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Maths · Ratio and Proportion

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Repeated Percentage Change and Compound Interest

Using multipliers for percentage change, compound interest and depreciation, and working back to the original amount.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Use a multiplier to increase or decrease an amount by a percentage in one step.
  2. 2Solve repeated percentage change problems, including compound interest and depreciation.
  3. 3Compare simple interest with compound interest.
  4. 4Find the original amount after a percentage change (Higher tier).

Percentages that build on each other

A single percentage change is easy, but real life often repeats it: a bank adds interest every year, and a car loses value every year. The important idea is that each change is applied to the new amount, not the original, so the changes cannot just be added together. A multiplier turns the whole calculation into repeated multiplication, which is quick and tidy, even without a calculator when the numbers are chosen well.

Percentage multipliers

A multiplier is the single number you multiply by to apply a percentage change in one step.

  • Increase

    Add the percentage to 100% and write it as a decimal. An increase of 15% gives \(115\% = 1.15\).

  • Decrease

    Subtract the percentage from 100% and write it as a decimal. A decrease of 15% gives \(85\% = 0.85\).

  • Some to know

    \(+10\% = 1.1\), \(+25\% = 1.25\), \(+5\% = 1.05\), \(-10\% = 0.9\), \(-20\% = 0.8\), \(-2\% = 0.98\).

  • Why it works

    Increasing by 15% means you keep 100% and gain a further 15%, which is 115% of the original amount.

A single change with a multiplier

A phone costs £400. In a sale the price is reduced by 15%. Work out the sale price.

Show the solutionHide the solution
  1. 1 Find the multiplier \(100\% - 15\% = 85\%\), so the multiplier is 0.85.
  2. 2 Multiply \(400 \times 0.85\).
  3. 3 Calculate \(400 \times 0.85 = 340\).
  4. 4 Write the answer £340.

Answer£340

Repeated percentage change

To apply the same percentage change \(n\) times, raise the multiplier to the power \(n\).

  • The formula

    New amount \(=\) original amount \(\times\) multiplier\(^n\).

  • Growth

    A multiplier above 1, such as 1.05, gives growth. This is compound interest, or population increase.

  • Decay

    A multiplier below 1, such as 0.8, gives decay. This is depreciation of a car, or a falling population.

  • Different percentages

    If the percentage changes each year, multiply by a different multiplier each time: \(\times 1.05 \times 1.03\).

Compound interest

Sam invests £2000 at 5% compound interest per year. Work out the value of the investment after 2 years.

Show the solutionHide the solution
  1. 1 Find the multiplier \(100\% + 5\% = 105\%\), so the multiplier is 1.05.
  2. 2 Apply it twice \(2000 \times 1.05^2\).
  3. 3 Work out the power \(1.05^2 = 1.1025\), so \(2000 \times 1.1025 = 2205\).
  4. 4 Or do it year by year \(2000 \times 1.05 = 2100\), then \(2100 \times 1.05 = 2205\).
  5. 5 Write the answer £2205.

Answer£2205

Depreciation

A car is worth £12 000. Its value falls by 20% each year. Work out its value after 2 years.

Show the solutionHide the solution
  1. 1 Find the multiplier \(100\% - 20\% = 80\%\), so the multiplier is 0.8.
  2. 2 Apply it twice \(12\,000 \times 0.8 = 9600\), then \(9600 \times 0.8 = 7680\).
  3. 3 Write the answer £7680.
  4. 4 Check The car is worth less each year, and the fall in the second year (£1920) is smaller than in the first (£2400), as it should be.

Answer£7680

Simple and compound interest

There are two ways a bank can pay interest, and the difference is what the interest is calculated on.

  • Simple interest

    The same amount is added every year, because it is always a percentage of the original amount. 3% of £2000 is £60 a year.

  • Compound interest

    Interest is added to the account and then earns interest itself, so the amount added grows each year.

  • Which is bigger

    After the first year they are the same. After that compound interest is always larger.

  • Reading questions

    Unless the question says "simple", assume compound interest.

Finding the original amount (Higher tier)

Sometimes you are told the amount after a change and asked for the amount before. You work backwards by dividing by the multiplier.

  • Divide, do not subtract

    If an amount after a 20% increase is £90, the original is \(90 \div 1.2 = 75\), not \(90 - 20\%\) of 90.

  • Same for repeated change

    After two 10% increases the amount is \(1.1^2 = 1.21\) times the original, so divide by 1.21.

  • Check by going forwards

    Multiply your answer by the multiplier and you should get the amount in the question.

Working back to the original

(Higher tier) After a 25% increase, a house is worth £150 000. Work out its value before the increase.

