A-Level Percentages Revision Guide
This guide covers the percentage topics that appear in A-Level Maths and Further Maths, pitched at Year 12 and Year 13 students. It builds from the compound interest formula through to exponential models, continuous compounding, the relative rate of change in calculus, and index numbers. All worked examples are written in the style expected in A-Level assessments.
Topics on this page
- Compound interest: the formula
- Exponential growth and decay
- Continuous compound interest
- Percentage change in differentiation
- Index numbers and percentage change
- A-Level exam-style worked examples
Compound interest: the formula
At A-Level, compound interest is generalised to allow for compounding frequencies other than annually. The full formula is:
A = P(1 + r/n)nt
- A = final amount (principal plus interest)
- P = principal (initial amount invested or borrowed)
- r = annual interest rate expressed as a decimal (e.g. 4.5% = 0.045)
- n = number of times interest is compounded per year
- t = time in years
When interest is compounded annually, n = 1 and the formula reduces to the GCSE version A = P(1 + r)t. At A-Level you are expected to work with monthly (n = 12), quarterly (n = 4), and other compounding frequencies.
Worked Example 1
£5,000 is invested at 4.5% per annum compounded monthly for 3 years. Calculate the value of the investment at the end of this period.
P = 5000, r = 0.045, n = 12, t = 3
A = 5000 × (1 + 0.045/12)12 × 3
A = 5000 × (1.00375)36
A = 5000 × 1.14423...
A = £5,721.15 (to the nearest penny)
Worked Example 2: Effective Annual Rate
A savings account pays 3.2% per annum compounded quarterly. Calculate the effective annual rate (EAR).
The EAR is the equivalent annual rate that gives the same return as the given rate with its compounding frequency.
EAR = (1 + r/n)n - 1
EAR = (1 + 0.032/4)4 - 1
EAR = (1.008)4 - 1
EAR = 1.032386... - 1
EAR = 3.239% (to 3 d.p.)
The effective annual rate is slightly higher than the nominal rate of 3.2% because of the quarterly compounding effect.
For more practice with the compound interest formula, see the compound interest worksheet.
Exponential growth and decay
Exponential models describe quantities that increase or decrease by a fixed percentage per unit time. These appear throughout A-Level Maths in the contexts of population growth, radioactive decay, depreciation, and biological models.
Growth: A = P(1 + r)t
Decay: A = P(1 - r)t
where r is the fractional rate of change per unit time and t is the number of time periods
Example: Population Growth
A town has a population of 45,000 and is growing at 2.3% per year. Estimate the population after 6 years.
P = 45,000, r = 0.023, t = 6
A = 45,000 × (1.023)6
A = 45,000 × 1.14529...
A ≈ 51,538 (to the nearest whole number)
Example: Radioactive Decay with Half-Life
A radioactive substance halves every 5 years. Find the percentage decay rate per year, correct to 3 significant figures.
If the substance halves every 5 years:
P(1 - r)5 = 0.5P
(1 - r)5 = 0.5
1 - r = 0.51/5 = 0.50.2
1 - r = 0.87055...
r = 0.12945...
Annual decay rate ≈ 12.9% per year
Example: Car Depreciation
A car is purchased for £22,000 and depreciates at 15% per year. Find its value after 4 years and the percentage of the original value remaining.
A = 22,000 × (1 - 0.15)4
A = 22,000 × (0.85)4
A = 22,000 × 0.52200625
A = £11,484.14 (to the nearest penny)
Percentage remaining = (0.85)4 × 100 = 52.2% (to 3 s.f.)
Continuous compound interest
As the compounding frequency n increases without limit, the formula A = P(1 + r/n)nt approaches a limit involving Euler's number e. This gives the continuous compounding formula:
A = Pert
- e = Euler's number ≈ 2.71828
- r = continuously compounded annual rate (as a decimal)
- t = time in years
Continuous compounding gives the theoretical upper bound for interest growth at a given annual rate. It is used in financial modelling, options pricing (such as the Black-Scholes model), and theoretical economics. In practice, most banks compound daily rather than continuously.
Derivation note
The limit limn→∞ (1 + r/n)n = er follows from the standard definition of e. This means that as compounding becomes infinitely frequent, the annual growth factor approaches er rather than (1 + r).
Worked Example
£10,000 is invested at 5% per annum continuously compounded for 10 years. Calculate the final value, and compare this with annual compounding at the same rate.
Continuous compounding:
A = 10,000 × e0.05 × 10 = 10,000 × e0.5
A = 10,000 × 1.64872...
A = £16,487.21
Annual compounding (for comparison):
A = 10,000 × (1.05)10 = 10,000 × 1.62889...
A = £16,288.95
Continuous compounding gives £198.26 more over 10 years in this case, illustrating that the difference, while real, is modest at everyday interest rates.
Percentage change in differentiation
In A-Level calculus, the relative rate of change of a function f(t) is defined as:
Relative rate of change = (1/f) × (df/dt)
This gives the fractional (or proportional) rate of change per unit time. Multiplying by 100 converts it to a percentage rate of change per unit time.
For exponential functions of the form N = N0ekt, the derivative is dN/dt = kN0ekt = kN. Therefore:
(1/N) × (dN/dt) = k
The relative rate of change is the constant k. If k = 0.03, the quantity grows at 3% per unit time. This is a key result for exponential growth models.
