Question

(c) A recurring deposit account in which
2500 is invested per month at 12%
p.a. has a maturity amount of 49275.
Find the time for which the account is
operated in years.
​

Answer 1

Solution!!

The concept of Recurring deposit account has to be used here. The principal, rate of interest and maturity value is given. We have to find the time. Just one formula is enough to solve this question. Let's do it!

Maturity value (MV) = Rs 49275

Principal (P) = Rs 2500

Rate of interest (R) = 12%

Time (n) = ?

$\sf \bold{MV=P\times n+P\times \dfrac{n(n+1)}{2\times 12}\times \dfrac{R}{100}}$

$\sf 49275=2500\times n+2500\times \dfrac{n(n+1)}{2\times 12}\times \dfrac{12}{100}$

$\sf 49275=2500n+25\times \dfrac{n(n+1)}{2}$

$\sf 49275=2500n+\dfrac{25n(n+1)}{2}$

$\sf 49275=2500n+\dfrac{25n^{2}+25n}{2}$

$\sf 49275=\dfrac{5000n+25n^{2}+25n}{2}$

98550 = 25n² + 5025n

3942 = n² + 201n

n² + 201n - 3942 = 0

n² + 219n - 18n - 3942 = 0

n(n + 219) -18(n + 219) = 0

(n - 18)(n + 219) = 0

n - 18 = 0

n = 18

or

n + 219 = p

n = -219

Since time cannot be negative, we reject n = -219

n = 18 months = 1.5 years