KnowledgeBoat Logo
|

Mathematics

Deepa has a 4-year recurring deposit account in a bank and deposits ₹ 1800 per month. If she gets ₹ 108450 at the time of maturity, find the rate of interest.

Banking

12 Likes

Answer

Let rate of interest be x%.

Given,

P = ₹ 1800, n = (4 × 12) = 48 months, r = x%.

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

∴I=₹1800×48×492×12×x100=₹18×98x=₹1764x\therefore I = ₹ 1800 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 18 \times 98x \\[1em] = ₹ 1764x

Sum deposited = ₹ 1800 × 48 = ₹ 86400

Interest = Maturity value - Sum deposited = ₹ 108450 - ₹ 86400 = ₹ 22050.

⇒1764x=22050⇒x=220501764⇒x=12.5%.\Rightarrow 1764x = 22050 \\[1em] \Rightarrow x = \dfrac{22050}{1764} \\[1em] \Rightarrow x = 12.5\%.

Hence, the rate of interest is 12.5% per annum.

Answered By

5 Likes

Related Questions