Rekha Invested a sum of `Rs. 12000` at `5%` per annum compound interest. She received an amount of `Rs.13230` after nyears. Find n.

Options

A. `2.8 years `
B. `3.0 years`
C. `2.5 years`
D. `2.0 years`

Answer

To find the number of years \( n \) that Rekha invested her money, we can use the formula for compound interest. The formula for the amount \( A \) after \( n \) years is given by:

\[ A = P \left(1 + \frac{r}{100}\right)^n \]

Where:
- \( A \) is the amount received after \( n \) years (in this case, Rs. 13230),
- \( P \) is the principal amount (initial investment, Rs. 12000),
- \( r \) is the rate of interest per annum (5%),
- \( n \) is the number of years.

### Step 1: Substitute the known values into the formula

We have:
- \( A = 13230 \)
- \( P = 12000 \)
- \( r = 5 \)

Substituting these values into the formula, we get:

\[ 13230 = 12000 \left(1 + \frac{5}{100}\right)^n \]

### Step 2: Simplify the equation

Calculate \( 1 + \frac{5}{100} \):

\[ 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \]

Now substitute this back into the equation:

\[ 13230 = 12000 \times (1.05)^n \]

### Step 3: Divide both sides by 12000

\[ \frac{13230}{12000} = (1.05)^n \]

Calculating the left side:

\[ \frac{13230}{12000} = 1.1025 \]

So, we have:

\[ 1.1025 = (1.05)^n \]

### Step 4: Take the logarithm of both sides

To solve for \( n \), we take the logarithm of both sides:

\[ \log(1.1025) = n \cdot \log(1.05) \]

### Step 5: Solve for \( n \)

Now, we can isolate \( n \):

\[ n = \frac{\log(1.1025)}{\log(1.05)} \]

Using a calculator to find the logarithms:

\[ \log(1.1025) \approx 0.0414 \quad \text{and} \quad \log(1.05) \approx 0.0212 \]

Now substitute these values into the equation:

\[ n \approx \frac{0.0414}{0.0212} \approx 1.95 \]

### Step 6: Round to the nearest option

Since the options provided are 2.8 years, 3 years, 2.5 years, and 2 years, we can see that \( n \approx 2 \) years is the closest option.

### Final Answer

Thus, the number of years \( n \) is approximately **2 years**.

---