Question

Akash deposited ₹2000 per month in a recurring deposit account. If rate of interest on recurring deposit is 24% per annum and he gets ₹27120 on maturity, find the time (in months) of his recurring deposit account.

• A. 12

Answer 1

The correct option is A 12

Given P = 2000
r = 10%
Let number of months be 'n '
Amount deposited P $\times$ n = 2000 n
Interest = $\frac{P\times n(n+1)}{2\times 12}\times \left(\frac{r}{100}\right)$
I = $\frac{2000\times n(n+1)}{2\times 12}\times \left(\frac{24}{100}\right)=20\times n(n+1)=20(n^2+n)$
Maturity value = Amount deposited + Interest
27120 = $2000n+20(n^2+n)$
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 or n = 12

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