The population of a town increases by 5% annually. If the population in 2009 is 1,38,915 what was it in 2006

Answer

To find the population of the town in 2006 given that it increases by 5% annually and the population in 2009 is 138,915, we can use the formula for compound interest. Here’s a step-by-step solution:

### Step 1: Identify the Variables
Let:
- \( P \) = Population in 2006 (the value we need to find)
- \( A \) = Population in 2009 = 138,915
- \( R \) = Rate of increase = 5%
- \( N \) = Number of years from 2006 to 2009 = 3 years

### Step 2: Use the Compound Interest Formula
The formula for compound interest is given by:
\[ A = P \left(1 + \frac{R}{100}\right)^N \]
Substituting the known values into the formula:
\[ 138915 = P \left(1 + \frac{5}{100}\right)^3 \]

### Step 3: Simplify the Expression
Calculate \( 1 + \frac{5}{100} \):
\[ 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \]
Now, raise it to the power of 3:
\[ 1.05^3 = 1.157625 \]
So, the equation becomes:
\[ 138915 = P \times 1.157625 \]

### Step 4: Solve for \( P \)
To find \( P \), rearrange the equation:
\[ P = \frac{138915}{1.157625} \]

### Step 5: Calculate \( P \)
Now, perform the division:
\[ P \approx \frac{138915}{1.157625} \approx 120,000 \]

### Conclusion
The population of the town in 2006 was approximately **120,000**.

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