Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for 4½ years at 11% p.a. (simple interest). If he gets ₹101418.75 at the time of maturity, find the monthly installment.

Here,
n = number of months for which the money is deposited = 4 x 12 + 6 = 54,
r = interest rate per annum = 11

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{54 \times 55}{2 \times 12} \times \dfrac{11}{100} \Big) \\[0.5em] \enspace\medspace = ₹\dfrac{1089}{80}x$

Total money deposited by Mr. Chaturvedi = ₹54x

∴ The amount of maturity = total money deposited + interest

$= ₹54x + ₹\dfrac{1089}{80}x \\[0.5em] = ₹\dfrac{5409}{80}x$

According to the given,

$\dfrac{5409}{80}x = 101418.75 \\[0.5em] \Rightarrow \dfrac{5409}{80}x = \dfrac{10141875}{100} \\[0.5em] \Rightarrow x = \dfrac{10141875 \times 80}{100 \times 5409} \\[0.5em] \Rightarrow x = 1500 \\[0.5em]$

∴ The monthly installment = ₹1500