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GCSE Maths · EdexcelChapter 5 · Percentages
GCSE Maths · Percentages · Foundation · Lesson 5.8

GCSE Maths: Compound interest, growth and decay (Foundation)

Part of the GCSE Maths percentages chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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5.8

Compound interest, growth and decay

Edexcel 1MA1: R9, R16
Grade 4-9

Simple interest pays the same amount every year. Most savings accounts work differently. With compound interest the interest is added to the account each year, and next year’s interest is worked out on the new, bigger amount. So you earn interest on your interest. Look at £1,000 at 5% a year.

Year 1: 5% of £1,000 = £50, so you have £1,050. Year 2: 5% of £1,050 = £52.50, so you have £1,102.50. With simple interest you would have had only £1,100.

YearsSimple interest totalCompound interest total
0£1,000.00£1,000.00
1£1,050.00£1,050.00
2£1,100.00£1,102.50
3£1,150.00£1,157.63
4£1,200.00£1,215.51
5£1,250.00£1,276.28
years1234567891011201260140015401680compound (orange) grows faster than simple (blue)Amount (£)
The orange curve (compound) bends upwards; the blue line (simple) is straight.

The multiplier shortcut

Each year the amount is multiplied by 1.05. After 2 years it has been multiplied by 1.05 twice: 1.05 × 1.05 = 1.052. After n years it is multiplied by 1.05n. So

final amount = starting amount × multipliern

where n is the number of years. On a calculator use the power key (xy or ^). Do not round in the middle of a calculation; round the final money answer to the nearest penny.

Repeated percentage change: growth and decay

The same idea works for anything that changes by a fixed percentage again and again. A population that grows 3% each year, bacteria that increase 20% every hour: use a multiplier above 1 (growth). A car that loses 15% of its value each year, or a medicine that falls by 30% each hour in the blood: use a multiplier below 1 (decay). A fall in value is called depreciation.

years123456300060009000120001500018000Value (£)
A car worth £18,000 loses 15% of its value each year. The fall is fastest at first.

Key ideas

  • Compound interest: interest is added each year and then earns interest itself.
  • Final amount = start × multipliern, with n = number of years (or other periods).
  • Growth by p%: multiplier = 1 + p/100.   Decay by p%: multiplier = 1 − p/100.
  • Compound interest earned = final amount − starting amount.
  • Do not round until the end. Money: nearest penny. Population: nearest whole number.
  • Compound interest is always at least as much as simple interest, and more after year 1.
Worked example 5.8.1
Maya invests £2,000 for 3 years at 5% compound interest per year. How much does she have at the end? Work year by year.

Key words. Compound interest is interest paid on the original amount and on the interest already added. Depreciation is a fall in value.

Method. Compound interest means the interest is added each year, so next year’s interest is worked out on the bigger amount. Use the multiplier 1.05 once for each year.

  1. Write down what we know. The starting amount, the percentage change each time, and how many times it happens.
    Start = £2,000,  increase = 5% per year,  number of years = 3
  2. Find the multiplier for one year. Each year the amount grows by 5%, so it becomes 105% of the previous amount. The multiplier is 105% as a decimal.
    105% = 105 ÷ 100 = 1.05
  3. Apply the multiplier once for each year. Each year we multiply the amount we have by 1.05.
    After year 1: £2,000 × 1.05 = £2,100
    After year 2: £2,100 × 1.05 = £2,205
    After year 3: £2,205 × 1.05 = £2,315.25

Check. Estimate: with simple interest the total would be £2,300. Compound interest is a little more because the interest earns interest, and £2,315.25 is a little above £2,300. ✓

Watch out. Use the new amount each year as the starting point for the next year.

Worked example 5.8.2
Work out the total amount when £5,000 is invested for 4 years at 3% compound interest per year.

Key words. Compound interest is interest paid on the original amount and on the interest already added. Depreciation is a fall in value.

Method. Compound interest means the interest is added each year, so next year’s interest is worked out on the bigger amount. Use the multiplier 1.03 once for each year.

