Straight-line and declining-balance depreciation for HSC Maths Standard 2

Syllabus dot point

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

HSCMaths Standard 2Year 12: Financial Mathematics11 min readUpdated 20 Jun 2026

Quick answer

Straight-line depreciation drops a fixed dollar amount each year, $V = P - Dn$ with $D = (P - S)/n_{\text{end}}$ ; declining-balance drops a fixed percentage, $V = P(1 - r)^n$ , identical in form to compound interest with the multiplier less than $1$ .

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What this dot point is asking
The answer
Exam-style questions
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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

The two methods differ in one thing: the shape of the value-against-time graph. That shape decides every calculation. Straight-line depreciation subtracts the same dollar amount each year. Its graph is a straight line falling from the purchase price to the salvage value, the value the asset is assumed to have left at the end. Declining-balance depreciation removes the same percentage of the current value each year. Its graph is an exponential-decay curve: it drops steeply at first, then flattens, and approaches zero without ever reaching it. The chart below puts both methods on the same $56000 vehicle so you can see how declining balance starts steeper, crosses below the straight line, and never lands on a salvage value.

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).

Key fact

Straight-line: $V = P - Dn$ , with $D = \dfrac{P - S}{n_{\text{end}}}$ when a salvage value $S$ is reached after $n_{\text{end}}$ years. The graph is a straight line and it stops at the salvage value. Declining-balance: $V = P(1 - r)^n$ , using the multiplier $1 - r$ , never the rate $r$ alone. The graph is exponential decay and it never reaches zero. Both give book value: the depreciation in any single year is the drop in $V$ across that year.

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.

The declining-balance schedule, stage by stage

Declining-balance value can be read straight off $V = P(1 - r)^n$ , but a question may instead ask you to complete a year-by-year schedule, and seeing the schedule built makes the formula concrete. The key move is that each year's closing value becomes the next year's opening value: you take $25\%$ off, carry the remainder down, and repeat. The four panels below build the schedule for the $56000 Hilux at a $25\%$ declining-balance rate.

Stage 1, the first year. Start at the purchase price $56000. One year's depreciation is $25\%$ of $56000 = 14000$ , i.e. $14000, leaving a closing book value of $42000. Equivalently, $56000 \times 0.75 = 42000$ , i.e. $42000.

Stage 2, carry the value down. The $42000 closing value becomes year $2$ 's opening value. Take $25\%$ off that ($10500) to leave $31500. Notice the dollar depreciation is smaller than year $1$ , because $25\%$ is now taken of a smaller base.

Stage 3, repeat again. Carry $31500 down, remove $25\%$ ($7875), and the closing value is $23625. The depreciation amount keeps shrinking, which is exactly why the curve flattens.

Stage 4, the fourth year. Carry $23625 down, remove $25\%$ ($5906.25), giving $17718.75. This is the closed-form answer too: $56000 \times 0.75^4 = 56000 \times 0.31641 = 17718.75$ , i.e. $17718.75. The schedule and the formula agree exactly.

The schedule is the long way and the formula $V = P(1 - r)^n$ is the shortcut; they must agree, so if a "complete the table" question and a "find the value after $n$ years" question both appear, use one to check the other.

Solving for the rate or the time

Depreciation questions are not always "find the value". Two rearrangements cover the rest:

Find the declining-balance rate from two known values. If an asset worth $P$ is worth $V$ after $n$ years, then $(1 - r)^n = V/P$ , so $1 - r = (V/P)^{1/n}$ and $r = 1 - (V/P)^{1/n}$ . This is the same $n$ th-root step as solving compound interest for its rate.
Find the time to reach a target value. Set $P(1 - r)^n = V$ , divide, and take logarithms: $n = \dfrac{\log(V/P)}{\log(1 - r)}$ . Because $\log(1 - r)$ is negative, the algebra still gives a positive $n$ .

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. In the opening chart the two cross between years $3$ and $4$ : before the crossover declining balance is higher, after it straight-line is higher.

Reading the question for the method and the unknown

Standard 2 depreciation questions always state which method to use, so read the wording first. A fixed dollar amount each year, or a salvage value over a set number of years, means straight-line. A percentage per year means declining balance. Then decide what is unknown. The question may give the price and rate and ask for a later value, which you get by substituting straight in. Or it may give two values and ask for the rate or the time. Pin down the method and the unknown before you substitute, so you do not reach for the wrong formula. Always check two things: a straight-line value should not be pushed below its salvage value, and a declining-balance multiplier uses $1 - r$ , not $r$ .

