How to Calculate Percentage Increase and Decrease

Percentage increase and decrease is a key Edexcel IGCSE Maths skill used for discounts, price changes, growth, and depreciation. In this guide you will learn three reliable methods (the multiplier method, the additive method, and the 10 percent shortcut), how to handle multi step changes, and how to work backwards to find an original value. Worked examples and our auto-marked practice rooms below give instant feedback.

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Prior Knowledge

You should be confident finding a percentage of an amount before starting. See Percentage Change if you need a refresher first.

How to Increase or Decrease by a Percentage

There are three methods, all giving the same answer: the multiplier method (fastest, and the one to use in exams), the additive method (good for showing working), and the 10% shortcut (handy for mental or non-calculator work).

  1. Find the multiplier. For an increase of \(p\%\), the multiplier is \(1 + \dfrac{p}{100}\). For a decrease, use \(1 - \dfrac{p}{100}\).
  2. Multiply. New value \(=\) Original \(\times\) multiplier.
  3. Round to the required degree of accuracy.

📐 Multiplier Method (use this in exams)

\[ \text{New value} = \text{Original} \times \left(1 \pm \frac{p}{100}\right) \]

Use \(+\) for an increase, \(-\) for a decrease. The multiplier for a 20% increase is \(1.20\); for a 15% decrease it is \(0.85\).

📐 Additive Method (useful for showing working)

\[ \text{Change} = \frac{p}{100} \times \text{Original} \] \[ \text{New value} = \text{Original} \pm \text{Change} \]

Find the change first, then add or subtract from the original.

📐 The 10% Shortcut (mental or non-calculator)

Build the percentage from easy chunks, then add it on (increase) or take it off (decrease):

  • 10% of an amount: divide by 10.
  • 5%: halve the 10%.   1%: divide by 100.
  • Combine the chunks to make the percentage you need.

Example: increase £340 by 15%. 10% of 340 is 34, and 5% is 17, so 15% is \(34 + 17 = 51\). New value \(= 340 + 51 = 391\), which matches the multiplier answer of \(340 \times 1.15\).

The multiplier decides increase or decrease

decrease increase 0.70 -30% 0.85 -15% 1.00 no change 1.15 +15% 1.30 +30%

A multiplier above 1 increases a value; below 1 decreases it. For a change of p%, increase uses 1 + p/100 and decrease uses 1 - p/100. So a 15% rise multiplies by 1.15: £200 becomes £230.

Core Ideas

The Multiplier

A multiplier bundles the whole calculation into one step. Increase 8% → multiply by \(1.08\). Decrease 8% → multiply by \(0.92\).

Multiplier + Decrease = <1

A multiplier less than 1 always means a decrease. \(0.75\) means a 25% decrease, not a 75% decrease.

Two-Step Changes

Apply multipliers one at a time in sequence. A 10% increase then a 10% decrease does not return to the original value.

Working Backwards

If a value after a change is given, divide by the multiplier to find the original. This is inverse percentages.

Percentage vs Amount

The question asks for the new value, not the change itself. Always check what is being asked before rounding.

Rounding

Only round at the very end. Carrying extra decimal places through working avoids rounding errors in multi-step problems.

Worked Examples

💡 Example 1: Increase £340 by 15%

  1. Multiplier for a 15% increase: \(1 + 0.15 = 1.15\)
\[ 340 \times 1.15 = 391 \]

New value = £391

💡 Example 2: Decrease £520 by 8%

  1. Multiplier for an 8% decrease: \(1 - 0.08 = 0.92\)
\[ 520 \times 0.92 = 478.40 \]

New value = £478.40

💡 Example 3: Salary of £28 000 rises by 6.5%

  1. Multiplier: \(1.065\)
\[ 28\,000 \times 1.065 = 29\,820 \]

New salary = £29 820

💡 Example 4: Two-step increase 10% then decrease 5%

  1. After +10%: \(£200 \times 1.10 = £220\)
  2. After −5%: \(£220 \times 0.95 = £209\)
\[ 200 \times 1.10 \times 0.95 = 209 \]

Note: this is not the same as a net 5% increase.

💡 Example 5: Additive method, increase 72 by 25%

  1. Find 25% of 72: \(\dfrac{25}{100} \times 72 = 18\)
  2. Add to original: \(72 + 18 = 90\)
\[ 72 + 18 = 90 \]

💡 Example 6: Car worth £14 500 depreciates by 12%

  1. Multiplier: \(1 - 0.12 = 0.88\)
\[ 14\,500 \times 0.88 = 12\,760 \]

Value after one year = £12 760

🔑 Key Points

  • Increase multiplier \(= 1 + \dfrac{p}{100}\); decrease multiplier \(= 1 - \dfrac{p}{100}\).
  • A multiplier less than 1 always represents a decrease.
  • For two-step changes, multiply the multipliers together.
  • Only round your final answer, not intermediate steps.
  • Always re-read: the question usually asks for the new value, not the amount of change.

⚠️ Common Pitfalls

  • Using the percentage as the multiplier (e.g. multiplying by 0.20 instead of 1.20 for a 20% increase).
  • Assuming a 10% increase then 10% decrease returns to the original (it doesn't).
  • Combining two percentages into one step (e.g. treating +10% then +5% as +15%).
  • Rounding too early in multi-step problems, causing a loss of accuracy.
  • Confusing percentage change with percentage of the original.
⇩ Jump to Practice Questions ⇩

Ready to practise? Use the auto-marked rooms below to build speed and accuracy, then move on when you are confident.

Next: Prime Factorisation →

Percentage Increase & Decrease: Practice Room

Practise percentage increase and decrease across five auto-marked rooms: one-step changes, real-life scenarios, two-step changes, working backwards to an original value, and a mixed challenge. Each room has 16 questions that get harder from left to right. Round money to 2 decimal places (e.g. 122.50), give counts as whole numbers, and write multipliers like 1.15. Currency and percent symbols are ignored when marking.

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