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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.

  1. For the declining-balance method, the salvage value of the machine after `n` years is given by the formula
  2.     `S=V_(0)xx(0.80)^(n),`
  3. where `S` is the salvage value and `V_(0)` is the initial value of the asset.
  4.  i. What is the annual rate of depreciation used in this formula?  (1 mark)
  5. ii. Calculate the salvage value of the machine after 3 years, based on the given formula.  (1 mark)
  6. For the straight-line method, the value of the machine is depreciated at a rate of 12.2% of the purchase price each year.
  7. 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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  1.  i. `20text{%}`
  2. ii. `$25\ 600`
  3. `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.

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

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

  3. 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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  1. `$24\ 000`
  2. `$7593.75`
  3. `text{See Worked Solutions}`
Show Worked Solution

i.  `text(Using)\ \ S = V_0 – Dn`

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

 

ii.  `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)`

 

iii.

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)`

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.

  1. 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)

  2. 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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  1. `$1200`
  2. `text(See Worked Solutions.)`
Show Worked Solution
i.    `S` `= V_0-Dn`
  `0` `= 3600-D xx 3`
  `3D` `= 3600`
  `D` `= 3600/3`
    `= 1200`

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

 

♦ Mean mark 45%
ii   `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`