  1. Write down what we know. The starting amount, the percentage change each time, and how many times it happens.
    Start = £5,000,  increase = 3% per year,  number of years = 4
  2. Find the multiplier for one year. Each year the amount grows by 3%, so it becomes 103% of the previous amount. The multiplier is 103% as a decimal.
    103% = 103 ÷ 100 = 1.03
  3. Raise the multiplier to the power 4. The same increase happens 4 times, so we multiply by 1.03 4 times. That is the multiplier to the power 4.
    1.034 = 1.125509…
  4. Multiply the starting amount by 1.034. This gives the amount after all the changes in one calculation.
    £5,000 × 1.125509… = £5,627.54405
  5. Round the answer. Money is rounded to the nearest penny (2 decimal places). Round only at the very end.
    £5,627.54405 = £5,627.54

Check. Estimate: with simple interest the total would be £5,600. Compound interest is a little more because the interest earns interest, and £5,627.54 is a little above £5,600. ✓

Watch out. Do not multiply by 1.03 and then by 4. Use the multiplier to the power 4.

Worked example 5.8.3
Work out the compound interest earned when £4,000 is invested for 2 years at 2.5% per year.

Key words. Compound interest is interest paid on the original amount and on the interest already added. Depreciation is a fall in value.

Method. Compound interest means the interest is added each year, so next year’s interest is worked out on the bigger amount. Use the multiplier 1.025 once for each year.

  1. Write down what we know. The starting amount, the percentage change each time, and how many times it happens.
    Start = £4,000,  increase = 2.5% per year,  number of years = 2
  2. Find the multiplier for one year. Each year the amount grows by 2.5%, so it becomes 102.5% of the previous amount. The multiplier is 102.5% as a decimal.
    102.5% = 102.5 ÷ 100 = 1.025
  3. Raise the multiplier to the power 2. The same increase happens 2 times, so we multiply by 1.025 2 times. That is the multiplier to the power 2.
    1.0252 = 1.050625
  4. Multiply the starting amount by 1.0252. This gives the amount after all the changes in one calculation.
    £4,000 × 1.050625 = £4,202.50
  5. Subtract the original amount to find the interest. The interest is only the extra money, not the total.
    £4,202.50 − £4,000 = £202.50

Check. Estimate: with simple interest the total would be £4,200. Compound interest is a little more because the interest earns interest, and £4,202.50 is a little above £4,200. ✓

Watch out. Do not multiply by 1.025 and then by 2. Use the multiplier to the power 2.

Worked example 5.8.4
Compare the total from £1,000 invested for 3 years at 4% per year with (a) simple interest and (b) compound interest. How much more does compound interest give?

Method. Work out each total, then subtract.

  1. Simple interest: interest for one year. 4% of £1,000.
    4% of £1,000 = £40
  2. Simple interest: total. Three years of £40 added to the £1,000.
    £1,000 + 3 × £40 = £1,120
  3. Compound interest: the multiplier. 104% as a decimal.
    1.04
  4. Compound interest: total. Multiply by 1.04 three times.
    £1,000 × 1.043 = £1,000 × 1.124864 = £1,124.864 = £1,124.86
  5. Find the difference. Subtract the simple total from the compound total.
    £1,124.86 − £1,120 = £4.86

Check. Compound interest in year 2 is 4% of £1,040 = £41.60 (more than £40), and in year 3 it is bigger again. ✓

Watch out. Compound interest only beats simple interest from year 2 onwards.

Worked example 5.8.5
A car is bought for £14,000. It loses 15% of its value every year. Work out its value after 3 years.

Key words. Depreciation means the value falls each year. Decay is a repeated percentage decrease.

Method. Repeated percentage decrease: multiply by 0.85 once for each year.

  1. Write down what we know. The starting amount, the percentage change each time, and how many times it happens.
    Start = £14,000,  decrease = 15% per year,  number of years = 3
  2. Find the multiplier for one year. Each year the amount falls by 15%, so it becomes 85% of the previous amount. The multiplier is 85% as a decimal.
    85% = 85 ÷ 100 = 0.85
  3. Raise the multiplier to the power 3. The same decrease happens 3 times, so we multiply by 0.85 3 times. That is the multiplier to the power 3.
    0.853 = 0.614125
  4. Multiply the starting amount by 0.853. This gives the amount after all the changes in one calculation.
    £14,000 × 0.614125 = £8,597.75

Check. Estimate: losing 15% of the original each time would leave £7,700. Each cut is smaller than that because it is taken from a smaller amount, so the answer should be a bit more than £7,700. ✓

Watch out. Do not multiply by 0.85 and then by 3. Use the multiplier to the power 3.

