Rs. 16000 invested at 10% per annum compounded semi-annually amounts to Rs. 18522. The period of investment is

Options

A. 16 yr
B. 3 yr
C. `1 (1)/(2)` yr
D. 2 yr

Answer

To solve the problem of finding the period of investment for Rs. 16,000 compounded semi-annually at a rate of 10% per annum that amounts to Rs. 18,522, we can follow these steps:

### Step 1: Understand the formula for compound interest
The formula for compound interest is given by:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = principal amount (the initial amount of money).
- \( r \) = annual interest rate (decimal).
- \( n \) = number of times that interest is compounded per year.
- \( t \) = the number of years the money is invested or borrowed.

### Step 2: Identify the given values
From the question:
- \( A = 18522 \)
- \( P = 16000 \)
- \( r = 10\% = 0.10 \)
- Since the interest is compounded semi-annually, \( n = 2 \).

### Step 3: Substitute the values into the formula
We substitute the known values into the compound interest formula:
\[ 18522 = 16000 \left(1 + \frac{0.10}{2}\right)^{2t} \]

### Step 4: Simplify the equation
Calculate \( \frac{0.10}{2} \):
\[ \frac{0.10}{2} = 0.05 \]
Thus, the equation becomes:
\[ 18522 = 16000 \left(1 + 0.05\right)^{2t} \]
Which simplifies to:
\[ 18522 = 16000 \left(1.05\right)^{2t} \]

### Step 5: Divide both sides by 16000
To isolate the term with \( t \):
\[ \frac{18522}{16000} = \left(1.05\right)^{2t} \]
Calculating the left side:
\[ \frac{18522}{16000} \approx 1.157625 \]

### Step 6: Take logarithm of both sides
Now we take the logarithm of both sides:
\[ \log(1.157625) = \log\left((1.05)^{2t}\right) \]
Using the property of logarithms:
\[ \log(a^b) = b \cdot \log(a) \]
We can rewrite the equation as:
\[ \log(1.157625) = 2t \cdot \log(1.05) \]

### Step 7: Solve for \( t \)
Now, we can solve for \( t \):
\[ t = \frac{\log(1.157625)}{2 \cdot \log(1.05)} \]

### Step 8: Calculate the logarithms
Using a calculator:
- \( \log(1.157625) \approx 0.0697 \)
- \( \log(1.05) \approx 0.0212 \)

Substituting these values:
\[ t = \frac{0.0697}{2 \cdot 0.0212} \]
\[ t = \frac{0.0697}{0.0424} \approx 1.645 \]

### Step 9: Finalize the answer
Thus, the time period \( t \) is approximately:
\[ t \approx 1.645 \text{ years} \]

### Step 10: Convert to a fraction if necessary
Since \( 1.645 \) years can be expressed as \( \frac{3}{2} \) years or approximately \( 1 \frac{5}{8} \) years, we can round it to:
\[ t \approx 1.5 \text{ years} \]

### Final Answer:
The period of investment is approximately **1.5 years**.