Question

Priyanka has a recurring deposit account of Rs. $1000$ per month at $10\%$ per annum. If she gets Rs. $5550$ as interest at the time of maturity, find the total time for which the account was held.

Answer 1

Let the maturity period be $n$ months,

$P$ be the money deposited every month which is $Rs.1000$

and $r$ be the rate of interest per annum which is $10\%$.

Now, the interest $I$ received at the end of maturity is calculated by the following formula:

$I=P\times \frac{n(n+1)}{24}\times \frac{r}{100}$

We are given $I=5500$, so

substituting the values, we get

$\Rightarrow 5550=1000\times \frac{n(n+1)}{24}\times \frac{10}{100}$

$\Rightarrow n(n+1)=\frac{5550\times 24}{100}=1332$

$\Rightarrow n^2+n-1332=0$

$\Rightarrow n^2+37n-36n-1332=0$

$\Rightarrow (n-36)(n+37)=0$

$\Rightarrow n=36,-37$

Since the maturity period cannot be negative, $n=36$.

Therefore, the total time for which the account was held was $36$ months which is the same as $3$ years. So the maturity period was $3$ years.