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.
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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.
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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.