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Financial Maths, STD1 F3 2025 HSC 27

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

\(\text{Find}\ r:\)

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

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

  
\(\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, 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)

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

Show Worked Solution

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

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

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

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

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

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

♦ Mean mark 40%.

Financial Maths, STD1 F3 2022 HSC 30

A car is purchased for $15 000. The graph shows the value of the car, `$V`, at time `t` years since it was purchased, using the declining-balance method of depreciation.
 


 

  1. When using the straight-line method of depreciation, the value of the car depreciates at a rate of $2500 per year.

  2. By first completing the table, plot on the grid above the value of the car for the first three years based on the straight-line method of depreciation.  (2 marks)

  3.      
  4. After how many years will the value of the car using the straight-line method of depreciation be equal to its value using the declining-balance method?  (2 marks)

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a.  
     

b.   
     

Values are equal when graphs intersect

→ after 4 years

Show Worked Solution

a.  
     

b.   
     

Values are equal when graphs intersect

→ after 4 years


♦♦ Mean mark part (a) 37%.
♦♦ Mean mark part (b) 31%.

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?
 

2015 10 mc1

2015 10 mc2

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`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.

  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 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?

 

2UG-2005-15abMC

2UG-2005-15cdMC

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`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 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?
 

VCAA 2007 12 mcii

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`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 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)

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  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 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)
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`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`