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Mathematics

Neema had a recurring deposit account in a bank and deposited ₹ 600 per month for 2122\dfrac{1}{2} years. If the rate of interest was 10% per annum, find the maturity value of this account.

Banking

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Answer

Given,

P = ₹600

n = 2122\dfrac{1}{2} years = 2.5 years = 24 months + 6 months = 30 months

r = 10%

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

∴I=600×30×312×12×10100I=600×93024×0.1I=600×38.75×0.1I=₹2,325\therefore I = 600\times \dfrac{30\times 31}{2 \times 12} \times \dfrac{10}{100} \\[1em] I = 600 \times \dfrac{930}{24} \times 0.1 \\[1em] I = 600 \times 38.75 \times 0.1 \\[1em] I = ₹2,325

Sum deposited = ₹600 x 30 = ₹18,000

Maturity value = Sum deposited + Interest = ₹18,000 + ₹2,325 = ₹20,325

Hence, Neema got ₹20,325 at the time of maturity.

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