A new electric vehicle costs $46 000 and is depreciated using the reducing balance method by 15% of its value each year.
Find its value after four years, giving your answer to the nearest dollar. (2 marks)
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A new electric vehicle costs $46 000 and is depreciated using the reducing balance method by 15% of its value each year.
Find its value after four years, giving your answer to the nearest dollar. (2 marks)
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`$24\ 012`
`V_o = 46\ 000,\ \ \ r = 15% = 0.15,\ \ \ n = 4`
`S` | `= V_0 (1-r)^n` |
`= 46\ 000 (1-0.15)^4` | |
`=46\ 000 xx 0.85^4` | |
`= $24\ 012\ \ \text{(nearest dollar)}` |
A new car costs $30 000 and is depreciated using the reducing balance method by 20% of its value each year.
After three years its value is
`C`
`V_0= 30\ 000,\ \ \ r = 20% = 0.2,\ \ \ n = 3`
`S` | `= V_0 (1-r)^n` |
`= 30\ 000 (1-0.2)^3` | |
`=30\ 000 xx 0.8^3` | |
`= $15\ 360` |
`=> C`
Peter installed his new pool fence on 1 January 2013 at a cost of $12 000.
On 1 January of each year after 2013 its value is depreciated by 15% using the reducing balance method.
The value of the pool fence will be below $4000 for the first time on 1 January of what year? Show all working. (3 marks)
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`text{Test the values of different years using:}\ \ S=V_0(1-r)^n`
`S \text{(Jan17)} = 12\ 000(1-0.15)^4= 5000 xx 0.85^4= $2560`
`S \text{(Jan19)} = 12\ 000 xx 0.85^6= $4525.79`
`S \text{(Jan20)} = 12\ 000 xx 0.85^7= $3846.93`
`\text{2020 is the 1st year its value is below $4000 on 1 January.}`
`r=15% = 0.15`
`text{Test the values of different years using:}\ \ S=V_0(1-r)^n`
`S \text{(Jan17)} = 12\ 000(1-0.15)^4= 5000 xx 0.85^4= $2560`
`S \text{(Jan19)} = 12\ 000 xx 0.85^6= $4525.79`
`S \text{(Jan20)} = 12\ 000 xx 0.85^7= $3846.93`
`\text{2020 is the 1st year its value is below $4000 on 1 January.}`
A new air-conditioning unit was purchased for $5000 on 1 January 2017.
On 1 January of each year after 2017 its value is depreciated by 20% using the reducing balance method.
The value of the air conditioner will be below $1500 for the first time on 1 January
`D`
`r=20% = 0.2`
`text{Test the values of given options using:}\ \ S=V_0(1-r)^n`
`S \text{(Jan20)} = 5000(1-0.2)^3= 5000 xx 0.8^3= $2560`
`S \text{(Jan22)} = 5000 xx 0.8^5= $1638.40`
`S \text{(Jan23)} = 5000 xx 0.8^6= $1310.72`
`=> D`
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.
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a. `text{Depreciation rate}\ = 1-0.8=0.2=20text{%}`
b. `text{Find}\ \ S\ \ text{when}\ \ n=3:`
`S` | `=V_0 xx (0.80)^n` | |
`=50\ 000 xx (0.80)^3` | ||
`=$25\ 600` |
Alan bought a light aircraft for $76 500. It will depreciate at 14% per annum.
Using the declining balance method, what will be the salvage value of the light aircraft after 6 years, to the nearest dollar?
`$30\ 949`
`S` | `= V_0 (1-r)^n` |
`= 76\ 500 (1-14/100)^6` | |
`= 76\ 500 (0.86)^6` | |
`= $30\ 949.39` | |
`=$30\ 949\ \ text{(nearest dollar)}` |
Marnus bought a cricket bowling machine two years ago that cost $3400. Its value has depreciated by 10% each year, based on the declining-balance method.
What is the salvage value today, to the nearest dollar? (2 marks)
`$2754`
`S` | `= V_0 (1-r)^n` |
`= 3400 (1-0.10)^2` | |
`= 3400 (0.90)^2` | |
`= $2754` |
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?
`D`
`S` | `= V_0 (1-r)^n` |
`= 2467 (1-0.15)^3` | |
`= 2467 (0.85)^3` | |
`= $1515` |
`=> D`
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`
`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))`
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))`
`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))` |
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?
`C`
`S` | `= V_0 (1-r)^n` |
`= 19\ 990 (1-18/100)^3` | |
`= 19\ 990 (0.82)^3` | |
`= $11\ 021.85` |
`=> C`