The difference between the population of a city 1 year and 2 year ago is 5000. If increment in population every year is `10%` then find the present population of city .

Options

A. 66000
B. 65550
C. 56660
D. 60500

Answer

To solve the problem step by step, we need to find the present population of the city given the difference in population over the past two years and the annual growth rate.

### Step 1: Define Variables
Let:
- \( P_2 \) = Population of the city 2 years ago
- \( P_1 \) = Population of the city 1 year ago
- \( P \) = Present population of the city

### Step 2: Establish the Relationship
According to the problem, the difference between the population of the city 1 year ago and 2 years ago is 5000. Therefore, we can write:
\[ P_1 - P_2 = 5000 \]

### Step 3: Express Population in Terms of Growth Rate
Since the population increases by 10% each year, we can express the populations as:
\[ P_1 = P_2 \times (1 + 0.10) = P_2 \times 1.10 \]

### Step 4: Substitute and Solve for \( P_2 \)
Now, we substitute \( P_1 \) in the difference equation:
\[ P_2 \times 1.10 - P_2 = 5000 \]
\[ 0.10 P_2 = 5000 \]
\[ P_2 = \frac{5000}{0.10} = 50000 \]

### Step 5: Calculate Present Population \( P \)
Now that we have \( P_2 \), we can find the present population \( P \):
\[ P = P_1 = P_2 \times 1.10 = 50000 \times 1.10 = 55000 \]

### Step 6: Final Calculation for Present Population
To find the present population, we can also calculate it directly from \( P_2 \):
\[ P = P_2 \times (1 + 0.10)^2 = 50000 \times (1.10)^2 \]
Calculating \( (1.10)^2 \):
\[ (1.10)^2 = 1.21 \]
Thus,
\[ P = 50000 \times 1.21 = 60500 \]

### Final Answer
The present population of the city is **60500**.

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