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ICSE Class 10 Mathematics Question 2 of 28

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Question 51

A man opened a recurring deposit account in a branch of PNB. The man deposits certain amount of money per month such that after 2 years, the interest accumulated is equal to his monthly deposits. Find the rate of interest per annum that the bank was paying for the recurring deposit account.

Answer

Given,

Time (n) = 2 years or 24 months

Rate = r% (let)

P = ₹ x/month

I = ₹ x

By formula,

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

Substituting values we get :

⇒x=x×24×(24+1)2×12×r100⇒x=x×24×2524×r100⇒1=r4⇒r=4\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1 = \dfrac{r}{4} \\[1em] \Rightarrow r = 4%.

Hence, rate of interest = 4%.