If 4000 becomes 4410at the rate of 10% com pound interest in certain time period then find the time period interest is compounded semi-annually?

Options

A. 6 month
B. 1 year
C. 1.5 year
D. 2 year

Answer

To solve the problem, we need to find the time period for which the principal amount of 4000 becomes 4410 at a compound interest rate of 10%, compounded semi-annually.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Principal (P) = 4000
- Amount (A) = 4410
- Rate (R) = 10% per annum
- Time (T) = ? (in years)

2. **Convert the Rate for Semi-Annual Compounding**:
Since the interest is compounded semi-annually, we need to adjust the rate and the time:
- Semi-annual rate (r) = R/2 = 10%/2 = 5% = 0.05 (as a decimal)
- The number of compounding periods per year (n) = 2

3. **Use the Compound Interest Formula**:
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Here, we can simplify it to:
\[ A = P \left(1 + r\right)^{nt} \]
Substituting the known values:
\[ 4410 = 4000 \left(1 + 0.05\right)^{2T} \]

4. **Simplify the Equation**:
\[ 4410 = 4000 \left(1.05\right)^{2T} \]
Divide both sides by 4000:
\[ \frac{4410}{4000} = \left(1.05\right)^{2T} \]
\[ 1.1025 = \left(1.05\right)^{2T} \]

5. **Take the Logarithm of Both Sides**:
To solve for \(2T\), take the logarithm:
\[ \log(1.1025) = 2T \cdot \log(1.05) \]

6. **Calculate the Logarithms**:
Using a calculator:
\[ \log(1.1025) \approx 0.0414 \]
\[ \log(1.05) \approx 0.0212 \]

7. **Substitute the Values**:
\[ 0.0414 = 2T \cdot 0.0212 \]

8. **Solve for T**:
\[ 2T = \frac{0.0414}{0.0212} \approx 1.95 \]
\[ T \approx \frac{1.95}{2} \approx 0.975 \text{ years} \]

### Final Answer:
The time period for which the interest is compounded semi-annually is approximately **0.975 years** or about **11.7 months**.