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hscscience Maths Std · Y11
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Module 3 · L13 of 14 ~50 min MST-12-S2-02 ⚡ +90 XP available

Depreciation

Assets lose value over time, but does a car lose the same dollar amount each year, or the same percentage? Straight-line depreciation uses a fixed dollar amount ($S = V_0 - Dn$); declining balance uses a fixed percentage ($S = V_0(1-r)^n$). Learn both, compare them, and find unknown rates.

Today's hook, The moment you drive a new car off the lot, it loses roughly 10–15% of its value. By the end of the first year it might be worth 20% less than you paid. Does an asset lose the same dollar amount each year, or the same percentage? The answer depends on which depreciation method you use, and the two methods give very different results.
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Get oriented

Recall what you already know, meet both formulas and settle the key 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 is purchased for $\$30,000$. After one year it might be worth only $\$24,000$, a loss of $\$6,000$ in year one. Does it lose another $\$6,000$ in year two, or something different? Does an asset lose the same dollar amount each year, or the same percentage each year? Do you think one method is more realistic for describing how assets like cars actually lose value?

Without calculating write your gut feeling. We'll revisit this at the end of the lesson.

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02
The key formulas you need to own
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Depreciation in Maths Standard uses two formulas that mirror simple and compound interest, but in reverse (value decreases instead of grows).

Straight-line: $S = V_0 - Dn$ loses the same dollar amount $D$ each year, linear, like simple interest in reverse. Declining balance: $S = V_0(1-r)^n$ loses the same percentage each year, exponential decay, like compound interest but shrinking.

Straight-line vs declining-balance depreciation.
Total depreciation = $V_0 - S$ for both methods
Fixed $\$$ → straight-line
"Depreciates by $\$2,400$ per year", a dollar figure signals straight-line. Use $S = V_0 - Dn$.
Fixed % → declining balance
"Depreciates at 18% per annum", a percentage signals declining balance. Use $S = V_0(1-r)^n$. The multiplier is $(1-r)$, e.g. $1 - 0.18 = 0.82$.
Total depreciation
Always use $V_0 - S$. For declining balance, never multiply the annual percentage by $n$, the dollar amount changes each year.
03
What you'll master
Know

Key facts

  • Straight-line formula: $S = V_0 - Dn$
  • Declining balance formula: $S = V_0(1-r)^n$
  • How to find $D$ from initial value, salvage value, and time
  • That total depreciation = $V_0 - S$ for both methods
  • That declining balance never theoretically reaches zero
Understand

Concepts

  • Why the declining balance multiplier is $(1-r)$, not $(1+r)$
  • Why straight-line is linear and declining balance is exponential
  • How to identify which method from question wording
  • Why declining balance depreciates faster in early years
Can do

Skills

  • Apply $S = V_0 - Dn$ including solving for $D$ or $n$
  • Apply $S = V_0(1-r)^n$ for any number of periods
  • Compare both methods for the same asset
  • Find depreciation rate $r$ by rearranging $S = V_0(1-r)^n$
04
Key terms
DepreciationThe decrease in the value of an asset over time due to wear, tear, or obsolescence.
Straight-LineA depreciation method where the asset loses the same dollar amount each year: $S = V_0 - Dn$.
Declining BalanceA depreciation method where a fixed percentage of the current value is deducted each year: $S = V_0(1-r)^n$.
Salvage Value ($S$)The estimated value of an asset at any point in time (also called book value or residual value).
Initial Value ($V_0$)The purchase price of the asset at time $n = 0$.
Total DepreciationThe total loss in value: $V_0 - S$. Use this for both methods, not $D \times n$ for declining balance.

Depreciate in a straight line

Work through S = V0 - Dn, where the asset loses the same amount every year.

05
Straight-Line Depreciation, $S = V_0 - Dn$
core concept

Straight-line depreciation reduces an asset's value by the same fixed dollar amount $D$ each period, producing a linear decrease in value over time, identical in structure to simple interest, but going down instead of up.

