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Depreciation (SACE Stage 2 General Mathematics)

Syllabus dot point

“Model depreciation using flat-rate, reducing-balance and unit-cost methods, and find the value of an asset over time.”

8 min read

Quick answer

Depreciation models an asset losing value. Flat-rate (straight-line) subtracts a fixed amount each year, reducing-balance multiplies by a fixed percentage each year, and unit-cost depreciates by an amount per unit of use. Each gives the value after a chosen time.

Jump to a section
  1. What this dot point is asking
  2. Flat-rate (straight-line) depreciation
  3. Reducing-balance depreciation
  4. Unit-cost depreciation
  5. Depreciation as negative growth
  6. Finding the time to a target value
  7. Choosing and comparing methods
  8. Exam-style questions

What this dot point is asking

You must apply all three depreciation methods, find an asset's value over time, and recognise which method suits a situation.

Flat-rate (straight-line) depreciation

The asset loses the same amount each year, often a fixed percentage of the original cost. If V0V_0 is the purchase price and DD is the annual loss, the value after nn years is:

Vn=V0−DnV_n = V_0 - Dn

Flat-rate depreciation

The asset's value decreases by a constant amount each period, producing a straight-line graph. The annual loss is often quoted as a percentage of the original purchase price, so the dollar loss stays the same every year.

Worked example

A van costs $40 000\$40\,000 and depreciates by 15% of its original value each year (flat-rate). Find its value after 4 years, and determine in which year the value first falls below $10 000\$10\,000.

Step 1: Calculate the fixed annual loss in dollars

In flat-rate depreciation the percentage is applied to the original cost every year, so the dollar loss is the same each year. Multiplying the purchase price by the depreciation rate gives:

D=0.15×40 000=6000.D = 0.15 \times 40\,000 = 6000.

The van loses $6000\$6000 of value every year regardless of its current worth.

Step 2: Find the value after 4 years using the flat-rate formula

Because the same $6000\$6000 is subtracted each year, after nn years the value is Vn=V0−DnV_n = V_0 - Dn. Substituting n=4n = 4:

V4=40 000−6000×4=40 000−24 000=16 000.V_4 = 40\,000 - 6000 \times 4 = 40\,000 - 24\,000 = 16\,000.

Step 3: Find when the value first falls below $10 000\$10\,000

We set up the inequality Vn<10 000V_n < 10\,000 and solve for nn:

40 000−6000n<10 000  ⟹  6000n>30 000  ⟹  n>5.40\,000 - 6000n < 10\,000 \implies 6000n > 30\,000 \implies n > 5.

At n=5n = 5 the value is exactly $10 000\$10\,000, and at n=6n = 6 it falls to $4000\$4000, so the value first drops below $10 000\$10\,000 during year 6.

Final answer: After 4 years the van is worth $16 000\$16\,000. The value first falls below $10 000\$10\,000 in year 6.

Reducing-balance depreciation

The asset loses a fixed percentage of its current value each year, so the dollar loss shrinks over time. With depreciation rate rr (decimal), the value after nn years is:

Vn=V0(1−r)nV_n = V_0(1 - r)^n

Reducing-balance depreciation

Vn=V0(1−r)nV_n = V_0(1 - r)^n

where V0V_0 is the initial value, rr is the annual depreciation rate as a decimal, and nn is the number of years. This is the compound interest formula with a negative growth rate.

Worked example

A laptop costs $2000\$2000 and depreciates at 25% per year on a reducing-balance basis. Find its value after 3 years.

Step 1: Recognise the depreciation method and write down the formula

The phrase "25% of its current value" signals reducing-balance depreciation: each year the laptop keeps a fixed fraction of its value at the start of that year. This is different from flat-rate, where the dollar loss is always the same. The formula is:

Vn=V0(1−r)n,V_n = V_0(1 - r)^n,

where V0V_0 is the purchase price, rr is the annual depreciation rate as a decimal, and nn is the number of years.

Step 2: Identify the values and substitute

Here V0=2000V_0 = 2000, r=0.25r = 0.25 (so each year the laptop retains 1−0.25=0.751 - 0.25 = 0.75 of the previous year's value), and n=3n = 3:

V3=2000(1−0.25)3=2000(0.75)3=2000×0.421875=843.75.V_3 = 2000(1 - 0.25)^3 = 2000(0.75)^3 = 2000 \times 0.421875 = 843.75.

Step 3: Interpret the result

The laptop has lost more than half its value in three years, which is typical of reducing-balance at a high rate like 25%. Each year the dollar loss is smaller than the last: $500\$500 in year 1, $375\$375 in year 2, $281.25\$281.25 in year 3.

Final answer: After 3 years the laptop is worth $843.75\$843.75.

