Question 4(i)Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years total time for which the account was held.

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Let time be n months.By formula,I = P×n(n+1)2×12×r100P \times \dfrac{n(n + 1)}{2\times 12} \times \dfrac{r}{100}P×2×12n(n+1)​×100r​Substituting values we get :⇒1040=2000×n(n+1)24×8100⇒1040=2000×n(n+1)300⇒n(n+1)=1040×3002000⇒n(n+1)=156⇒n2+n−156=0⇒n2+13n−12n−156=0⇒n(n+13)−12(n+13)=0⇒(n−12)(n+13)=0⇒n−12=0 or n+13=0⇒n=12 or n=−13.\Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} \[1em] \Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{300} \[1em] \Rightarrow n(n + 1) = \dfrac{1040 \times 300}{2000} \[1em] \Rightarrow n(n + 1) = 156 \[1em] \Rightarrow n^2 + n - 156 = 0 \[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \[1em] \Rightarrow (n - 12)(n + 13) = 0 \[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \[1em] \Rightarrow n = 12 \text{ or } n = -13.⇒1040=2000×24n(n+1)​×1008​⇒1040=2000×300n(n+1)​⇒n(n+1)=20001040×300​⇒n(n+1)=156⇒n2+n−156=0⇒n2+13n−12n−156=0⇒n(n+13)−12(n+13)=0⇒(n−12)(n+13)=0⇒n−12=0 or n+13=0⇒n=12 or n=−13.Since, no. of months cannot be negative.∴ n = 12.Hence, total time for which the account was held = 12 months.

Solved 2024 Question Paper ICSE Class 10 Mathematics — Question 1

Question 4(i)