Reverse Percentages
Questions designed to stretch thinking, reveal misconceptions, and spark mathematical reasoning.
Convince Me That…
Students must construct a mathematical argument for why each statement is true.
If the original price was £48, then a 25% increase means adding 25% of £48: \( 0.25 \times 48 = 12 \). So the new price is £48 + £12 = £60. ✓ The correct method is to divide by the multiplier: £60 ÷ 1.25 = £48.
A common mistake is to subtract 25% of £60 instead: 25% of £60 = £15, giving £60 − £15 = £45. But this is the “percentage of the wrong amount” error — it treats £60 as if it were 100% of the original, when it’s actually 125% of the original. The 25% was calculated on the original amount, not on £60. Since the original is smaller than £60, 25% of it must be less than £15.
Start with £100. A 10% increase gives \( 100 \times 1.1 = 110 \). Now decrease £110 by 10%: 10% of £110 = £11, so £110 − £11 = £99. We end up with £99, not £100. The price has dropped by 1%.
The two 10%s are calculated on different amounts. The increase adds 10% of the original, but the decrease subtracts 10% of the larger amount. So the decrease removes more than the increase added. This is the “percentage changes are reversible” misconception — students assume that doing and undoing the same percentage cancels out. In fact, multiplying by 1.1 then 0.9 gives 0.99, not 1.
A 30% reduction means the customer pays 70% of the original. So £42 represents 70% of the original price. To find the original, divide by 0.7: \( 42 \div 0.7 = 60 \). Check: \( 60 \times 0.7 = 42 \). ✓
Many students try to add 30% of £42 back on: 30% of £42 = £12.60, giving £42 + £12.60 = £54.60. But this is wrong because the 30% discount was calculated on the original price (£60), not on the sale price (£42). The 30% of the original (£18) is larger than 30% of the sale price (£12.60), which is exactly why the “percentage of the wrong amount” method always undershoots.
If the original value was \( x \), then after a 15% increase it becomes \( x \times 1.15 = 2300 \). This is a multiplication equation, and the inverse of multiplication is division: \( x = 2300 \div 1.15 = 2000 \). Check: \( 2000 \times 1.15 = 2300 \). ✓ The multiplier 1.15 represents 100% + 15% = 115% = 1.15.
The key insight is that £2300 IS 115% of the original — not 100%. Once students see that £2300 = 1.15 × original, dividing both sides by 1.15 becomes the obvious step. This is why you can’t simply subtract 15% of £2300: that would treat £2300 as 100%, but it’s actually 115%. This is the “confusing which value is 100%” misconception.
We can test the original price working forwards: a 20% increase on £50 gives \( 50 \times 1.2 = 60 \). Then, a 10% decrease on £60 gives \( 60 \times 0.9 = 54 \). We get £54, so the original must be £50. ✓
Alternatively, we can work backwards systematically by chaining the inverse operations. To reverse the 10% decrease, divide by 0.9: \( 54 \div 0.9 = 60 \). To reverse the 20% increase, divide by 1.2: \( 60 \div 1.2 = 50 \). This tests whether students know that compound reverse percentages can be solved by dividing by both multipliers in sequence: \( 54 \div 0.9 \div 1.2 = 50 \).
Give an Example Of…
Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.
Example: New price £72, 20% increase (\( 60 \times 1.2 = 72 \), so \( 72 \div 1.2 = 60 \) ✓)
Another: New price £90, 50% increase (\( 60 \times 1.5 = 90 \) ✓)
Creative: New price £60.60, 1% increase (\( 60 \times 1.01 = 60.60 \)). Or new price £120, 100% increase.
Trap: “New price £48 after a 20% decrease.” This correctly gets back to £60 (\( 48 \div 0.8 = 60 \)), but the question specifically asks for a percentage increase! A student might ignore the condition and just look for any combination that works.
Example: 10% off, sale price £180 (\( 200 \times 0.9 = 180 \), so \( 180 \div 0.9 = 200 \) ✓)
Another: 25% off, sale price £150 (\( 200 \times 0.75 = 150 \) ✓)
Creative: 0.5% off, sale price £199 (\( 200 \times 0.995 = 199 \) ✓). Or 80% off, sale price £40 (\( 200 \times 0.2 = 40 \) ✓).
Trap: “20% off, sale price £200” — a student might think the original is £200, but if £200 is the sale price after 20% off, the original = \( 200 \div 0.8 = 250 \), not £200. The trap exploits confusing which value is the original and which is the sale price.
