Depreciation (AA SL)
A car, a laptop, or a piece of machinery loses a fixed percentage of its value every year - which makes depreciation the mirror image of compound interest: the same geometric structure, just shrinking instead of growing. This page covers the depreciation formula, how to spot the common ratio, and the wording that most often trips students up. It's part of the broader Financial Maths topic.
11 questions on this sub-topic.
The depreciation formula
Covered under IB syllabus reference SL1.4: financial applications of geometric sequences and series, which includes annual depreciation alongside compound interest.
Depreciation
\[V=V_0(1-r)^{n}\]
A geometric sequence with common ratio \(1-r\), where \(V_0\) is the original value, \(r\) is the annual depreciation rate as a decimal, and \(n\) is the number of years.
Not in the formula booklet - same structure as compound interest with a negative rateNeed the compound interest and annuity formulas, or a full GDC finance-solver walkthrough? See Financial Maths.
Worked examples
A laptop worth $1200 loses 20% of its value each year.
Find its value after 2 years.
Worked solution
Losing 20% multiplies by \(0.80\): \(V=1200(0.80)^2.\) M1
\(=1200(0.64)=$768.\) A1
A car worth $24 000 depreciates by 15% each year.
Find its value after 3 years, to the nearest dollar.
Worked solution
Losing 15% each year multiplies by \(1-0.15=0.85\): \(V=24000(0.85)^n.\) M1
\(V=24000(0.85)^3=24000(0.614125)\) A1 \(\approx$14739.\) A1
A car bought for $35000 is valued at $19000 after 5 years, having depreciated by the same percentage each year.
Find the annual rate of depreciation, correct to 3 significant figures.
Worked solution
Let \(r\) be the annual depreciation rate (as a decimal): \(19000=35000(1-r)^5.\) M1
\((1-r)^5=\dfrac{19000}{35000}=0.542857\ldots\) A1
\(1-r=(0.542857)^{1/5}.\) M1
\(1-r\approx0.88499,\) so \(r\approx0.11501.\) A1
The annual rate of depreciation is \(\approx11.5\%\) (3 s.f.). A1
A printing press purchased for $60000 depreciates at 18% per year.
Find the least number of complete years after which its value first falls below $15000.
Worked solution
\(1-0.18=0.82\), so the value after \(n\) years is \(V(n)=60000(0.82)^n.\) We require \(V(n)<15000.\) M1
Use the GDC (table or graph) to test successive integer values of \(n\): M1
\(V(6)=60000(0.82)^6\approx$18\,240\), still above $15000. A1
\(V(7)=60000(0.82)^7\approx$14\,957\), below $15000. A1
The least number of complete years is \(n=7.\) A1
Common mistakes
- Forgetting depreciation uses a ratio below 1. A machine depreciating \(15\%\) per year has common ratio \(1-0.15=0.85\), not \(0.15\) itself.
- Reading "loses 20% of its value" as the remaining fraction. "Loses 20%" means \(80\%\) is left, so the ratio is \(0.80\) - not \(0.20\), which is the amount lost, not the amount kept.
- Entering a positive rate into the GDC's TVM solver. The finance app models depreciation with a negative \(I\%\) - leaving it positive turns a shrinking value into a growing one.
Ready to practise properly?
11 depreciation questions, marked instantly like the real exam.
Quick answers
What is the formula for depreciation?
\(V=V_0(1-r)^{n}\), where \(V_0\) is the original value, \(r\) is the annual depreciation rate as a decimal, and \(n\) is the number of years. It's the same geometric structure as compound interest, but with a common ratio below 1.
Is the depreciation formula in the IB formula booklet?
No - it isn't listed separately, because it has exactly the same structure as compound interest with a negative rate. Most GDC finance solvers handle it by entering the interest rate as a negative percentage.