The big idea: Many financial situations change by the same percentage each period. If the value keeps multiplying by the same factor, the model is compound growth or compound depreciation.
| Situation | What happens each period? | Type |
|---|---|---|
| Savings account earning 4% yearly | Multiply by 1.04 | Compound growth |
| Car losing 15% of value each year | Multiply by 0.85 | Depreciation |
| Price rising by 3% each year | Multiply by 1.03 | Compound growth |
| Phone worth 20% less each year | Multiply by 0.80 | Depreciation |
Add or subtract the percentage from 100%: A rise of 6% means you keep 100% and gain 6% → multiply by 106% = 1.06. A fall of 12% means you keep 88% → multiply by 0.88.
Quick recognition example
A tablet is worth $800 and depreciates by 10% each year. What is the multiplier?
Step by step
- A 10% decrease means the value keeps 90% each year.
- Convert 90% to a decimal multiplier.
Final answer
The multiplier is 0.90.
Common mistake: Students often use 0.10 instead of 0.90 for a 10% depreciation. The multiplier is what the value becomes after the change, not the size of the change itself.
The model: If a value changes by the same percentage each year, the future amount is found by multiplying by the same factor over and over again.
| Type | Formula | Meaning |
|---|---|---|
| Compound growth | A = P(1 + r/100)n | Start with P, grow by r% for n years |
| Depreciation | A = P(1 - r/100)n | Start with P, lose r% for n years |
Here, P is the starting value, A is the value after n years, and r is the yearly percentage rate.
Worked example — write the model
A machine costs $12 000 and depreciates by 8% each year. Write a formula for its value V after n years.
Step by step
- Start value P = 12 000.
- Depreciation of 8% means multiplier 1 - 8/100.
- Use the depreciation model.
Final answer
V = 12000(0.92)n
Read the wording carefully: If the question says 'after 5 years', then n = 5. If it gives you the value at the start and asks for the value after each year, the exponent counts the number of percentage changes.
Learn what examiners really want
See exactly what to write to score full marks. Our AI shows you model answers and the key phrases examiners look for.
The big idea: Once the model is built, IB usually asks for the value after a certain number of years. The mathematics is straightforward, but the final answer must still be written in context with sensible rounding.
Worked example — compound growth
A savings plan starts with $5 000 and grows by 4% per year. Find the value after 6 years.
Step by step
- Use the compound-growth model.
- Calculate.
Final answer
After 6 years, the savings plan is worth about $6 326.60.
Worked example — depreciation
A car is bought for $18 000 and depreciates by 12% per year. Find its value after 4 years.
Step by step
- Use the depreciation model.
- Calculate.
Final answer
After 4 years, the car is worth about $10 794.52.
Money answers should look like money: For financial values, it is usually sensible to round to the nearest cent unless the question says otherwise. IB will also accept appropriate rounding in dollars when clearly justified.
| Student mistake | Why it is wrong | Correct fix |
|---|---|---|
| Using 0.07 for a 7% growth multiplier | 0.07 is the increase only, not the full new value | Use 1.07 |
| Using 0.15 for 15% depreciation | 0.15 is the loss, not the remaining value | Use 0.85 |
| Forgetting the power n | The percentage change happens repeatedly | Raise the multiplier to n |
| Giving only the number | IB finance questions need context | State what the value represents |
Worked example — choosing the correct model
A house increases in value by 3% each year. A student writes V = 250000(0.03)n. Explain the mistake.
Step by step
- 0.03 represents the increase only, not the new yearly multiplier.
- A 3% increase means keep 100% and add 3%.
- Write the correct model.
Final answer
The correct model is V = 250000(1.03)n.
What the examiner wants: In these questions, the examiner is really checking whether you can convert words like 'increases by 4%' or 'depreciates by 10%' into the correct multiplier before calculating.