Ramesh wants to get ₹ 6,050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is:

1. ₹ 5,500

2. ₹ 5,000

3. ₹ 5,600

4. ₹ 5,800

Given,

A = ₹ 6,050

R = 10%

n = 2 years

Let amount he needs to keep be ₹ P.

$A = P\Big(1 + \dfrac{r}{100}\Big)^{n}$

Substituting values we get :

$\Rightarrow 6050 = P \Big(1 + \dfrac{10}{100}\Big)^{2} \\[1em] \Rightarrow 6050 = P \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P (1.1)^2 \\[1em] \Rightarrow 6050 = P \times 1.21 \\[1em] \Rightarrow \dfrac{6050}{1.21} = P \\[1em] \Rightarrow P = ₹ 5,000. \\[1em]$

Hence, option 2 is correct option.