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Mathematics

Puneet has a recurring deposit account in the Bank of Baroda and deposits ₹ 140 per month for 4 years. If he gets ₹ 8092 on maturity, find the rate of interest given by the bank.

Banking

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Answer

Let rate of interest given by bank = x%.

So,

P = ₹ 140, n = (4 × 12) = 48 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=₹140×48×492×12×x100=₹140×98x100=₹13720x100\therefore I = ₹ 140 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 140 \times \dfrac{98x}{100} \\[1em] = ₹ \dfrac{13720x}{100}

Maturity value = Sum deposited + Interest

= ₹140×48+₹13720x100=₹6720+₹13720x100₹ 140 \times 48 + ₹\dfrac{13720x}{100} = ₹6720 + ₹\dfrac{13720x}{100}

Given, maturity value = ₹ 8092.

∴6720+13720x100=8092⇒13720x100=8092−6720⇒13720x100=1372⇒x=1372×10013720=10.\therefore 6720 + \dfrac{13720x}{100} = 8092 \\[1em] \Rightarrow \dfrac{13720x}{100} = 8092 - 6720 \\[1em] \Rightarrow \dfrac{13720x}{100} = 1372 \\[1em] \Rightarrow x = \dfrac{1372 \times 100}{13720} = 10.

Hence, the rate of interest given by bank is 10%.

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