smc-6925-50-Declining Balance vs Straight-line

Financial Maths, STD2 F4 EQ-Bank 33

Yolanda purchased a motorcycle for $30 000. She explores two options for predicting the value of the motorcycle after four years.

Option 1:

For the first two years, the value of the motorcycle is depreciated by 10% per annum using flat rate depreciation. For the next two years, the value of the motorcycle is depreciated by 10% per annum using reducing balance depreciation.

Option 2:

The value of the motorcycle is depreciated using reducing balance depreciation with a constant depreciation rate per annum for four years.

For both options to predict the same value after four years, determine the rate per annum used for Option 2, giving your answer as a percentage correct to one decimal place. (3 marks)

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$10.3\%$

Show Worked Solution

$\text{Let \(V_n$ = Value after $n$ years}\)

$\text{Option 1:}$

$V_1=30\,000-(10\% \times 30\,000)=27\,000,\ \ V_2=27\,000-3000=24\,000$

$V_3=24\,000 \times 0.9=21\,600,\ \ V_4=21\,600 \times 0.9 = 19\,440$

$\text{Option 2 (predicting the same value):}$

$30\,000(1-r)^{4}$	$=19\,440$
$(1-r)^{4}$	$=\dfrac{19\,440}{30\,000}$
$r$	$=\left(\dfrac{19\,440}{30\,000}\right)^{\frac{1}{4}}-1$
	$= 0.1027…$
	$=10.3\%\ \text{(to 1 d.p.)}$

Financial Maths, STD2 F4 2024 HSC 29

The graph shows the decreasing value of an asset.

For the first 4 years, the value of the asset depreciated by $1500 per year, using a straight-line method of depreciation.

After the end of the 4th year, the method of depreciation changed to the declining-balance method at the rate of 35% per annum.

What is the total depreciation at the end of 10 years? (4 marks)

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$\text{Total depreciation}\ =$46\,681.57$

Show Worked Solution

$\text{Depreciation after 4 years}\ = 4 \times 1500 = $6000$

$\text{Value after 4 years}\ = 50\,000-6000=44\,000$

$\text{Declining balance used for the next 6 years:}$

$V_0=$44\,000, r=0.35, n=6$

$S$	$=V_0(1-r)^n$
	$=44\,000(1-0.35)^6$
	$=$3318.43$

$\therefore\ \text{Total depreciation}\ =50\,000-3318.43=$46\,681.57$

Financial Maths, STD2 F4 2023 HSC 28

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \textit{Rate of depreciation} \\
\hline
\rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$3500 per annum} \\
\hline
\rule{0pt}{2.5ex} \text{Declining balance method} \rule[-1ex]{0pt}{0pt} & \text{12% per annum} \\
\hline
\end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations. (3 marks)

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`text{Straight-line method:}`

`S`	`=V_0-Dn`
	`=60\ 000-3500×3`
	`=$49\ 500`

`text{Declining-balance method:}`

`S`	`=V_0(1-r)^n`
	`=60\ 000(1-0.12)^3`
	`=60\ 000(0.88)^3`
	`=$40\ 888.32`

`text{Salvage value is lower for the declining-balance method.}`

Show Worked Solution

`text{Straight-line method:}`

`S`	`=V_0-Dn`
	`=60\ 000-3500×3`
	`=$49\ 500`

`text{Declining-balance method:}`

`S`	`=V_0(1-r)^n`
	`=60\ 000(1-0.12)^3`
	`=60\ 000(0.88)^3`
	`=$40\ 888.32`

`text{Salvage value is lower for the declining-balance method.}`

Financial Maths, STD2 F4 2022 HSC 27

A company purchases a machine for $50 000. The two methods of depreciation being considered are the declining-balance method and the straight-line method.