For example, a machine worth $\$24,000$ that depreciates by $\$3,000$ per year will be worth $24,000 − ($3,000 × 5) = $9,000 after 5 years. An asset depreciated to zero is "fully depreciated."

If the initial value, final salvage value, and time are all known, calculate $D = (V_0 - S_n) \div n$.

Must do: Identify $V_0$, $D$, and $n$ clearly before substituting. $V_0$ is the purchase price, $D$ is the annual depreciation amount in dollars (not a percentage), and $n$ is the number of years.
Common error: Straight-line depreciation uses a fixed dollar amount, not a percentage. If $D = \$2{,}500$ per year, the asset loses $2,500 every year regardless of its current value. Applying a percentage to the current value each year is declining balance depreciation, a different formula.
Straight-line depreciation falls by a fixed $3,000/yr; declining balance falls by 18% of the current value.

Declining balance depreciates faster in early years; straight-line is constant.

Straight-line depreciation: S = V₀ − Dn, where V₀ = purchase price, D = annual depreciation amount, n = number of years. The asset loses an equal dollar amount each year; S is the salvage value after n years.

Pause, copy S = V₀ − Dn with definitions: V₀ = purchase price, D = fixed annual depreciation amount, n = number of years, S = salvage value, and note D = (V₀ − S_final) / n when the scrap value is known into your book.

Quick check: An asset depreciates at 18% per annum declining balance. What multiplier should be used each year?

Depreciate on the declining balance

Work through S = V0(1-r)^n, where the asset loses a percentage of what is left.

06
Declining Balance Depreciation, $S = V_0(1-r)^n$
core concept

Straight-line depreciation S = V₀ − Dn removes the same dollar amount D every year, giving a linear decline. Many assets, vehicles, electronics, lose a larger dollar amount when new and smaller amounts later. Declining balance (reducing value) depreciation S = V₀(1 − r)^n models this: it removes the same percentage r of the current value each year, producing exponential decay that mirrors compound interest but in the falling direction.

Declining balance depreciation reduces the asset's value by a fixed percentage of its current value each period, so the dollar amount of depreciation decreases as the asset becomes worth less.

The formula $S = V_0(1-r)^n$ mirrors the compound interest formula structurally, except instead of multiplying by $(1+r)$ to grow, we multiply by $(1-r)$ to shrink. For example, a vehicle purchased for $\$32,000$ depreciating at 18% per annum after 4 years: $S = \$32{,}000 \times (0.82)^4 = \$32{,}000 \times 0.45212 = \$14{,}468$.

Must do: Convert the depreciation rate to a decimal and subtract from 1. $r = 18\%$ means $(1-r) = 0.82$. Label this multiplier explicitly in your working, it prevents sign mistakes.
Insight: $S = V_0(1-r)^n$ is structurally identical to $A = P(1+r)^n$. Same calculator process, same care with $r$ and $n$, just subtraction instead of addition inside the bracket. If you can do compound interest, you can do declining balance.
Common error: Under declining balance, an asset never theoretically reaches zero, because each year's depreciation is a percentage of the remaining value. Unlike straight-line, the model does not predict the asset reaching $0.

To find total depreciation: always use $V_0 - S$, do not multiply the annual percentage by $n$, as the dollar amount changes each year.

Declining balance (reducing value) depreciation: S = V₀(1 − r)^n, where r = depreciation rate per period. The asset loses a fixed percentage of its current value each year, so losses are large early and small later.

Pause, copy S = V₀(1 − r)^n with definitions (r = annual depreciation rate as decimal), and note the structural parallel: compound growth uses (1 + r)^n; declining balance uses (1 − r)^n, same formula, opposite direction into your book.

True or false: Under declining balance depreciation, the dollar amount lost each year stays the same.

Work four depreciation examples

Follow salvage value, the depreciation rate, the useful life and a method comparison.

