QLDGeneral MathematicsSyllabus dot point

Topic 2: Growth and decay in sequences - how do we model the loss in value of an asset over time?

Model depreciation of an asset using flat-rate (straight-line), reducing-balance and unit-cost methods with recurrence relations and rules, compute book value and scrap value, and compare the methods over the life of the asset

A focused answer to the QCE General Mathematics Unit 3 dot point on depreciation. Covers flat-rate (straight-line), reducing-balance and unit-cost depreciation as recurrence relations and rules, book value and scrap value, and how the three methods differ over the life of an asset, with arithmetic-verified worked examples for IA2 and the external assessment.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to model how an asset such as a car, machine or computer loses value over time, using the three standard methods: flat-rate (straight-line), reducing-balance and unit-cost. Each method is a sequence, so you express it as a recurrence relation and a rule, compute the book value at any time, and recognise when an asset reaches its scrap value. You should also be able to say how the methods behave differently. This is a distinct application of Unit 3 Topic 2 sequences and is a reliable source of marks in IA2 and the external assessment.

The answer

Book value, scrap value and depreciation

The book value is the current worth of an asset after depreciation. Depreciation is the amount of value lost. The scrap value (or salvage value) is the value at which the asset is written off. Each method below starts from the purchase price $V_0$ and reduces it.

Flat-rate (straight-line) depreciation

The asset loses the same fixed dollar amount $d$ each period. This is linear (arithmetic) decay.

V_{n+1} = V_n - d, \qquad V_n = V_0 - dn.

If depreciation is quoted as a flat-rate percentage, that percentage is always taken of the original price, so $d$ is constant. A graph of book value against time is a straight line sloping down.

Reducing-balance depreciation

The asset loses a fixed percentage $r$ of its current value each period. This is geometric decay, the same structure as compound interest but shrinking.

V_{n+1} = V_n (1 - r), \qquad V_n = V_0 (1 - r)^n.

Because the percentage is taken of the falling balance, the dollar loss is largest early and shrinks over time. A graph curves downward, steep at first then levelling off, and the value approaches but never reaches zero.

Unit-cost depreciation

The asset loses value in proportion to its use rather than time: a fixed amount per unit of output (per kilometre driven, per item produced). If the asset depreciates by $c$ dollars per unit and has produced $u$ units in total,

V = V_0 - c u.

This is linear in the number of units used, not in time, so the value falls faster in heavy-use periods.

Comparing the methods

Flat-rate gives a constant dollar loss and a straight-line graph.
Reducing-balance gives a large early loss that tapers, and the value never quite hits zero, so a scrap value is reached by a set time rather than by the formula.
Unit-cost ties the loss to actual usage, so two identical assets depreciate at different rates if used differently.

Worked example

Flat-rate versus reducing-balance after three years

A machine is bought for $\$ 30,000 $. Compare its book value after$ 3 $years under flat-rate depreciation of$ \ $4\,000$ per year and reducing-balance depreciation of $15$ percent per year.

Step 1: Flat-rate. Subtract $\$ 4,000 $each year, or use$ V_n = V_0 - dn$.

V_3 = 30000 - 4000 \times 3 = 30000 - 12000 = 18000.

Verify: $4000 \times 3 = 12000$ and $30000 - 12000 = 18000$ . Book value is $\$ 18,000$.

Step 2: Reducing-balance. Multiply by $1 - 0.15 = 0.85$ each year, or use $V_n = V_0(1 - r)^n$ .

V_3 = 30000 \times 0.85^3.

Step 3: Evaluate the power.

0.85^2 = 0.7225, \qquad 0.85^3 = 0.7225 \times 0.85 = 0.614125.

Step 4: Final value.

30000 \times 0.614125 = 18423.75.

Verify: $30000 \times 0.614125 = 18423.75$ . Book value is $\$ 18,423.75$.

Step 5: Compare. After three years reducing-balance leaves a slightly higher value ( $\$ 18,423.75 $versus$ \ $18\,000$ ) because its dollar loss shrinks each year, whereas flat-rate keeps removing the full $\$ 4,000$.

Common traps

Taking the reducing-balance percentage of the original price: Reducing-balance depreciation is a percentage of the current book value, not the purchase price. Using the original price every year turns it into flat-rate by mistake.
Using time instead of usage for unit-cost: Unit-cost depreciation depends on units produced or distance travelled, not on years elapsed. Multiplying by the number of years is wrong.
Forgetting the scrap value floor: Reducing-balance value never reaches zero algebraically, so an asset is retired when its book value first falls to or below the stated scrap value, not when the formula gives zero.
Mixing up $(1 - r)$ and $(1 + r)$: Depreciation shrinks the value, so the multiplier is $1 - r$ . Using $1 + r$ models growth instead.
Dropping the $n$ in flat-rate: Flat-rate after $n$ years subtracts $dn$ , not $d$ . Subtracting only one period of depreciation understates the loss.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 QCAA4 marksA farmer bought a tractor for

45 100 at the start of 2012. It depreciates by

2700 each year. Identify and use a mathematical model to determine the value of the tractor at the start of 2021.

Show worked answer →

Step 1 - identify the model (1 mark). The asset loses a fixed dollar amount ($2700) each year, so this is flat-rate (straight-line) depreciation, which behaves like an arithmetic sequence with first term t_1 = 45 100 and common difference d = -2700.

Step 2 - identify the parameters (1 mark). From the start of 2012 to the start of 2021 is 9 years, so n = 9 full years of depreciation. Using the arithmetic rule t_(n+1) = t_1 + nd, the value after 9 years is the term with n = 9.

Step 3 - substitute (1 mark): value = 45 100 - 2700 x 9 = 45 100 - 24 300.

Step 4 - evaluate and state (1 mark): value = $20 800. The tractor is worth$ 20 800 at the start of 2021.

The trap is the number of years: start of 2012 to start of 2021 is exactly 9 steps, not 10. Equivalently you can use the linear rule V = 45 100 - 2700t with t = 9.