Percentage Increase and Decrease

A 20% increase means you keep the original 100% AND add 20% more. That’s 100% + 20% = 120% of the original. As a decimal, 120% = 120 ÷ 100 = 1.2. So multiplying by 1.2 gives you the original plus the increase in one step.

You can verify with £50: finding 20% gives £10, and £50 + £10 = £60. Alternatively, £50 × 1.2 = £60. Both approaches give the same answer — but the multiplier method does it in a single calculation, which is why it’s so powerful.

If you decrease by 35%, you’re removing 35% and keeping whatever’s left. Since the whole amount is 100%, the part you keep is 100% − 35% = 65%. So “decrease by 35%” and “find 65%” are just two ways of describing the same operation: multiply by 0.65.

Test it with £200: finding 35% gives £70, and £200 − £70 = £130. Alternatively, 65% of £200 = £200 × 0.65 = £130. The answers match. This also explains why the decimal multiplier for a percentage decrease is always less than 1 — you’re finding less than 100% of the original.

Start with £100. A 50% increase adds £50, giving £150. Now decrease £150 by 50%: that’s 50% of £150 = £75 removed, leaving £75. You started with £100 but ended with £75 — a loss of £25. The two “50% changes” don’t cancel out.

The key insight is that the decrease is applied to the LARGER amount (£150), not the original (£100). So 50% of £150 = £75 is bigger than the £50 that was added. In multiplier terms: ×1.5 × ×0.5 = ×0.75, not ×1. This works for any percentage: increasing and then decreasing by the same percentage always results in a net loss, because \( (1 + p)(1 – p) = 1 – p^2 \), which is always less than 1.

12.5% means 12.5 out of 100, which simplifies to \( \frac{1}{8} \) (since 12.5 ÷ 100 = 0.125 = \( \frac{1}{8} \)). So finding 12.5% of an amount is exactly the same as dividing by 8. Adding that on to the original gives the 12.5% increase.

For example, if a coat costs £64: one-eighth is £64 ÷ 8 = £8. Adding on: £64 + £8 = £72. Using the decimal multiplier: £64 × 1.125 = £72. They match. This is a powerful mental strategy — any percentage equivalent to a simple fraction (25% = \( \frac{1}{4} \), 50% = \( \frac{1}{2} \), 12.5% = \( \frac{1}{8} \), 20% = \( \frac{1}{5} \)) can be calculated using division, which is often easier in your head.

If you increase by 150%, you keep the original 100% and add 150% more. This means 100% + 150% = 250% of the original value. As a decimal, 250% = 250 ÷ 100 = 2.5. So, multiplying by 2.5 gives you the original amount plus the increase.

Many students confuse a 150% increase with finding 150%. Finding 150% is just multiplying by 1.5 (a 50% increase). But to increase by 150%, the multiplier must be 2.5.

Example: £75 increased by 20% → 75 × 1.2 = £90

Another: £60 increased by 50% → 60 × 1.5 = £90

Creative: £80 increased by 12.5% → 80 × 1.125 = £90 — uses the fraction link (12.5% = \( \frac{1}{8} \), and 80 ÷ 8 = 10, so 80 + 10 = 90).

Trap: £100 decreased by 10% = £90 — this does give £90, but it uses a percentage decrease, not an increase as the question asks. A student focused on reaching the target of £90 might not notice they’ve used the wrong direction of change.

Example: ×0.9 — this represents a 10% decrease (keeping 90% of the original).

Another: ×0.75 — this represents a 25% decrease (keeping 75%).

Creative: ×0.995 — this is a 0.5% decrease, showing that decimal multipliers can be very close to 1 for tiny percentage changes. Or ×0.01 — a 99% decrease, showing the multiplier can be very small.

Trap: ×0.05 — a student might write this thinking “it’s a 5% decrease” because of the digits. But ×0.05 means you KEEP 5% of the original, which is actually a massive 95% decrease! For a 5% decrease, you keep 95%, so the correct multiplier is ×0.95.

Example: Increase £40 by 20% → 40 × 1.2 = £48.

Another: Decrease £80 by 25% → 80 × 0.75 = £60.

Creative: Increase £25 by 12% → 25 × 1.12 = £28. This uses a less “standard” percentage and still gives a clean two-digit result.

Trap: Increase £70 by 50% → 70 × 1.5 = £105. This starts as a two-digit number, but the result is £105 — a three-digit number. A student might assume that a percentage change on a two-digit number always gives a two-digit result, but large percentage increases (or starting near 100) can push the answer past 99.

Example: ×0.3 — this represents a 70% decrease (keeping 30%).

Another: ×0.1 — this represents a 90% decrease (keeping 10%).

Creative: ×0.49 — this represents a 51% decrease, only just satisfying the condition. Or ×0.001 — a 99.9% decrease, showing how close the multiplier can get to zero.

Trap: ×0.6 — a student might think “0.6 means a 60% decrease.” But ×0.6 means you KEEP 60% of the original, which is only a 40% decrease — less than 50%. For a decrease of more than 50%, the multiplier must be less than 0.5. This is the classic “finding x%” vs “decreasing by x%” confusion.

Example: Final amount £60, percentage increase 20%. (Because £50 × 1.2 = £60).

Another: Final amount £75, percentage increase 50%. (Because £50 × 1.5 = £75).

Creative: Final amount £50.50, percentage increase 1%. (Because £50 × 1.01 = £50.50).

