GCSE Maths Practice: percentages
A town has a population of 5000. It increases by 8% each year. Work out the population after 4 years, giving your answer to the nearest whole number.
Compound increases multiply each year’s total by (1 + rate). Use brackets carefully and round only at the end.
Understanding Compound Growth in Context
In GCSE Higher Tier Maths, compound percentage questions often appear in real-life settings such as population change, bank interest, or inflation. Unlike simple percentage problems, where growth happens once, compound growth applies the increase repeatedly each year — the new total becomes the base for the next calculation.
The Compound Growth Formula
The standard formula is:
\[ N = P(1 + r)^t \]
- \( N \) = final amount
- \( P \) = initial amount (starting population or balance)
- \( r \) = rate of change as a decimal
- \( t \) = number of time periods (years)
In this example: \( P = 5000, r = 0.08, t = 4. \)
Substitute and calculate:
\[ N = 5000(1.08)^4 = 5000 \times 1.36049 = 6802.45. \]
So, after four years, the population is approximately 6802 people.
Why the Growth is Compounded
Each year, the increase is based on the most recent total — not the original 5000. This is why the growth accelerates. After the first year, the population becomes \( 5000 \times 1.08 = 5400. \) In the next year, the 8% is taken from 5400, not 5000. That creates the compounding effect, similar to earning interest on both your savings and the previous interest added.
Step-by-Step Yearly Calculation
| Year | Population |
|---|---|
| 0 | 5000 |
| 1 | 5000 × 1.08 = 5400 |
| 2 | 5400 × 1.08 = 5832 |
| 3 | 5832 × 1.08 = 6298.56 |
| 4 | 6298.56 × 1.08 ≈ 6802.45 |
This table clearly shows how growth compounds each year.
Real-World Interpretation
Population growth rates vary globally, and compound models provide realistic projections. For instance, if a town’s population grows by 8% annually, this represents a consistent, steady increase — similar to how bacteria multiply or how investments grow. Over several years, even modest percentages can lead to substantial growth due to compounding.
Common Higher-Tier Variations
- Compound decay: The same formula applies, but with a minus sign: \( (1 - r)^t. \)
- Mixed changes: e.g., grow by 8% for three years, then shrink by 5%.
- Reverse problems: e.g., “After 4 years the population is 6948. What was the original population?”
Being able to rearrange the formula algebraically is essential for higher-level marks.
Checking with Estimation
Estimation provides a quick reality check. If the population grows by 8% per year for 4 years, that’s roughly 32% total simple growth, or \( 5000 + 0.32 \times 5000 = 6600. \) The compound result of around 6800 is slightly higher, which makes sense because each year’s increase builds on the previous year’s total.
Realistic Applications
- Finance: Compound interest on £5000 at 8% gives £6948 after 4 years — same principle as population growth.
- Environment: A forest growing 8% annually doubles its biomass in about 9 years — a useful approximation using the rule of 72 (72 ÷ 8 ≈ 9).
- Data science: Exponential models describe spread of information, population growth, and even disease progression.
Common Mistakes
- Adding 8% four times instead of compounding — that’s simple, not compound growth.
- Using 1.8 instead of 1.08 — remember the decimal conversion.
- Rounding too early; always round only the final result.
Summary
Compound growth means repeated percentage increases on a changing base. The formula \( N = P(1 + r)^t \) captures this perfectly. Substituting \( P = 5000, r = 0.08, t = 4 \) gives \( N ≈ 6802. \) This realistic town population example illustrates how percentages describe long-term growth patterns. Whether applied to investments, bacteria, or human populations, compound percentages reveal how small changes accumulate dramatically over time.