KnowledgeBoat Logo
|

Mathematics

David opened a recurring deposit account in a bank and deposited ₹300 per month for two years. If he received ₹7,725 at the time of maturity, find the rate of interest per annum.

Banking

8 Likes

Answer

Given,

P = ₹300

n = 2 years = 24 months

Maturity Value = ₹7,725

Let the rate of interest be 'r' % per annum.

Sum deposited = P × n = 300 × 24 = ₹7,200

Maturity Value = Sum deposited + Interest=7,200+I

I = Maturity Value - Sum deposited

I = ₹7,725 - ₹7,200

I= ₹525

I=P×n(n+1)2×12×r100I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}

∴I=300×24×2524×r100I=300×25×r100I=7500×r100525=75rr=52575r=7%\therefore I = 300 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100}\\[1em] I = 300 \times 25 \times \dfrac{r}{100}\\[1em] I = 7500 \times \dfrac{r}{100}\\[1em] 525 = 75r\\[1em] r = \dfrac{525}{75} \\[1em] r=7 \%

Hence, the rate of interest per annum is 7%p.a.

Answered By

1 Like

Related Questions