1, 10, 100, 1000 Common Difference? Difference
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Part A
Purchase Value of Anna's Car in 2015 = $36,000
Depreciated Value of Anna's Car in 2017 = $25,000
We know that the value of the car is depreciating linearly.
From the above data, we can see that the value of the car has reduced by $36,000-$25,000 = $11,000 over 2 years.
⇒ Per year the value has been depreciated by
⇒ Per year the value has been depreciated by $5,500
So if the value of a car in 2015 is V₀ then value of the car after t years can be determined by the below function:
V (t) = V₀ - (t*5,500), where t indicates the years passed since 2015
OR V (t) = 36,000 - (t*5,500), where t indicates the years passed since 2015
Part B
Suggested value of Anna's car for trade in option in 2018 = $15,000
In 2018, 3 years would have been passed since 2015. Using the value of t = 3 in the function determined in Part A:
V(3) = V₀ - (3*5,500)
⇒ V(3) = 36,000 - (16,500)
⇒ V(3) = $19,000
From the above calculations, we can see that the value of the car in 2018 should be $19,000, however the suggested value for trade in is indicated as $15,000 which is lower than what the value of the car should be. Hence, basis linear depreciation, the $15,000 value is not fair for the car in 2018.