Example: 60% discount, sale price £40 → \( 40 \div 0.4 = 100 \). Since £100 > 2 × £40 = £80. ✓
Another: 75% discount, sale price £25 → \( 25 \div 0.25 = 100 \). Since £100 > 2 × £25 = £50. ✓
Creative: 95% discount, sale price £3 → \( 3 \div 0.05 = 60 \). Since £60 > 2 × £3 = £6. ✓ Extreme discounts produce dramatic multiplier effects.
Trap: 40% discount, sale price £60 → \( 60 \div 0.6 = 100 \). But £100 < 2 × £60 = £120, so the original is NOT more than double. A student might think “40% is a big reduction, so the original must be more than double,” but you actually need a discount greater than 50% for the original to exceed double the sale price.
Example: 200% increase to £150 → multiplier = 1 + 2 = 3. Original = \( 150 \div 3 = 50 \). Check: \( 50 \times 3 = 150 \). ✓
Another: 150% increase to £250 → multiplier = 2.5. Original = \( 250 \div 2.5 = 100 \). ✓
Creative: 900% increase to £1000 → multiplier = 10. Original = \( 1000 \div 10 = 100 \). The multiplier is a whole number, making the division trivial — students rarely consider that a percentage problem can reduce to simple division by 10.
Trap: “100% increase to £200” → multiplier = 2. Original = \( 200 \div 2 = 100 \). But 100% is NOT greater than 100%, it’s equal. A student might include this thinking “100% counts,” but the question asks for greater than 100%.
Always, Sometimes, Never
Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.
The decrease is always calculated on the larger amount produced by the increase, so it removes more than the increase added. For example: start with £100, increase by 20% → £120, decrease by 20% → \( 120 \times 0.8 = 96 \), not £100. Algebraically, multiplying by \( (1 + p) \) then \( (1 – p) \) gives \( (1 – p^2) \), which is less than 1 for any non-zero \( p \).
Students often think percentage changes work like adding and subtracting the same number. But percentages are multiplicative, not additive — the base changes after the first operation, so the second operation acts on a different amount. (Strictly, a 0% increase followed by a 0% decrease does return to the original — but that’s a trivial case of “no change at all.” For any non-zero percentage, the result is always less than the original.)
Think of percentages like time travel. A percentage is bound to the specific moment in time it was created. If you calculate 20% on Monday’s price, you can’t subtract 20% from Tuesday’s completely different price and expect to end up back on Monday. The ‘size’ of 1% changes as the base value changes.
It depends on whether the percentage change was an increase or a decrease. If the value increased (e.g. 20% increase to £120), the original (£100) is less than the given value — TRUE. If the value decreased (e.g. 20% decrease to £80), the original (£100) is greater than the given value — FALSE.
This catches students who only practise reverse percentages with increases. They need to think about the direction of the change: increases make the original smaller than the new amount; decreases make the original larger.
A percentage increase of \( p\% \) means the new value is \( (100 + p)\% \) of the original, which corresponds to a multiplier of \( 1 + \frac{p}{100} \). Since \( p \) is positive, this multiplier is always greater than 1. To reverse the operation, you divide by this multiplier — and dividing by something greater than 1 always gives a smaller result, which makes sense because the original must be less than the increased value.
Students who confuse the multipliers might try dividing by 0.8 to “undo” a 20% increase (using the decrease multiplier). This would give a larger result, which contradicts the fact that the original should be smaller than the increased amount.
For percentage increases, this is true: a 10% increase uses multiplier 1.1, a 30% increase uses 1.3, and \( 1.3 > 1.1 \). Larger percentage → larger divisor. But for percentage decreases, it’s the opposite: a 10% decrease uses multiplier 0.9, a 30% decrease uses 0.7, and \( 0.7 < 0.9 \). Larger percentage → smaller divisor.
This highlights the asymmetry between increases and decreases. For increases, the multiplier grows with the percentage; for decreases, the multiplier shrinks. Students who don’t recognise this may apply a blanket rule that “bigger percentage = bigger number to divide by,” which fails for decreases.
Odd One Out
Which is the odd one out? Can you make a case for each one? There’s no single right answer!
Explain the Mistake
Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?
Answer: £112.50
Reasoning: “I found 25% of £150, which is £37.50, and subtracted it. £150 − £37.50 = £112.50.”
