Question

A man has a recurring deposit account and deposits ₹ 2000 per month at an interest rate of 24%. If he got ₹ 27,120 after the end of the maturity period, Find the time (in months) for which account was held.

Answer 1

Given P = ₹ 2000
r = 10%
Let number of months be 'n'.

Amount deposited = P \(\times\) n = 2000 n

Interest = \(\frac{P \times n (n+1)}{2 \times 12} \times (\frac{r}{100})\)
I = \(\frac{2000 \times n (n+1)}{2 \times 12} \times (\frac{24}{100})= 20\times n(n+1) = 20(n^2 + n)\)

Maturity value = Amount deposited + Interest

\(27120 = 2000 n + 20(n^2 + n)\)
\(1356 = 100n + n^2 + n\)
\(n^2 + 101n - 1356 = 0\)

\(n^2 + 113n -12n - 1356 = 0\)

\(n(n+113) - 12(n+ 113) = 0\)

\(n = -113\ \text{ or} \ n = 12\)

Since time period cannot be negative, so n = 12 months.

Hence, the account was held for 12 months.