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Financial Maths, STD1 F3 2024 HSC 5 MC

A car is valued at $25 000 when new. Its value depreciates by 25% per annum.

Which of the following best describes the change in value of the car after one year?

  1. Decrease of $1000
  2. Increase of $1000
  3. Decrease of $6250
  4. Increase of $6250
Show Answers Only

\(C\)

Show Worked Solution
\(S\) \(=V_0(1-r)^n\)
  \(=25000(1-0.25)^1\)
  \(=$18\,750\)

 
\(\therefore\ \text{Decrease in value}\ = $25\,000-$18\,750=$6250\)
 

\(\Rightarrow C\)

Financial Maths, STD1 F3 2024 HSC 30

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)

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

♦ Mean mark 40%.

Financial Maths, STD1 F3 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?

  1. $952
  2. $1110
  3. $1357
  4. $1515
Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 50%.
`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?

  1.  $1749.01
  2.  $4182.12
  3.  $4343.88
  4.  $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, STD1 F3 2019 HSC 21

A new car is bought for $24 950. Each year the value of the car depreciates by 14%.

Using the declining-balance method, calculate the salvage value of the car at the end of 10 years.  (2 marks)

Show Answers Only

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

Show Worked Solution

`V_0 = 24\ 950, \ r = 0.14, \ n = 10`

`S` `= V_0(1 – r)^n`
  `= 24\ 950(1 – 0.14)^10`
  `= $5521.47\ \ (text(nearest cent))`

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)

Show Answers Only

`$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)

Show Answers Only

`$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 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)

Show Answers Only

`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 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.

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

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

Show Answers Only
  1. `$2600`
  2. `$1404`
Show Worked Solution
i.  `text(Tax deduction)` `= 40 text(%) xx $6500`
  `= $2600`

 

ii. `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 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? 

  1. $8968
  2. $9195
  3. $11 022
  4. $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`

 
`=>  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.
 

 2011 28b

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

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

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

  4. 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)

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

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

  2. 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)

Show Answers Only
  1. `text(Gradient) =-4000`
  2. `text(The amount the tractor depreciates each year.)`
  3. `S = 60\ 000\-4000n`
  4. `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.)`
  5. `text(After 14 years, the tractor is worth $2638.83)`
  6. `text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases but remains)>0.`
  7.  
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 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`