Question

A machine depreciates every year at the rate of $20\%$ of its value at the beginning of the year. The machine was purchased for Rs. $2,50,000$ when new and the scrap value realized when sold was Rs. $1,28,000$. Find the number of years that the machine was used.

Answer 1

Hint: We will first start by writing down the definition of depreciation and then we will jot down all the values that are given in the question. We will then use the formula for rate of depreciation that is, \[\text{Rate of Depreciation}=1{{\left( \dfrac{s}{c} \right)}^{\dfrac{1}{n}}}\times 100\text{ }\]. We will put all the values in this equation and we will have only 1 unknown left that is $n$ . We will find its value and that will be our answer.

Complete step-by-step answer:
First of all, let us understand what is depreciation? It is defined as a continuing, permanent and gradual decrease in the book value or original value of an asset.
\[\begin{align}
  & \text{Depreciation}=\text{Original Value}\times \dfrac{\text{Rate of Depreciation}}{100} \\
 & \text{Rate of Depreciation}=1{{\left( \dfrac{s}{c} \right)}^{\dfrac{1}{n}}}\times 100\text{ }..........\text{ Equation 1}\text{.} \\
\end{align}\]
Where, n = useful life ; s = scrap value ; c = cost price
We are given the rate of depreciation = $20\%$
Cost price of the machine = $c$ = Rs. $2,50,000$
Selling price of the machine = $s$ = Rs. $1,28,000$
Let the required number of years be $n$ years,
On putting all the above equation in equation 1, we will have:
\[\begin{align}
  & \Rightarrow \text{Rate of Depreciation}=1{{\left( \dfrac{s}{c} \right)}^{\dfrac{1}{n}}}\times 100 \\
 & \Rightarrow 20=1-{{\left( \dfrac{128000}{250000} \right)}^{\dfrac{1}{n}}}\times 100\Rightarrow \dfrac{20}{100}=1-{{\left( \dfrac{128}{250} \right)}^{\dfrac{1}{n}}} \\
 & \Rightarrow 1-\dfrac{1}{5}={{\left( \dfrac{128}{250} \right)}^{\dfrac{1}{n}}}\Rightarrow \dfrac{4}{5}={{\left( \dfrac{64}{125} \right)}^{\dfrac{1}{n}}} \\
\end{align}\]
Now taking the powers of n on both sides of the equation:
\[\begin{align}
  & \Rightarrow \dfrac{4}{5}={{\left( \dfrac{64}{125} \right)}^{\dfrac{1}{n}}}\Rightarrow {{\left( \dfrac{4}{5} \right)}^{n}}={{\left( {{\left( \dfrac{64}{125} \right)}^{\dfrac{1}{n}}} \right)}^{n}} \\
 & \Rightarrow {{\left( \dfrac{4}{5} \right)}^{n}}=\left( \dfrac{64}{125} \right)\Rightarrow {{\left( \dfrac{4}{5} \right)}^{n}}={{\left( \dfrac{4}{5} \right)}^{3}} \\
\end{align}\]
On simplifying we get, $n=3$ . Therefore the number of years that machine was used was 3 years.

Note: At last we equated the powers because the bases are the same. Elongated method can be seen below:
$\begin{align}
  & \Rightarrow {{\left( \dfrac{4}{5} \right)}^{n}}={{\left( \dfrac{4}{5} \right)}^{3}}\Rightarrow \log {{\left( \dfrac{4}{5} \right)}^{n}}=\log {{\left( \dfrac{4}{5} \right)}^{3}} \\
 & \Rightarrow n\log \left( \dfrac{4}{5} \right)=3\log \left( \dfrac{4}{5} \right)\Rightarrow n=3 \\
\end{align}$
Under the rate of depreciation formula, the machine can be depreciated up to the net scrap value or zero value.