# What percentage of population in 2001 were males given the following conditions?

In a town the population of males decreased by $25$% from 2001 to 2002.The population of females increased by $20$% in this period.If females formed $44$ . $4$/$9$ % of the population in 2002,what percentage of the population in 2001 were males?

MyApproach

From $2001$ to $2002$, population of males decreases from $1$/$4$ to $3$/$4$.

and From $2001$ to $2002$, population of females increases from $1$/$5$ to $6$/$5$.

In 2002, $4$/$9$ . x=females population.($44$ . $4$/$9$ %)=$400$/$900$ I could solve this problem till here.

In 2001 we have $m + f$ many males and females.

In 2002 we have $\frac{3}{4}m + \frac{6}{5}f$ as the population.

We know that the percentage of women in 2002 can be written as $100\frac{\frac{6}{5}f}{\frac{3}{4}m + \frac{6}{5}f}$, which should equal $44 \frac{4}{9}$.

The question is now, what is $100 \frac{m}{m+f}$ ?

The easiest is to compute $z = \frac{f}{m}$. The percentage condition yields an equation in this that should be solvable, and then the asked for percentage can also be expressed in it.

• What do you mean by f/$3$/$4$ . m +$6$/$5$ .f ?
– Jack
Commented Nov 6, 2015 at 13:53
• The males have decreased etc. So I multiply them with the right fraction. Commented Nov 6, 2015 at 13:55