NSWMaths Standard 2Syllabus dot point

How is depreciation calculated using the straight-line and declining-balance methods?

Use the straight-line and declining-balance methods to calculate the value of a depreciating asset over time

A focused answer to the HSC Maths Standard 2 dot point on depreciation. Both straight-line and declining-balance formulas, how they differ, salvage value, and worked Australian examples for cars, equipment and electronics.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to apply the two standard depreciation methods (straight-line and declining-balance) to estimate the value of an asset over time, and to compare the two methods.

The answer

Straight-line depreciation

The asset loses a fixed dollar amount each year. The value after $n$ years is

V = P - D n,

where $P$ is the purchase price and $D$ is the constant annual depreciation.

If a salvage value $S$ is given for $n_{\text{end}}$ years, then

D = \frac{P - S}{n_{\text{end}}}.

The value falls linearly from $P$ to $S$ over $n_{\text{end}}$ years. After $n_{\text{end}}$ years the asset is assumed to have value $S$ (often $0$ ).

Declining-balance depreciation

The asset loses a fixed percentage of its current value each year. After $n$ years:

V = P(1 - r)^n,

where $r$ is the annual depreciation rate (as a decimal). $1 - r$ is the per-year multiplier; this is identical in form to compound interest with a negative rate.

The value falls quickly at first then slowly, asymptotically approaching zero (never actually reaching it).

Choosing between methods

Straight-line is used for tax purposes when the asset wears out at a steady rate (office equipment with a known useful life, buildings).
Declining-balance is more common for assets that lose value quickly when new (cars, computers, machinery). The Australian Taxation Office allows both for many asset classes.

Book value vs market value

These formulas give the book value (what the asset is recorded as on the balance sheet). Market value (resale value) can differ, and the question will usually be explicit about which is asked.

Comparing values at a given year

Plot or tabulate the two methods side by side. Declining balance gives higher book value early on but never reaches zero. Straight-line gives a constant drop and hits salvage on schedule.

Worked example

Vehicle depreciation (Australian context)

A new Toyota Hilux costs $\$ 56000 $on the road. ATO declining-balance rate for new utes is approximately$ 25%$ per year.

After $1$ year: $V = 56000 \times 0.75 = \$ 42000$.

After $3$ years: $V = 56000 \times (0.75)^3 = 56000 \times 0.4219 \approx \$ 23625$.

After $5$ years: $V = 56000 \times (0.75)^5 = 56000 \times 0.2373 \approx \$ 13289$.

After $8$ years: $V = 56000 \times (0.75)^8 = 56000 \times 0.1001 \approx \$ 5608$.

The asset never reaches zero on declining balance.

Same vehicle on straight-line

Estimate useful life of $8$ years with salvage value $\$ 8000$.

$D = \frac{56000 - 8000}{8} = \$ 6000$ per year.

After $3$ years: $V = 56000 - 18000 = \$ 38000$.

After $5$ years: $V = 56000 - 30000 = \$ 26000$.

After $8$ years: $V = 56000 - 48000 = \$ 8000$ (salvage).

Straight-line gives a higher book value in the middle years and exactly $\$ 8000 $at year$ 8 $. Declining-balance is higher in year$ 1 $but lower from year$ 4$ onwards.

Office computer

A $\$ 2400 $laptop on a$ 40%$ declining-balance rate.

After $2$ years: $V = 2400 \times 0.60^2 = 2400 \times 0.36 = \$ 864$.

After $4$ years: $V = 2400 \times 0.60^4 = 2400 \times 0.1296 \approx \$ 311$.

A common ATO rule of thumb is to write off the laptop entirely once book value falls below a threshold ( $\$ 300$ small-asset threshold for some categories).

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q133 marksA van is bought for

\

48000

. Using the straight-line method, the salvage value after

years is estimated at

8000

. Find the annual depreciation and the value after

5

years.

Show worked answer →

Annual depreciation: $D = \frac{48000 - 8000}{8} = \frac{40000}{8} = \$ 5000$ per year.

Value after $5$ years: $V = 48000 - 5 \times 5000 = 48000 - 25000 = \$ 23000$.

Markers reward the depreciation calculation, the linear formula $V = P - D t$ , and the value with units.

2021 HSC Q193 marksMachinery worth

\

25000

depreciates at

18\%

per annum using the declining-balance method. Find its value after

4$ years.

Show worked answer →

Declining balance: $V = P(1 - r)^n = 25000(0.82)^4$ .

$(0.82)^4 \approx 0.4521$ .

$V \approx 25000 \times 0.4521 \approx \$ 11302.81$.

Markers reward the multiplier $0.82$ , the correct exponent, and the answer rounded to cents. Half a mark for using $1.18$ instead of $0.82$ .