The compound interest on ₹40,000 at 6% per annum is ₹4,944. What is the period (in years) for which the amount is invested?

Options

A. 5 years
B. 4 years
C. 3 years
D. 2 years

Answer

To solve the problem of finding the period for which the amount is invested, we can use the formula for compound interest. The formula for the amount (A) after time (T) with principal (P) and rate (R) is given by:

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

Where:
- \( A \) = Total amount after time T
- \( P \) = Principal amount (initial investment)
- \( R \) = Rate of interest per annum
- \( T \) = Time in years

Given:
- Principal (P) = ₹40,000
- Rate (R) = 6%
- Compound Interest (CI) = ₹4,944

First, we need to find the total amount (A) after the interest is added to the principal:

\[ A = P + CI \]
\[ A = 40,000 + 4,944 \]
\[ A = 44,944 \]

Now we can substitute the values into the compound interest formula:

\[ 44,944 = 40,000 \left(1 + \frac{6}{100}\right)^T \]

Next, simplify the equation:

\[ 44,944 = 40,000 \left(1 + 0.06\right)^T \]
\[ 44,944 = 40,000 \left(1.06\right)^T \]

Now, divide both sides by 40,000:

\[ \frac{44,944}{40,000} = (1.06)^T \]
\[ 1.1236 = (1.06)^T \]

To solve for T, we can take the logarithm of both sides:

\[ \log(1.1236) = T \cdot \log(1.06) \]

Now, calculate the logarithms:

1. Calculate \( \log(1.1236) \) and \( \log(1.06) \):
- \( \log(1.1236) \approx 0.0513 \)
- \( \log(1.06) \approx 0.0253 \)

Now substitute these values back into the equation:

\[ 0.0513 = T \cdot 0.0253 \]

Now solve for T:

\[ T = \frac{0.0513}{0.0253} \]
\[ T \approx 2.03 \]

Since T must be a whole number, we round it to the nearest whole number, which gives us:

\[ T = 2 \text{ years} \]

Thus, the period for which the amount is invested is **2 years**.