It is observed that the population of a city increases at the rate of 8% per annum. If the present population of the city is 125000, then the population of the city after 2 years will be

Options

A. `145000`
B. `145800`
C. `154000`
D. `154800`

Answer

To find the population of the city after 2 years with an annual increase of 8%, we can use the formula for compound interest, which is also applicable for population growth:

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

Where:
- \( A \) = the amount of population after \( n \) years
- \( P \) = the present population (initial amount)
- \( r \) = rate of increase (as a decimal)
- \( n \) = number of years

### Step 1: Identify the values
- Present population \( P = 125000 \)
- Rate of increase \( r = 8\% = \frac{8}{100} = 0.08 \)
- Number of years \( n = 2 \)

### Step 2: Substitute the values into the formula
Now we substitute the values into the formula:

\[ A = 125000(1 + 0.08)^2 \]

### Step 3: Calculate \( (1 + 0.08) \)
First, calculate \( (1 + 0.08) \):

\[ 1 + 0.08 = 1.08 \]

### Step 4: Raise \( 1.08 \) to the power of \( 2 \)
Now calculate \( (1.08)^2 \):

\[ (1.08)^2 = 1.08 \times 1.08 = 1.1664 \]

### Step 5: Multiply by the present population
Now multiply this result by the present population:

\[ A = 125000 \times 1.1664 \]

### Step 6: Calculate the final population
Now perform the multiplication:

\[ A = 125000 \times 1.1664 = 145800 \]

### Final Answer
The population of the city after 2 years will be approximately **145800**.

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