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Mathematics

Kavita has a cumulative time deposit account in a bank. She deposits ₹800 per month and gets ₹16,700 as maturity value. If the rate of interest be 5% per annum, find the total time for which the account was held. (Hint: x² + 481x − 10020 = 0 ⇒ x² + 501x − 20x − 10020 = 0)

Banking

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Answer

Given,

P = ₹800

Maturity Value = ₹16,700

r = 5%

Let the number of months be 'x'.

Sum deposited = P × x = 800x

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

I=800×x(x+1)24×5100I=800×5x(x+1)2400=4000x(x+1)2400=53x(x+1)I = 800 \times \dfrac{x(x+1)}{24} \times \dfrac{5}{100}\\[1em] I = 800 \times \dfrac{5x(x+1)}{2400} \\[1em] = \dfrac{4000x(x+1)}{2400} \\[1em] = \dfrac{5}{3} x(x+1)

Maturity Value = Sum deposited + Interest

16700=800x+53x(x+1)50100=2400x+5x(x+1)50100=2400x+5x2+5x5x2+2405x50100=05(x2+481x10020)=0x2+481x10020=0x2+501x20x10020=0x(x+501)20(x+501)=0(x20)(x+501)=0x=20 or x=501\Rightarrow 16700 = 800x + \dfrac{5}{3} x(x+1) \\[1em] \Rightarrow 50100 = 2400x + 5x(x+1) \\[1em] \Rightarrow 50100 = 2400x + 5x^2 + 5x \\[1em] \Rightarrow 5x^2 + 2405x - 50100 = 0 \\[1em] \Rightarrow 5(x^2 + 481x - 10020) = 0 \\[1em] \Rightarrow x^2 + 481x - 10020 = 0 \\[1em] \Rightarrow x^2 + 501x - 20x - 10020 = 0 \\[1em] \Rightarrow x(x + 501) - 20(x + 501) = 0 \\[1em] \Rightarrow (x - 20)(x + 501) = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -501

Since the number of months cannot be negative

∴ x = 20 months

Hence, total time for which the account was held is 20 months.

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