KnowledgeBoat Logo
|

Mathematics

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.

Banking

134 Likes

Answer

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×n(n+1)2×12×r100P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

∴I=₹2500×24×252×12×x100=₹2500×x4=₹625x\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%.

Answered By

47 Likes

Related Questions