Rs. 1600 at `10%` per annum compound interest compounded half yearly amount to Rs. 1944·81 in-

Options

A. 2 years
B. 3 years
C. `(1)1/2 `years
D. `(2)1/2 `years

Answer

To solve the problem of finding the time period for Rs. 1600 to amount to Rs. 1944.81 at a compound interest rate of 10% per annum compounded half-yearly, we can follow these steps:

### Step 1: Identify the given values
- Principal (P) = Rs. 1600
- Amount (A) = Rs. 1944.81
- Rate of interest (R) = 10% per annum
- Compounding frequency = Half-yearly

### Step 2: Convert the annual interest rate to half-yearly
Since the interest is compounded half-yearly, we need to divide the annual interest rate by 2:
- Half-yearly rate (r) = R / 2 = 10% / 2 = 5%

### Step 3: Determine the number of compounding periods
Let the number of years be \( t \). Since the interest is compounded half-yearly, the number of compounding periods (n) will be:
- n = 2 * t (because there are 2 half-year periods in a year)

### Step 4: Use the compound interest formula
The formula for compound interest is given by:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]

Substituting the known values into the formula:
\[ 1944.81 = 1600 \left(1 + \frac{5}{100}\right)^{2t} \]

### Step 5: Simplify the equation
First, calculate \( 1 + \frac{5}{100} \):
\[ 1 + \frac{5}{100} = 1.05 \]

Now substitute this back into the equation:
\[ 1944.81 = 1600 \cdot (1.05)^{2t} \]

### Step 6: Divide both sides by 1600
\[ \frac{1944.81}{1600} = (1.05)^{2t} \]
\[ 1.21550625 = (1.05)^{2t} \]

### Step 7: Take the logarithm of both sides
To solve for \( t \), take the logarithm:
\[ \log(1.21550625) = 2t \cdot \log(1.05) \]

### Step 8: Solve for \( t \)
Now, isolate \( t \):
\[ t = \frac{\log(1.21550625)}{2 \cdot \log(1.05)} \]

### Step 9: Calculate the values
Using a calculator:
- \( \log(1.21550625) \approx 0.0833 \)
- \( \log(1.05) \approx 0.0212 \)

Now substitute these values:
\[ t = \frac{0.0833}{2 \cdot 0.0212} \]
\[ t \approx \frac{0.0833}{0.0424} \approx 1.96 \]

### Step 10: Conclusion
Thus, the time period \( t \) is approximately 1.96 years.

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