Question

A person invested a certain money in a scheme offering simple interest of 16% per annum and received Rs 736 as interest. Had he invested the amount for 8 more years, he would have got an interest of Rs 3680. What is the amount obtained if the same sum is invested at 9% per annum for 2 years, interest compounded annually?

• A. Rs. 2714
• B. Rs. 2720.75
• C. Rs. 2732.63
• D. Rs. 2845.26
• E. Rs. 2987.43

Hint:

For compound interest, remember the formula \( A = P \left(1 + \frac{R}{100}\right)^T \). Ensure to apply it correctly for different time periods and rates.

Answer 1

The Correct Option is C

Step 1: Understanding the given information.
Let the principal amount be \( P \). The interest is given for 16% per annum and for a period of \( t \) years. The formula for simple interest is: \[ I = \frac{P \times R \times T}{100} \] where: - \( I \) is the interest, - \( P \) is the principal, - \( R \) is the rate of interest, - \( T \) is the time in years. Step 2: Calculate the principal.
The person receives Rs. 736 as interest at a rate of 16% per annum. The time period is not given initially. But, the question states that if the time were 8 years more, the interest would have been Rs. 3680. The interest for 8 more years is: \[ 3680 - 736 = 2944 \] Since the rate is 16% and the extra time is 8 years, we can use the formula for interest: \[ 2944 = \frac{P \times 16 \times 8}{100} \] \[ 2944 = \frac{128P}{100} \] \[ 2944 \times 100 = 128P \] \[ P = \frac{294400}{128} = 2300 \] Step 3: Calculate the compound interest.
Now, we need to find the amount when the same principal is invested at 9% per annum for 2 years with compound interest. The formula for compound interest is: \[ A = P \left(1 + \frac{R}{100}\right)^T \] Substitute the values \( P = 2300 \), \( R = 9 \), and \( T = 2 \): \[ A = 2300 \left(1 + \frac{9}{100}\right)^2 \] \[ A = 2300 \left(1.09\right)^2 \] \[ A = 2300 \times 1.1881 = 2732.63 \] Final Answer: \[ \boxed{2732.63} \]