Question

Kalyan purchased an old bike for Rs. $12000$. If its cost after $2$ years is Rs. $11524.80$, find the rate of depreciation.
A. \[1\%\] per annum
B. \[4\%\] per annum
C. \[3\%\] per annum
D. \[2\%\] per annum

Answer 1

Hint: First, we will find the formula for simple interest using amount and principal that is $\text{Simple Interest}=\text{ }\left( \text{Amount - Principal Amount} \right)$ after that we will compare it with $S.I.=\dfrac{P\times R\times T}{100}$and put all the values and take \[R\] as negative as it is the rate of depreciation and finally get the answer

Complete step-by-step answer:
It is given that the purchase amount or principal amount of the bike is Rs. $12000$now it is given that the time period is \[2\] years and the final amount is Rs. $11524.80$.
So, we know that: $\text{Simple Interest}=\text{ }\left( \text{Amount - Principal Amount} \right)$
So, the Simple Interest = \[11524.80-12000=-475.2\] ,
Now, we know that: $S.I.=\dfrac{P\times R\times T}{100}$
Where, $S.I.$ is simple interest, $P$ is the principal amount, $R$ is the rate of interest and $T$ is the time period.
Now, since we are given the rate of depreciation therefore we will be taking $R$negative. Therefore, the formula will become:
$S.I.=\dfrac{P\times \left( -R \right)\times T}{100}$
Now, we will put the values in the above formula of simple interest,
$\begin{align}
  & \Rightarrow -475.2=\dfrac{12000\times \left( -R \right)\times 2}{100}\Rightarrow \dfrac{475.2}{120\times 2}=R \\
 & \Rightarrow 1.98=R \\
 & \Rightarrow R=2\% \\
\end{align}$

So, the correct answer is “Option D”.

Note: We can approach this question by a different formula, we know that the amount is calculated as follows: $A=P{{\left( 1+\dfrac{R}{100} \right)}^{2}}$ , now since $R$ is the rate of depreciation therefore we will take it as negative , therefore, the formula will become: : $A=P{{\left( 1-\dfrac{R}{100} \right)}^{2}}$ , we will now put the values in this formula:
\[\begin{align}
  & \Rightarrow A=P{{\left( 1-\dfrac{R}{100} \right)}^{2}} \\
 & \Rightarrow 11524.80=12000{{\left( 1-\dfrac{R}{100} \right)}^{2}} \\
 & \Rightarrow \dfrac{11524.80}{12000}={{\left( 1-\dfrac{R}{100} \right)}^{2}} \\
 & \Rightarrow \dfrac{2401}{2500}={{\left( 1-\dfrac{R}{100} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{49}{50} \right)}^{2}}={{\left( 1-\dfrac{R}{100} \right)}^{2}} \\
 & \Rightarrow \left( \dfrac{49}{50} \right)=\left( 1-\dfrac{R}{100} \right) \\
 & \Rightarrow \dfrac{R}{100}=\left( \dfrac{50-49}{50} \right) \\
 & \Rightarrow \dfrac{R}{100}=\dfrac{1}{50} \\
 & \Rightarrow R=2 \\
\end{align}\]

Hence, the rate of depreciation is $2\%$ .