Question

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:

• A. Rs. 16,000
• B. Rs. 18,000
• C. Rs. 20,000
• D. Rs. 25,000
• E. None of the above

Hint:

Remember to use the compound interest formula and solve using the relationship between the two accumulated amounts at different points in time.

Answer 1

The Correct Option is C

Let the principal amount be \( P \) and the annual interest rate be \( r \). The formula for compound interest is: \[ A = P(1 + r)^t \] where: - \( A \) is the amount after \( t \) years, - \( P \) is the principal, - \( r \) is the rate of interest, - \( t \) is the number of years. In 2007, the accumulated interest is Rs. 10,000. The amount at the beginning of 2007 is: \[ A_1 = P + 10000 \] Using the formula for compound interest from 2004 to 2007 (3 years), we have: \[ P(1 + r)^3 = P + 10000 \quad \text{(Equation 1)} \] In 2010, the accumulated interest is Rs. 25,000. The amount at the beginning of 2010 is: \[ A_2 = P + 25000 \] Using the compound interest formula from 2004 to 2010 (6 years), we have: \[ P(1 + r)^6 = P + 25000 \quad \text{(Equation 2)} \] Now, divide Equation 2 by Equation 1: \[ \frac{P(1 + r)^6}{P(1 + r)^3} = \frac{P + 25000}{P + 10000} \] This simplifies to: \[ (1 + r)^3 = \frac{P + 25000}{P + 10000} \] Substitute the values from the question: \[ (1 + r)^3 = \frac{25000 + P}{10000 + P} \] Solve this equation to find \( P = 20000 \). \[ \boxed{P = 20000} \]