In a town,the population was 8000.In one year,male population increased by 10% and female population increased by 8% but the total population increased by 9% .The number of males in the town was:

Options

A. 4000
B. 4500
C. 5000
D. 6000

Answer

To solve the problem step by step, we can follow this approach:

1. **Identify the initial population**:
The total population of the town is given as 8000.

2. **Determine the percentage increase in population**:
- The male population increases by 10%.
- The female population increases by 8%.
- The total population increases by 9%.

3. **Set up the equations**:
Let the initial male population be \( M \) and the initial female population be \( F \). Thus, we have:
\[ M + F = 8000 \]

4. **Calculate the new populations after the increase**:
- The new male population after a 10% increase is \( M + 0.10M = 1.10M \).
- The new female population after an 8% increase is \( F + 0.08F = 1.08F \).

5. **Set up the equation for the total population after the increase**:
The total population after the increase can be expressed as:
\[ 1.10M + 1.08F = 8000 + 0.09 \times 8000 = 8000 + 720 = 8720 \]

6. **Now we have two equations**:
- \( M + F = 8000 \) (1)
- \( 1.10M + 1.08F = 8720 \) (2)

7. **Substitute \( F \) from equation (1) into equation (2)**:
From equation (1), we can express \( F \) as:
\[ F = 8000 - M \]
Substitute this into equation (2):
\[ 1.10M + 1.08(8000 - M) = 8720 \]

8. **Simplify the equation**:
\[ 1.10M + 8640 - 1.08M = 8720 \]
Combine like terms:
\[ 0.02M + 8640 = 8720 \]

9. **Isolate \( M \)**:
\[ 0.02M = 8720 - 8640 \]
\[ 0.02M = 80 \]
\[ M = \frac{80}{0.02} = 4000 \]

10. **Conclusion**:
The number of males in the town is \( 4000 \).