In 2 years at simple interest the principal in creases by 8%. What will be the compound interest earned (in Rs) on Rs 10 lakhs in 2 years at the same rate?

Options

A. 86000
B. 81600
C. 90000
D. 94000

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understand the Problem**: We know that in 2 years at simple interest, the principal increases by 8%. We need to find the compound interest earned on Rs 10 lakhs in 2 years at the same rate.

2. **Identify the Simple Interest Formula**: The formula for simple interest (SI) is given by:
\[ SI = \frac{P \times R \times T}{100} \]
where \( P \) is the principal, \( R \) is the rate of interest, and \( T \) is the time in years.

3. **Set Up the Equation**: Since the principal increases by 8% in 2 years, we can express this as:
\[ SI = 8\% \text{ of } P = \frac{8P}{100} \]
For our case, we can write:
\[ \frac{P \times R \times 2}{100} = \frac{8P}{100} \]

4. **Cancel Out P**: Since \( P \) is common in both sides, we can cancel it out (assuming \( P \neq 0 \)):
\[ \frac{R \times 2}{100} = \frac{8}{100} \]

5. **Solve for R**: Rearranging the equation gives:
\[ R \times 2 = 8 \implies R = \frac{8}{2} = 4\% \]
So, the rate of interest \( R \) is 4%.

6. **Calculate Compound Interest**: Now we will use the compound interest formula:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]
Substituting \( P = 10,00,000 \), \( R = 4\% \), and \( T = 2 \):
\[ CI = 10,00,000 \left(1 + \frac{4}{100}\right)^2 - 10,00,000 \]

7. **Calculate \( \left(1 + \frac{4}{100}\right)^2 \)**:
\[ = 10,00,000 \left(1 + 0.04\right)^2 = 10,00,000 \times (1.04)^2 \]
\[ = 10,00,000 \times 1.0816 \]

8. **Final Calculation**:
\[ = 10,00,000 \times 1.0816 - 10,00,000 \]
\[ = 10,81,600 - 10,00,000 = 81,600 \]

9. **Conclusion**: The compound interest earned on Rs 10 lakhs in 2 years at the same rate is Rs 81,600.

### Final Answer:
The compound interest earned is **Rs 81,600**.