Growth and Decay with Multipliers Revision Notes | GCSE AQA Higher Maths

1. Why Growth and Decay Matters

Your savings account, the value of a car, the spread of a virus — all of these change by a percentage of whatever they are right now, not a fixed amount. That one difference changes everything about how you calculate them.

If a car loses £500 every year, that is simple decay — straightforward subtraction. But if it loses 15% of its current value every year, the loss gets smaller each year because the car is worth less. That is exponential decay, and it needs a different method.

The same logic applies to growth. A savings account paying 3% compound interest earns more in year two than year one, because the interest from year one is now also earning interest. This compounding effect is what makes the topic so powerful — and so frequently examined.

2. The Core Idea: Multipliers

The key tool for every growth and decay problem is the multiplier. Instead of calculating the percentage change and then adding or subtracting it, you combine both steps into one multiplication. This is faster, less error-prone, and essential for repeated percentage changes.

Think of it this way: if a quantity increases by 20%, you keep 100% of what you had and gain 20% more — so you end up with 120% of the original. As a decimal, that is a multiplier of 1.20. For a decrease of 20%, you keep 80%, giving a multiplier of 0.80.

Here is where it gets interesting: if you apply the same multiplier repeatedly over $n$ time periods, you raise it to the power $n$. That single idea — repeated multiplication — is the engine behind every compound interest and exponential decay calculation you will meet at GCSE.

Before moving to the formula, check your multiplier instinct. This is where most marks are dropped — not in the arithmetic, but in setting up the wrong multiplier from the start.

3. The Growth and Decay Formula

Now that the multiplier idea is clear, we can write it as a single compact formula. This formula handles both growth and decay — the only difference is whether the multiplier is greater than 1 (growth) or less than 1 (decay).

The compound interest formula you will see in exam questions is the same structure, just written with different letters. Recognising that both formulas are the same idea will save you time and confusion in the exam.

Notice that $\left(1 + \frac{r}{100}\right)$ is precisely the multiplier $m$ from the general formula. The two formulas are identical in structure — the compound interest version just makes the percentage rate explicit.

3a. Worked Example: Compound Interest

Theory is straightforward, but exam questions can be tricky. Let's walk through a standard compound interest question step by step, paying attention to exactly where marks are awarded.

Worked example: Compound Interest Calculation

Priya invests £4,500 in a savings account that pays 3.5% compound interest per year. Calculate the total amount in the account after 4 years. Give your answer to the nearest penny. [3 marks]

1. $P = 4500$, $r = 3.5$, $n = 4$ — Read the question carefully. The principal is £4,500, the rate is 3.5%, and the time is 4 years. [M1 — correct identification of values]
2. $m = 1 + \frac{3.5}{100} = 1 + 0.035 = 1.035$ — This is a growth problem, so the multiplier is greater than 1. Writing this step explicitly earns a method mark.
3. $A = 4500 \times 1.035^4$ — [M1 — correct substitution into the compound interest formula]
4. $1.035^1 = 1.035$ $1.035^2 = 1.035 \times 1.035 = 1.071225$ $1.035^3 = 1.071225 \times 1.035 = 1.108717875$ $1.035^4 = 1.108717875 \times 1.035 = 1.147522990...$ — On a calculator-allowed exam question, use the power key directly: $1.035^4$. Never round this intermediate result — carry the full decimal through.
5. $A = 4500 \times 1.147522990... = 5163.85...$ — Carry the unrounded power result into this multiplication. Only round at the very end.
6. $A = £5163.85$ — [A1 — correct final answer to the nearest penny, with £ sign]. Always include the currency symbol.

Answer: £5163.85

This question is suitable for the

3b. Worked Example: Exponential Decay

Decay problems follow exactly the same structure — the only change is that the multiplier is less than 1. A common exam context is the depreciation of a car or the decline of a population.

