Depreciation
The moment you drive a new car off the lot, it loses value. But how much? And does it keep losing the same amount every year, or does the loss accelerate? In this lesson you will learn the two mathematical models for depreciation — flat rate and reducing balance — and discover why accountants and insurers choose one over the other.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A new car costs $40,000. It loses 20% of its original value every year.
After 5 years, is it worth $0? More than $0? Or can it go negative? What does your intuition tell you, and what might the mathematics reveal?
There are exactly two methods you need for HSC depreciation. Both start from the same initial value $V_0$ — but they calculate the loss differently each year.
Flat rate subtracts the same fixed dollar amount each year — the graph is a straight line. Reducing balance multiplies by $(1-r)$ each year — the graph is exponential decay.
Key facts
- Flat rate formula: $S = V_0 - Dn$
- Reducing balance formula: $S = V_0(1-r)^n$
- Flat rate can produce negative values; reducing balance cannot
Concepts
- Why flat rate is linear and reducing balance is exponential decay
- When each method is appropriate (accounting, tax, insurance)
- The salvage value concept and why assets rarely depreciate to zero
Skills
- Calculate book value using both methods
- Find the depreciation rate given initial and final values
- Compare total depreciation across methods
- Identify which method a scenario describes
For the same asset and same nominal rate, the two methods produce very different book-value curves. Below: a $45,000 van depreciating at 15% p.a.
Reducing balance preserves more book value early; flat rate reaches zero (then stops) while reducing balance never mathematically hits zero.
After 4 years on the same $45,000 van at 15%:
- Flat rate: $S = 45{,}000 - 6{,}750 \times 4 = \$18{,}000$
- Reducing balance: $S = 45{,}000(0.85)^4 = \$23{,}490$
The reducing balance value is higher because the annual loss shrinks as the book value shrinks.
Flat rate: $S = V_0 - Dn$, where $D = rV_0$. Same dollar amount lost each year.; Reducing balance: $S = V_0(1-r)^n$. Same as compound interest but with $-r$.
Pause — copy the two depreciation formulas: flat rate $S = V_0 - Dn$ where $D = rV_0$ (constant dollar loss each year) and reducing balance $S = V_0(1-r)^n$ (constant percentage loss) into your book.
Did you get this? True or false: reducing balance depreciation can produce a negative book value if enough years pass.
Worked examples · 3 in a row, reveal as you go
A delivery van is purchased for $45,000. It depreciates at 15% p.a. flat rate. Find its value after 4 years and determine after how many years it would theoretically reach zero.
The same $45,000 van now depreciates at 15% p.a. reducing balance. Find its value after 4 years and compare with the flat rate result.
$S = 45{,}000 \times 0.52200625 = \$23{,}490.28$
Industrial machinery purchased for $120,000 has a book value of $28,000 after 6 years using reducing balance depreciation. Find the annual depreciation rate $r$.
$(1-r)^6 = \dfrac{28{,}000}{120{,}000} = 0.2\overline{3}$
Quick check: A laptop purchased for $2,400 depreciates at 25% p.a. flat rate. Its value after 3 years is:
Common errors · the 3 traps that cost marks
Tell the logic: Why does reducing balance depreciation give a higher book value than flat rate after several years, even when both use the same depreciation rate?
Quick-fire practice · 4 depreciation calculations
$V_0 = \$60{,}000$, flat rate 12%, after 8 years. Find $S$.
$V_0 = \$60{,}000$, reducing balance 12%, after 8 years. Find $S$.
$V_0 = \$30{,}000$, flat rate 25%, after 5 years. What happens?
Machinery: $V_0 = \$80{,}000$, $S = \$32{,}000$ after 3 years, reducing balance. Find $r$.
Fill in the blanks: Flat rate depreciation produces a [blank] graph, while reducing balance produces an [blank] decay curve. Only the flat rate method can produce a [blank] book value.
Match each formula with its description:
At 20% flat rate on $40,000: $S = 40{,}000 - 8{,}000n$. After 5 years: $S = 40{,}000 - 40{,}000 = \$0$. So the car is worth exactly $0 after 5 years using flat rate. But this is unrealistic — a 5-year-old car still has salvage value. Using reducing balance: $S = 40{,}000(0.8)^5 = \$13{,}107.20$. This reveals why reducing balance is preferred for vehicles: it better matches reality, where assets retain some value even after significant age.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A commercial vehicle was purchased for $35,000. It depreciates at 18% p.a. flat rate. (a) Find its value after 4 years. (b) Find the total dollar depreciation over those 4 years. (3 marks)
Q2. A piece of machinery was purchased for $65,000 and has a book value of $18,000 after 5 years using reducing balance depreciation. Find the annual depreciation rate $r$ to 2 decimal places. (3 marks)
Q3. A piece of technology (initial value $80,000) is depreciated at 20% p.a. (a) Find the value after 3 years using flat rate. (b) Find the value after 3 years using reducing balance. (c) Explain why reducing balance is a more appropriate model for technology assets. (4 marks)
Comprehensive answers (click to reveal)
Drill 1: $D = 0.12 \times 60{,}000 = \$7{,}200$/yr. $S = 60{,}000 - 7{,}200 \times 8 = 60{,}000 - 57{,}600 = \$2{,}400$
Drill 2: $S = 60{,}000(0.88)^8 = 60{,}000 \times 0.41579 = \$24{,}947.40$
Drill 3: $S = 30{,}000 - 7{,}500 \times 5 = -\$7{,}500$. Negative — impossible. Asset fully depreciated before year 5 (at exactly year 4).
Drill 4: $32{,}000 = 80{,}000(1-r)^3 \Rightarrow (1-r)^3 = 0.4 \Rightarrow 1-r = 0.7368 \Rightarrow r = 0.2632 = 26.32\%$
Q1 (3 marks): $D = 0.18 \times 35{,}000 = \$6{,}300$/year [0.5]. After 4 years: $S = 35{,}000 - 6{,}300 \times 4 = \$9{,}800$ [1.5]. Total depreciation $= 35{,}000 - 9{,}800 = \$25{,}200$ [1].
Q2 (3 marks): $18{,}000 = 65{,}000(1-r)^5$ [0.5]. $(1-r)^5 = 18{,}000/65{,}000 = 0.27692$ [0.5]. $1-r = (0.27692)^{0.2} = 0.76847$ [1]. $r = 1 - 0.76847 = 0.23153 \approx 23.15\%$ [1].
Q3 (4 marks): (a) Flat rate: $D = 0.20 \times 80{,}000 = \$16{,}000$. $S = 80{,}000 - 16{,}000 \times 3 = \$32{,}000$ [1]. (b) Reducing balance: $S = 80{,}000(0.80)^3 = 80{,}000 \times 0.512 = \$40{,}960$ [1]. (c) Technology loses the most value in its first 1–2 years due to rapid obsolescence — reducing balance, with its steep early decline, better models this real-world pattern [1]. Flat rate would overstate the value in early years and understate it later [1].
Five timed questions on flat rate and reducing balance depreciation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms using depreciation, flat rate, and reducing balance. Pool: lessons 1–4.
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