How exam questions ask about depreciation

Translate the wording into the method and the unknown:

"Using the straight-line method, find the value after $t$ years." Compute $D$ (from salvage if given), then $V = P - Dt$ . Stop at the salvage value; do not go below it.
"Find the annual depreciation / how much it loses each year (straight-line)." That is $D = \frac{P - S}{n_{\text{end}}}$ , or $\frac{P}{n_{\text{end}}}$ if salvage is zero.
"Depreciating at $r\%$ per annum (reducing/diminishing/declining balance), find the value after $n$ years." Use the multiplier: $V = P(1 - r)^n$ .
"Complete the depreciation schedule / table." Year by year, take the percentage off the opening value and carry the closing value down, as in the four panels above.
"At what rate does it depreciate?" given two values. $r = 1 - (V/P)^{1/n}$ .
"After how many years is it worth less than $X?" or "when does it reach the salvage value?" Solve for $n$ with logs (declining balance) or by $\frac{P - V}{D}$ (straight-line), and round to whole years as the wording demands.
"Which method gives the higher value at year $k$ ?" or "compare the two methods." Compute both and state the difference; mention that declining balance is higher early and lower later, and never reaches zero.

Vehicle depreciation: declining balance vs straight-line, and a laptop write-off

Apply the declining-balance formula to a $56000 Toyota Hilux at $25\%$ per year, then compare with straight-line over $8$ years, and finally apply a $40\%$ declining-balance rate to a $2400 laptop.

Toyota Hilux: declining balance versus straight-line

Identify the declining-balance multiplier: Under declining-balance (also called diminishing value), a fixed percentage of the current book value is removed each year. The key is to use $1 - r$ as the multiplier, not $r$ itself. For a $25\%$ rate, the vehicle retains $75\%$ of its value each year, so the multiplier is $0.75$ .

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

Calculate the book value at selected years under declining balance: Substituting into $V = P(1-r)^n$ for each required year gives the value directly. Each result should look reasonable: roughly halving the previous few-year figure as the exponent grows.

V_1 = 56000 \times 0.75 = 42000

V_3 = 56000 \times (0.75)^3 = 56000 \times 0.4219 \approx 23625

V_5 = 56000 \times (0.75)^5 = 56000 \times 0.2373 \approx 13289

V_8 = 56000 \times (0.75)^8 = 56000 \times 0.1001 \approx 5608

In dollars, these book values are $42000, $23625, $13289 and $5608.

Note: under declining balance the asset never reaches exactly zero.

Find the annual straight-line depreciation charge: Straight-line spreads the net cost (purchase price less salvage value) evenly over the asset's useful life. The annual charge is constant, which makes it easy to track but gives a slower deduction in the early years compared to declining balance.

Estimated useful life: $8$ years. Salvage value: $8000.

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

i.e. $6000 per year.

Calculate straight-line book values and compare: Subtracting cumulative depreciation from the purchase price gives the book value at each year. The value falls in a straight line and reaches exactly the salvage value at year $8$ .

V_3 = 56000 - 3 \times 6000 = 56000 - 18000 = 38000

V_5 = 56000 - 5 \times 6000 = 56000 - 30000 = 26000

V_8 = 56000 - 8 \times 6000 = 56000 - 48000 = 8000 \text{ (salvage)}

In dollars, these are $38000, $26000 and the $8000 salvage value. 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, exactly as the opening chart shows.

Declining-balance write-off of a laptop

The method is identical regardless of asset type: identify the multiplier ( $1 - 0.40 = 0.60$ ) and substitute into the formula. A $2400 laptop depreciates at $40\%$ per year on a declining-balance rate.

V_2 = 2400 \times 0.60^2 = 2400 \times 0.36 = 864

V_4 = 2400 \times 0.60^4 = 2400 \times 0.1296 \approx 311

In dollars, the laptop is worth $864 after $2$ years and about $311 after $4$ years. A common ATO rule of thumb is to write off the laptop entirely once book value falls below a threshold (the $300 small-asset threshold for some categories).

Final answer: Hilux book values under declining balance: $42000 (year 1), $23625 (year 3), $13289 (year 5), $5608 (year 8). Under straight-line: $38000 (year 3), $26000 (year 5), $8000 (year 8, salvage). Laptop book value: $864 after 2 years, approximately $311 after 4 years.