Example: Population Model
A population is modelled by N = 1000e0.03t, where t is the time in years. Find the annual percentage growth rate and verify it using the relative rate of change.
Method 1: Direct reading
The model is of the form N = N0ekt with k = 0.03.
Annual percentage growth rate = k × 100 = 3% per year.
Method 2: Differentiation
dN/dt = 1000 × 0.03 × e0.03t = 0.03N
(1/N) × (dN/dt) = 0.03
Relative rate of change = 3% per year. ✓
Note: this is the instantaneous percentage rate of change. The actual percentage increase over a full year is slightly different (e0.03 - 1 ≈ 3.045%) due to continuous compounding.
Index numbers and percentage change
An index number expresses the value of a variable relative to its value in a chosen base year, which is set to 100. Subsequent years are expressed as a percentage of the base year value. Index numbers allow meaningful comparison of quantities over time without needing to compare raw values.
Index = (Current value / Base year value) × 100
% change = ((Index - 100) / 100) × 100 = Index - 100 (when base is 100)
More generally: % change between two index values = ((New index - Old index) / Old index) × 100
In the UK, the most commonly referenced index numbers are:
- CPI (Consumer Prices Index): measures the change in prices paid by households for a basket of goods and services. Used as the ONS's primary measure of inflation.
- RPI (Retail Prices Index): an older measure of inflation that includes housing costs such as mortgage interest payments. Still used for some wage agreements and index-linked bonds.
- GDP deflator: measures price changes across the whole economy, used to convert nominal GDP into real GDP.
Worked Example: CPI Percentage Change
The UK CPI index stood at 100 in 2015 and 128.4 in 2023. Calculate the percentage increase in prices over this period.
% change = ((128.4 - 100) / 100) × 100
% change = (28.4 / 100) × 100
% change = 28.4%
Prices, on average, were 28.4% higher in 2023 than in 2015 according to this index.
Calculating percentage change between two non-base years
To find the percentage change between two years where neither is the base year, use the general formula. For example, if the CPI was 112.3 in 2019 and 128.4 in 2023:
% change = ((128.4 - 112.3) / 112.3) × 100
= (16.1 / 112.3) × 100 = 14.3% (to 3 s.f.)
A-Level exam-style worked examples
Question 1
A bank account pays 3.6% per annum compounded monthly. Rosalind invests £8,000. Calculate the value of the account after 5 years, giving your answer to the nearest penny. Hence find the effective annual rate, giving your answer correct to 3 significant figures.
Solution:
P = 8000, r = 0.036, n = 12, t = 5
A = 8000 × (1 + 0.036/12)60
A = 8000 × (1.003)60
A = 8000 × 1.19668...
A = £9,573.44 [2 marks]
EAR = (1.003)12 - 1 = 1.03660... - 1 = 0.03660...
EAR = 3.66% [2 marks]
Question 2
A radioactive isotope decays exponentially. Its mass M grams after t years satisfies M = M0(0.88)t, where M0 is the initial mass.
(a) Find the annual percentage decay rate. [1 mark]
(b) Find the half-life of the isotope, giving your answer correct to 3 significant figures. [3 marks]
(c) Find the percentage of the original mass remaining after 20 years. [1 mark]
Solution:
(a) The multiplier per year is 0.88 = 1 - 0.12, so the annual decay rate is 12% per year.
(b) For the half-life, set M = 0.5M0:
(0.88)t = 0.5
t ln(0.88) = ln(0.5)
t = ln(0.5) / ln(0.88) = (-0.69315...) / (-0.12783...) = 5.4218...
Half-life = 5.42 years (to 3 s.f.)
(c) (0.88)20 × 100 = 0.08264... × 100 = 8.26% remaining.
Question 3
The size of a bacterial colony is modelled by N = 500e0.04t, where N is the number of bacteria and t is the time in hours.
(a) State the percentage growth rate of the colony per hour. [1 mark]
(b) Find the size of the colony after 24 hours, giving your answer to the nearest whole number. [2 marks]
(c) By differentiating, verify that the relative rate of change is 4% per hour. [2 marks]
(d) Find the time at which the colony first reaches 2,000 bacteria. Give your answer in hours correct to 3 significant figures. [1 mark]
Solution:
(a) k = 0.04, so the percentage growth rate is 4% per hour.
(b) N = 500e0.04 × 24 = 500e0.96 = 500 × 2.61170... = 1,306 bacteria.
(c) dN/dt = 500 × 0.04 × e0.04t = 0.04N
(1/N) × (dN/dt) = 0.04 = 4% per hour. ✓
(d) 500e0.04t = 2000
e0.04t = 4
0.04t = ln 4
t = ln 4 / 0.04 = 1.38629... / 0.04 = 34.7 hours (to 3 s.f.)
Further practice and resources
Compound Interest Worksheet
Printable worksheet with exam-standard questions and a full mark scheme.
Reverse Percentage Calculator
Find the original value before a percentage change was applied.
A-Level Practice Questions
20 A-Level standard questions with full worked solutions covering all topics above.
A-Level Percentage Worksheet
Free printable A-Level worksheet with 20 questions and a full mark scheme.