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Mathematics

Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹2,500 per month for 2 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

Banking

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Answer

Given,

P = ₹2,500

n = 2 years = 24 months

Maturity Value = ₹67,500

Sum deposited = ₹2,500 × 24 = ₹60,000

Maturity value = Sum deposited + Interest

Interest = Maturity value - Sum deposited

∴ I = ₹67,500 − ₹60,000 = ₹7,500

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×r1007500=62500×r100r=7500×10062500r=12%\therefore I = 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] 7500 = 62500 \times \dfrac{r}{100} \\[1em] r = \dfrac{7500 \times 100}{62500} \\[1em] r=12\%

Hence, (i)Mr. Gupta earned ₹7,500 as interest.(ii)The rate of interest was 12% per annum.

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