Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for repeated percentage change in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Repeated percentage change topic page.
When you see 'per year for years' (interest, depreciation, population growth), use one multiplier and raise it to the power .
When the rate changes between years, or you must find how many years until a value passes a target, work one period at a time.
A 10% increase uses 1.1. A 10% decrease uses 0.9. Apply it once for each year.
£200 increases by 10% each year for 2 years ends at 242.
£200 decreases by 10% each year for 2 years is 200×0.9×0.9 = 162.
A third year multiplies once more. £1000 increases by 10% each year for 3 years is 1331.
Sometimes the question asks for the yearly multiplier, not the final amount.
The amount increases by 10% each year. It is not a one-off add of a tenth of the start.
Write the multiplier for each change before you touch the money. A 10% rise is × 1.1. A 10% fall is × 0.9. A 25% rise is × 1.25. A 20% fall is × 0.8. Then multiply in the order the story happens. £200 up 10% is 200 × 1.1 = 220. Down 10% from there is 220 × 0.9 = 198. The final amount is £198.00.
The second percentage uses the latest amount. 10% of £220.00 is £22.00, so the fall is larger in pounds than the £20.00 rise that came first. That is why the result is £198.00 and not £200.00. One combined multiplier is 1.1 × 0.9 = 0.99. The chain is a 1% fall overall, which on £200 is a £2.00 drop.
Different rises do not add either. £250 up 10% then up 20% is 250 × 1.1 = 275, then 275 × 1.2 = 330. One 30% rise would be 250 × 1.3 = 325. The chain is £5.00 higher because the 20% was taken on £275.00, not on £250.00. Add the percentages only when the question says simple interest, or a single change.
Method: 1. Write a multiplier for every rise or fall. 2. Multiply the multipliers in the order of the changes. 3. Multiply the starting amount by that combined multiplier.
On £1000 at 10% for two years, simple interest adds £100.00 twice. The amount is 1000 + 200 = 1200, so £1200.00. The compound chain, same rate twice, is 1000 × 1.1 = 1100 and 1100 × 1.1 = 1210. Compound finishes £10.00 higher. The extra £10.00 is 10% of the first year's interest.
Use that comparison as a check when the rates happen to be equal. When the rates change, there is no single 'simple' percentage to add, and the chain is the only correct method. A 25% rise followed by a 20% fall has multipliers 1.25 and 0.8. Their product is 1.25 × 0.8 = 1. The amount returns to the start. That is a property of these two multipliers, not a reason to skip the working.
On £4800, a 25% rise gives £6000.00, and a following 20% fall gives 6000 × 0.8 = 4800. A further 10% rise then gives 4800 × 1.1 = 5280. The middle step landed back on the start, and the last step moved off it. Write every year's amount. A single overall percentage is hard to see until the product of the multipliers is known.
Method: 1. Simple interest adds the same amount each year, using the original sum. 2. Compound interest multiplies the new amount each year. 3. Compare the two final amounts, not just the percentages.
If the final amount is given, divide by the multipliers in reverse order. A bike is worth £576 after a 10% fall and then a 20% fall. The last change was × 0.8, so the price just before that fall was 576 ÷ 0.8 = 720. The change before that was × 0.9, so the original was 720 ÷ 0.9 = 800. The original price was £800.00.
The order of undoing matters because the multipliers are different. Dividing by 0.9 first would answer a story in which the 10% fall happened last. Follow the sentence from the end. 'After a 10% fall then a 20% fall' means the 20% fall is the last multiplication, so it is the first division.
Check by going forwards from your original. 800 × 0.9 = 720 and 720 × 0.8 = 576. If the forward chain misses £576.00, one of the divisors is the wrong multiplier. A 20% fall is undone by dividing by 0.8, not by dividing by 1.2 and not by adding 20% of £576.
Method: 1. Write the multipliers that were applied, in order. 2. Divide the final amount by those multipliers, last one first. 3. The result is the original amount.
Keep every stage exact, and round to the nearest penny only at the end if a stage actually produces parts of a penny. The chains in this lesson are built so each stage is a whole number of pounds: £220.00, £198.00, £275.00, £330.00, £5280.00, £800.00. If a calculator shows a long decimal, do not round between stages. The next multiplier has to see the full amount.