The student subtracted 25% of the new price instead of the original price. This is the “percentage of the wrong amount” misconception — the most common error in reverse percentages. The 25% increase was calculated on the original, not on £150. Since £150 represents 125% of the original, we divide: \( 150 \div 1.25 = 120 \).
A quick check exposes the error: 25% of £120 = £30, and £120 + £30 = £150. ✓ But 25% of £112.50 = £28.125, and £112.50 + £28.125 = £140.625 ≠ £150. ✗ The student’s answer doesn’t reverse back to the given price, which should always be the first check.
Answer: £200,000 ✓
Reasoning: “I tried subtracting 10% of £220,000, which is £22,000, giving £198,000. That didn’t feel right because it’s not a round number. So I guessed £200,000 and checked: 10% of £200,000 is £20,000, and £200,000 + £20,000 = £220,000. So the answer must be £200,000.”
The answer is correct, but the student has no reliable method. Their first instinct — subtracting 10% of £220,000 — is the classic wrong approach and gave £198,000. They only found the correct answer (£200,000) by guessing a round number and checking it. This guess-and-check approach fails when the answer isn’t a round number: try reversing a 10% increase to £214,500 by guessing!
The systematic method is \( 220{,}000 \div 1.1 = 200{,}000 \). Students must understand that £220,000 represents 110% of the original, so dividing by 1.1 always works — regardless of whether the answer is “round.” Getting the right answer through guessing gives false confidence and masks a fundamental misunderstanding of the method.
Answer: £288.89 (to 2 d.p.)
Reasoning: “To reverse a 35% change, I divided by 1.35. £390 ÷ 1.35 = £288.89.”
The student used the increase multiplier (1.35) instead of the decrease multiplier (0.65). This happens when students memorize a rigid rule like “for reverse percentages, always divide by 1 point something” without understanding why. A 35% decrease means the customer pays 65% of the original, so the multiplier is 0.65, not 1.35. The correct calculation is \( 390 \div 0.65 = 600 \). Check: \( 600 \times 0.65 = 390 \). ✓
The student’s answer of £288.89 is less than £390, which should be an immediate red flag — if the price was reduced in a sale, the original must be higher than the sale price, not lower. Dividing by a number greater than 1 always makes the result smaller, but after a reduction, we need the result to be larger. This sense-check catches this error every time.
Answer: £864
Reasoning: “20% increase means I multiply by 1.2. £720 × 1.2 = £864.”
The student correctly identified 1.2 as the relevant multiplier but applied it in the wrong direction — they multiplied when they should have divided. This error often comes from impulsive reading. A student sees “£720” and “20% increase”, stops reading, and simply calculates a 20% increase on the number in front of them, ignoring the phrase “Find the original price”.
Multiplying by 1.2 applies a 20% increase, but we need to undo a 20% increase, which requires the inverse operation: division. The correct answer is \( 720 \div 1.2 = 600 \). Check: \( 600 \times 1.2 = 720 \). ✓ Whenever students know the multiplier but get confused about the direction, they should ask: “Is my answer bigger or smaller than the amount given — and does that make sense given the direction of the change?”
Answer: £39.13 (to 2 d.p.)
Reasoning: “£45 ÷ 1.15 = £39.13.”
The student assumed that £45 was the new total price (representing 115%), but £45 actually represents the 15% increase itself. The problem states the price “goes up by £45”, not “goes up to £45”.
Because £45 equals 15% of the original, we should divide by 0.15 (the decimal for 15%) rather than 1.15. The correct calculation is \( 45 \div 0.15 = 300 \). Check: 15% of £300 is £45. ✓ Students frequently confuse a problem that gives the difference with one that gives the new total.
Answer: £20,571.43
Reasoning: “Total drop is 10% + 20% = 30%. So the multiplier is 0.7. £14,400 ÷ 0.7 = £20,571.43.”
The student added the percentages together to make 30%. However, you cannot add percentages that are calculated from different base amounts. The 20% drop was taken from a smaller value at the end of year one, not the original value.
The changes must be reversed sequentially by dividing by each multiplier in turn. The correct calculation is \( 14{,}400 \div 0.8 \div 0.9 = 20{,}000 \). (Alternatively, finding the combined multiplier: \( 0.9 \times 0.8 = 0.72 \), then \( 14{,}400 \div 0.72 = 20{,}000 \)). Check: \( 20{,}000 \times 0.9 \times 0.8 = 14{,}400 \). ✓