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Mathematics

A man deposits ₹ 900 per month in a recurring account for 2 years. If he gets ₹ 1800 as interest at the time of maturity, find the rate of interest.

Banking

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Answer

Let rate of interest be r%.

Given,

P = ₹ 900

Time (n) = 2 years = 24 months.

By formula,

Interest = P × n(n+1)2×12×r100\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Substituting values we get :

⇒1800=900×24×2524×r100⇒1800=225r⇒r=1800225⇒r=8%.\Rightarrow 1800 = 900 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1800 = 225r \\[1em] \Rightarrow r = \dfrac{1800}{225} \\[1em] \Rightarrow r = 8\%.

Hence, the rate of interest = 8%.

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