The big idea: Depreciation is compound interest in reverse: a value loses the same percentage each year, so it multiplies by r = 1 − rate (with 0 < r < 1).
For example, a car losing 15% a year → multiply by 0.85 each year: value = start × 0.85ⁿ.
- the depreciated value
- the original value
- the annual depreciation rate (percent)
- the number of years
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Multiply by (1 − rate) each year: Find the decay ratio r = 1 − rate, then value = start × rⁿ.
IB-style question — depreciated value
A car is worth $24 000. It depreciates by 12% each year.
Find its value after 5 years, to the nearest dollar.
Step by step
- Decay ratio.
- Value after 5 years = start × r⁵.
- Work it out.
Final answer
≈ $12 666.
Decay never quite reaches zero: Each year keeps a fixed percentage (88% here), so the value shrinks but never hits 0.
Round only at the very end.
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The multiplier tells you the rate: A decay model V = V₀ × bᵗ has yearly multiplier b, so the depreciation rate is (1 − b) as a percent.
For example, V = 5000(0.92)ᵗ loses 8% a year.
IB-style question — read and use a model
The value of a van is V = 30 000(0.85)ᵗ dollars, t years after purchase.
(a) Write down the annual rate of depreciation. (b) Find the value after 4 years, to the nearest dollar.
Step by step
- (a) The multiplier is 0.85, so the rate lost is 1 − 0.85.
- (b) Substitute t = 4.
- Work it out.
Final answer
(a) 15% per year. (b) ≈ $15 660.
Common mistakes
- Reading the multiplier (0.85) as the rate.
- Using 1 + rate for decay (that's growth).
- Subtracting a flat percentage each year (simple).
Do this instead
- Rate lost = 1 − multiplier (1 − 0.85 = 15%).
- Decay multiplies by (1 − rate), 0 < r < 1.
- Multiply by the same ratio each year (compound).
When the first year is different: Sometimes the value drops by one rate in the first year and a different rate every year after. Handle it in two stages:
Year 1: value = start × (1 − first rate). Each later year: multiply by (1 − second rate) once more.
So after n years: value = start × (1 − rate₁) × (1 − rate₂)ⁿ⁻¹.
IB-style question — a machine that loses value
A company buys a printing machine for $48 000. It loses 20% of its value in the first year, then 12% in each year after that.
(a) Find the value after 1 year.
(b) Find the value after 8 years, to the nearest dollar.
(c) Find the least number of complete years for the value to fall below 15% of $48 000.
Step by step
- (a) The first year loses 20%, so multiply by 0.80.
- (b) After year 1, multiply by 0.88 once for each further year. After 8 years that is 7 more ×0.88.
- Work it out on the GDC.
- (c) 15% of $48 000 is $7 200. We need the first whole year where the value is below 7200.
- Read it from the GDC table (below): the value first drops below 7200 at n = 15.
Final answer
(a) $38 400 (b) ≈ $15 693 (c) n = 15 years
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
“How many years?” on Paper 2: Type the value formula into Y₁, then press 2nd → GRAPH for the table. Scroll until the value first passes the target — the X on that row is your answer.
“Complete years” means take the first whole X that works (here X = 15).