Question

A car is purchased for 26000. Each year it loses 30% of its value. After how many years will the car be worth 2200 or less?

Answer 1

Answer:

The smallest possible number of years for the depreciation value of car is 7 years .

Step-by-step explanation:

Given as :

The purchased price of car = $26000

The depreciation rate of car every year = r = 30%

The reduced price of car after n years = $2200 or less

Let The time period for the depreciation value = n years

Now, According to question

The reduced price of car after n years = The purchased price of car × $(1-\dfrac{\textrm rate}{100})^{\textrm time}$

Or, $2200 = $26000 × $(1-\dfrac{\textrm r}{100})^{\textrm n}$

Or, $\dfrac{2200}{26000}$ = $(1-\dfrac{\textrm 30}{100})^{\textrm n}$

Or, 0.084615 = $(0.7)^{n}$

Now, Taking Log both side

So, $Log_{10}$0.084615 =  $Log_{10}$( $(0.7)^{n}$)

Or, -1.07255 = n $Log_{10}$0.7

Or, -1.07255 = n ( - 0.15490 )

∴ n = $\dfrac{1.07255}{0.15490}$

I.e n = 6.92 ≈ 7 years

So, Time period for depreciation of car value = n = 7 years

Hence The smallest possible number of years for the depreciation value of car is 7 years . Answer