Depreciation | Year 10 | Ascent Library

01 / LEARN

The idea, step by step

Two depreciation structures

Straight-line depreciation subtracts a fixed amount: $V = P - d n$ . Reducing-balance depreciation multiplies current value by a factor: $V = P {(1 - r)}^{n}$ . For the decreasing, positive-value model used here, $P > 0$ , $0 < r < 1$ and $n \geq 0$ .

Under reducing balance, the absolute loss gets smaller as the value decreases. Do not repeatedly subtract the same percentage of the original principal unless that is the stated model.

Keep the interpretation sensible

A straight-line model can eventually become negative. Restrict or revise it if a negative value has no meaning for the asset being modelled.

A supplied value model is not a guarantee of resale price. State the duration, rounding and assumptions before interpreting its answer.

A common trap

Add annual percentage losses until their sum is 100%.

Instead: Each reducing-balance percentage acts on a different base. The model decreases towards zero without reaching it in a finite number of periods.

02 / WORKED EXAMPLES

See the thinking

Example 1

Reducing balance

Equipment starts at 2400 dollars and loses 15% of current value annually. Find its year-2 model value.

The retained factor is $0.85$ .
$V = 2400 {(0.85)}^{2} = 1734$ .

1734 dollars.

Example 2

Straight-line value

An asset starts at 2400 dollars and loses 300 dollars annually. Find its year-5 value and when the formula reaches zero.

At year 5, $V = 2400 - 300 (5) = 900$ .
$2400 - 300 n = 0$ gives $n = 8$ .

900 dollars after 5 years; zero after 8 years.

Check your understanding

Find the retained factor for an 18% annual loss.

Show answer

0.82

$1 - 0.18 = 0.82$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the value of a 5000-dollar model losing 8% annually after 3 years.
2 marks
A device loses 90 dollars annually from 720 dollars. Find its value after 4 years.
2 marks
For $V = 18000 {(0.84)}^{t}$ , state the initial value and annual percentage loss.
2 marks
A value halves every 3 years from 6400 dollars. Find its value after 6 years.
2 marks
Compare year-2 values from 1000 dollars with fixed loss 100 annually versus 10% reducing-balance loss.
3 marks
Give a sensible nonnegative time interval for $V = 1500 - 250 t$ , with $t$ in years.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the value of a 5000-dollar model losing 8% annually after 3 years.
2 marks
$5000 {(0.92)}^{3} = 3893.44$ .
3893.44 dollars
A device loses 90 dollars annually from 720 dollars. Find its value after 4 years.
2 marks
$720 - 90 (4) = 360$ .
360 dollars
For $V = 18000 {(0.84)}^{t}$ , state the initial value and annual percentage loss.
2 marks
The coefficient is the initial value; $1 - 0.84 = 0.16$ .
18000 dollars; 16% loss.
A value halves every 3 years from 6400 dollars. Find its value after 6 years.
2 marks
Two halving periods give $6400 {(1 / 2)}^{2} = 1600$ .
1600 dollars
Compare year-2 values from 1000 dollars with fixed loss 100 annually versus 10% reducing-balance loss.
3 marks
$1000 - 2 (100) = 800$ and $1000 {(0.9)}^{2} = 810$ .
800 dollars fixed; 810 dollars reducing balance.
Give a sensible nonnegative time interval for $V = 1500 - 250 t$ , with $t$ in years.
3 marks
Both time and value are nonnegative. $1500 - 250 t \geq 0$ gives $t \leq 6$ .
$0 \leq t \leq 6$

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Simple versus compound growth; compound interest; depreciation; repeated percentage change; savings/loan scenarios with supplied models; changing assumptions and checking results; spreadsheets/digital models.

Official curriculum source ↗

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