A sum of money when invested for a year at the rate of 10% interest per ammum compounded half-yearly becomes ₹ 44,100 at maturity. The sum invested was _____ .
A)40000
B)40500
C)39800
D)40250

Options

A. 40000
B. 40500
C. 39800
D. 40250

Answer

To solve the problem, we need to determine the original sum of money (principal) that, when invested at a rate of 10% per annum compounded half-yearly, grows to ₹44,100 after one year.

### Step-by-Step Solution:

1. **Understand the Compounding Period**:
- The interest rate is given as 10% per annum.
- Since the interest is compounded half-yearly, we need to divide the annual interest rate by 2.
- Thus, the half-yearly interest rate is:
\[ \text{Half-yearly rate} = \frac{10\%}{2} = 5\% \]

2. **Determine the Number of Compounding Periods**:
- Since the investment is for one year and compounded half-yearly, the number of compounding periods is:
\[ \text{Number of periods} = 2 \text{ (half-years)} \]

3. **Use the Compound Interest Formula**:
- The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
- Where:
- \(A\) = Amount after time \(n\)
- \(P\) = Principal amount (the sum invested)
- \(r\) = Rate of interest per period
- \(n\) = Number of periods

4. **Substitute the Known Values**:
- We know \(A = 44,100\), \(r = 5\%\), and \(n = 2\).
- Plugging these values into the formula gives:
\[ 44100 = P \left(1 + \frac{5}{100}\right)^2 \]
\[ 44100 = P \left(1 + 0.05\right)^2 \]
\[ 44100 = P \left(1.05\right)^2 \]
\[ 44100 = P \times 1.1025 \]

5. **Solve for the Principal \(P\)**:
- Rearranging the equation to solve for \(P\):
\[ P = \frac{44100}{1.1025} \]
- Calculating \(P\):
\[ P = 40000 \]

### Conclusion:
The sum invested was **₹40,000**.