Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹2500 per month for two years. At the time of maturity he got ₹67500. Find:

(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.

(i) Here,
P = money deposited per month = ₹2500,
n = number of months for which the money is deposited = 2 x 12 = 24

∴ Total money deposited by Mr. Gupta = ₹(2500 x 24) = ₹60000

Money Mr. Gupta gets at the time of maturity = ₹67500

∴ Total interest earned by Mr. Gupta = ₹67500 - ₹60000 = ₹7500

(ii) Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] 7500 = \Big( 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] 7500 = 625r \\[0.5em] \Rightarrow r = \dfrac{7500}{625} \\[0.5em] \Rightarrow r = 12$

∴ Rate of (simple) interest = 12% p.a.