Show the solutionHide the solution
  1. 1 Find the multiplier \(100\% + 25\% = 125\%\), so the multiplier is 1.25.
  2. 2 Divide by it \(150\,000 \div 1.25\).
  3. 3 Calculate \(150\,000 \div 1.25 = 120\,000\).
  4. 4 Check \(120\,000 \times 1.25 = 150\,000\). Correct.

Answer£120 000

What the AQA exam gives you

AQA gives very few formulae in the exam, so most of the formulae in this lesson have to be remembered.

  • Compound interest is not given

    AQA expects you to know total accrued \(= P\left(1 + \dfrac{r}{100}\right)^n\), where \(P\) is the amount invested, \(r\) is the percentage rate and \(n\) is the number of times it is compounded, such as the number of years.

  • Multipliers do the same job

    Writing \(P \times 1.05^n\) is the same formula, so learn the multiplier method and you have the formula.

  • Also not given

    Simple interest and percentage change are not on the page, so learn how to do them.

Test yourself

  1. 1

    What is the multiplier for an increase of 8%?

    Show answerHide answer

    1.08.

  2. 2

    What is the multiplier for a decrease of 12%?

    Show answerHide answer

    0.88.

  3. 3

    How do you apply a 5% increase three times?

    Show answerHide answer

    Multiply by \(1.05^3\).

  4. 4

    What is the difference between simple and compound interest?

    Show answerHide answer

    Simple interest is always calculated on the original amount, while compound interest is calculated on the amount at the start of each year.

  5. 5

    How do you find the original amount after a percentage increase?

    Show answerHide answer

    Divide by the multiplier.

Exam technique: repeated percentage change

Show the multiplier and the power, because a final answer alone can score no marks if it is wrong.

  • Write the multiplier

    "\(\times 1.05\)" is a mark in itself.

  • Keep every decimal until the end

    Rounding the multiplier too early changes the answer.

  • Read the wording

    "Compound" and "each year" mean repeat the multiplier, and "simple" means the same amount each year.

  • Give money to 2 decimal places

    A final answer of £2205.0 or £2205.123 loses the mark.

Summary and exam focus

  • A multiplier applies a percentage change in one step: 1.15 for +15% and 0.85 for \(-15\%\).
  • Repeated change uses multiplier\(^n\), and each change is applied to the new amount.
  • Compound interest grows faster than simple interest after the first year.
  • To find the original amount after a change, divide by the multiplier (Higher tier).

Exam focus

Aisha invests £2000 for 2 years at 4% per year compound interest. Work out the total interest she earns. (3 marks) (3 marks)

Use \(2000 \times 1.04^2 = 2163.20\) and then subtract the £2000 to get £163.20. The question asks for the interest, not the total value, so the last subtraction is worth a mark. Show the multiplier clearly.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Multiplier
The number you multiply by to apply a percentage change in one step.
Compound interest
Interest calculated on the original amount plus any interest already added.
Simple interest
Interest calculated only on the original amount, giving the same amount each year.
Depreciation
The fall in value of an item, such as a car, over time.
Growth
An increase over time, with a multiplier greater than 1.
Decay
A decrease over time, with a multiplier less than 1.
Principal
The amount of money first invested or borrowed.
Original amount
The starting value before a percentage change is applied.
Per annum
Each year.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Work out 2 marks Easier

Increase \(\pounds 60\) by 15%.

Mark scheme — 2 marks available

  • 15% of 60 \(= 9\), or \(60 \times 1.15\) — M1
  • \(\pounds 69\) — A1

Model answer

10% of 60 is 6 and 5% is 3, so 15% is 9. The new amount is \(60 + 9 = \pounds 69\).

2. Exam question Work out 3 marks Core

\(\pounds 4000\) is invested at 3% per year compound interest. Work out the value of the investment after 2 years.

Mark scheme — 3 marks available

  • \(4000 \times 1.03\) or \(4000 \times 1.03^2\) — M1
  • \(4120 \times 1.03\) — M1
  • \(\pounds 4243.60\) — A1

Model answer

After year 1: \(4000 \times 1.03 = 4120\). After year 2: \(4120 \times 1.03 = 4243.60\). The value is \(\pounds 4243.60\).

3. Exam question Work out 3 marks Core

A tractor is worth \(\pounds 30\,000\). Each year its value falls by 20% of its value at the start of that year. Work out the value of the tractor after 2 years.

Mark scheme — 3 marks available

  • \(30\,000 \times 0.8\) — M1
  • \(24\,000 \times 0.8\) — M1
  • \(\pounds 19\,200\) — A1

Model answer

After year 1: \(30\,000 \times 0.8 = 24\,000\). After year 2: \(24\,000 \times 0.8 = 19\,200\). The value is \(\pounds 19\,200\).

4. Exam question Work out 4 marks Stretch

Zoe invests \(\pounds 500\) for 2 years at 10% per year simple interest. Yan invests \(\pounds 500\) for 2 years at 10% per year compound interest. How much more money does Yan have than Zoe at the end of the 2 years?

