Depreciation (AI SL)
Depreciation is compound decrease: an asset - a car, a machine, a laptop - loses a fixed percentage of its current value every year, so the amount it loses shrinks each time too. It's really the compound interest model run with a negative rate, and it's part of the broader Financial Maths topic.
9 questions on this sub-topic.
The depreciation model
Covered under IB syllabus reference SL1.4 - financial applications of geometric sequences and series, including annual depreciation and compound interest calculated yearly, half-yearly, quarterly, or monthly, often using a GDC's built-in financial package.
Depreciation
\[FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}\]
The same compound-growth formula as compound interest, but \(r\) is entered as a negative rate. \(k\) is the number of compounding periods per year - for straightforward annual depreciation, \(k=1\).
✓ In the formula bookletMany depreciation questions are easier to picture as a retention factor: if a value falls by \(8\%\) each year, it keeps \(92\%\), so you multiply by \(0.92\) once per year. Need the full syllabus wording and formula-booklet reference table? See Financial Maths.
Worked examples
A car worth $18,000 depreciates at 8% per year.
Find its value after 4 years, correct to 2 decimal places.
Worked solution
Losing 8% each year means keeping 92%, so the multiplier is \(0.92\) per year. M1
\(V = 18000(0.92)^4.\) \((0.92)^4 \approx 0.7164\)
, so \(V \approx 18000(0.7164) = $12,895.07.\) A1
A \($15{,}000\) machine depreciates at \(10\%\) per year. Find the first whole year its value falls below \($9000\).
Worked solution
\(15000(0.9)^n<9000\Rightarrow n=5.\) M1
\(n=5.\) A1
A car bought for \($24\,000\) is worth \($12\,288\) after 3 years of depreciation at a constant annual rate. Find the annual depreciation rate.
Worked solution
\(12288=24000(1-r)^3.\) M1
\((1-r)^3=\dfrac{12288}{24000}=0.512\Rightarrow1-r=\sqrt[3]{0.512}=0.8.\) M1
\(r=1-0.8=0.20=20\%.\) A1
A \($15\,000\) machine depreciates at \(10\%\) per year. Find the first whole year its value falls below \($9000\).
Worked solution
\(15000(0.9)^n<9000\Rightarrow n=5.\) M1 - Attempt to set up 15000(0.9)^n<9000 and solve for n
\(n=5.\) A1
Common mistakes
- Adding instead of subtracting. Depreciation is negative growth - the multiplier is \(\left(1-\tfrac{r}{100}\right)\), not \(\left(1+\tfrac{r}{100}\right)\). Using the compound-interest sign on a depreciation question overstates the value badly.
- Misreading "depreciates at 8%" as "worth 8% of its value". The asset loses 8% and keeps the other 92% - it isn't reduced to 8% of the original price.
- Forgetting \(n\) must be a whole number of years. For "find the first year the value falls below..." questions, solving the inequality gives a decimal - you then round to the nearest whole year that actually satisfies the inequality, which usually means rounding up rather than to the nearest integer.
Ready to practise properly?
10 depreciation questions, marked instantly like the real exam.
Quick answers
How do you calculate depreciation on the IB Applications & Interpretation syllabus?
Depreciation uses the same compound-growth model as compound interest, but with a negative rate: \(V = PV\left(1+\tfrac{r}{100}\right)^n\), where \(r\) is negative (or you can multiply by the retention factor, \(1\) minus the depreciation rate, once per year).
What is a retention factor?
If an asset depreciates by \(r\%\) per year, it keeps \((100-r)\%\) of its value each year. That fraction, written as a decimal, is the retention factor you raise to the power of the number of years.