KnowledgeBoat Logo
|

Mathematics

Sajal invests ₹600 per month for 2122\dfrac{1}{2} years in a recurring deposit scheme of Oriental Bank of Commerce. If the bank pays simple interest at 6236\dfrac{2}{3} % per annum, find the amount received by him on maturity.

Banking

1 Like

Answer

Given,

P = ₹600

n = 2122\dfrac{1}{2} years = 2.5 years = 24 months + 6 months = 30 months

r = 6 23\dfrac{2}{3} % = 203\dfrac{20}{3}

I = P×n(n+1)2×12×r100P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}

I=600×30×312×12×203100I=600×93024×20300I=600×38.75×0.67I=1,550\therefore I = 600\times \dfrac{30\times 31}{2 \times 12} \times \dfrac{\dfrac{20}{3}}{100}\\[1em] I = 600 \times \dfrac{930}{24} \times \dfrac{20}{300}\\[1em] I = 600 \times 38.75 \times 0.67\\[1em] I = ₹1,550

Sum deposited = ₹600 x 30 = ₹18,000

Maturity value = Sum deposited + Interest = ₹18,000 + ₹1,550 = ₹19,550

Hence, Sajal got ₹19,550 at the time of maturity.

Answered By

3 Likes


Related Questions