Question

The difference between the compound interest and simple interest on an amount of Rs. $18000$ in 2 years is Rs. $405$. Find the rate of interest.

Answer 1

Hint: First we write down all the given information and calculate the simple interest and compound interest by using their respective formula. As given in the question the difference between compound interest and simple interest, we subtract simple interest from compound interest to get the desired answer. The formula for simple interest and compound interest is –
When $P=$ Principal amount
$R=$ Rate of interest and
 $T=$ Time period
 Then, $S.I.=\dfrac{P\times R\times T}{100}$
Compound Interest = Amount – Principal and \[\text{Amount =}P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]

Complete step by step solution:
We have given that,
 Principal $P=18000$
Time period $T=2$ years
We have to find the Rate of interest.
First, we calculate the simple interest by using the formula $S.I.=\dfrac{P\times R\times T}{100}$, we get
$S.I.=\dfrac{18000\times R\times 2}{100}$
Now, we calculate the compound interest by using the formula Compound Interest = Amount – Principal
We get Compound Interest \[=P{{\left( 1+\dfrac{R}{100} \right)}^{T}}-P\]
When we put values and solve further we get
$\begin{align}
  & C.I.=18000{{\left( 1+\dfrac{R}{100} \right)}^{2}}-18000 \\
 & C.I.=18000\left[ {{\left( 1+\dfrac{R}{100} \right)}^{2}}-1 \right] \\
\end{align}$
Now, we have given that the difference between the compound interest and simple interest is Rs. $405$.
So, we have $18000\left[ {{\left( 1+\dfrac{R}{100} \right)}^{2}}-1 \right]-\dfrac{18000\times R\times 2}{100}=405$
Now, we use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to solve ${{\left( 1+\dfrac{R}{100} \right)}^{2}}$, we get
$18000\left[ \left( {{1}^{2}}+{{\left( \dfrac{R}{100} \right)}^{2}}+2\times 1\times \dfrac{R}{100} \right)-1 \right]-\dfrac{18000\times R\times 2}{100}=405$
$\begin{align}
  & \Rightarrow 18000\left[ {{1}^{2}}+{{\left( \dfrac{R}{100} \right)}^{2}}+2\times 1\times \dfrac{R}{100}-1 \right]-\dfrac{18000\times R\times 2}{100}=405 \\
 & \Rightarrow 18000\left[ {{\left( \dfrac{R}{100} \right)}^{2}}+\dfrac{2R}{100}-\dfrac{2R}{100} \right]=405 \\
 & \Rightarrow 18000{{\left( \dfrac{R}{100} \right)}^{2}}=405 \\
 & \Rightarrow {{\left( \dfrac{R}{100} \right)}^{2}}=\dfrac{405}{18000} \\
 & \Rightarrow {{\left( \dfrac{R}{100} \right)}^{2}}=\dfrac{81}{3600} \\
 & \Rightarrow {{\left( \dfrac{R}{100} \right)}^{2}}={{\left( \dfrac{9}{60} \right)}^{2}} \\
 & \Rightarrow \dfrac{R}{100}=\dfrac{9}{60} \\
 & \Rightarrow R=\dfrac{9\times 100}{60} \\
 & \Rightarrow R=15 \\
\end{align}$
The required Rate of interest is $15%$.

Note: The simple interest is based on the principle of a deposit or a loan but the compound interest is based on the principal and the interest which accumulates every year. The fact is that the value of compound interest is always greater than the simple interest. That’s why we subtract simple interest from compound interest. If you subtract compound interest from simple interest, you will get the wrong answer.