Question

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 msles in town was.

Answer 1

4000 males in that town

Answer 2

Answer:

4000

Step-by-step explanation:

Total population = 8000

we are given that the total population increased by 9%

New population = $8000+\frac{9}{100} \times 8000=8720$

let no. of males be x

So, No. of females = 8000-x

Male population increased by 10%

New Male population = $x+\frac{10}{100}x=1.1x$

Female population increased by 8%

New Female population = $(8000-x)+\frac{8}{100}(8000-x)=8640-1.08x$

New Population = New Males + new females = 8640-1.08x +1.1x=8640+0.02x

$8640+0.02x=8720$

$0.02x=8720-8640$

$x=\frac{8720-8640}{0.02}$

$x=4000$

Hence the number of males in town was 4000