Depreciation | Year 10 | Ascent Library

01 / LEARN

Understand the idea.

Two depreciation structures

Straight-line depreciation subtracts a fixed dollar amount each period: $V = P - d n$ . Reducing-balance depreciation removes a fixed percentage of current value: $V = P {(1 - r)}^{n}$ . Identify whether the question specifies dollars or a percentage.

For a positive decreasing reducing-balance model, $P > 0$ , $0 < r < 1$ and $n \geq 0$ . The retained factor is $1 - r$ . As current value decreases, the same percentage removes a smaller dollar amount each period.

Keep the interpretation sensible

A straight-line formula reaches zero after $\frac{P}{d}$ periods when $d > 0$ , and becomes negative afterwards. If negative asset value is meaningless, restrict its domain or revise the model rather than reporting a negative resale value.

A reducing-balance model stays positive at every finite period under these assumptions and approaches zero. Neither model guarantees an actual sale price; interpret the answer as the value predicted by the supplied rule.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Reducing balance

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

Worked solution

1
Find the retained annual factor.
$b = 1 - 0.15 = 0.85$
A loss of $15 %$ retains $85 %$ of current value.
2
Apply two annual reductions.
$V = 2400 {(0.85)}^{2}$
The second-year percentage acts on the first-year reduced value.
3
Evaluate the model value.
$V = 2400 (0.7225) = 1734 dollars$
The exponent counts two complete annual periods.

Answer

1734 dollars.

EXAMPLE 02

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.

Worked solution

1
Write the straight-line value rule.
$V = 2400 - 300 n$
The same $300$ dollars is subtracted each year.
2
Substitute five years.
$V (5) = 2400 - 300 (5) = 900 dollars$
Five fixed annual losses total $1500$ dollars.
3
Solve for zero value.
$2400 - 300 n = 0 ⟹ n = \frac{2400}{300} = 8$
Zero value occurs when accumulated fixed losses equal the starting value.
4
State the sensible time limit.
$0 \leq n \leq 8$
Within this interval the straight-line model gives nonnegative ordinary asset values.

Answer

900 dollars after 5 years; zero after 8 years.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the retained factor for an 18% annual loss.

Show answer & reasoning

0.82

An $18 %$ annual loss leaves $82 %$ of current value, so the retained factor is $1 - 0.18 = 0.82$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 $6400 {(\frac{1}{2})}^{2} = 1600$ .
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 $t \leq 6$ .
$0 \leq t \leq 6$

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Two depreciation structures

Keep the interpretation sensible

See the method in action.

Reducing balance

Find the retained annual factor.

Apply two annual reductions.

Evaluate the model value.

Straight-line value

Write the straight-line value rule.

Substitute five years.

Solve for zero value.

State the sensible time limit.

Try it yourself.

Check your understanding

Keep practising