Question

A certain amount of money is invested at the rate of $10\%$ per annum compound interest, the interest compounded annually. If the difference between the interests of third year and first year is Rs. 1105. Find the sum invested.

Answer 1

Hint: We assume the sum invested as a variable. We find out the individual interests of third year and first year at the rate of $10\%$ per annum compound interest, the interest compounded annually. We make the conditions as a linear equation of the variable. We solve it to get the solution of the problem.

Complete step by step answer:
Let’s assume the invested money is p. It's invested at the rate of $10\%$ per annum compound interest, the interest compounded annually.
We find the interests of that particular amount in individual years.
As the interest is compounded annually, we can take the formula of interest for 1 year as $I=\dfrac{pr}{100}$ and the total invested sum after every year as $p+\dfrac{pr}{100}$. Here r is the rate of interest. So, $r=10$. The formula of total investment after n years would be $p{{\left( 1+\dfrac{r}{100} \right)}^{n}}$.
The interest in the first year is \[{{I}_{1}}=\dfrac{p\times 10}{100}=\dfrac{p}{10}\].
The total sum invested after 3 years which is at the start of the fourth year is $p{{\left( 1+\dfrac{10}{100} \right)}^{3}}=1.331p$.
The interest in the third year would be \[{{I}_{3}}=1.331p-p=0.331p\].
We have been given that the difference between the interests of third year and first year is Rs. 1105.
So, we get that \[{{I}_{3}}-{{I}_{1}}=1105\].
We place the values to get the linear equation of p
\[\begin{align}
  & {{I}_{3}}-{{I}_{1}}=1105 \\
 & \Rightarrow 0.331p-\dfrac{p}{10}=1105 \\
\end{align}\]
We solve the equation to find the value of p using binary operations
\[\begin{align}
  & 0.331p-\dfrac{p}{10}=1105 \\
 & \Rightarrow 0.331p-0.1p=1105 \\
 & \Rightarrow 0.231p=1105 \\
 & \Rightarrow p=\dfrac{1105}{0.231}=4783.55 \\
\end{align}\]

The amount of sum invested is 4783.55 Rs. (approx.)

Note: We need to remember that the difference between the interests of third year and first year is equal to the difference between the total sum invested at the end of third year and first year as the invested sum gets cancelled each time. So, the difference can be expressed as $p{{\left( 1+\dfrac{r}{100} \right)}^{3}}-p{{\left( 1+\dfrac{r}{100} \right)}^{1}}$ where r is 10.