Reducing-balance depreciation in WACE Mathematics Applications Unit 3

Syllabus dot point

“Model reducing-balance depreciation with a recurrence relation, compare it with flat-rate depreciation, and find book value and scrap-value timing.”

WACEMathematics ApplicationsUnit 37 min readUpdated 2 Jun 2026

Quick answer

Reducing-balance depreciation removes a fixed percentage of an asset's current value each year, so it is geometric decay: $V_{n+1} = R V_n$ with $R = 1 - i$ . The explicit form is $V_n = V_0 R^n$ . Unlike flat-rate (straight-line) depreciation, which subtracts the same dollar amount each year, reducing-balance loses less in dollars each year as the value shrinks, giving a curve rather than a straight line.

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What this dot point is asking
The reducing-balance model
Reducing-balance versus flat-rate
Finding the rate from two values
Choosing a depreciation method
Unit-cost (kilometre) depreciation
Exam-style questions

What this dot point is asking

You must model reducing-balance depreciation with a recurrence, contrast it with flat-rate, find a book value, and find when an asset reaches a scrap value.

The reducing-balance model

Each year the asset loses a percentage of whatever it is currently worth, so it multiplies by a fixed factor.

Reducing-balance depreciation

$V_{n+1} = R V_n$ , where $R = 1 - i$ and $i$ is the annual depreciation rate as a decimal. Explicit form: $V_n = V_0 R^n$ , with $V_0$ the purchase price.

A $12\%$ annual depreciation gives $R = 0.88$ . Because $0 < R < 1$ , the value decays towards (but never reaches) zero, flattening as it falls.

Reducing-balance versus flat-rate

The two methods produce very different value curves.

Feature	Flat-rate (straight-line)	Reducing-balance
Loss each year	Same dollar amount	Same percentage
Model	Arithmetic, $V_n = V_0 - dn$	Geometric, $V_n = V_0 R^n$
Graph	Straight line	Decaying curve
Early years	Loses less	Loses more

Reducing-balance loses most early

Reducing-balance depreciation removes the largest dollar amounts in the first years (when the value is high) and tapers off later. Flat-rate removes equal dollars throughout. This is why a new car loses value fastest in its first years under a reducing-balance model.

Worked example

Book value and scrap timing

A printing machine costs $\$120\,000$ and depreciates by $25\%$ of its value each year. Find its book value after $3$ years and the first year it falls below a scrap value of $\$30\,000$ .

Step 1: Set up the depreciation model

A $25\%$ annual loss means $75\%$ of the value survives each year, so the common ratio is $R = 0.75$ and the purchase price gives $V_0 = 120000$ . The recurrence is $V_{n+1} = 0.75 V_n$ and the explicit rule is $V_n = 120000 \times 0.75^n$ .

Step 2: Calculate the book value after 3 years

Substituting $n = 3$ into the explicit rule gives the value at that point in time.

$V_3 = 120000 \times 0.75^3 = 120000 \times 0.421875 = 50625$ , so $\$50\,625$ .

Step 3: Find the first year the value falls below the scrap value

To find when $V_n$ first drops below $\$30\,000$ , rearrange to $0.75^n < 0.25$ and test successive whole-number years. Testing, $0.75^4 = 0.3164$ and $0.75^5 = 0.2373$ , so the value first falls below $\$30\,000$ in year $5$ .

Step 4: Interpret the results

After $3$ years the machine is worth $\$50\,625$ on the books, and it reaches its scrap value early in year $5$ , the point at which it would typically be sold or replaced.

Finding the rate from two values

If you know the value at two times, find the ratio. If $V_0 = 40000$ and $V_3 = 23000$ , then $40000 R^3 = 23000$ , so $R^3 = 0.575$ and $R = \sqrt[3]{0.575} = 0.8315$ , giving an annual depreciation rate of about $16.9\%$ .

Choosing a depreciation method

Businesses choose between the two methods for tax and accounting reasons, and SCSA questions often ask you to compare them. Flat-rate depreciation writes off the same dollar amount each year, so its straight-line graph is simple to budget for and the asset reaches zero in a fixed number of years ( $\dfrac{\text{cost}}{\text{annual loss}}$ ). Reducing-balance depreciation writes off more in the early years, which better matches assets that lose value fastest when new (cars, technology), and it never quite reaches zero because each year removes a percentage of a shrinking value.

A useful check is the total depreciation. Under flat-rate, after $n$ years the total lost is $n \times d$ dollars. Under reducing-balance, the total lost is $V_0 - V_0 R^n = V_0(1 - R^n)$ . Comparing these for the same asset shows reducing-balance losing more early and flat-rate catching up later, the trade-off that decides which method suits a given asset. SCSA comparison questions reward computing both book values at the same year and stating which method leaves the higher value and why (the percentage loss shrinks each year under reducing-balance).

Unit-cost (kilometre) depreciation

A related model charges depreciation per unit of use rather than per year, for example cents per kilometre driven. The value after travelling $k$ kilometres is $V = V_0 - (\text{rate per km}) \times k$ , an arithmetic (flat-rate) model in $k$ rather than in time. Read whether the question depreciates by time or by usage, because the variable in the explicit rule changes accordingly.

Common errors

Using the rate as the ratio: A $15\%$ depreciation has ratio $R = 0.85$ , not $0.15$ .
Confusing flat-rate with reducing-balance: Flat-rate subtracts a fixed dollar amount (arithmetic); reducing-balance multiplies by a fixed ratio (geometric). Read which the question states.
Rounding mid-calculation: Keep full precision in $R^n$ and round the final dollar value.
Off-by-one on the year: When finding the first year below a threshold, round the comparison up to the next whole year, then state that year.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks

Equipment is bought for

\$80\,000

and depreciates by

18\%

of its value each year (reducing balance). (a) Write a recurrence relation for the book value

V_n

. (b) Find the book value after

5

years. (c) Find the first year the value falls below the scrap value of

\$25\,000

Show worked answer →

Reducing-balance loss of $18\%$ keeps $82\%$ each year.

(a) Common ratio $R = 1 - 0.18 = 0.82$ , so $V_{n+1} = 0.82 V_n$ , with $V_0 = 80000$ . (2 marks)

(b) $V_n = 80000 \times 0.82^n$ , so $V_5 = 80000 \times 0.82^5 = 80000 \times 0.37074 = 29659$ , that is $\$29\,659$ . (2 marks)

(c) Solve $80000 \times 0.82^n < 25000$ , so $0.82^n < 0.3125$ . Testing, $0.82^5 = 0.3707$ and $0.82^6 = 0.3040$ , so the value first falls below $\$25\,000$ in year $6$ . (2 marks)

Markers reward the ratio $0.82$ , the explicit substitution, and finding the first whole year below the scrap value.

WACE 2024-style5 marks

A van costs

\$48\,000

. Under flat-rate depreciation it loses

\$6000

per year; under reducing-balance it loses

15\%

per year. Compare the two book values after

4

years and state which method gives the higher value.

Show worked answer →

Compute each method's value at year $4$ .

Flat-rate: $V_4 = 48000 - 6000 \times 4 = 48000 - 24000 = 24000$ , that is $\$24\,000$ . (2 marks)

Reducing-balance: $R = 0.85$ , $V_4 = 48000 \times 0.85^4 = 48000 \times 0.52200 = 25056$ , that is $\$25\,056$ . (2 marks)

Reducing-balance gives the higher value after $4$ years ( $\$25\,056$ versus $\$24\,000$ ), because the percentage loss shrinks each year as the value falls. (1 mark)

Markers reward both correct book values and the comparison.