Worked examples · 4 problems, reveal step by step

PROBLEM 1 · STRAIGHT-LINE, VALUE AND INEQUALITY

A piece of industrial equipment was purchased for $\$85,000$. It depreciates by $\$7,400$ per year using the straight-line method. (a) What is its value after 6 years? (b) After how many complete years will its value first fall below $30,000?

1
Part (a)
$S = V_0 - Dn = \$85{,}000 - (\$7{,}400 \times 6) = \$40{,}600$
Substitute directly into the straight-line formula.
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PROBLEM 2 · DECLINING BALANCE DEPRECIATION

A car is purchased new for $38,500 and depreciates at 22% per annum declining balance. Calculate: (a) its value after 3 years, and (b) the total depreciation over this period.

1
Part (a), salvage value
$r = 0.22$, so $(1-r) = 0.78$
$S = \$38{,}500 \times (0.78)^3$
$(0.78)^3 = 0.474552$
$S = \$38{,}500 \times 0.474552 = \$18{,}270.25$
Convert 22% to decimal, subtract from 1 to get the multiplier, raise to power 3, then multiply by initial value.
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PROBLEM 3 · COMPARING BOTH METHODS

Office furniture is purchased for $\$16,000$. Method A: straight-line depreciation of $\$1,800$ per year. Method B: declining balance at 14% per annum. (a) Find the value under each method after 5 years. (b) Which method gives a higher salvage value after 5 years, and by how much?

1
Method A, straight-line
$S_A = \$16{,}000 - (\$1{,}800 \times 5) = \$7{,}000.00$
Fixed dollar amount: $1,800 × 5 = $9,000 total depreciation.
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PROBLEM 4 · FINDING THE DEPRECIATION RATE

A machine was purchased for $\$28,000$ and is worth $\$15,950$ after 4 years under declining balance depreciation. Find the annual depreciation rate, correct to two decimal places.

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Set up the equation
$15{,}950 = 28{,}000(1-r)^4$
Start with the declining balance formula and substitute the known values.
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Fill the gap: A vehicle worth $\$45,000$ depreciates at 25% per annum declining balance. After 2 years its value is $\$$ (use $S = 45000 \times (0.75)^2$).

Avoid the depreciation traps

Fix the method, rate and units slips that cost marks.

Common errors · the 3 traps that cost marks

Trap 01
Using the wrong method
"Depreciates by $3,000 per year" → straight-line. "Depreciates at 15% per annum" → declining balance. Read the wording before choosing a formula.
Trap 02
Using $(1+r)$ instead of $(1-r)$
Depreciation shrinks the value, so the multiplier is $(1-r)$. Using $(1+r)$ gives compound interest growth instead. $r = 18\%$ → multiplier = $0.82$, not $1.18$.
Trap 03
Multiplying $r \times n$ for declining balance total depreciation
Under declining balance, the dollar depreciation changes each year. Total depreciation = $V_0 - S$. Never use $r \times n \times V_0$ for declining balance.

Match each description to the correct model:

  • depreciates by $2,400 per year
  • depreciates at 18% per annum
  • same amount each year
  • same percentage each year
  • Straight-line
  • Declining balance
  • S = V₀ − Dn
  • S = V₀(1−r)^n

Drill it and revisit

Run the drill, model both methods in a spreadsheet, then name what has changed.

Quick-fire practice · 5 calculations

1

A printer costing $\$6,200$ depreciates by $\$540$ per year using straight-line depreciation. Find its value after 5 years.

2

A boat was purchased for $64,000 and depreciates at 12% per annum declining balance. Find its value after 3 years.

3

An asset is purchased for $\$50,000$ and has a salvage value of $\$18,000$ after 8 years of straight-line depreciation. What is the annual depreciation amount?

4

A machine originally worth $\$2,800$ depreciates by $\$420$ per year (straight-line). What is its value after 4 years?

5

An asset was bought for $\$24,000$. Under declining balance it depreciates at 10% per annum. Compare the salvage value after 4 years with straight-line at $\$2,200$ per year.

