Mathematics Advanced • Year 12 • Module 7 • Lesson 4
Depreciation
Apply flat-rate and reducing-balance depreciation to vehicles, technology, machinery and tax planning.
Problem 1 — New car drives off the lot (the "Think First" scenario)
A new car is purchased for $40,000. It depreciates at 20% per year.
Set up: What are we solving for?
(i) Using flat-rate depreciation, find the book value after 5 years. Is the figure realistic? 2 marks
(ii) Using reducing-balance depreciation, find the book value after 5 years. 2 marks
(iii) A used-car dealer offers $14,500 for the car after 5 years. Comment in one sentence on whether that price is fair for each depreciation model. 2 marks
Stuck? Revisit lesson § Think First and § Reducing Balance.Problem 2 — Delivery van for a small business
A small business purchases a delivery van for $45,000. The accountant models depreciation at 15% per year.
Set up: What are we solving for?
(i) Find the book value at the end of year 4 under flat rate and reducing balance. 2 marks
(ii) Find the total dollar depreciation over the 4-year period under each method. 2 marks
(iii) Explain in 1-2 sentences why a delivery business might prefer the reducing-balance method for tax purposes. 2 marks
Problem 3 — Industrial machinery (find the rate)
A factory purchased industrial machinery for $120,000. After 6 years its book value is $28,000 under reducing balance.
Set up: What are we solving for?
(i) Find the annual depreciation rate r to two decimal places. 2 marks
(ii) Project the book value at year 10 using your rate from (i). 2 marks
(iii) If the factory wants to dispose of the machine when its book value first falls below $10,000, use logarithms to find the year (to nearest whole). 3 marks
Stuck? Revisit lesson § Worked Example.Problem 4 — Office IT fleet (technology obsolescence)
A company buys $80,000 of laptops. Technology depreciates at 30% p.a. reducing balance (rapid obsolescence). The same fleet under flat rate would depreciate at 20% p.a.
Set up: What are we solving for?
(i) Find the book value at the end of years 1, 3 and 5 under reducing balance. 3 marks
(ii) Find the book value at the same years under flat rate (20% p.a.). At which year do the two methods cross over (i.e. give the same book value)? 3 marks
(iii) Recommend which method best models technology obsolescence and justify in one sentence. 1 mark
Problem 5 — Salvage value and end of life
An accountant uses reducing-balance depreciation for a $50,000 piece of equipment at 15% p.a., with a salvage value of $5,000 (the lowest value the asset can have on the books).
Set up: What are we solving for?
(i) Compute the book value at year 5, 10 and 15. 3 marks
(ii) Use logarithms to find the first year in which the unadjusted book value drops below the salvage value $5,000. 2 marks
(iii) Explain in one sentence why reducing balance approaches but never reaches zero, and why salvage values are imposed in practice. 2 marks
Stuck? Solve V₀(1 − r)ⁿ < salvage and take logs.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Car at 20% per year
Set up. We are valuing the car under each model and judging which matches a real dealer offer.
(i) Flat: S = 40,000 − 0.20 × 40,000 × 5 = 40,000 − 40,000 = $0. Not realistic — a 5-year-old car still has trade-in value.
(ii) RB: S = 40,000(0.80)⁵ = 40,000 × 0.32768 = $13,107.20.
(iii) The dealer's $14,500 offer is generous against reducing balance ($13,107) and absurdly generous against flat rate ($0) — only the RB model gives a price comparison that makes sense for vehicles.
Problem 2 — Delivery van at 15% per year
Set up. We are comparing the dollar depreciation under each method and explaining the tax incentive.
(i) Flat: D = 0.15 × 45,000 = $6,750/yr; S4 = 45,000 − 27,000 = $18,000. RB: S4 = 45,000(0.85)⁴ = 45,000 × 0.52201 = $23,490.28.
(ii) Total depreciation: flat = $27,000; RB = 45,000 − 23,490.28 = $21,509.72. Flat rate "expenses" more in total over 4 years.
(iii) Reducing balance front-loads the depreciation expense (largest dollar amount in year 1), which reduces taxable income sooner — and a tax saving today is worth more than the same saving in the future (time value of money).
Problem 3 — Machinery (find rate then project)
Set up. We are inferring r from two book values, projecting forward, and inverting with logs.
(i) (1 − r)⁶ = 28,000/120,000 = 0.23333; 1 − r = 0.233331/6 = 0.79263; r = 0.20737 ≈ 20.74% p.a.
(ii) S10 = 120,000(0.79263)¹⁰ = 120,000 × 0.09995 ≈ $11,994.
(iii) Solve 120,000(0.79263)ⁿ < 10,000 ⇒ (0.79263)ⁿ < 0.0833 ⇒ n > ln 0.0833 / ln 0.79263 = −2.485 / −0.23253 = 10.69 years. First year below $10,000 is year 11.
Problem 4 — IT fleet (technology obsolescence)
Set up. We are tracing two depreciation paths to find the crossover and recommend a method.
(i) RB at 30%: yr 1 = 80,000 × 0.7 = $56,000; yr 3 = 80,000 × (0.7)³ = 80,000 × 0.343 = $27,440; yr 5 = 80,000 × (0.7)⁵ = 80,000 × 0.16807 = $13,445.60.
(ii) Flat at 20%: yr 1 = 80,000 − 16,000 = $64,000; yr 3 = 80,000 − 48,000 = $32,000; yr 5 = 80,000 − 80,000 = $0. Crossover: Year 5 flat hits $0, while RB still at $13,446 — flat falls below RB between year 4 (flat $16,000 vs RB $19,208) and year 5 (flat $0 vs RB $13,446). They cross at roughly year 4.4.
(iii) Reducing balance better models technology because the largest loss in value is in the first year (a 1-year-old laptop is worth far less than a new one) — exactly the steep early decline of exponential decay.
Problem 5 — Equipment with salvage cap
Set up. We are valuing an asset under RB and finding when the unadjusted curve crosses the salvage floor.
(i) S = 50,000(0.85)ⁿ. yr 5 = 50,000 × 0.4437 = $22,184.99; yr 10 = 50,000 × 0.1969 = $9,843.74; yr 15 = 50,000 × 0.0874 = $4,367.86.
(ii) Solve 50,000(0.85)ⁿ < 5,000 ⇒ (0.85)ⁿ < 0.10 ⇒ n > ln 0.10 / ln 0.85 = −2.3026 / −0.16252 = 14.17. First year below salvage is year 15 (unadjusted $4,367.86); after that the books are capped at the $5,000 salvage value.
(iii) Reducing balance approaches but never reaches zero because each year multiplies by a factor (1 − r) < 1 — the curve is asymptotic; in real accounting a non-zero salvage value is imposed to reflect the practical floor (scrap or resale value) and to stop the books carrying assets at unrealistically small positive values forever.