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hscscience Maths Adv · Y12
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Module 7 · L4 of 20 ~35 min ⚡ +95 XP available

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.

Today's hook, A new car costs $\$40,000$. It loses 20% of its original value every year. After 5 years, is it worth exactly $\$0$? More than $0? The mathematics of depreciation has a surprising answer depending on which method you use.
0/5QUESTS

Get oriented

Recall what you already know, meet the key ideas and settle the terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

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?

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02
The two depreciation models
+5 XP to read

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.

Comparison of flat rate vs reducing balance depreciation formulas
$$S = V_0(1-r)^n \quad \text{or} \quad S = V_0 - Dn$$
Flat Rate
$S = V_0 - Dn$ where $D = rV_0$. Same amount lost every year. Can go negative if held too long, needs a salvage cap.
Reducing Balance
$S = V_0(1-r)^n$. Percentage of current value lost each year. Never goes negative asymptotically approaches zero.
Key connection
Reducing balance is just compound interest with $-r$ instead of $+r$: $S = V_0(1+(-r))^n$. Same structure, opposite direction.
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What you'll master
Know

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
Understand

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
Can do

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
04
Key terms
Initial value ($V_0$)The purchase price or starting book value of the asset.
Book value / salvage value ($S$)The value of the asset after $n$ periods of depreciation.
Flat rate depreciationA fixed dollar amount $D = rV_0$ subtracted each year. Linear decline.
Reducing balance depreciationA fixed percentage $r$ of the current book value removed each year. Exponential decline.
Depreciation rate ($r$)The percentage by which the asset loses value per period, expressed as a decimal.
Salvage capIn flat rate, depreciation stops when the book value reaches a specified minimum (usually $0).

Compare flat rate and reducing balance

See what each depreciation method does to the same asset.

05
Flat rate vs reducing balance, visual comparison
core concept

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.

Flat Rate vs Reducing Balance Depreciation Flat Rate (linear) Reducing Balance (exponential) Years (n) Book Value ($) 0 5 10 $45k reaches $0 $23,490 $18,000

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.

Why this matters for tax. Reducing balance gives larger depreciation expenses in early years, reducing taxable income sooner. A dollar saved on tax today is worth more than a dollar saved in the future (time value of money), so many businesses prefer reducing balance for tax purposes.

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.

Work the examples

Follow three worked problems across both depreciation methods.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FLAT RATE DEPRECIATION

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.

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$D = 0.15 \times 45{,}000 = \$6{,}750$ per year
Find the fixed annual depreciation amount $D = rV_0$.
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PROBLEM 2 · REDUCING BALANCE DEPRECIATION

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.

1
$S = 45{,}000(1 - 0.15)^4 = 45{,}000(0.85)^4$
Use the reducing balance formula $S = V_0(1-r)^n$ with $r = 0.15$, $n = 4$.
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PROBLEM 3 · FIND THE DEPRECIATION RATE

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
$28{,}000 = 120{,}000(1-r)^6$
$(1-r)^6 = \dfrac{28{,}000}{120{,}000} = 0.2\overline{3}$
Substitute into $S = V_0(1-r)^n$ and isolate $(1-r)^6$.
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Quick check: A laptop purchased for $2,400 depreciates at 25% p.a. flat rate. Its value after 3 years is:

Avoid the traps

Fix the method and rate slips that cost marks.

Common errors · the 3 traps that cost marks

Trap 01
Mixing up the formulas
Using $V_0(1-r)^n$ when you should use $V_0 - rV_0 n$ (or vice versa). The key: if the loss is described as a "flat" or "straight-line" amount, use $V_0 - Dn$. If it's a percentage "of the current value", use $(1-r)^n$.
Trap 02
Confusing depreciation with interest
Some students write $(1+r)^n$ instead of $(1-r)^n$, the compounding formula, and the asset grows instead of shrinking! Always check: depreciation means the sign inside the bracket is negative: $(1\mathbf{-}r)^n$.
Trap 03
Accepting a negative book value
If flat rate gives a negative answer, don't write it, the book value floors at zero (or the salvage value). A negative book value is a signal that the formula has gone beyond its valid range, not a real-world answer.

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?

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Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

Quick-fire practice · 4 depreciation calculations

1

$V_0 = \$60{,}000$, flat rate 12%, after 8 years. Find $S$.

2

$V_0 = \$60{,}000$, reducing balance 12%, after 8 years. Find $S$.

3

$V_0 = \$30{,}000$, flat rate 25%, after 5 years. What happens?

4

Machinery: $V_0 = \$80{,}000$, $S = \$32{,}000$ after 3 years, reducing balance. Find $r$.

Fill in the blanks: Flat rate depreciation produces a graph, while reducing balance produces an decay curve. Only the flat rate method can produce a book value.

Match each formula with its description:

  • S = V₀ − Dn
  • S = V₀(1−r)ⁿ
  • r = 1 − (S/V₀)^(1/n)
  • S = V₀(1+r)ⁿ
  • Find the depreciation rate from given values
  • Compound interest growth (NOT depreciation)
  • Flat rate, same dollar loss each year
  • Reducing balance, percentage of current value
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Revisit your initial thinking

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.

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Practise depreciation

Write full short-answer responses, then check them against the model answers.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 43 marks

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)

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ApplyBand 43 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)

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AnalyseBand 54 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)

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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].