How to Solve Direct and Inverse Proportion Problems

Direct and inverse proportion are core topics in IGCSE and GCSE Maths that appear in almost every exam series. In direct proportion, two quantities increase together at a constant rate (when one doubles, so does the other), modelled by the formula y = kx. In inverse proportion, one quantity increases as the other decreases (when one doubles, the other halves), modelled by xy = k. This page explains how to recognise each type, find the constant of proportionality k, and solve exam-style proportion problems step by step using the unitary method and the constant ratio method. Worked examples, key points, and a set of randomly generated practice questions are included so you can build accuracy and confidence before your Edexcel IGCSE exam.

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Prior Knowledge

You should be comfortable with substituting into formulae and ratio and percentage change before starting.

⇩ Jump to Practice Questions ⇩

Direct and Inverse Proportion: the key difference

Both types describe a relationship between two quantities, but they behave in opposite ways:

📈 Direct Proportion  \(y = kx\)

Both quantities change in the same direction.

  • Double one → the other doubles
  • Halve one → the other halves
  • Graph: straight line through the origin
  • Symbol: \(y \propto x\)

📉 Inverse Proportion  \(y = \dfrac{k}{x}\)

Quantities change in opposite directions.

  • Double one → the other halves
  • Treble one → the other becomes a third
  • Graph: rectangular hyperbola
  • Symbol: \(y \propto \dfrac{1}{x}\)
Key Formulas
Direct Proportion \(\displaystyle y = kx \quad \Rightarrow \quad k = \dfrac{y}{x}\)
Inverse Proportion \(\displaystyle y = \dfrac{k}{x} \quad \Rightarrow \quad k = xy\)

\(k\) is the constant of proportionality, always find it first using the given pair of values.

Step-by-Step Method (both types)

  1. Identify the type: do the quantities increase together (direct) or does one increase as the other decreases (inverse)?
  2. Write the formula: \(y = kx\) for direct, or \(y = \dfrac{k}{x}\) for inverse.
  3. Find \(k\): substitute the known pair of values into your formula.
  4. Solve: substitute the new value and calculate the answer.
  5. Check: does the answer make sense? (direct → both should go the same way; inverse → opposite).

Core Ideas

Find k first, always
Every proportion question gives you one pair of values to find \(k\). Never skip this step, \(k\) is the engine of the whole calculation.
Unitary method works too
For direct proportion you can find the value for 1 unit first, then scale. Both methods give the same answer, use whichever feels cleaner.
Spot inverse by context
Workers and time, speed and journey time, pipes and fill time, if more of one means less of the other, it's inverse proportion.
Check your answer's direction
If you have more workers the job takes fewer days, not more. Always ask: does my answer go in the right direction before moving on?

Worked Examples

💡 Example 1: Direct Proportion (cost)

A 6 km taxi ride costs £9. How much does a 10 km ride cost?

Method A: Unitary

\[ \begin{aligned} \text{Cost per km} &= \frac{9}{6} = 1.50 \\[6pt] \text{Cost for 10 km} &= 10 \times 1.50 \\ &= \mathbf{£15} \end{aligned} \]
What's happening?
  • Divide to find the cost of 1 km.
  • Multiply by the new distance.
  • Works whenever the rate is constant.

Method B: Constant of proportionality

\[ \begin{aligned} k &= \frac{y}{x} = \frac{9}{6} = 1.50 \\[6pt] y &= kx = 1.50 \times 10 \\ &= \mathbf{£15} \end{aligned} \]
What's happening?
  • Identify \(y = kx\) (direct proportion).
  • Substitute the known pair to find \(k\).
  • Use \(k\) to find the unknown.

💡 Example 2: Direct Proportion (scaling)

300 g of flour makes 12 cupcakes. How much flour for 20 cupcakes?

\[ \begin{aligned} k &= \frac{y}{x} = \frac{300}{12} = 25 \\[6pt] y &= kx = 25 \times 20 \\ &= \mathbf{500 \text{ g}} \end{aligned} \]
What's happening?
  • Flour is proportional to the number of cupcakes.
  • \(k = 25\) g per cupcake.
  • Scale up: \(25 \times 20 = 500\) g.

Tip: always state your units, "g per cupcake", to make the method clear to an examiner.

💡 Example 3: Inverse Proportion (workforce)

8 workers complete a job in 5 days. How many days for 4 workers?

\[ \begin{aligned} k &= x_1 \times y_1 = 8 \times 5 = 40 \\[6pt] y_2 &= \frac{k}{x_2} = \frac{40}{4} \\[4pt] &= \mathbf{10 \text{ days}} \end{aligned} \]
What's happening?
  • Fewer workers → more days, so this is inverse.
  • Find \(k\) by multiplying the known pair.
  • Divide \(k\) by the new number of workers.

Sense check: halving the workers should double the time. \(5 \times 2 = 10\) ✓

💡 Example 4: Inverse Proportion (speed)

At 60 km/h a journey takes 2 hours. How long at 80 km/h?

\[ \begin{aligned} k &= 60 \times 2 = 120 \\[6pt] t &= \frac{k}{v} = \frac{120}{80} \\[4pt] &= 1.5 \text{ h} \\ &= \mathbf{1 \text{ h } 30 \text{ min}} \end{aligned} \]
What's happening?
  • Same distance: faster speed → less time (inverse).
  • \(k\) = distance (speed × time = distance).
  • Divide \(k\) by the new speed.
  • Convert 1.5 h to hours and minutes if asked.

Here \(k\) has real meaning, it's the distance of the journey (120 km).

🔑 Key Points

  • Direct: \(y = kx\), graph is a straight line through the origin.
  • Inverse: \(y = \dfrac{k}{x}\), graph is a rectangular hyperbola.
  • Find \(k\) first using the pair of values given in the question.
  • Unitary method: find value for 1 unit, then scale, quick and reliable for direct proportion.
  • Sense check: always verify your answer goes in the right direction before writing it down.

⚠️ Common Pitfalls

  • Using the direct formula (\(y=kx\)) when the relationship is inverse, always identify the type first.
  • Calculating \(k\) incorrectly by adding instead of multiplying (inverse) or dividing the wrong way (direct).
  • Forgetting to convert the answer, e.g. leaving time as 1.5 hours when the question asks for hours and minutes.
  • Not showing the \(k\) step, examiners award a method mark for finding \(k\) even if the final answer is wrong.
⇩ Jump to Practice Questions ⇩

Confident with direct and inverse proportion? Next, explore percentage increase and decrease, another key topic where multipliers and proportional reasoning combine.

Next up → Percentage Change

Direct & Inverse Proportion: Practice Rooms

Practise direct proportion (\(y = kx\)) and inverse proportion (\(xy = k\)) with auto-marked questions. Numbers refresh on every load. Where the answer is not a whole number, round to 2 decimal places.

✓ Correct 0
✗ Re-attempts 0
🔥 Streak 0
🏆 Best 0

Type the number only (e.g. 24 or 3.50). Decimals accepted, round to 2 d.p. if needed.