For the declining-balance method, the salvage value of the machine after `n` years is given by the formula
`S=V_(0)xx(0.80)^(n),`
where `S` is the salvage value and `V_(0)` is the initial value of the asset.
i. What is the annual rate of depreciation used in this formula? (1 mark)
ii. Calculate the salvage value of the machine after 3 years, based on the given formula. (1 mark)
For the straight-line method, the value of the machine is depreciated at a rate of 12.2% of the purchase price each year.
When will the value of the machine, using this method, be equal to the salvage value found in part (a) (ii)? (2 marks)

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i. `20text{%}`
ii. `$25\ 600`
`text{4 years}`

Show Worked Solution

a.i. `text{Depreciation rate}\ = 1-0.8=0.2=20text{%}`

a.ii. `text{Find}\ \ S\ \ text{when}\ \ n=3:`

`S`	`=V_0 xx (0.80)^n`
	`=50\ 000 xx (0.80)^3`
	`=$25\ 600`

b. `text{Using the SL method}`

`S_n`	`= 50\ 000-(0.122 xx 50\ 000)xxn`
	`=50\ 000-6100n`

`text{Find}\ \ n\ \ text{when}\ \ S_n=$25\ 600`

`25\ 600`	`=50\ 000-6100n`
`6100n`	`=24\ 400`
`n`	`=(24\ 400)/6100`
	`=4\ text{years}`

♦♦ Mean mark (a.i.) 24%.
COMMENT: A poor State result in part (a.i.) that warrants attention.

♦ Mean mark part (b) 38%.

Financial Maths, STD2 F4 2006 HSC 27c

Kai purchased a new car for $30 000. It depreciated in value by $2000 per year for the first three years.

After the end of the third year, Kai changed the method of depreciation to the declining balance method at the rate of 25% per annum.

Calculate the value of the car at the end of the third year. (1 mark)

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Calculate the value of the car seven years after it was purchased. (2 marks)

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Without further calculations, sketch a graph to show the value of the car over the seven years.

Use the horizontal axis to represent time and the vertical axis to represent the value of the car. (3 marks)

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a. `$24\ 000`

b. `$7593.75`

c. `text{See Worked Solutions}`

Show Worked Solution

a. `text(Using)\ \ S = V_0-Dn`

`S= 30\ 000-(2000 xx 3)= $24\ 000`

b. `text(Using)\ \ S = V_0(1-r)^n`

`text(where)\ V_0= 24\ 000,\ r=0.25,\ n=4`

`S= 24\ 000(1-0.25)^4= $7593.75`

`:.\ text(The value of the car after 7 years is $7593.75)`

Financial Maths, STD2 F4 2011 HSC 28b

Norman and Pat each bought the same type of tractor for $60 000 at the same time. The value of their tractors depreciated over time.

The salvage value `S`, in dollars, of each tractor, is its depreciated value after `n` years.

Norman drew a graph to represent the salvage value of his tractor.

Find the gradient of the line shown in the graph. (1 mark)

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What does the value of the gradient represent in this situation? (1 mark)

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Write down the equation of the line shown in the graph. (1 mark)

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Find all the values of `n` that are not suitable for Norman to use when calculating the salvage value of his tractor. Explain why these values are not suitable. (2 marks)

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Pat used the declining balance formula for calculating the salvage value of her tractor. The depreciation rate that she used was 20% per annum.

What did Pat calculate the salvage value of her tractor to be after 14 years? (2 marks)

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Using Pat’s method for depreciation, describe what happens to the salvage value of her tractor for all values of `n` greater than 15. (1 mark)

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a. `text(Gradient) =-4000`

b. `text(The amount the tractor depreciates each year.)`

c. `S = 60\ 000\-4000n`

d. `text(It is unsuitable to use)`

`n<0\ text(because time must be positive)`

`n>15\ text(because the tractor has no more value after 15 years and)`

`text(therefore can’t depreciate further.)`

e. `text(After 14 years, the tractor is worth $2638.83)`

f. `text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases but remains)>0.`

Show Worked Solution

♦♦♦ Mean mark 14%
COMMENT: The intercepts of both axes provide points where the gradient can be quickly found.

i.	`text(Gradient)`	`= text(rise)/text(run)`
		`= (-60\ 000)/15`
		`=-4000`

♦ Mean mark 37%

ii. `text(The amount the tractor depreciates each year)`

♦♦ Mean mark 28%
COMMENT: Using the general form `y=mx+b` is quick here because you have the gradient (from part (i)) and the `y`-intercept is obviously `60\ 000`.

iii.	`text(S)text(ince)\ \ S = V_0\-Dn`
	`:.\ text(Equation of graph:)`
	`S = 60\ 000-4000n`

iv. `text(It is unsuitable to use)`

♦♦♦ Mean mark 20%

`n<0,\ text(because time must be positive:)`

`n>15,\ text(because it has no more value after 15)`

`text(years and therefore can’t depreciate further.)`

v.	`text(Using)\ S = V_0 (1-r)^n\ \ text(where)\ r = text(20%,)\ n = 14`

`S`	`= 60\ 000 (1-0.2)^14`
	`= 60\ 000 (0.8)^14`
	`= 2\ 638.8279…`

`:.\ text(After 14 years, the tractor is worth $2638.83`

♦ Mean mark 37%

vi.	`text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases)`
	`text(but remains > 0.)`

Financial Maths, STD2 F4 2013 HSC 28e

Zheng has purchased a computer for $5000 for his company. He wants to compare two different methods of depreciation over two years for the computer.

Method 1: Straight-line with $1250 depreciation per annum.

Method 2: Declining balance with 35% depreciation per annum.

Which method gives the greatest depreciation over the two years? Justify your answer with suitable calculations. (3 marks)

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`text(Method 2)`

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`text(Method 1)`

`text(Depreciation over 2 years)`	`=2xx 1250`
	`= $2500`

`text(Method 2)`

`text(Depreciation (Year 1) )`	`=35text(%) xx 5000`
	`=$1750`

`text(Depreciation (Year 2) )`	`=35text(%) xx (5000-1750)`
	`=$1137.50`

`text(Depreciation over 2 years)`	`=1750 + 1137.50`
	`=$2887.50`

`:.\ text(Method 2 gives the greater depreciation.)`

Financial Maths, STD2 F4 2009 HSC 24e

Jay bought a computer for $3600. His friend Julie said that all computers are worth nothing (i.e. the value is $0) after 3 years.

Find the amount that the computer would depreciate each year to be worth nothing after 3 years, if the straight line method of depreciation is used. (1 mark)

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Explain why the computer would never be worth nothing if the declining balance method of depreciation is used, with 30% per annum rate of depreciation. Use suitable calculations to support your answer. (2 marks)

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a. `$1200`

b. `text(See Worked Solutions.)`

Show Worked Solution

a.	`S`	`= V_0-Dn`
	`0`	`= 3600-D xx 3`
	`3D`	`= 3600`
	`D`	`= 3600/3`
		`= 1200`

`:.\ text(Annual depreciation = $1200`

♦ Mean mark 45%

b.	`text(Using)\ \ S = V_0 (1-r)^n`
	`text(where)\ r = text(30%)\ \ text(and)\ \ V_0 = 3600`

`S`	`=3600 (1-30/100)^n`
	`= 3600 (0.7)^n`

`(0.7)^n > 0\ text(for all)\ n`

`:.\ text(Salvage value is always)\ >0`

\(30\,000(1-r)^{4}\)	\(=19\,440\)
\((1-r)^{4}\)	\(=\dfrac{19\,440}{30\,000}\)
\(r\)	\(=\left(\dfrac{19\,440}{30\,000}\right)^{\frac{1}{4}}-1\)
	\(= 0.1027…\)
	\(=10.3\%\ \text{(to 1 d.p.)}\)

\(S\)	\(=V_0(1-r)^n\)
	\(=44\,000(1-0.35)^6\)
	\(=$3318.43\)