Worked example: Exponential Decay: Car Depreciation

A car is bought for £18,000. It depreciates at a rate of 15% per year. Find the value of the car after 5 years. Give your answer to the nearest pound. [3 marks]

1. $P = 18000$, $r = 15$, $n = 5$ — This is a decay problem because the car is losing value. [M1 — correct identification]
2. $m = 1 - \frac{15}{100} = 1 - 0.15 = 0.85$ — For decay, subtract the decimal from 1. The multiplier 0.85 means the car retains 85% of its value each year.
3. $A = 18000 \times 0.85^5$ — [M1 — correct formula with decay multiplier and correct power]
4. $0.85^2 = 0.85 \times 0.85 = 0.7225$ $0.85^3 = 0.7225 \times 0.85 = 0.614125$ $0.85^4 = 0.614125 \times 0.85 = 0.52200625$ $0.85^5 = 0.52200625 \times 0.85 = 0.4437053125$ — Each multiplication is carried out in full. Do not round until the final step.
5. $A = 18000 \times 0.4437053125 = 7986.6956...$ — Carry the full decimal value of $0.85^5$ into this step.
6. $A \approx £7987$ — [A1 — correct answer to the nearest pound with £ sign]

Answer: £7987

Notice that after 5 years the car has lost more than half its original value. This is the power of exponential decay — the percentage loss compounds, so the total effect is much larger than multiplying 15% by 5 years (which would give 75%, a much bigger loss than actually occurs). The car retains more value than a simple percentage calculation would suggest, because each year's loss is calculated on a smaller base.

4. Visualising Growth and Decay

Seeing the shape of these curves helps you check whether your answer is reasonable. Growth curves sweep upwards and get steeper over time; decay curves fall steeply at first and then level off, always approaching zero but never quite reaching it.

The blue curve shows £1,000 growing at 10% compound interest per year — it starts slowly but accelerates noticeably after year 7 or 8. The red curve shows £1,000 decaying at 15% per year — it drops quickly at first, then the rate of fall slows as the value gets smaller. Both curves are smooth and continuous, never straight lines.

Earlier we established that the multiplier determines whether a problem is growth or decay. You can now see that visually: a multiplier above 1 produces the rising blue curve, and a multiplier below 1 produces the falling red curve. If your graph looks like a straight line, something has gone wrong — check that you have used a power, not a multiplication.

5. Setting Up Problems from Context

The hardest part of growth and decay questions is not the arithmetic — it is reading the question carefully enough to set up the correct formula. Examiners deliberately phrase questions to test whether you can extract $P$, $r$, and $n$ from a real-world context.

Watch out for questions that give you the final amount and ask for the starting amount, or that ask how many years it takes to reach a target value. These problems require you to rearrange the formula or use trial and improvement.

Worked example: Finding the Number of Years

Aisha invests £5,000 at 4% compound interest per year. After how many complete years will her investment first exceed £6,000? [3 marks]

1. $5000 \times 1.04^n > 6000$ — We need the amount to exceed £6,000, so we write an inequality. The multiplier for 4% growth is $1.04$. [M1]
2. $1.04^4 = 1.04 \times 1.04 \times 1.04 \times 1.04$ $1.04^2 = 1.0816$ $1.04^4 = 1.0816 \times 1.0816 = 1.16985856$ $A = 5000 \times 1.16985856 = £5849.29$ — After 4 years: £5849.29. This is less than £6,000, so 4 years is not enough.
3. $1.04^5 = 1.16985856 \times 1.04 = 1.2166529024$ $A = 5000 \times 1.2166529024 = £6083.26$ — After 5 years: £6083.26. This exceeds £6,000. [M1 — correct trial and improvement method]
4. After 4 complete years: £5849.29 (below £6,000) After 5 complete years: £6083.26 (above £6,000) — [A1 — correct answer with clear comparison showing why 5 years is the first year the target is exceeded]

Answer: 5 years

This trial-and-improvement approach is the expected GCSE method for finding the number of periods. You do not need logarithms at this level — systematic substitution with clear working earns full marks.

6. Practice Problems

The best way to consolidate this method is to work through a range of question types. The problems below cover growth, decay, and interpretation — the three skills the examiner will test. Additionally, we will include extra items to ensure full coverage of the specification points.

7. Exam Traps and Mark-Winning Technique

Growth and decay questions are reliable mark-scorers if you avoid a small set of predictable errors. Examiners see the same mistakes every year — knowing them in advance puts you ahead of most candidates.

The most common trap is confusing compound and simple interest. Simple interest adds the same fixed amount each year; compound interest recalculates on the new total. If a question says 'compound', you must use the multiplier raised to a power — never multiply the rate by the number of years and add it on.

A second trap is rounding too early. If you round $1.035^4$ to 1.15 before multiplying by the principal, your final answer will be wrong and you will lose the accuracy mark. Keep the full decimal until the very last step.