Dear Student, Please find below the solution to the asked query: Given : Value of car = Rs 150000 And value depreciated at rate of 10 % per annum and value reduced to 109350 We know formula for compound interest when value depreciated annually : Amount = P 1 - r100t , Here , P = 150000 , r = 10 % and Time = t ( years ) and Amount = 109350 , So 109350 = 1500001 - 10100t⇒109350150000 = 1 - 110t⇒7291000 = 910t⇒93103 = 910tWe know : ambm = abm , So ⇒9103 = 910t Now we compare both hand side and get : t = 3 Therefore, In three years value of given car reduced to Rs 109350 ( Ans ) Hope this information will clear your doubts about topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

do 11 ths sum please

do 11 ths sum please If its present is J. birth rate and death rate is thousand. Its population after 2 year. 40 per S. the rate the in the first hour, decreases by in the and increases by in the third hour. the count of bacteria initially was what a with a capital of in 2010 First year his profit was but it to 20%. If each the ptoiit was on the previous year capital. find out his total profit price of a u•iling fan is If value decreases every year by 10%, then find 2 years. Gautam purchased a rww computer for t 17.5m Its value depreciates at the rate of What will value three .Veqrs hence? 10. had purchased •t bicycle three years back. Its value depreciated to 1728 at the rate Of annum. For what amount had Sonu purchasehd the value of a car north is depreciating at the rate of per annum. In how many years wu its value be reduced to present price ofa machine is If its value depreciates e.•ery year by then find its on• two years. 88 Series MATHEMATICS-8

Dear Student,

Please find below the solution to the asked query:

Given : Value of car = Rs. 150000

And value depreciated at rate of 10 % per annum and value reduced to 109350

We know formula for compound interest when value depreciated annually :

Amount = P

{(1 - \frac{r}{100})}^{t}

, Here , P = 150000 , r = 10 % and Time = t ( years ) and Amount = 109350 , So

109350 = 150000 {(1 - \frac{10}{100})}^{t} \Rightarrow \frac{109350}{150000} = {(1 - \frac{1}{10})}^{t} \Rightarrow \frac{729}{1000} = {(\frac{9}{10})}^{t} \Rightarrow \frac{9^{3}}{10^{3}} = {(\frac{9}{10})}^{t} We know : \frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}, So \Rightarrow {(\frac{9}{10})}^{3} = {(\frac{9}{10})}^{t}

Now we compare both hand side and get :

t = 3

Therefore,

In three years value of given car reduced to Rs. 109350 ( Ans )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

View Full Answer