The population of a village increases at the rate of 16% per annum. If the present population of the village is 403680, then what was the population of the village 2 years ago?

Options

A. 30000
B. 40000
C. 300000
D. 400000

Answer

To find the population of the village 2 years ago, we will use the formula for compound interest since the population increases at a certain percentage per annum.

Let's denote:
- Present population (P) = 403680
- Rate of increase (r) = 16% = 0.16
- Population 2 years ago (x)

### Step 1: Write the equation for the population after 1 year
The population after 1 year can be calculated using the formula:
\[ P_1 = x \times (1 + r) \]
Where \( P_1 \) is the population after 1 year.

Substituting the values:
\[ P_1 = x \times (1 + 0.16) = x \times 1.16 \]

### Step 2: Write the equation for the population after 2 years
The population after 2 years can be calculated similarly:
\[ P = P_1 \times (1 + r) \]
Substituting \( P_1 \):
\[ P = (x \times 1.16) \times (1 + 0.16) = x \times 1.16 \times 1.16 = x \times (1.16)^2 \]

Calculating \( (1.16)^2 \):
\[ (1.16)^2 = 1.3456 \]

Thus, we have:
\[ P = x \times 1.3456 \]

### Step 3: Set up the equation with the current population
Now, we know the present population \( P \) is 403680:
\[ 403680 = x \times 1.3456 \]

### Step 4: Solve for x
To find \( x \), we rearrange the equation:
\[ x = \frac{403680}{1.3456} \]

Calculating this:
\[ x = \frac{403680}{1.3456} \approx 299,000 \]

### Conclusion
The population of the village 2 years ago was approximately **299,000**.

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