Edge case: finding the rate from two values

A delivery van bought for $60000 is worth $30000 after $3$ years on a declining-balance basis. The rate is not "halved over $3$ years means about $17\%$ "; it is the third root: $1 - r = (30000/60000)^{1/3} = 0.5^{1/3} \approx 0.7937$ , so $r \approx 0.2063$ , about $20.6\%$ per year. Check: $60000 \times 0.7937^3 \approx 30000$ , i.e. $30000.

Exam technique

State which method you are using and write the formula before substituting; markers award method marks for $V = P - Dn$ or $V = P(1 - r)^n$ even if the arithmetic slips. For declining balance, write the multiplier $1 - r$ explicitly (for example $0.82$ for an $18\%$ rate) so you cannot accidentally use the rate. Round only the final value, to cents for money. Never let a straight-line value drop below the salvage value, and remember declining-balance values never reach zero. If asked for the depreciation in a particular year, give the drop across that year, not the book value.

Common traps

Using the rate, not the multiplier: Declining-balance uses $1 - r$ , not $r$ . $18\%$ depreciation gives base $0.82$ , not $0.18$ .
Forgetting salvage value in straight-line: Annual depreciation is $\frac{P - S}{n}$ , not $\frac{P}{n}$ . Skipping salvage gives a value that drops below the salvage at year $n$ .
Letting straight-line go negative: Beyond year $n_{\text{end}}$ , the asset is at salvage; do not extrapolate to negative values.
Mixing methods: Choose one method per question. The HSC will specify which is being used.
Confusing depreciation rate with interest rate: Declining-balance depreciation has the same formula as compound interest but with the multiplier less than $1$ .

Note

Depreciation just means something you bought loses value as it gets older, like a new car being worth less the day after you drive it home. The two methods are two stories about how that loss happens. The first takes off the same number of dollars every year, so the value drops in an even, steady way. The second takes off the same slice, like a quarter, of whatever it is worth right now, so the dollar loss is big at first and gets smaller each year. The everyday picture: a tool that wears out a little each year fits the steady method, while a phone that loses lots of value when new fits the percentage one.

Exam-style questions

Questions in the style of NESA 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.

2022 HSC-style3 marks

A van is bought for $48000. Using the straight-line method, the salvage value after

8

years is estimated at $8000. Find the annual depreciation and the value after

5

years.

Show worked answer →

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

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

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

2021 HSC-style3 marks

Machinery worth $25000 depreciates at

18\%

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

4

years.

Show worked answer →

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

$(0.82)^4 \approx 0.4521$ .

$V \approx 25000 \times 0.4521 \approx 11303.04$ , i.e. $11303.04.

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

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A tip truck is bought for $

85\,000

and depreciates by $

9000

each year using the straight-line method. Find its book value after

6

years.

Show worked solution →

Identify the method. A fixed dollar amount removed each year is straight-line depreciation, so use $V = V_0 - Dn$ with starting value $V_0 = 85\,000$ and annual depreciation $D = 9000$ .

Substitute $n = 6$ . Six years of depreciation is $6 \times 9000 = 54\,000$ , so

V = 85\,000 - 6 \times 9000 = 85\,000 - 54\,000 = 31\,000.

So the book value after $6$ years is $ $31\,000$ .

Check. The truck loses $ $9000$ a year, so over $6$ years it loses a little over half its $ $85\,000$ price, leaving $ $31\,000$ , which is reasonable.

Answer: $ $31\,000$ .

foundation2 marks

A farm quad bike is bought for $

9000

and depreciates at

20\%

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

3

years.

Show worked solution →

Identify the multiplier. Declining balance removes a fixed percentage of the current value each year, so use the multiplier $1 - r$ , not the rate. At $r = 0.20$ the quad bike keeps $80\%$ of its value each year, so the multiplier is $0.8$ .

Substitute into the formula. Using $V = V_0(1 - r)^n$ with $V_0 = 9000$ and $n = 3$ :

V = 9000 \times (0.8)^3 = 9000 \times 0.512 = 4608.

So the book value after $3$ years is $ $4608$ .

Check. Year by year the value goes $9000 \to 7200 \to 5760 \to 4608$ , each step keeping $80\%$ , which agrees with the formula.

Answer: $ $4608$ .

foundation3 marks

An office printer is bought for $4800. Using the straight-line method it is expected to have a salvage value of $600 after

6

years. (a) Find the annual depreciation. (b) Find the book value after

4

years.

