Ashish deposits a certain sum of money every month in a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12715 as the maturity value of this account, what sum of money did he pay every month?

Let Ashish deposits ₹ x per month.

So,

P = ₹ x, n = 12 months and r = 11%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{11}{100} \\[1em] = ₹ x \times \dfrac{11 \times 13}{200} \\[1em] = ₹ \dfrac{143x}{200}$

Maturity value = Sum deposited + Interest

$= x \times 12 + ₹\dfrac{143x}{200} \\[1em] = ₹\dfrac{2400x + 143x}{200} \\[1em] = ₹\dfrac{2543x}{200}$

Given, maturity value = ₹ 12715.

$\therefore \dfrac{2543x}{200} = 12715 \\[1em] \Rightarrow x = \dfrac{12715 \times 200}{2543} \\[1em] \Rightarrow x = \dfrac{2543000}{2543} = ₹ 1000.$

Hence, Ashish paid ₹ 1000 per month.