Question

Deepa has a 4-year recurring deposit account in a bank and deposits Rs 1,800 per month. If she gets Rs 1,08,450 at the time of maturity, find the rate of interest.

Answer 1

Instalment per month P= Rs. 1800
no of months n = 48
Let rate of interest r = r% p.a.

S.I = $P\times \frac{n(n+1)}{2\times 12}\times \frac{r}{100}$

$=1800\times \frac{48(48+1)}{2\times 12}\times \frac{r}{100}$

$=1800\times \frac{2352}{24}\times \frac{r}{100}$

$=1764r$

Maturity value $=(1800\times 48)+\left(1764\right)r$

given Maturity value = Rs.1,08,450

then

$(1800\times 48)+\left(1764\right)r=1,08,450$

$\left(1764\right)r=1,08,450-86,400$

$\left(1764\right)r=22050$

$r=\frac{22050}{1764}$

r = 12.5%

Answer:- the rate of interest is 12.5%