Show worked solution →

Identify the method and find the annual depreciation. Straight-line means a fixed dollar amount is removed each year. With a salvage value $S$ reached after $n_{\text{end}}$ years, the annual depreciation is

D = \frac{P - S}{n_{\text{end}}} = \frac{4800 - 600}{6} = \frac{4200}{6} = 700.

So the printer loses $700 each year.

Find the book value after $4$ years. Substitute into $V = P - Dn$ with $n = 4$ :

V = 4800 - 4 \times 700 = 4800 - 2800 = 2000.

So the book value after $4$ years is $2000.

Check. Carrying on to year $6$ gives $V = 4800 - 6 \times 700 = 4800 - 4200 = 600$ , which is exactly the stated salvage value, so the depreciation is correct.

foundation2 marks

A smartphone costs $1200 and depreciates at

30\%

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

2

years.

Show worked solution →

Identify the multiplier. Declining balance removes a fixed percentage of the current value each year, so use the multiplier $1 - r$ , not the rate. For $r = 0.30$ the phone keeps $70\%$ of its value each year, so the multiplier is $0.7$ .

Substitute into the formula. Using $V = P(1 - r)^n$ with $P = 1200$ and $n = 2$ :

V = 1200 \times (0.7)^2 = 1200 \times 0.49 = 588.

So the book value after $2$ years is $588.

Check. A second year of depreciation removes $30\%$ of the year- $1$ value $1200 \times 0.7 = 840$ : $840 \times 0.7 = 588$ . The recursion agrees, so the value is $588.

core3 marks

An office photocopier is bought for $

12\,000

. Using the straight-line method it depreciates to a salvage value of $

2000

over

8

years. (a) Find the annual depreciation. (b) Find the book value after

3

years. (c) Find the first whole year at the end of which the book value has fallen below $

5000

Show worked solution →

Find the annual depreciation. The straight-line charge spreads the net cost evenly over the useful life:

D = \frac{V_0 - S}{n_{\text{end}}} = \frac{12\,000 - 2000}{8} = \frac{10\,000}{8} = 1250.

So the photocopier loses $ $1250$ each year.

Find the book value after $3$ years. Substitute $n = 3$ into $V = V_0 - Dn$ :

V = 12\,000 - 3 \times 1250 = 12\,000 - 3750 = 8250.

So after $3$ years the book value is $ $8250$ .

Find when the value falls below $ $5000$ . Solve $12\,000 - 1250n < 5000$ :

7000 < 1250n \implies n > \frac{7000}{1250} = 5.6.

Depreciation is recorded at the end of each whole year, so round up to $n = 6$ . At year $5$ the value is $12\,000 - 5 \times 1250 = 5750$ , still above $ $5000$ , and at year $6$ it is $12\,000 - 6 \times 1250 = 4500$ , below $ $5000$ .

Check. The year- $6$ value $ $4500$ is above the $ $2000$ salvage, so the straight-line value has not been taken below salvage, and the first year below $ $5000$ is year $6$ .

Answer: (a) $ $1250$ per year; (b) $ $8250$ ; (c) year $6$ .

core4 marks

A farm tractor is bought for $90000. Using the straight-line method it depreciates to a salvage value of $18000 over

12

years. (a) Find the annual depreciation. (b) Find the book value after

7

years. (c) Find the first whole year at the end of which the book value has fallen below $40000.

Show worked solution →

Find the annual depreciation. The straight-line charge spreads the net cost evenly over the useful life:

D = \frac{P - S}{n_{\text{end}}} = \frac{90000 - 18000}{12} = \frac{72000}{12} = 6000.

So the tractor loses $6000 each year.

Find the book value after $7$ years. Substitute $n = 7$ into $V = P - Dn$ :

V = 90000 - 7 \times 6000 = 90000 - 42000 = 48000.

So after $7$ years the book value is $48000.

Find when the value falls below $40000. Solve $90000 - 6000n < 40000$ :

50000 < 6000n \implies n > \frac{50000}{6000} \approx 8.33.

Since depreciation is recorded at the end of each whole year, round up to $n = 9$ . Checking: at year $8$ the value is $90000 - 8 \times 6000 = 42000$ , still above $40000, and at year $9$ it is $90000 - 9 \times 6000 = 36000$ , below $40000.

Check. The year- $9$ value $36000 is above the $18000 salvage value, so the straight-line value has not been extrapolated below salvage, and the first year below $40000 is year $9$ .

core4 marks

A car is bought for $32000 and depreciates at

22\%

per annum on a declining-balance basis. (a) Find its book value after

4

years, to the nearest cent. (b) Find the depreciation during the third year, to the nearest cent.