Top 3 list: Name THREE things that distinguish declining balance depreciation from straight-line depreciation.

08
Modelling both methods in a spreadsheet
core concept

The two depreciation methods are hard to compare from formulas alone, because one is a subtraction and the other is a multiplication. Put them in adjacent columns and the comparison becomes obvious.

Take an asset worth $\$40\,000$, depreciated by $\$6000$ per year on the straight line, or by $20\%$ per year on the declining balance.

CellWhat goes in itFill down?
A2the year number, $0$yes, $=$A2+1
B2straight line, $=$40000-6000*A2yes
C2declining balance, $=$40000*0.8^A2yes

Filling down to year $5$ gives:

YearStraight lineDeclining balance
1$\$34\,000.00$$\$32\,000.00$
2$\$28\,000.00$$\$25\,600.00$
3$\$22\,000.00$$\$20\,480.00$
4$\$16\,000.00$$\$16\,384.00$
5$\$10\,000.00$$\$13\,107.20$

Graph both columns against the year. Straight line plots as a straight line, falling by the same $\$6000$ every year. Declining balance plots as a curve that falls steeply then flattens, because $20\%$ of a shrinking value is a shrinking amount.

Read where the two lines cross. Declining balance is below the straight line for the first three years and above it from year $4$ on. It writes value off faster early and slower late, which is why it suits assets like vehicles. A model where one curve never crosses the other has a wrong rate somewhere.

To model depreciation in a spreadsheet, put the year in one column and fill down. Straight line is $=V_0-D*n$ and plots as a straight line; declining balance is $=V_0*(1-r)^n$ and plots as a curve that falls fast then flattens. Graph both against the year to see where they cross.

Pause, copy the two cell formulas and one line on what the crossing point means, into your book.

09
Revisit your thinking

Look back at what you wrote in the Think First section. Under straight-line depreciation, the car loses the same dollar amount every year. Under declining balance, it loses a fixed percentage of its current value, so the dollar amount decreases over time. Declining balance is more realistic for assets like cars, which lose a large percentage of value early on.

What has changed? What did you get right? What surprised you?

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Practise depreciation

Answer the question bank, then write full short-answer responses.

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 32 marks

SA 1. A printer costing $\$6,200$ depreciates by $\$540$ per year using straight-line depreciation. Find its value after 5 years. (2 marks)

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ApplyBand 33 marks

SA 2. A boat was purchased for $64,000 and depreciates at 12% per annum declining balance. Find: (a) its value after 3 years, and (b) the total depreciation over this period. (3 marks)

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AnalyseBand 43 marks

SA 3. An asset was bought for $\$24,000$. Under straight-line depreciation it loses $\$2,200$ per year. Under declining balance it depreciates at 10% per annum. Compare the salvage value after 4 years. Which method gives the higher salvage value, and by how much? (3 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $S = 6200 - (540 \times 5) = \$3{,}500$ · 2: $(1-0.12)=0.88$; $S = 64000(0.88)^3 = \$43{,}614.21$ · 3: $D = (50000-18000) \div 8 = \$4{,}000$ · 4: $S = 2800 - (420 \times 4) = \$1{,}120$ · 5: SL: $S = 24000 - (2200 \times 4) = \$15{,}200$; DB: $S = 24000(0.90)^4 = \$15{,}746.40$; DB higher by $\$546.40$

SA 1 (2 marks): $S = 6200 - (540 \times 5) = 6200 - 2700$ [1] $= \$3{,}500.00$ [1].

SA 2 (3 marks): $(1-r) = 0.88$ [1]. $S = 64000(0.88)^3 = \$43{,}614.21$ [1]. Total depreciation $= 64000 - 43614.21 = \$20{,}385.79$ [1].

SA 3 (3 marks): SL: $S = 24000 - (2200 \times 4) = \$15{,}200$ [1]. DB: $S = 24000(0.90)^4 = \$15{,}746.40$ [1]. Declining balance gives the higher salvage value by $\$546.40$ [1].