Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for compound interest & depreciation 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 Compound interest & depreciation topic page.
When you see 'per year' or 'per annum' with a fixed percentage over several years, use one multiplier raised to a power.
When there are only 2 or 3 years, or the rate changes between years, work each year out separately.
An increase of 10% multiplies by 1.1. A decrease of 10% multiplies by 0.9.
The multiplier is used once for each year.
£100 increased by 10% is £110. The next year, 10% of £110 is £11, so the total is £121.
The second year’s interest is larger because it is calculated on the new amount.
Two years at 10% is × 1.1². Three years is × 1.1³.
Do not multiply the percentage by the number of years and add that once. That would be simple interest.
A car losing 10% a year multiplies by 0.9 each year.
After two years the factor is 0.9² = 0.81, so it keeps 81% of the start.
The amount in the account includes the original money.
Interest earned is the final amount minus the start.
Compound growth should be a little more than simple interest over the same years.
Depreciation should leave you with less than you started with.
Start from 100%, which is the multiplier 1. A rise is added on. 5% is 0.05, so a 5% rise multiplies by 1.05. A 20% rise multiplies by 1.20. A 10% rise multiplies by 1.10. Write the percentage as a decimal before you add it to 1. 5% is not the multiplier. 5% of the money is only the extra part.
A fall is taken off 100%. A 20% fall leaves 80%, so the multiplier is 0.80. A 15% fall leaves 85%, so the multiplier is 0.85. A 25% fall leaves 75%, so the multiplier is 0.75. Subtract the percentage from 100 first, then write that as a decimal. Using 1.15 for a 15% fall sends the value up.
Keep the multiplier exact through the years. 1.05 and 0.85 are already decimals, so you can type them straight in. If you are working on paper, multiply in stages and write each year's amount. Round money to the nearest penny only when the question asks for a final cash amount, and only after the last multiplication.
Method: 1. Turn the percentage into a decimal and add it to 1 for a rise. 2. For a fall, subtract the decimal from 1. 3. Use that single multiplier for every later year.
Each year uses the multiplier once, on the amount you have just found. For £800 at 5% compound, year one is 800 × 1.05 = 840. Year two is 840 × 1.05 = 882. The second multiplication uses £840.00, not £800.00. That is why the second year's interest is £42.00 rather than another £40.00.
When the rate stays the same, the years can be written as a power. Two years at 5% is × 1.05^2. Three years is × 1.05^3. The power counts the years. It does not replace the percentage. 1.05^2 means 1.05 × 1.05. On a calculator, find 1.05^2 and then multiply by the starting amount, or multiply by 1.05 once per year. Both routes match when the rate does not change.
Simple interest does not use a power. Two years of simple interest at 5% on £800 adds 2 × £40.00 = £80.00, and the amount is £880.00. Compound interest on the same deal ends at £882.00. Over more years the gap grows, because each new interest payment is calculated on a higher balance. If a question says simple, do not raise the multiplier to a power.
Method: 1. Multiply the starting amount by the multiplier once for year one. 2. Multiply the new amount by the same multiplier for each later year. 3. Do not add the interest onto the original amount again.
A fall uses a multiplier below 1, then the same year-by-year habit. A van worth £2000 loses 20% of its current value each year. Year one is 2000 × 0.8 = 1600. Year two is 1600 × 0.8 = 1280. After two years the van is worth £1280.00. Two years at the same rate is × 0.8^2. 0.8^2 = 0.64, and 2000 × 0.64 = 1280.
A 15% fall is the multiplier 0.85. On £1200 for two years, 1200 × 0.85 = 1020, then 1020 × 0.85 = 867. The value is £867.00. Notice £867.00 is not £1200 minus 30% of £1200. Subtracting 15% of the original twice would be simple depreciation, and it gives a different figure. Read the question: 'each year' or 'compound' or 'depreciates by 15% per annum' means the multiplier is used once a year.
The value lost is the start minus the end. The van fell from £2000.00 to £1280.00, so it lost £720.00. The laptop style of question that asks 'how much value is lost' wants that subtraction. The question that asks 'what is it worth' wants the final amount. Label which one you have found before you stop.
Method: 1. Build a multiplier below 1 from the percentage lost. 2. Multiply the current value by that multiplier once each year. 3. The value lost is the start minus the final value.