Show worked solution →

Identify the multiplier. A $22\%$ declining-balance rate means the car keeps $78\%$ of its value each year, so the multiplier is $1 - r = 0.78$ .

Find the book value after $4$ years. Substitute into $V = P(1 - r)^n$ with $P = 32000$ and $n = 4$ :

V_4 = 32000 \times (0.78)^4 = 32000 \times 0.37015056 = 11844.82.

So after $4$ years the book value is $11844.82.

Find the depreciation during the third year. This is the drop from the end of year $2$ to the end of year $3$ , not a book value. The year- $2$ value is

V_2 = 32000 \times (0.78)^2 = 32000 \times 0.6084 = 19468.80,

and the third year removes $22\%$ of that opening value:

0.22 \times 19468.80 = 4283.136 \approx 4283.14.

So the depreciation during the third year is $4283.14.

Check. The year- $3$ value is $V_3 = 32000 \times (0.78)^3 = 15185.66$ , and $V_2 - V_3 = 19468.80 - 15185.66 = 4283.14$ , matching the third-year depreciation.

core3 marks

A commercial dough mixer is bought for $50000. After

2

years its book value on a declining-balance basis is $32000. Find the annual depreciation rate as a percentage.

Show worked solution →

Set up the equation. Declining balance gives $V = P(1 - r)^n$ . With $P = 50000$ , $V = 32000$ and $n = 2$ :

50000 \times (1 - r)^2 = 32000.

Solve for the multiplier. Divide both sides by $50000$ :

(1 - r)^2 = \frac{32000}{50000} = 0.64.

Take the square root (the multiplier is positive):

1 - r = \sqrt{0.64} = 0.8.

Solve for the rate. Rearrange $1 - r = 0.8$ to get $r = 0.2$ , that is $20\%$ per annum.

Check. Substituting back, $50000 \times (0.8)^2 = 50000 \times 0.64 = 32000$ , which matches the given value, so the rate is $20\%$ per annum.

exam5 marks

A warehouse buys a forklift for $45000. Method A uses straight-line depreciation to a salvage value of $9000 over

8

years. Method B uses declining-balance depreciation at

20\%

per annum. (a) Find the book value after

5

years under each method. (b) State which method gives the higher book value at year

5

and by how much.

Show worked solution →

Find the straight-line value at year $5$ . The annual charge is

D = \frac{45000 - 9000}{8} = \frac{36000}{8} = 4500,

so after $5$ years, using $V = P - Dn$ :

V_{\text{A}} = 45000 - 5 \times 4500 = 45000 - 22500 = 22500.

Find the declining-balance value at year $5$ . The multiplier is $1 - 0.20 = 0.8$ , so using $V = P(1 - r)^n$ :

V_{\text{B}} = 45000 \times (0.8)^5 = 45000 \times 0.32768 = 14745.60.

So at year $5$ the book values are $22500 (straight-line) and $14745.60 (declining-balance).

Compare the two methods. Straight-line is higher at year $5$ . The difference is

22500 - 14745.60 = 7754.40.

So Method A (straight-line) gives the higher book value at year $5$ , by $7754.40.

Check. Declining balance is higher only in the very early years and falls faster afterwards, so by year $5$ the straight-line value being larger is expected, and both values lie above $0, consistent with the methods.

exam4 marks

A business buys laptops at $2500 each. Each laptop depreciates at

40\%

per annum on a declining-balance basis. The business writes a laptop off once its book value first falls below $300. After how many whole years is each laptop written off?

Show worked solution →

Set up the inequality. The multiplier is $1 - 0.40 = 0.6$ , so the book value after $n$ years is $V = 2500 \times (0.6)^n$ . We need the first whole $n$ with

2500 \times (0.6)^n < 300.

Rearrange and take logarithms. Divide both sides by $2500$ :

(0.6)^n < \frac{300}{2500} = 0.12.

Taking logarithms and dividing by $\log 0.6$ , which is negative, reverses the inequality:

n > \frac{\log 0.12}{\log 0.6} \approx \frac{-0.9208}{-0.2218} \approx 4.15.

Round to the next whole year. Since depreciation is applied at the end of each whole year, round up to $n = 5$ .

