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Mathematics

Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5% per annum simple interest. If he gets ₹8,325 as interest at the time of maturity, find:

(i) The monthly deposit,

(ii) The maturity value.

Banking

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Answer

Given,

n = 3 year = 36 months

r = 7.5%

I = ₹8,325

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

∴8325=P×36×372×12×7.51008325=P×133224×0.0758325=P×4.1625P=83254.1625I=₹2,000\therefore 8325 = P\times \dfrac{36\times 37}{2 \times 12} \times \dfrac{7.5}{100} \\[1em] 8325 = P \times \dfrac{1332}{24} \times 0.075 \\[1em] 8325 = P \times 4.1625 \\[1em] P=\dfrac{8325}{4.1625}\\[1em] I = ₹2,000

Sum deposited = ₹2,000 x 36 = ₹72,000

Maturity value = Sum deposited + Interest = ₹72,000 + ₹8,325 = ₹80,325

Hence, (i) Mr.Richard deposited ₹2,000 monthly (ii) Mr.Richard got ₹80,325 at the time of maturity.

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