Worked example 5.8.6
The population of a village is 12,500. It grows by 4% every year. Estimate the population after 5 years, to the nearest whole number.

Key words. Compound interest is interest paid on the original amount and on the interest already added. Depreciation is a fall in value.

Method. Compound interest means the interest is added each year, so next year’s interest is worked out on the bigger amount. Use the multiplier 1.04 once for each year.

  1. Write down what we know. The starting amount, the percentage change each time, and how many times it happens.
    Start = 12,500,  increase = 4% per year,  number of years = 5
  2. Find the multiplier for one year. Each year the amount grows by 4%, so it becomes 104% of the previous amount. The multiplier is 104% as a decimal.
    104% = 104 ÷ 100 = 1.04
  3. Raise the multiplier to the power 5. The same increase happens 5 times, so we multiply by 1.04 5 times. That is the multiplier to the power 5.
    1.045 = 1.216653…
  4. Multiply the starting amount by 1.045. This gives the amount after all the changes in one calculation.
    12,500 × 1.216653… = 15,208.16128
  5. Round the answer. Round to the nearest whole number.
    15,208.16128 = 15,208

Check. Estimate: with simple interest the total would be 15,000. Compound interest is a little more because the interest earns interest, and 15,208 is a little above 15,000. ✓

Watch out. Do not multiply by 1.04 and then by 5. Use the multiplier to the power 5.

Watch out

Do not work out the percentage once and multiply it by the number of years. 5% for 3 years is not 15%: the multiplier is 1.053 = 1.157625, which is a 15.76% increase. Also take care with the power: 1.053 means 1.05 × 1.05 × 1.05, not 1.05 × 3.

Exercise 5M  Compound interest, growth and decay

Calculator allowed
1
FWork out the total amount in each case. Give your answers to the nearest penny.
(a)£1,000 at 5% per year for 2 years
(b)£2,000 at 3% per year for 3 years
(c)£5,000 at 4% per year for 2 years
(d)£800 at 2% per year for 3 years
(e)£1,500 at 6% per year for 2 years
(f)£10,000 at 1.5% per year for 4 years
2
FWork out the total amount year by year in each case.
(a)£1,000 at 10% per year for 2 years
(b)£200 at 5% per year for 2 years
(c)£500 at 20% per year for 2 years
3
FWork out the compound interest earned in each case.
(a)£4,000 at 2.5% per year for 2 years
(b)£6,000 at 3% per year for 2 years
(c)£2,500 at 4% per year for 3 years
4
PSEve invests £3,000 for 3 years at 4% compound interest per year.
(a)Work out the total amount at the end of 3 years.
(b)Work out the compound interest she earns.
5
PSHiba invests £2,000 for 3 years at 4% per year. Work out how much more interest she gets if the interest is compound interest rather than simple interest.
6
PSA town has a population of 25,000. It grows by 3% each year. Work out the population after 4 years, to the nearest whole number.
7
PSA scientist starts with 500 bacteria. The number of bacteria increases by 20% every hour. Estimate the number after 5 hours, to the nearest whole number.
8
PSA car is worth £14,000 when new. It loses 15% of its value every year.
(a)Work out the value of the car after 3 years.
(b)Work out the value of the car after 5 years.
9
PSA new phone costs £800. It loses 25% of its value each year. Work out its value after 2 years. Work year by year.
10
RJoe says, “£1,000 at 10% compound interest for 2 years gives the same as 20% simple interest, because 10% + 10% = 20%.” Is he right? Work out both totals.
11
RDan wants to work out the value of £1,000 after 3 years at 5% compound interest. He writes: 1,000 × 1.15 = £1,150. Explain his mistake and work out the correct value.
12
R£1,000 is invested at 5% compound interest per year. Show that the interest earned in year 2 is £52.50. Explain why it is more than the interest in year 1.

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