4) Ravi opens a recurring deposit account of Rs.800 per month at 7% p.

1.	4) Ravi opens a recurring deposit account of Rs.800 per month at 7% p.a. If he receives Rs.20600 at the time of maturity, find the time in years for which the account was held.please give the correct answer, don't give unnecessary answer. Class 10 ICSE Banking.
Answer» <P> $\large\underline{\sf{Given- }}$ Ravi has an Recurring Deposit Account of Rs 800 per month. The rate of interest is 7 % p.a. He get Rs 20600 on the time of maturity. $\large\underline{\sf{To\:Find - }}$ Time in years. $\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}$ Amount on a certain sum of money of Rs P invested per month at the rate of r % per annum for n MONTHS is $\bold{ \red {\boxed{\text{Maturity value} = \text{nP} + \text{P} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{r}}{100} }}}$ $\red{\large\underline{\sf{Solution-}}}$ Given that, Ravi has an Recurring Deposit Account of Rs 800 per month. The rate of interest is 7 % p.a. He get Rs 20600 at the time of maturity. Let assume that the RD is deposited for n months. So, on SUBSTITUTING the values in the formula $\bold{ \red {\boxed{\text{Maturity value} = \text{nP} + \text{P} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{r}}{100} }}}$ We have, Sum deposited, P = Rs 800 Rate of interest, r = 7 % per annum Maturity Value = Rs 20600 $\rm :\longmapsto\:20600 = 800n + 800 \times \dfrac{n(n + 1)}{24} \times \dfrac{7}{100}$ $\rm :\longmapsto\:20600 = 800n + \dfrac{7}{3} n(n + 1)$ $\rm :\longmapsto\:61800 = 2400n + 7 {n}^{2} + 7n$ $\rm :\longmapsto\:7 {n}^{2} + 2407n - 61800 = 0$ Its a QUADRATIC equation in 'n', So, by using splitting of middle terms, we have $\rm :\longmapsto\:7 {n}^{2} + 2575n - 168n - 61800 = 0$ $\rm :\longmapsto\:n(7n + 2575) - 24(7n + 2575) = 0$ $\rm :\longmapsto\:(7n + 2575)(n - 24) = 0$ $\bf\implies \:n = 24$ So, Time = 2 years.