You can multiply the multipliers first. For up 10% then up 20%, the overall factor is 1.1 × 1.2 = 1.32, and 250 × 1.32 = 330. That is one multiplication instead of two. It is worth doing when you only need the final amount. If the question also asks for the amount after the first change, do the stages separately anyway.
Name what you have found. £330.00 is the final amount. It is not the increase. The increase on the £250 chain is 330 - 250 = 80, so £80.00. On the backwards bike, £800.00 is the original and £576.00 is the final value. The value lost is £224.00. Match the noun in the question, amount or change, before you box the answer.
Method: 1. Keep each stage exact until the end. 2. Round to the nearest penny only when a money answer needs it. 3. A combined multiplier above 1 is an overall rise, and below 1 is an overall fall.
When the same percentage change happens every year, do not multiply year by year. Use one multiplier raised to a power equal to the number of years. A rise of 5% each year has multiplier 1.05, so after 4 years the amount is the original times 1.05^4. A fall of 12% each year has multiplier 0.88, so after 3 years it is the original times 0.88^3. This is the idea behind compound interest, depreciation and population growth.
On a calculator, use the power key and press equals once at the end so that no rounding creeps in. Write a line such as "amount = 2500 × 1.02^2" because the method mark is for the multiplier and the power. Round only the final answer: money to the nearest penny, people or animals to the nearest whole number. A fall of 12% for three years is not a fall of 36%, because each year's 12% is taken from a smaller amount.
Method: 1. Write the yearly multiplier, for example 1.05 for a 5% rise or 0.88 for a 12% fall. 2. Raise it to the power equal to the number of years. 3. Multiply the original amount by that result in one calculator step. 4. Round at the end to the accuracy asked for.
Tip: Press equals once. Rounding at each year gives a final answer that is slightly wrong.
Sometimes the multiplier is the unknown. If £2500 grows to £2704 in 2 years at the same rate each year, divide to get the overall multiplier, 2704 ÷ 2500 = 1.0816, then take the square root to get the yearly multiplier 1.04. That is a rate of 4%. For 3 years you would take a cube root. A calculator can find a root with its root key, or by raising the number to the power tfrac13.
Other times the number of years is unknown. For a question such as "After how many whole years will the value first be more than £25 000?" work out the value at each year using the power key. Increase the number of years by one each time until the value passes the target. Write the trials in a short table, showing the last year below the target and the first year above it, and then state the answer in a sentence.
Method: 1. For an unknown rate, divide the final amount by the original and take the matching root. 2. Subtract 1 from the yearly multiplier and write the result as a percentage. 3. For an unknown number of years, try values of n in order and record each result. 4. State the answer in a sentence, and check it by running the calculation forwards.
Banks and shops often change the rate: 4% in the first year and 2% in every later year. Do not use a single power. Write each year's multiplier separately and multiply them: 1.04 × 1.02 × 1.02, which is 1.04 × 1.02^2. Use the same idea for a car that falls by 20% in the first year and by 10% each year after that. The order does not change the final product, but the number of times each multiplier appears must match the wording.
Comparison questions ask which account or option is better. Work out the final amount for each, to the nearest penny, write the two values side by side and state which is bigger. If you are asked for interest, subtract the amount invested from the final amount. Read the question to see whether it wants the total amount or the interest, because the two differ by the original sum.
Method: 1. Write a multiplier for every year, noting where the rate changes. 2. Group repeated multipliers as powers, such as 1.02^2. 3. Multiply the original amount by all the multipliers in one calculation. 4. Compare the final amounts and write a sentence naming the larger one.
Many marks go to reasoning questions. Take "Kai says £400 rising by 10% each year for 2 years rises by 20% in total, to £480. Is he correct?" Work out the real value 400 × 1.1^2 = 484, compare it with £480 and say no. The extra £4 is the interest earned on the first year's interest. A bare "no" earns nothing. The working and the comparison earn the marks.
In a show-that question you know the answer, so the working has to reach it: 1.15 × 0.85 = 0.9775 and 1 - 0.9775 = 0.0225, which is a decrease of 2.25%. Always write the multipliers, their product and a closing sentence saying what has been shown. Keep at least four decimal places in a product of multipliers so that the percentage you quote is accurate.
Method: 1. Write down each multiplier clearly. 2. Work out the exact value and compare it with the claim in the question. 3. Use a closing sentence such as "so he is not correct because 484 is not 480". 4. Give any percentage to the accuracy requested.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.