The differnce between compound interest compounded every 6 months and simple interest after 2 years is 248.10. The rate of interest is 10 precent Find the sum

Options

A. 12000
B. 14000
C. 16000
D. 18000

Explanation

To solve the problem, we need to find the principal amount (P) given that the difference between the compound interest (CI) compounded every 6 months and the simple interest (SI) after 2 years is 248.10, with an interest rate of 10% per annum.

### Step-by-Step Solution:

1. **Identify the Rate and Time Period**:
- Rate of interest (R) = 10% per annum
- Time period (T) = 2 years

2. **Calculate Compound Interest (CI)**:
- Since the interest is compounded every 6 months, the effective rate for each half-year is:
\[ R_{half} = \frac{10}{2} = 5\% \text{ per half-year} \]
- The number of compounding periods in 2 years (T) is:
\[ n = 2 \times 2 = 4 \text{ (half-year periods)} \]
- The formula for the amount (A) when compounded is:
\[ A = P \left(1 + \frac{R_{half}}{100}\right)^n \]
- Substituting the values:
\[ A = P \left(1 + \frac{5}{100}\right)^4 = P \left(1.05\right)^4 \]
- Calculate \( (1.05)^4 \):
\[ (1.05)^4 = 1.21550625 \approx 1.2155 \]
- Therefore, the amount after 2 years is:
\[ A \approx 1.2155P \]
- The compound interest (CI) is:
\[ CI = A - P = 1.2155P - P = 0.2155P \]

3. **Calculate Simple Interest (SI)**:
- The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]
- Substituting the values:
\[ SI = \frac{P \times 10 \times 2}{100} = \frac{20P}{100} = 0.2P \]

4. **Set Up the Equation**:
- According to the problem, the difference between CI and SI is given as:
\[ CI - SI = 248.10 \]
- Substituting the expressions for CI and SI:
\[ 0.2155P - 0.2P = 248.10 \]
- Simplifying the left side:
\[ (0.2155 - 0.2)P = 248.10 \]
\[ 0.0155P = 248.10 \]

5. **Solve for Principal (P)**:
- To find P, divide both sides by 0.0155:
\[ P = \frac{248.10}{0.0155} \]
- Calculating the right side:
\[ P \approx 16000 \]

### Final Answer:
The principal amount (P) is **16000**.