Question

Shiva invested a certain sum of money in a simple interest bond whose value grew to ₹300 at the end of 3 years and to ₹400 at the end of another 5 years. What was the rate of interest at which he invested his sum?

• A. 12.5%
• B. 12%
• C. 6.67%
• D. 8.33%

Hint:

In simple interest problems, use the relationships between amounts at different times to set up equations and solve for the rate of interest.

Answer 1

The Correct Option is D

Let the principal amount be \( P \) and the rate of interest be \( r \). - After 3 years, the value of the investment is ₹300.
- After 8 years (3 + 5 years), the value of the investment is ₹400.
Using the simple interest formula: \[ A = P + \frac{P \times r \times t}{100} \] For 3 years, the amount is ₹300, so: \[ 300 = P + \frac{P \times r \times 3}{100} \] This simplifies to: \[ 300 = P \left( 1 + \frac{3r}{100} \right) \quad \cdots (1) \] For 8 years, the amount is ₹400, so: \[ 400 = P + \frac{P \times r \times 8}{100} \] This simplifies to: \[ 400 = P \left( 1 + \frac{8r}{100} \right) \quad \cdots (2) \] Now, subtract equation (1) from equation (2): \[ 400 - 300 = P \left( 1 + \frac{8r}{100} \right) - P \left( 1 + \frac{3r}{100} \right) \] \[ 100 = P \left( \frac{8r}{100} - \frac{3r}{100} \right) \] \[ 100 = P \times \frac{5r}{100} \] \[ 100 = \frac{5P \times r}{100} \] \[ 100 \times 100 = 5P \times r \] \[ 10000 = 5P \times r \] \[ P \times r = 2000 \] Now substitute \( P \times r = 2000 \) into equation (1): \[ 300 = P \left( 1 + \frac{3r}{100} \right) \] \[ 300 = P + \frac{3P \times r}{100} \] \[ 300 = P + \frac{3 \times 2000}{100} \] \[ 300 = P + 60 \] \[ P = 240 \] Now that we have \( P = 240 \), substitute this into \( P \times r = 2000 \): \[ 240 \times r = 2000 \] \[ r = \frac{2000}{240} = 8.33% \] Thus, the rate of interest is 8.33%.