Mark scheme — 4 marks available

  • \(\pounds 600\) for simple interest — M1
  • \(500 \times 1.1\) or \(\pounds 550\) — M1
  • \(\pounds 605\) — A1
  • \(\pounds 5\) — A1

Model answer

Zoe: \(10\%\) of 500 is 50, so after 2 years she has \(500 + 2 \times 50 = \pounds 600\). Yan: \(500 \times 1.1 = 550\) and \(550 \times 1.1 = 605\). The difference is \(605 - 600 = \pounds 5\).

5. Exam question Work out 2 marks Stretch

After a 10% increase, the price of a meal is \(\pounds 44\). Work out the price before the increase.

Mark scheme — 2 marks available

  • \(44 \div 1.1\) — M1
  • \(\pounds 40\) — A1

Model answer

The multiplier is 1.1, so the original price is \(44 \div 1.1 = \pounds 40\).

6. Exam question Work out 3 marks Core

The number of rabbits on an island increases by 20% each year. There are 500 rabbits at the start of the first year. Work out the number of rabbits after 2 years.

Mark scheme — 3 marks available

  • \(500 \times 1.2\) — M1
  • \(600 \times 1.2\) or \(500 \times 1.44\) — M1
  • 720 — A1

Model answer

After year 1: \(500 \times 1.2 = 600\). After year 2: \(600 \times 1.2 = 720\). There are 720 rabbits.

7. Multiple choice 1 mark Easier

What is the multiplier for an increase of 15%?

  1. A 0.15
  2. B 1.15 Correct
  3. C 0.85
  4. D 1.015

Why: An increase of 15% gives \(100\% + 15\% = 115\%\), which is 1.15.

8. Multiple choice 1 mark Easier

What is the multiplier for a decrease of 12%?

  1. A 0.88 Correct
  2. B 0.12
  3. C 1.12
  4. D 0.012

Why: \(100\% - 12\% = 88\%\), which is 0.88.

9. Multiple choice 1 mark Easier

A phone costs \(\pounds 400\). The price is reduced by 15%. What is the new price?

  1. A \(\pounds 385\)
  2. B \(\pounds 60\)
  3. C \(\pounds 460\)
  4. D \(\pounds 340\) Correct

Why: \(400 \times 0.85 = 340\).

10. Multiple choice 1 mark Core

\(\pounds 1000\) increases by 10% each year for 2 years. What is it worth after 2 years?

  1. A \(\pounds 1200\)
  2. B \(\pounds 1100\)
  3. C \(\pounds 1210\) Correct
  4. D \(\pounds 1110\)

Why: \(1000 \times 1.1 = 1100\) and \(1100 \times 1.1 = 1210\). The second increase is 10% of the new amount.

11. Multiple choice 1 mark Core

\(\pounds 2000\) is invested at 5% compound interest per year. What is it worth after 2 years?

  1. A \(\pounds 2200\)
  2. B \(\pounds 2205\) Correct
  3. C \(\pounds 2100\)
  4. D \(\pounds 2210\)

Why: \(2000 \times 1.05^2 = 2000 \times 1.1025 = 2205\). The simple interest answer would be \(\pounds 2200\).

12. Multiple choice 1 mark Core

A car worth \(\pounds 12\,000\) loses 20% of its value each year. What is it worth after 2 years?

  1. A \(\pounds 7680\) Correct
  2. B \(\pounds 7200\)
  3. C \(\pounds 9600\)
  4. D \(\pounds 6144\)

Why: \(12\,000 \times 0.8 = 9600\) and \(9600 \times 0.8 = 7680\). Taking 40% off in one go would give \(\pounds 7200\).

13. Multiple choice 1 mark Easier

Which statement about interest is correct?

  1. A Simple interest grows faster than compound interest
  2. B Simple and compound interest are always equal
  3. C Compound interest adds the same amount every year
  4. D After the first year, compound interest is greater than simple interest Correct

Why: Compound interest is paid on interest already earned, so it grows faster than simple interest, which is always paid on the original amount.

14. Multiple choice 1 mark Stretch

After a 25% increase, a house is worth \(\pounds 150\,000\). What was it worth before the increase?

  1. A \(\pounds 112\,500\)
  2. B \(\pounds 187\,500\)
  3. C \(\pounds 120\,000\) Correct
  4. D \(\pounds 125\,000\)

Why: The multiplier is 1.25, so the original is \(150\,000 \div 1.25 = 120\,000\).

15. Multiple choice 1 mark Core

A population of 8000 falls by 10% each year. What is the population after 2 years?

  1. A 6400
  2. B 6480 Correct
  3. C 7200
  4. D 6600

Why: \(8000 \times 0.9^2 = 8000 \times 0.81 = 6480\).