The annual increase in the population of a town is 10%. If the present population of the town is 180000, then what will be its population after two years ?

Answer

To find the population of the town after two years with an annual increase of 10%, we can use the formula for compound interest, which is also applicable for population growth. The formula is:

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

Where:
- \( A \) = the amount (or population after N years)
- \( P \) = the present population
- \( R \) = the rate of increase (in percentage)
- \( N \) = the number of years

### Step-by-step Solution:

1. **Identify the given values:**
- Present population \( P = 180,000 \)
- Rate of increase \( R = 10\% \)
- Time \( N = 2 \) years

2. **Convert the rate of increase into decimal form:**
\[ R = \frac{10}{100} = 0.10 \]

3. **Substitute the values into the formula:**
\[ A = 180,000 \left(1 + 0.10\right)^2 \]

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

5. **Raise \( 1.10 \) to the power of 2:**
\[ (1.10)^2 = 1.21 \]

6. **Multiply the present population by \( 1.21 \):**
\[ A = 180,000 \times 1.21 \]

7. **Calculate \( 180,000 \times 1.21 \):**
\[ A = 217,800 \]

8. **Final Answer:**
The population after 2 years will be **217,800**.