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 ₹ 67,500. Find :

(i) the total interest earned by Mr. Gupta

(ii) the rate of interest per annum.

(iii) how much more interest will Mr. Gupta get, if he deposits ₹ 100 more per month at the same rate and for the same time ?

Sum deposited = ₹ 2500 × 24 = ₹ 60000.

(i) Interest = Maturity value - Sum deposited = ₹ 67500 - ₹ 60000 = ₹ 7500.

Hence, the total interest earned ₹ 7500.

(ii) Let rate of interest be x%.

Given,

P = ₹ 2500, n = (2 × 12) = 24 months and r = x%

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

Substituting values we get :

$\therefore I = ₹ 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 2500 \times \dfrac{x}{4} \\[1em] = ₹ 625x$

As, Interest = ₹ 7500

⇒ 625x = 7500

⇒ x = 12.

Hence, the rate of interest is 12%.

(iii) New monthly deposit be ₹ 2500 + ₹ 100 = ₹ 2600.

P = ₹ 2,600, n = (2 × 12) = 24 months and r = 12%

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

Substituting values we get :

$\Rightarrow I = 2600 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times \dfrac{600}{24} \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times 25 \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times 3 \\[1em] \Rightarrow I = ₹ 7,800.$

Difference in interest earned = ₹ 7,800 - ₹ 7,500 = ₹ 300.

Hence, Mr. Gupta will get ₹ 300 more as interest.