Unit-cost depreciation

Here the value falls by a fixed amount for each unit of use (kilometres driven, items produced, hours run), rather than per year. If the asset loses cc dollars per unit and has been used uu units:

V=V0−c×uV = V_0 - c \times u

Worked example

Unit-cost depreciation

A machine costs $50 000 and depreciates by $0.40 for each item it produces. Find its value after producing 60 000 items.

V=50000−0.40×60000=50000−24000=26000V = 50000 - 0.40 \times 60000 = 50000 - 24000 = 26000

The machine is worth $26 000 after 60 000 items.

Depreciation as negative growth

Reducing-balance depreciation is mathematically the compound-interest formula with a negative rate: where an investment multiplies by (1+i)(1 + i) each period, a depreciating asset multiplies by (1−r)(1 - r). This is why a reducing-balance asset never reaches exactly zero - each year it keeps a fixed fraction of its value, so the curve approaches the axis without touching it. Flat-rate depreciation, by contrast, is a linear model with a negative gradient and will eventually reach zero (and would go negative if not stopped at the scrap value). Recognising these as the linear and exponential models from earlier topics ties the financial work back to the modelling strand.

Finding the time to a target value

A common task is to find how long until an asset reaches a given value. For flat-rate, set V0−DnV_0 - Dn equal to the target and solve the linear equation for nn. For reducing-balance, set V0(1−r)nV_0(1 - r)^n equal to the target; solving for nn needs logarithms or a financial solver. For unit-cost, set V0−cuV_0 - cu equal to the target and solve for the number of units uu. In every case, round to a sensible whole number of years or units and check the answer makes practical sense in the asset's context.

Choosing and comparing methods

Flat-rate gives a straight-line decline and never quite reflects how assets lose value fastest when new. Reducing-balance drops quickly at first then levels off, matching items like cars and electronics, and never reaches exactly zero. Unit-cost suits assets whose wear depends on use rather than time.

Scrap value and useful life

The scrap (or salvage) value is the value at the end of an asset's useful life. To find when an asset reaches a given scrap value, set VnV_n equal to that value and solve for nn, the number of years or units.

Identify the method from the wording

A fixed dollar or fixed-percentage-of-original loss each year is flat-rate. A percentage "of its current value" each year is reducing-balance. A loss "per kilometre" or "per item" is unit-cost. Spotting the keyword tells you which formula to use.

Common mistake

Do not confuse flat-rate and reducing-balance when a percentage is given. "15% of the original value" each year is flat-rate (a constant dollar loss), while "15% of its current value" each year is reducing-balance (a shrinking dollar loss). Using the wrong one gives very different long-term values.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2023-style3 marks
Calculator-assumed. A van costs 40000 dollars and depreciates by 15% of its original value per year (flat-rate). (a) Find its value after 4 years. (b) Find when its value first falls below 10000 dollars.
Show worked answer →

(a) Annual loss D=0.15×40000=6000D = 0.15 \times 40000 = 6000. Value after 4 years: V4=40000−6000(4)=16000V_4 = 40000 - 6000(4) = 16000 dollars. (2 marks)

(b) Set 40000−6000n<1000040000 - 6000n < 10000, so 30000<6000n30000 < 6000n, giving n>5n > 5. The value first falls below 10000 dollars during year 6 (at n=6n = 6, V=40000−36000=4000V = 40000 - 36000 = 4000; at n=5n = 5, V=10000V = 10000). (1 mark)

Marks: two for the value after 4 years, one for solving the inequality.

SACE 2022-style2 marks
Calculator-assumed. A laptop costs 2000 dollars and depreciates at 25% per year on a reducing-balance basis. Find its value after 3 years.
Show worked answer →

Use Vn=V0(1−r)nV_n = V_0(1 - r)^n with V0=2000V_0 = 2000, r=0.25r = 0.25, n=3n = 3:

V3=2000(0.75)3=2000×0.421875=843.75 dollars.V_3 = 2000(0.75)^3 = 2000 \times 0.421875 = 843.75 \text{ dollars}.

Marks: one for the reducing-balance formula, one for the value of about 843.75 dollars.

SACE 2021-style2 marks
Calculator-assumed. A machine costs 50000 dollars and depreciates by 0.40 dollars for each item it produces (unit-cost). Find its value after producing 60000 items.
Show worked answer →

Use V=V0−c×uV = V_0 - c \times u with V0=50000V_0 = 50000, c=0.40c = 0.40, u=60000u = 60000:

V=50000−0.40×60000=50000−24000=26000 dollars.V = 50000 - 0.40 \times 60000 = 50000 - 24000 = 26000 \text{ dollars}.

Marks: one for the unit-cost set-up, one for the value of 26000 dollars.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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