Trap: Final amount £60, percentage increase 16.6% (repeating). A student working backwards knows the difference is £10, but they might mistakenly calculate the percentage out of the final amount (£10 ÷ £60 = 16.6%) rather than the original amount (£10 ÷ £50 = 20%). This exposes a deep misunderstanding of which value represents the “base” 100%.

A percentage increase means adding a positive percentage to the original 100%. The multiplier is \( \frac{100 + p}{100} \), where \( p \) is the percentage increase. Since \( p \) is positive, this is always greater than \( \frac{100}{100} = 1 \). For example, a 5% increase uses ×1.05, a 50% increase uses ×1.5, and even a 0.1% increase uses ×1.001 — all greater than 1.

Students might be tempted to say SOMETIMES if they confuse increase multipliers with decrease multipliers. The multiplier for a decrease is less than 1 (e.g. ×0.85 for a 15% decrease), but the question asks specifically about increases.

Multiplying by 0.3 finds 30% of the value — it doesn’t decrease by 30%. Finding 30% means keeping only 30% of the original, which is actually a 70% decrease. To decrease by 30%, you need to keep 70%, which means multiplying by 0.7.

For example, 0.3 × £100 = £30 (you’ve lost £70, a 70% decrease). But 0.7 × £100 = £70 (you’ve lost £30, the intended 30% decrease). The misconception arises because students see “30%” and “0.3” and assume they go together for a decrease, without thinking about whether 0.3 represents what you lose or what you keep.

If you increase by \( a\% \) and then by \( b\% \), the combined multiplier is \( (1 + \frac{a}{100})(1 + \frac{b}{100}) = 1 + \frac{a}{100} + \frac{b}{100} + \frac{ab}{10000} \). The overall percentage increase is \( a + b + \frac{ab}{100} \), which is always more than \( a + b \) (since \( \frac{ab}{100} \) is positive when both percentages are positive).

For example, two successive 10% increases: ×1.1 × ×1.1 = ×1.21, which is a 21% increase — not 20%. A 10% increase followed by a 20% increase: ×1.1 × ×1.2 = ×1.32, which is a 32% increase — not 30%. The “extra” comes from the fact that the second percentage increase acts on the already-increased amount, not just the original.

It depends on the size of the decrease. A 60% decrease means keeping 40% of the original, which IS less than half. For example, a 60% decrease on £200: £200 × 0.4 = £80, which is less than £100 (half of £200). But a 20% decrease means keeping 80%, which is MORE than half: £200 × 0.8 = £160, which is more than £100.

The boundary is exactly 50%. A decrease of more than 50% always gives a result less than half the original; a decrease of less than 50% always gives more than half. This question probes whether students connect the size of the percentage decrease to the proportion remaining — a 30% decrease keeps 70% (more than half), while a 70% decrease keeps 30% (less than half).

The student has found 30% OF £80, not increased £80 BY 30%. Multiplying by 0.3 calculates the percentage itself (the bit to add on), but the student stopped there instead of adding it back to the original. This is the “finding vs increasing” misconception.

The correct method is either: (a) find 30% then add: £80 × 0.3 = £24, then £80 + £24 = £104; or (b) use the decimal multiplier directly: £80 × 1.3 = £104. The multiplier 1.3 works because 100% + 30% = 130% = 1.3. The student’s answer of £24 is actually LESS than the starting amount — an increase should always give a result larger than the original.

The answer is correct, but the reasoning is dangerously flawed. The “cross out the last digit” rule happens to work for 90 because 90 ends in zero, and removing the zero is equivalent to dividing by 10. But this rule breaks down for numbers not ending in zero.

Even when numbers do end in zero, “crossing it out” is pedagogically dangerous because it masks the underlying place-value shifting. For example, 10% of £85 is £8.50, not £8 (which is what “crossing out the 5” would give). And for £124, “crossing out the 4” gives 12, but 10% of £124 = £12.40. The correct method is to divide by 10 (or multiply by 0.1), which moves the digits one place to the right — it doesn’t simply remove a digit. Getting the right answer by flawed reasoning builds false confidence that fails on the very next problem.

The student has used the multiplier for a 15% increase instead of a 15% decrease. This is the “wrong direction multiplier” misconception. ×1.15 adds 15% on top of the original, making the value go up — the opposite of what was asked.

For a decrease, you keep 100% − 15% = 85% of the original, so the correct multiplier is ×0.85: £60 × 0.85 = £51. Notice the student’s answer of £69 is greater than the starting value of £60 — a clear red flag that something has gone wrong, since a decrease should always give a smaller result. Students should build the habit of checking whether their answer is bigger or smaller than the original to catch this error.

The student has applied the “reversing by subtracting the same percentage” misconception. They correctly calculate 30% less than £91, but £91 is the INCREASED price — subtracting 30% of the increased price is not the reverse of adding 30% of the original price.

The correct approach: if the original was increased by 30%, then £91 = original × 1.3. To find the original, divide: £91 ÷ 1.3 = £70. Check: £70 × 1.3 = £91 ✔. The student’s answer of £63.70 is wrong because 30% of £91 (= £27.30) is larger than 30% of the true original £70 (= £21). This misconception is particularly important in real-world contexts like VAT: to find the pre-tax price, you must divide by the increase multiplier, not subtract the tax percentage.

The student has treated compound change as simple, additive change. They assumed the 10% is calculated on the original value every year. In reality, each 10% increase is calculated on the new, already-increased value from the previous year.

Because the base amount grows each year, the amount added also grows each year. The correct overall multiplier is found by multiplying the individual multipliers: \( 1.1 \times 1.1 \times 1.1 = 1.1^3 = 1.331 \). This represents an overall increase of 33.1%, not 30%.