The population of a city is 7.26 lakh today. If the population has been increasing at the rate of 10% per year, then two years ago, the poplulation would have been

Options

A. 6 lakh
B. 5.5 lakh
C. 5 lakh
D. 4.5 lakh

Answer

To find the population of the city two years ago, we can use the formula for compound interest, which can also be applied to population growth. The formula is:

\[ A = P \times (1 + r)^t \]

Where:
- \( A \) is the amount (current population),
- \( P \) is the principal amount (population two years ago),
- \( r \) is the rate of interest (population growth rate),
- \( t \) is the time in years.

### Step-by-Step Solution:

1. **Identify the known values**:
- Current population \( A = 7.26 \) lakh
- Growth rate \( r = 10\% = 0.10 \)
- Time \( t = 2 \) years

2. **Rearrange the formula to find \( P \)**:
\[ P = \frac{A}{(1 + r)^t} \]

3. **Substitute the known values into the formula**:
\[ P = \frac{7.26}{(1 + 0.10)^2} \]

4. **Calculate \( (1 + 0.10)^2 \)**:
\[ (1 + 0.10)^2 = (1.10)^2 = 1.21 \]

5. **Now substitute this back into the equation for \( P \)**:
\[ P = \frac{7.26}{1.21} \]

6. **Perform the division**:
\[ P = 6.0 \text{ lakh} \]

### Final Answer:
The population of the city two years ago was **6 lakh**.