Question

Two equal sums of money were lent at simple interest at 11% p.a. for \[3\dfrac{1}{2}\] years and $4\dfrac{1}{2}$ years respectively. If the difference in the interest for two periods was Rs. 412.50, find the each sum.

Answer 1

Hint: To solve this question, we will first assume both the sums of the money to be x as they have been given to be equal. Then we will calculate the rate of interests at the end of both the time periods by the formula given as:
Simple interest I on a principal sum P after time T at the rate of interest R is given as:
$I=\dfrac{P\times R\times T}{100}$
Then, we will keep their difference equal to the given difference, i.e. 412.50 and this will give us an equation in x. Then, we will solve that equation and hence we will get our answer.

Complete step by step answer:
We need to find each sum of money but they have been given to us to be equal hence we can assume both of them to be x. thus, we need to find the value of x.
Now, we have been given that the one sum for lent for $4\dfrac{1}{2}$ years and the other for $3\dfrac{1}{2}$ years at the same amount of simple interest at 11% p.a.
We have also been given that the difference between their interests is Rs. 412.50. If we take the interest at the end of $3\dfrac{1}{2}$ years to be ${{I}_{1}}$ and that after the end of $4\dfrac{1}{2}$ years to be ${{I}_{2}}$ , then we can say that:
${{I}_{2}}-{{I}_{1}}=412.50$ …..(i)
Now, we know that simple interest I on a principal sum P after time T at the rate of interest R is given as:
$I=\dfrac{P\times R\times T}{100}$
Thus, for ${{I}_{1}}$, we have:
P=x
R=11%
T=$3\dfrac{1}{2}=\dfrac{7}{2}$
Hence, we get ${{I}_{1}}$ as:
$\begin{align}
  & {{I}_{1}}=\dfrac{x\times 11\times \dfrac{7}{2}}{100} \\
 & \Rightarrow {{I}_{1}}=\dfrac{77x}{200} \\
\end{align}$
Now, for ${{I}_{2}}$, we have:
P=x
R=11%
T=$4\dfrac{1}{2}=\dfrac{9}{2}$
Hence, we get ${{I}_{2}}$ as:
$\begin{align}
  & {{I}_{2}}=\dfrac{x\times 11\times \dfrac{9}{2}}{200} \\
 & \Rightarrow {{I}_{2}}=\dfrac{99x}{100} \\
\end{align}$
Thus, putting the values of ${{I}_{1}}$ and ${{I}_{2}}$ in equation (i) we get:
$\begin{align}
  & {{I}_{2}}-{{I}_{1}}=412.50 \\
 & \Rightarrow \dfrac{99x}{200}-\dfrac{77x}{200}=412.50 \\
\end{align}$
Solving this equation we get:
$\begin{align}
  & \dfrac{99x}{200}-\dfrac{77x}{200}=412.50 \\
 & \Rightarrow \dfrac{22x}{200}=412.50 \\
 & \Rightarrow \dfrac{11x}{100}=412.50 \\
 & \Rightarrow 11x=41250 \\
 & \Rightarrow x=\dfrac{41250}{11}=3750 \\
\end{align}$

Thus, each sum of money is Rs. 3750.

Note: Be careful with the difference between simple interest and compound interest. They both are very different and any confusion between them may result in a wrong answer.
Simple interest uses the formula:
Simple interest I on a principal sum P after time T at the rate of interest R is given as:
$I=\dfrac{P\times R\times T}{100}$
Compound interest uses the formula:
Final amount A on principal sum P at rate of interest r with n as number of times interest applied per time period and t as number of time periods elapsed is given by:
$A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$