Question

Mr. Dhruv deposits Rs. 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.

Answer 1

Step 1: Given data:

Instalment per month, $\left(P\right)=\mathrm{Rs}.600$

Rate of Interest, $\left(R\right)=10\%\mathrm{p}.\mathrm{a}.$

Number of years, $\left(n\right)=5$

Convert years into months.

$$\begin{gathered} n=5 \\ =5\times 12\left(\because 1\mathrm{year}=12\mathrm{months}\right) \\ =60 \\ \mathrm{So},n=60\mathrm{months} \end{gathered}$$

Step 2: Calculate the maturity amount:

$$\begin{gathered} \mathrm{Maturity}\mathrm{amount}=\mathit{[}1\mathit{+}\frac{\left(n\mathit{+}\mathit{1}\right)R}{2400}\mathit{]}\times n\times P \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}[1+\frac{\left(60+1\right)\times 10}{2400}]\times 60\times 600 \\ =[1+\frac{61\times 10}{2400}]\times 60\times 600 \\ =[1+\frac{610}{2400}]\times 60\times 600 \\ =\left[\frac{2400+610}{2400}\right]\times 36000 \\ =\left[\frac{3010}{2400}\right]\times 36000 \\ \mathit{=}\mathrm{Rs}.45150 \end{gathered}$$

Hence, the amount received by Mr. Dhruv at the time of maturity is $\mathrm{Rs}.45150$.