The compound interest on a certain sum for 2 years at 15% per annum is ₹3,641, when the interest is compounded 8 monthly. The sum is:

Options

A. ₹9,000
B. ₹9,600
C. ₹11,000
D. ₹10,000

Answer

To find the principal sum for which the compound interest for 2 years at 15% per annum compounded every 8 months is ₹3,641, we can follow these steps:

### Step 1: Understand the given data
- Compound Interest (CI) = ₹3,641
- Rate of Interest (R) = 15% per annum
- Time (T) = 2 years
- Compounding frequency = every 8 months

### Step 2: Calculate the effective rate of interest for 8 months
Since the interest is compounded every 8 months, we need to convert the annual interest rate to the rate for 8 months.

1. **Convert annual rate to 8-month rate:**
\[ \text{Rate for 8 months} = \frac{15\%}{12} \times 8 = 10\% \]

### Step 3: Calculate the number of compounding periods
2. **Determine the number of compounding periods in 2 years:**
- 2 years = 24 months
- Number of 8-month periods in 24 months = \( \frac{24}{8} = 3 \)

### Step 4: Use the compound interest formula
3. **Apply the compound interest formula:**
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
Where:
- \( A \) = Total amount after interest
- \( P \) = Principal amount (initial sum)
- \( r \) = Rate of interest per compounding period
- \( n \) = Number of compounding periods

Since CI = A - P, we can express it as:
\[ CI = P \left(1 + \frac{r}{100}\right)^n - P \]
Rearranging gives:
\[ CI = P \left[\left(1 + \frac{r}{100}\right)^n - 1\right] \]

### Step 5: Substitute the known values
4. **Substituting the values into the formula:**
- CI = ₹3,641
- \( r = 10\% \)
- \( n = 3 \)

Thus, we have:
\[ 3641 = P \left[\left(1 + \frac{10}{100}\right)^3 - 1\right] \]
Simplifying further:
\[ 3641 = P \left[\left(1.1\right)^3 - 1\right] \]

### Step 6: Calculate \( (1.1)^3 \)
5. **Calculate \( (1.1)^3 \):**
\[ (1.1)^3 = 1.331 \]
Therefore:
\[ 3641 = P \left[1.331 - 1\right] \]
\[ 3641 = P \times 0.331 \]

### Step 7: Solve for P
6. **Rearranging to find P:**
\[ P = \frac{3641}{0.331} \]
Calculating this gives:
\[ P \approx 11,000 \]

### Conclusion
The principal sum is approximately ₹11,000.

---