stays constant.

The relationship is written \(y = \frac{k}{x}\), so the graph is a reciprocal curve rather than a straight line. It never touches either axis.

Everyday examples make the idea concrete. More workers means less time to finish a job; a faster speed means a shorter journey time. In each case the product — total work, or total distance — stays the same.

Revision notes

The relationship

If \(y\) is inversely proportional to \(x\), then \(y = \frac{k}{x}\), or equivalently \(xy = k\).

The product of the two quantities is constant, which is often the quickest way to work with these questions.

Finding the constant

Multiply a known pair of values together.

If \(y = 8\) when \(x = 3\), then \(k = 24\), so \(y = \frac{24}{x}\).

Recognising inverse proportion

Look for one quantity increasing while the other decreases, with the product staying the same.

If \(4\) workers take \(6\) hours, the job needs \(24\) worker-hours, so \(8\) workers take \(3\) hours.

Key points

  • Inverse proportion means \(y = \frac{k}{x}\).
  • The product \(xy\) is constant.
  • The graph is a reciprocal curve.
  • Doubling one quantity halves the other.
  • Find \(k\) by multiplying a known pair.
  • More workers means less time.

Worked examples

Example 1

\(y\) is inversely proportional to \(x\), and \(y = 6\) when \(x = 4\). Find \(k\).

Working

\[k = xy\]the product is constant
\[k = 4 \times 6\]substitute the known values
\[= 24\]state the constant

Example 2

Using \(y = \dfrac{24}{x}\), find \(y\) when \(x = 8\).

Working

\[y = \frac{24}{8}\]substitute into the formula
\[= 3\]state the answer

Example 3

\(5\) workers take \(12\) hours. How long would \(4\) workers take?

Working

\[5 \times 12 = 60 \text{ worker-hours}\]find the constant product
\[60 \div 4\]divide by the new number of workers
\[= 15 \text{ hours}\]fewer workers means more time

Common mistakes

  • Treating it as direct proportion.

    In inverse proportion one goes up as the other goes down, so you divide rather than multiply.

  • Adding the constant instead of multiplying.

    k comes from multiplying the pair, not adding them.

  • Expecting a straight-line graph.

    Inverse proportion gives a reciprocal curve, not a line.

  • Getting the direction wrong.

    Fewer workers means more time, so the answer should be larger.

Exam tips

  • Multiply a known pair to find k immediately.
  • Sense-check the direction: one up means the other down.
  • Use the constant product for worker-and-time questions.
  • Remember the graph is a curve, never a straight line.

Key terms

Inverse proportion
A relationship where \(y = \frac{k}{x}\).
Constant product
The fixed value of \(xy\).
Reciprocal
One divided by a value.
Worker-hours
The total work, found by multiplying workers by time.

Related topics

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