1. Simple interest
How it works
- Interest is money added to a sum at regular intervals, increasing savings or increasing a debt
- Simple interest is worked out on the starting amount alone, so the same payment recurs every period
- The balance grows by equal steps, which is what separates it from compound interest
Calculating it
- Find the stated percentage of the starting amount to get one period's interest
- Multiply by the number of periods for the total interest
- Add that total to the starting amount only if the question asks for the final balance
Worked example
Simple interest over several years
£4500 is invested at 3% simple interest per year for 7 years. Find the total interest earned and the final balance.
Solution:
- Simple interest is worked out on the starting amount every year, so each year earns exactly the same
- One year's interest is 3% of £4500
- 3% as a decimal is 0.03, so the interest is 0.03 × 4500 = £135
- That same £135 is added in each of the 7 years, so the interest never grows
- Total interest = 135 × 7 = £945
- The balance is the original sum plus the interest it has earned
- Final balance = 4500 + 945 = £5445
- Read the question carefully: the interest (£945) and the balance (£5445) are different answers, and each is asked for separately
- Both answers are exact, so neither needs rounding
2. Compound interest
How it differs
- With compound interest each period's payment is worked out from the balance as it then stands, so earlier interest goes on to earn interest of its own
- The balance therefore grows by increasing steps rather than equal ones
- Over the same rate and more than one period, compound interest always exceeds simple interest
Calculating it
- Applying a percentage increase repeatedly means multiplying by the same multiplier repeatedly
- Raising the multiplier to the power of the number of periods does this in one step
- A balance P at r% for n periods becomes P × (1 + r/100)ⁿ
- The result is the final balance, not the interest, so subtract the original amount when the interest is what was asked for
Changing rates
- Where the rate changes partway through, treat each stretch as its own calculation
- Carry the balance from the end of one stretch into the start of the next
Worked example
Compound interest over several years
£4500 is invested at 3% compound interest per year for 7 years. Find the final balance and the interest earned.
Solution:
- Compound interest is worked out on the balance at the time, so each year's interest earns interest of its own afterwards
- An increase of 3% leaves 103% of the amount, so the multiplier is 1.03
- Applying that multiplier once per year for 7 years means raising it to the power 7
- Balance = 4500 × 1.03⁷
- 1.03⁷ = 1.2298739…, and keeping the full figure in the calculator matters here
- Balance = £5534.43 to the nearest penny
- The interest earned is the growth above the original sum, not the balance itself
- Interest = 5534.43 − 4500 = £1034.43
- Compare that with the £945 simple interest at the same rate over the same 7 years: the extra £89.43 is what compounding adds
- Rounding 1.03⁷ to 1.23 partway through would give £5535, which is out by more than half a pound
3. Losing value over time
Repeated percentage decrease
- An item losing a fixed percentage of its value each year behaves exactly like compound interest with the sign reversed
- Use a multiplier below 1 and raise it to the power of the number of years
- A yearly loss of 12% has multiplier 0.88
- A value P falling by r% for n years becomes P × (1 − r/100)ⁿ
- Where the question asks how much value was lost, subtract the final value from the starting value
Worked example
Value falling each year
A machine bought for £7500 loses 12% of its value each year. Find its value after 4 years, and the value lost.
Solution:
- A repeated percentage decrease works the same way as compound interest, with a multiplier below 1
- Losing 12% leaves 100 − 12 = 88% of the value, so the multiplier is 0.88
- Using 0.12 instead would give the value lost in one year, not the value remaining, which is the usual slip
- Applying the multiplier once per year for 4 years: 7500 × 0.88⁴
- 0.88⁴ = 0.59969536
- Value after 4 years = £4497.72 to the nearest penny
- The value lost is the fall from the starting price, so subtract
- Value lost = 7500 − 4497.72 = £3002.28
- A check on the scale: after 4 years a little under 60% remains, and £4497.72 is a little under 60% of £7500
4. Exponential growth and decay
The model
- A quantity changing by a fixed percentage each period grows or decays exponentially
- The same relationship covers both directions: B = A × kⁿ, where A is the starting amount, B the amount after n periods, and k the multiplier
- A value of k above 1 gives growth, and a value between 0 and 1 gives decay
- k is never negative, since a multiplier below zero would flip the sign of the quantity
- The period need not be a year: questions can also use other periods, such as days
Where it appears
| Growth | Decay |
|---|---|
| Population size | Radioactive mass |
| Bacterial cultures | A cooling liquid |
| An investment compounding | An asset losing value |
Worked example
Exponential growth of a population
A colony of 500 bacteria grows by 8% every hour. Find the population after 10 hours.
Solution:
- Growth of a fixed percentage each period is exponential, so the same multiplier is applied once per period
- Growing by 8% leaves 108% of the population, so the multiplier is 1.08
- The rate is per hour and the time is in hours, so the power is 10: always match the power to the period the rate is quoted for
- Population = 500 × 1.08¹⁰
- 1.08¹⁰ = 2.158925…, so the population has slightly more than doubled
- 500 × 2.158925… = 1079.46
- Bacteria are counted in whole organisms, so the answer is about 1079
- The context fixes the form of the answer here: 1079.46 bacteria is not a possible population, so give a whole number
Rearranging the model
- Where the starting amount is unknown, divide the final amount by kⁿ
- Where the multiplier is unknown, divide the final amount by the starting amount and take the nth root
- Where the number of periods is unknown, test whole-number values until the target is first passed, since n counts completed periods
Worked example
Finding the yearly rate of compound interest
£5000 is invested at r% per year compound interest, and is worth £6100 after 7 years. Find r.
Solution:
- Let k be the yearly multiplier, so 5000 × k⁷ = 6100
- Divide both sides by 5000: k⁷ = 6100 ÷ 5000 = 1.22
- Take the 7th root: k = ⁷√1.22 = 1.028814…
- The multiplier is 1 + r/100, so r/100 = 0.028814…
- r = 2.88 to 3 significant figures
- Check: 5000 × 1.0288⁷ = 6099.39…, which is £6100 to the nearest pound
- Writing 1.0288 as the answer would give the multiplier, not the rate
Exam tip
Round a decimal number of periods up
If your working gives a decimal number of periods, such as n = 10.48, the answer is the next whole number, 11, even though 10.48 would normally round to 10: the target is first passed during the eleventh period. The schemes gave a decimal n every mark except the last, so finish with the whole number the question asks for.
Worked example
Whole hours until a quantity falls below a level
The mass of a drug in a patient's blood is 400 mg, and it falls by 18% each hour. After how many complete hours is the mass first below 50 mg?
Solution:
- Losing 18% leaves 82%, so the hourly multiplier is 0.82
- After n hours the mass is 400 × 0.82ⁿ mg
- A first trial: 400 × 0.82⁵ = 148.29… mg, still well above 50, so try larger values of n
- n = 10: 400 × 0.82¹⁰ = 54.98… mg, still above 50
- n = 11: 400 × 0.82¹¹ = 45.08… mg, now below 50
- 10 and 11 are the neighbouring values either side of 50 mg, so the mass is first below 50 mg after 11 hours
- Rounding down to 10 is wrong: after 10 hours the mass is still 54.98 mg