In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs. 10,000 and in the beginning of 2010, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is:

Let the principal amount = $$P$$ and rate of interest = $$r \%$$

Interest accumulated from 2004 to 2007 is Rs.10,000 and from 2004 to 2010 is Rs.25,000

Using, $$C.I. = P[(1 + \frac{R}{100})^T - 1]$$

=> $$P[(1 + \frac{r}{100})^3 - 1] = 10,000$$ ----------Eqn(I)

and $$P[(1 + \frac{r}{100})^6 - 1] = 25,000$$ -----------Eqn(II)

Dividing eqn(II) from (I), we get : 

=> $$\frac{P[(1 + \frac{r}{100})^6 - 1]}{P[(1 + \frac{r}{100})^3 - 1]} = \frac{5}{2}$$

Let $$(1 + \frac{r}{100})^3 = x$$

=> $$\frac{x^2 - 1}{x - 1} = \frac{5}{2}$$

=> $$2x^2 - 5x + 3 = 0$$

=> $$(2x - 3) (x - 1) = 0$$

=> $$x = \frac{3}{2} , 1$$    $$(x \neq 1)$$ because then, r = 0

=> $$(1 + \frac{r}{100})^3 = \frac{3}{2}$$

Substituting it in eqn(I)

=> $$P[\frac{3}{2} - 1] = 10,000$$

=> $$P = 10,000 \times 2 = 20,000$$