Question

A two wheeler depreciates at 20 % of its value every year.
If present value of the same be 90000, its depreciated
value would be 36864 after how many years?
​

Answer 1

Answer:

Present cost= Co

Rate = 20%

Time = 2 years

Let after 2 years cost, C=208000

Thus, C=Co(1+100R)T

208000=Co(1−10020)2

Co=325000

Thus present cost is 325000 Rs.

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Answer 2

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{Original_{(price)}, P = Rs \: 90000} \\ &\sf{Depreciated_{(price)}, A = Rs \: 36864}\\ &\sf{Rate_{(depreciated)}, R = 20\% \: per \: annum} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{Time \: period, \: n \: years}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

↝Given that

$\: \: \: \: \: \: \: \: \: \bull \: \sf \:Original_{(price)}, P = Rs \: 90000$

$\: \: \: \: \: \: \: \: \: \bull \: \sf \:Depreciated_{(price)}, A = Rs \: 36864$

$\: \: \: \: \: \: \: \: \: \bull \: \sf \:Rate_{(depreciated)}, R = 20\% \: per \: annum$

$\rm \: Let \: time \: in \: which \: value \: depreciates \: be \: n \: years$

We know,

↝The depreciation amount is given by

$\rm \:Depreciated_{(price)} = Original_{(price)}{\bigg( 1 - \dfrac{Rate_{(depreciated)}}{100} \bigg)}^{n}$

↝On substituting all the above values, we get

$\rm :\longmapsto\:36864 = 90000 {\bigg( 1 - \dfrac{20}{100} \bigg)}^{n}$

$\rm :\longmapsto\:\dfrac{36864}{90000} = {\bigg( \dfrac{100 - 20}{100} \bigg)}^{n}$

$\rm :\longmapsto\:\dfrac{256}{625} = {\bigg( \dfrac{80}{100} \bigg)}^{n}$

$\rm :\longmapsto\: {\bigg( \dfrac{4}{5} \bigg)}^{4} = {\bigg( \dfrac{4}{5} \bigg)}^{n}$

$\bf\implies \:n \: = \: 4 \: years$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \: price \: depreciates \: in \: 4 \: years}}}$