Check. At year $4$ the value is $2500 \times (0.6)^4 = 2500 \times 0.1296 = 324$ , which is still above $300, and at year $5$ it is $2500 \times (0.6)^5 = 2500 \times 0.07776 = 194.40$ , which is below $300. So each laptop is written off after $5$ years.

exam6 marks

A cafe buys an espresso machine for $18000 that depreciates at

15\%

per annum on a declining-balance basis. (a) Complete a declining-balance schedule for the first

3

years, giving the depreciation and closing book value each year. (b) Use the formula to find the book value after

6

years, to the nearest cent. (c) If instead the machine were depreciated by the straight-line method to a salvage value of $3000 over

10

years, find the book value after

6

years, and state which method gives the higher book value at year

6

Show worked solution →

Build the schedule year by year. The multiplier is $1 - 0.15 = 0.85$ . Each year remove $15\%$ of the opening value and carry the closing value down.

Year $1$ : opening $18000, depreciation $0.15 \times 18000 = 2700$ , closing $18000 - 2700 = 15300$ .

Year $2$ : opening $15300, depreciation $0.15 \times 15300 = 2295$ , closing $15300 - 2295 = 13005$ .

Year $3$ : opening $13005, depreciation $0.15 \times 13005 = 1950.75$ , closing $13005 - 1950.75 = 11054.25$ .

So the closing book values are $15300, $13005 and $11054.25, with depreciation $2700, $2295 and $1950.75.

Find the year- $6$ value from the formula. Using $V = P(1 - r)^n$ with $n = 6$ :

V_6 = 18000 \times (0.85)^6 = 6788.69.

So after $6$ years the declining-balance book value is $6788.69.

Find the straight-line value at year $6$ and compare. The straight-line charge is

D = \frac{18000 - 3000}{10} = \frac{15000}{10} = 1500,

so $V = 18000 - 6 \times 1500 = 18000 - 9000 = 9000$ . The straight-line value $9000 is higher than the declining-balance value $6788.69 at year $6$ .

Check. The schedule closing values fall by a shrinking amount each year ( $2700$ , then $2295$ , then $1950.75$ ), exactly as declining balance should, and the straight-line value $9000 sits above its $3000 salvage, so both methods are applied correctly and straight-line is higher at year $6$ .

exam6 marks

A vineyard buys a grape harvester for $

240\,000

. (a) Under declining-balance depreciation at

30\%

per annum, find the book value after

3

years, to the nearest cent. (b) Under straight-line depreciation to a salvage value of $

30\,000

over

10

years, find the book value after

3

years. (c) State which method gives the higher book value at year

3

and by how much. (d) Using the declining-balance method, find the first whole year at the end of which the book value has fallen below $

60\,000

Show worked solution →

Find the declining-balance value at year $3$ . A $30\%$ rate keeps $70\%$ of the value each year, so the multiplier is $1 - 0.30 = 0.7$ . Using $V = V_0(1 - r)^n$ with $n = 3$ :

V_{\text{DB}} = 240\,000 \times (0.7)^3 = 240\,000 \times 0.343 = 82\,320.

So the declining-balance book value after $3$ years is $ $82\,320.00$ .

Find the straight-line value at year $3$ . The annual charge is

D = \frac{240\,000 - 30\,000}{10} = \frac{210\,000}{10} = 21\,000,

so with $V = V_0 - Dn$ and $n = 3$ :

V_{\text{SL}} = 240\,000 - 3 \times 21\,000 = 240\,000 - 63\,000 = 177\,000.

So the straight-line book value after $3$ years is $ $177\,000$ .

Compare the two methods. Straight-line is higher at year $3$ . The difference is

177\,000 - 82\,320 = 94\,680.

So straight-line gives the higher book value at year $3$ , by $ $94\,680.00$ .

Find when declining balance falls below $ $60\,000$ . Solve $240\,000 \times (0.7)^n < 60\,000$ . Dividing by $240\,000$ gives $(0.7)^n < 0.25$ . Taking logarithms and dividing by $\log 0.7$ , which is negative, reverses the inequality:

n > \frac{\log 0.25}{\log 0.7} \approx 3.89,

so round up to $n = 4$ . At year $3$ the value is $ $82\,320.00$ , still above $ $60\,000$ , and at year $4$ it is $240\,000 \times (0.7)^4 = 57\,624$ , below $ $60\,000$ .

Check. Declining balance falls faster early on, so it sitting well below straight-line by year $3$ is expected, and the year- $4$ value $ $57\,624.00$ confirms the threshold year.

Answer: (a) $ $82\,320.00$ ; (b) $ $177\,000$ ; (c) straight-line, by $ $94\,680.00$ ; (d) year $4$ .

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA