Depreciation

Financial Maths, STD2 F4 2025 HSC 36

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year? (4 marks)

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$$1476.40$

Show Worked Solution

$\text{Find}\ r:$

$\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}$

$S$	$=V_0(1-r)^n$
$44\ 000$	$=55\ 000(1-r)^1$
$\dfrac{44\ 000}{55\ 000}$	$=1-r$
$1-r$	$=0.8$
$r$	$=1-0.8=0.20$

♦ Mean mark 47%.

$\text{Find \(S$ when}\ \ n=9\ \ \text{and}\ \ n=10:\)

$S_9=55\ 000(1-0.20)^{9}=$7381.97504$

$S_{10}=55\ 000(1-0.20)^{10}=$5905.5800$

$S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}$

$\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}$

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)

Show Answers Only

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, STD1 F1 2021 HSC 19

Yin purchased a car for $20 000. The value of the car decreases according to a linear model. The graph shows the value of the car, $$V$, against the time, $t$ months, since it was purchased.

By how much does the value of the car decrease every 10 months? (1 mark)

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Find the value of the car after 5 years. (1 mark)

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Identify ONE problem with using this model to determine the value of Yin’s car over time. (1 mark)

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a. $$2000$

b. $$8000$

c. $\text{Car will have a negative value after 100 months.}$

Show Worked Solution

a. $\text{Decrease (10 months)}= 20\,000-18\,000= $2000$

b. $\text{5 years} = 5 \times 12 = 60\ \text{months}$

$\text{At}\ \ t=60\ \text{(from graph)}:$

$V = $8000$

c. $\text{Car will have a negative value after 100 months.}$

♦ Mean mark (c) 11%.

Financial Maths, STD2 F1 2021 HSC 19

Adam purchased some office furniture five years ago. It depreciated by $2300 each year based on the straight-line method of depreciation. The salvage value of the furniture is now $7500.

Find the initial value of the office furniture. (2 marks)

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`$19\ 000`

Show Worked Solution

`text{Find initial value}\ (V_0):`

`S`	`=V_0-Dn`
`7500`	`=V_0-2300 xx 5`
`V_0`	`=7500 + 11\ 500`
	`=$19\ 000`

Financial Maths, STD2 F4 2021 HSC 4 MC

Three years ago an appliance was valued at $2467. Its value has depreciated by 15% each year, based on the declining-balance method.

What is the salvage value today, to the nearest dollar?

$952
$1110
$1357
$1515

Show Answers Only

`D`

Show Worked Solution

`S`	`= V_0 (1-r)^n`
	`= 2467 (1-0.15)^3`
	`= 2467 (0.85)^3`
	`= $1515`

`=> D`

Financial Maths, STD2 F4 2020 HSC 11 MC

An asset is depreciated using the declining-balance method with a rate of depreciation of 8% per half year. The asset was bought for $10 000.

What is the salvage value of the asset after 5 years?

$1749.01
$4182.12
$4343.88
$6590.82

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 43%.
COMMENT: 8% depreciation is applicable every 6 months here (n=10). Read carefully!

`V_0 = 10\ 000 \ , \ r = 0.08 \ , \ n = 10`

`S`	`= V_0 (1-r)^n`
	`= 10\ 000 (1-0.08)^10`
	`= 10\ 000 (0.92)^10`
	`= $ 4343.88`

`=> \ C`

Financial Maths, STD2 F4 2019 HSC 37

A new car is bought for $24 950. Each year the value of the car is depreciated by the same percentage.

The table shows the value of the car, based on the declining-balance method of depreciation, for the first three years.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{End of year}\rule[-1ex]{0pt}{0pt} & \textit{Value}\\
\hline
\rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & \$21\ 457.00 \\
\hline
\rule{0pt}{2.5ex}2\rule[-1ex]{0pt}{0pt} & \$18\ 453.02 \\
\hline
\rule{0pt}{2.5ex}3\rule[-1ex]{0pt}{0pt} & \$15\ 869.60 \\
\hline
\end{array}

What is the value of the car at the end of 10 years? (3 marks)

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`$5521.47`

Show Worked Solution

`text(Find the depreciation rate:)`

`S`	`= V_0(1-r)^n`
`21\ 457`	`= 24\ 950(1-r)^1`
`1-r`	`= (21\ 457)/(24\ 950)`
`1-r`	`= 0.86`
`r`	`= 0.14`

`:.\ text(Value after 10 years)`

`= 24\ 950(1-0.14)^10`

`= 5521.474…`

`= $5521.47\ \ (text(nearest cent))`

Financial Maths, STD2 F4 2018 HSC 26h

A car is purchased for $23 900.

The value of the car is depreciated by 11.5% each year using the declining-balance method.

What is the value of the car after three years? (2 marks)

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`$16\ 566\ \ (text(nearest dollar))`

Show Worked Solution

`S`	`= V_0(1-r)^n`
	`= 23\ 900(1-0.115)^3`
	`= 23\ 900(0.885)^3`
	`= 16\ 566.383…`
	`= $16\ 566\ \ (text(nearest dollar))`

Financial Maths, STD2 F1 EQ-Bank 19

Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.

By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation? (1 mark)

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Determine the annual flat rate of depreciation correct to one decimal place. (1 mark)

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a. `$900`

b. `5.3text{% (1 d.p.)}`

Show Worked Solution

a. `text(Depreciation)= 17\ 000-3500= $13\ 500`

`:.\ text(Annual depreciation)= (13\ 500)/15= $900`

b. `:.\ text(Flat rate of depreciation )`

`= 900/(17\ 000) xx 100text(%)`

`= 5.29 …= 5.3text{% (1 d.p.)}`

Financial Maths, STD2 F1 EQ-Bank 18

Khan paid $900 for a printer.

This price includes 10% GST (goods and services tax).

Determine the price of the printer before GST was added.

Write your answer correct to the nearest cent. (2 marks)

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Khan is able to depreciate the full $900 purchase price of his printer for taxation purposes.

Under flat rate depreciation the printer will be valued at $300 after five years.

Calculate the annual depreciation in dollars. (1 mark)

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a. `$818.18`

b. `$120`

Show Worked Solution

a. `text(Let)\ \ $P = text(price ex-GST)`

COMMENT: Reverse GST questions regularly cause problems for many students.

`:. P + 10text(%) xx P`	`= 900`
`1.1P`	`= 900`
`P`	`= 900/1.1`
	`= 818.181…`
	`= $818.18\ \ text(nearest cent)`

b. `text(Annual depreciation)= ((900-300))/5= $120`

Financial Maths, STD2 F1 EQ-Bank 17

A company purchased a machine for $60 000.

For taxation purposes the machine is depreciated over time using the straight line depreciation method.

The machine is depreciated at a flat rate of 10% of the purchase price each year.

By how many dollars will the machine depreciate annually? (1 mark)

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Calculate the value of the machine after three years. (1 mark)

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After how many years will the machine be $12 000 in value? (1 mark)

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a. `$6000`

b. `$42\ 000`

c. `8\ text(years)`

Show Worked Solution

a. `text(Annual depreciation)= 10text(%) xx 60\ 000= $6000`

b. `text(After 3 years,)`

`text(Value)`	`=V_0-Dn`
	`= 60\ 000-(3 xx 6000)`
	`= $42\ 000`

c. `text(Find)\ n\ text(when value = $12 000)`

`12\ 000`	`= 60\ 000-6000 xx n`
`6000n`	`= 48\ 000`
`:.n`	`=(48\ 000)/6000`
	`= 8\ text(years)`

Financial Maths, STD2 F1 EQ-Bank 16

Hugo is a professional bike rider.

The value of his bike will be depreciated over time using the flat rate method of depreciation.

The graph below shows his bike’s initial purchase price and its value at the end of each year for a period of three years.

What was the initial purchase price of the bike? (1 mark)

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Use calculations to show that the bike depreciates in value by $1500 each year. (1 mark)

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Assume that the bike’s value continues to depreciate by $1500 each year. Determine its value five years after it was purchased. (1 mark)

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a. `$8000`

b. `text(See Worked Solutions)`

c. `$500`

Show Worked Solution

a. `$8000`

b. `text(Value after 1 year) = $6500\ \ \ text{(from graph)}`

`:.\ text(Annual depreciation)= 8000-6500= $1500`

c. `text(After 5 years:)`

`S`	`=V_0-Dn`
	`=8000-5 xx 1500`
	`=$500`

Financial Maths, STD2 F1 EQ-Bank 3 MC

Rae paid $40 000 for new office equipment at the start of the 2013 financial year.

At the start of each following financial year, she used flat rate depreciation to revalue her equipment.

At the start of the 2016 financial year she revalued her equipment at $22 000.

The annual flat rate of depreciation she used, as a percentage of the purchase price, was

11.25%
15%
17.5%
35%

Show Answers Only

`B`

Show Worked Solution

`text(Depreciation over 3 years)`

♦ Mean mark 50%.

`=40\ 000-22\ 000=$18\ 000`

`:.\ text(Annual depreciation) = (18\ 000)/3 = $6000`

`:.\ text(Depreciation rate) = 6000/(40\ 000) = 0.15 = 15text(%)`

`=> B`

Financial Maths, STD2 F1 2017 HSC 11 MC

A new car was bought for $19 900 and one year later its value had depreciated to $16 300.

What is the approximate depreciation, expressed as a percentage of the purchase price?

Show Answers Only

`A`

Show Worked Solution

`text(Net Depreciation)= 19\ 900-16\ 300= $3600`

`:. %\ text(Depreciation)= 3600/(19\ 900) xx 100= 18.09…text(%)`

`=>A`

Financial Maths, STD2 F4 2015 HSC 10 MC

A piece of machinery, initially worth $56 000, depreciates at 8% per annum.

Which graph best shows the salvage value of this piece of machinery over time?

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`text(A depreciation of 8% per annum depreciates the largest)`

`text(amount in year 1 and then gradually depreciates less each)`

`text(subsequent year.)`

`:.text(Cannot be)\ C\ text(or)\ D`

`text(Consider when)\ t`	`= 5`
`text(Salvage Value)`	`= V_0(1-r)^n`
	`= 56\ 000(1-0.08)^5`
	`= 36\ 908.5\ …`

`text(Graph B depreciates too quickly)`

`:.text(Cannot be)\ B`

`⇒ A`

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 2005 HSC 26a

A sports car worth $150 000 is bought in December 2005.

In December each year, beginning in 2006, the value of the sports car is depreciated by 10% using the declining balance method of depreciation.

In which year will the depreciated value first fall below $120 000? (2 marks)

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`text(The value falls below $120 000 in the third year)`

`text{which will be during 2008.}`

Show Worked Solution

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

`text(where)\ \ V_0 = 150\ 000, r = text(10%)`

`text(If)\ \ n = 2,`

`S= 150\ 000(1-0.1)^2= $121\ 500`

`text(If)\ \ n= 3,`

`S= 150\ 000(1-0.1)^3= $109\ 350`

`:.\ text(The value falls below $120 000 in the third year)`

`text{which will be during 2008.}`

Financial Maths, STD2 F4 2005 HSC 15 MC

A car bought for $50 000 is depreciated using the declining balance method.

Which graph best represents the salvage value of the car over time?

Show Answers Only

`D`

Show Worked Solution

`text(Declining Balance Method means that the salvage value)`

`text(of the car drops the most value in the 1st year and then)`

`text(drops less value each following year.)`

`=> D`

Financial Maths, STD2 F4 2004 HSC 25a

Tai uses the declining balance method of depreciation to calculate tax deductions for her business. Tai’s computer is valued at $6500 at the start of the 2003 financial year. The rate of depreciation is 40% per annum.

Calculate the value of her tax deduction for the 2003 financial year. (1 mark)

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What is the value of her computer at the start of the 2006 financial year? (2 marks)

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a. `$2600`

b. `$1404`

Show Worked Solution

a. `text(Tax deduction)`

`= 40 text(%) xx $6500= $2600`

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

`text(Value at the start of 2006 FY)`

`= 6500(1-0.4)^3= $1404`

Financial Maths, STD2 F1 2007 HSC 26b

Myles is in his third year as an apprentice film editor.

Myles purchased film-editing equipment for $5000.

After 3 years it has depreciated to $3635 using the straight-line method.