The amount in the account is the original money plus every bit of interest. Interest earned is only the extra. For the £800 at 5% compound for two years, the amount is £882.00 and the interest earned is 882 - 800 = 82, so £82.00. Quoting £882.00 when the question asks for the interest is a full-sum error. Read the last line of the question again.
A sensible check for growth: the compound amount should be a little more than the simple amount over the same years, as long as the rate is positive. £360.00 compound against £350.00 simple, on £250 at 20% for two years, fits that pattern. A sensible check for depreciation: the final value must be below the start, and a second year of loss must leave it even lower.
If the rate changes from year to year, do not use a single power. Multiply by the first year's multiplier, then by the next year's multiplier, and so on. That chain of different rates is repeated percentage change. This lesson keeps one rate so the power form is fair. Either way, round pounds and pence at the end, and pick a check that uses the original multiplier forwards.
Method: 1. The amount in the account is the original money plus the interest. 2. Interest on its own is the final amount minus the money invested. 3. Check a short case by writing the year-by-year amounts.
For more than two years, repeated multiplication is slow and risks a slip. Use a power instead. If the yearly multiplier is 1.03 and the time is 3 years, the final amount is the starting amount times 1.03^3. On a calculator, press the power key, then the number of years. For £1 200 at 3% for 3 years, 1200 × 1.03^3 = 1311.2724, which is £1311.27.
Do not round the multiplier or any middle value. Keep all the digits in the calculator and round once, at the end, to the nearest penny. Rounding after each year moves the answer by a penny or more. Write the multiplier, the power and the unrounded product, then the rounded answer. If a question asks for the nearest pound, round to a whole number only at the last step.
Method: 1. Write the yearly multiplier, such as 1.03 for 3% growth. 2. Raise it to the power of the number of years. 3. Multiply by the starting amount without rounding. 4. Round once at the end, to the nearest penny or as the question says.
Tip: Check the sign of the change: growth gives a multiplier above 1, decay gives a multiplier below 1.
When the rate changes, you cannot use a power. Write a separate multiplier for each year and multiply them. A car that loses 20% in the first year and 10% in each later year has multipliers 0.8, 0.9 and 0.9 over three years. Its value is the price times 0.8 × 0.9 × 0.9. Where the rate repeats, a power can be used for that part: 0.8 × 0.9^2.
Comparison questions give two offers, for example Bank X at 3% every year and Bank Y at 1% then 5%. Work out the final amount for each and state which is larger and by how much. Notice that the order of the multipliers does not change the final result, because multiplication can be done in any order. The one thing that changes the result is the size of the rates themselves.
Method: 1. List the yearly rates and convert each into a multiplier. 2. Multiply the starting amount by each multiplier in turn, or multiply the multipliers together first. 3. Keep the full calculator value and round once at the end. 4. For a comparison, work out both amounts and find the difference.
To find the starting amount, divide the final amount by the overall multiplier. After two years of 4% growth the multiplier is 1.04^2 = 1.0816, so an account holding £2 704 started with 2704 ÷ 1.0816 = £2500. Taking the overall growth of 8.16% off 2704 does not give 2500, because that growth was a percentage of the smaller starting amount.
To find the rate, use the first year. If £3 200 grows to £3 296 in one year, the multiplier is 3296 ÷ 3200 = 1.03 and the rate is 3%. You then apply that rate for any further years. If the final amount is given after two years, the multiplier for one year is the square root of the overall multiplier, for example √(1.1236) = 1.06, which is a 6% rate.
Method: 1. Write the overall multiplier, or find it as new amount divided by old amount. 2. To find the start, divide the end amount by the overall multiplier. 3. To find the rate for one year, take the square root for two years, or use the first year directly. 4. Check by applying the multiplier forwards to your answer.
A harder question puts a growing amount against a shrinking one and asks when one overtakes the other. Make a table with a column for each year. A bike worth £1 200 losing 25% then 10% a year is £900, £810 and so on. A coin worth £800 growing at 8% is £864, then £933.12. After year 1 the bike is still ahead. After year 2 the coin is ahead.
To "show that" something happens in a given year, you must show the values at the end of the year before and the end of the year itself. State the comparison and a sentence that answers the question. The final mark is for the sentence, for example "so the coin is worth more than the bike during the second year". For a question that asks how many years, try n = 1, 2, 3, dots until the target is passed.
Method: 1. Set up a table with one row for each item and one column for each year. 2. Calculate each year using the multiplier, keeping full values. 3. Find the year where the order of the values changes. 4. Write a sentence that states the conclusion and quotes the two values.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.