Question

Mr. Chaturvedi has a recurring deposit account in a Grindlay's bank for $4\frac{1}{2}$years at 11% interest p.a.(Simple interest). If he gets Rs. 101418.75 at the time of maturity, find the monthly instalment.

Answer 1

Step 1: Given data

Maturity amount received by Mr. Chaturvedi$=\mathrm{Rs}.101418.75$

Rate of Interest, $\left(R\right)=11\%\mathrm{p}.\mathrm{a}.$

Number of years, $\left(n\right)=4\frac{1}{2}$

Step 2:Convert years into months

$$\begin{gathered} n=4\frac{1}{2} \\ =\frac{9}{2} \\ =\frac{9}{2}\times 12\left(\because 1\mathrm{year}=12\mathrm{months}\right) \\ =54 \\ \mathrm{So},n=54\mathrm{months} \end{gathered}$$

Step 3:Calculate the monthly instalment

Let the monthly instalment be $\mathrm{Rs}.P$

Then,

$$\begin{gathered} \mathrm{Maturity}\mathrm{amount}=\left[1+\frac{\left(n+1\right)R}{2400}\right]\times n\times P \\ 101418.75=\left[1+\frac{\left(54+1\right)11}{2400}\right]\times 54\times P \\ 101418.75=\left[1+\frac{55\times 11}{2400}\right]\times 54\times P \\ 101418.75=\left[1+\frac{605}{2400}\right]\times 54\times P \\ 101418.75=\left[\frac{2400+605}{2400}\right]\times 54\times P \\ 101418.75=\left[\frac{3005}{2400}\right]\times 54\times P \\ \Rightarrow P=\frac{101418.75\times 2400}{3005\times 54} \\ \therefore P=\mathrm{Rs}.1500 \end{gathered}$$

Hence, the monthly instalment per month is $\mathrm{Rs}.1500$.