Calculate the rate of depreciation per year as a percentage. (2 marks)

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Myles earns $800 per week. Calculate his taxable income for this year if the only allowable deduction is the amount of depreciation of his film-editing equipment in the third year of use. (1 mark)

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Use this tax table to calculate Myles’s tax payable. (2 marks)

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a. `text(9.1%)`

b. `$41\ 145`

c. `$8443.50`

Show Worked Solution

a.	`S = V_0-Dn`
	`S = $3635,\ \ \ V_0 = 5000,\ \ \ n = 3`

`3635`	`= 5000-D xx 3`
`3D`	`= 1365`
`D`	`= $455`

`:.\ text(Rate of depreciation per year)`

`= 455/5000 xx 100`

`= 9.1 text(%)`

b. `text(Income per year)= 52 xx 800= $41\ 600`

`text(Taxable income)`	`=\ text(Income – Deductions)`
	`= 41\ 600-455= $41\ 145`

c.	`text(Tax payable)`	`= 4500 + 0.3(41\ 145-28\ 000)`
		`= 4500 + 3943.50= $8443.50`

Financial Maths, STD2 F4 2007 HSC 12 MC

The value of a car is depreciated using the declining balance method.

Which graph best illustrates the value of the car over time?

Show Answers Only

`C`

Show Worked Solution

`text(Declining balance depreciates quicker in absolute)`

`text(terms in the early stages, and slower as time goes)`

`text(on and the balance owing decreases.)`

`=> C`

Financial Maths, STD2 F4 2008 HSC 27c

A plasma TV depreciated in value by 15% per annum. Two years after it was purchased it had depreciated to a value of $2023, using the declining balance method.

What was the purchase price of the plasma TV? (2 marks)

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`$2800`

Show Worked Solution

`S = V_0 (1-r)^n`

`2023`	`= V_0 (1-0.15)^2`
`2023`	`= V_0 (0.85)^2`
`V_0`	`= 2023/0.85^2`
	`= 2800`

`:.\ text(The purchase price) = $2800`

Financial Maths, STD2 F4 2014 HSC 9 MC

A car is bought for $19 990. It will depreciate at 18% per annum.

Using the declining balance method, what will be the salvage value of the car after 3 years, to the nearest dollar?

$8968
$9195
$11 022
$16 392

Show Answers Only

$C$

Show Worked Solution

$S$	$= V_0 (1-r)^n$
	$= 19\ 990 (1-18/100)^3$
	$= 19\ 990 (0.82)^3$
	$= $11\ 021.85\approx $11\ 022$

$\Rightarrow C$

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)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Method 2)`

Show Worked Solution

`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)

--- 3 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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`

Financial Maths, STD2 F1 2010 HSC 11 MC

Which of the following graphs shows the lowest rate of depreciation over the given time period?

Show Answers Only

`D`

Show Worked Solution

`text(The lowest rate of depreciation will occur when)`

`text(an item retains value for the longest time.)`

`=> D`

Financial Maths, STD2 F4 2012 HSC 26b

Jim buys a photocopier for $22 000.

Its value is depreciated using the declining balance method at the rate of 15% per annum.

What is its value at the end of 3 years? (2 marks)

Show Answers Only

`$13\ 510.75`

Show Worked Solution

`S`	`= V_0 (1-r)^n`
	`= 22\ 000 (1-0.15)^3`
	`= 22\ 000 (0.85)^3`
	`= 13\ 510.75`

`:.\ text(After 3 years, it is worth)\ $13\ 510.75`

Financial Maths, STD2 F4 2012 HSC 16 MC

A machine was bought for $25 000.

Which graph best represents the salvage value of the machine over 10 years using the declining balance method of depreciation?

(A)			(B)

(C)			(D)

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`B\ \ text(and)\ \ D\ \ text(represent straight line depreciation.)`

`C\ \ text(incorrectly has no salvage value after 10 years)`

`=>A`

\(S\)	\(=V_0(1-r)^n\)
\(44\ 000\)	\(=55\ 000(1-r)^1\)
\(\dfrac{44\ 000}{55\ 000}\)	\(=1-r\)
\(1-r\)	\(=0.8\)
\(r\)	\(=1-0.8=0.20\)

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

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

\(S\)	\(= V_0 (1-r)^n\)
	\(= 19\ 990 (1-18/100)^3\)
	\(= 19\ 990 (0.82)^3\)
	\(= $11\ 021.85\approx $11\ 022\)