Question

A car costs 240000 rupees. Its value depreciates at the rate of 5% every year. If after some time period, its value is 205770 rupees, find the time period.

Answer 1

Hint: We use the compound depreciation formula $A=P{{\left( 1-\dfrac{r}{100} \right)}^{n}}$ where $P$ is the initial amount or the fair value , $n$ is the time period in years is , $r$ is the rate of decrease in percentage , $A$ is the amount after time period or the depreciated value. We put the known values given in the question in the formula and then find the unknown $n$.

Complete step-by-step solution:
Depreciation is the decrease in the fair value of an asset over a certain time period because of the loss of utility of the asset.
We have to use here compound depreciation formula. We know from compound depreciation formula that if the initial amount or the fair value is $P$, the time period in years is $n$, the rate of decrease in percentage is $r$ then the amount after period or the depreciated value as $A$ is given by
\[A=P{{\left( 1-\dfrac{r}{100} \right)}^{n}}\]
We are given here that the initial amount or the fair value is $P=240000$ rupees , the rate of decrease in percentage is $r=5\%$, the amount after certain time period or the depreciated value $A=205770$ . We are asked to find , the time period in years is $n$. So let us put the given values in the compound depreciation formula and get,
\[\begin{align}
  & 205770=240000{{\left( 1-\dfrac{5}{100} \right)}^{n}} \\
 & \Rightarrow \dfrac{205770}{240000}={{\left( \dfrac{100-5}{100} \right)}^{n}} \\
 & \Rightarrow \dfrac{6859}{8000}={{\left( \dfrac{95}{100} \right)}^{n}}={{\left( \dfrac{19}{20} \right)}^{n}} \\
 & \Rightarrow {{\left( \dfrac{19}{20} \right)}^{3}}={{\left( \dfrac{19}{20} \right)}^{n}} \\
\end{align}\]
We equate the exponents for the same base at the left and right hand side of the equation and get $n=3$. So the time period is 3 years.

Note: We can alternatively solve the problem taking logarithm both side of the equation $A=P{{\left( 1-\dfrac{r}{100} \right)}^{n}}$. The compound interest is analogues to compound depreciation and the simple interest is analogues to straight-line deprecation formula which is given by $A=P\left( 1-\dfrac{rn}{100} \right)$.