The male population of a city has increased by 20% from the previous year, while the female population has increased by 42.5%. If the total population of the city has increased by 30%, then the female population of the city was what percentage of male population in the previous year?

Let's assume the total population of a city in previous year is 100y.

Let's assume the male and female population of a city in previous year is 'M' and 'F' respectively.

M+F = 100y

M = (100y-F)    Eq.(i)

The male population of a city has increased by 20% from the previous year, while the female population has increased by 42.5%. If the total population of the city has increased by 30%.

M of (100+20)% + F of (100+42.5)% = 100y of (100+30)%

$$M\ of120\%+F\ of\ 142.5\%=100y\ of\ 130\%$$

$$M\times\frac{120}{100}+F\times\frac{142.5}{100}=100y\times\frac{130}{100}$$

$$M\times120+F\times142.5=100y\times130$$

$$1.2M+1.425F=130y$$
Put Eq.(i) in the above equation.

$$1.2(100y-F)+1.425F=130y$$

$$120y-1.2F+1.425F=130y$$

$$-1.2F+1.425F=130y-120y$$

$$0.225F=10y$$

$$F=\frac{10y}{0.225}$$

$$F=\frac{10000y}{225}$$

$$F=\frac{400y}{9}$$    Eq.(ii)

Put Eq.(ii) in Eq.(i).

$$M = (100y-\frac{400y}{9})$$

$$M = (\frac{900y-400y}{9})$$

$$M = (\frac{500y}{9})$$    Eq.(iii)

Female population of the city was what percentage of male population in the previous year = $$\frac{Eq.(ii)}{Eq.(iii)}\times\ 100$$

= $$\frac{\frac{400y}{9}}{\frac{500y}{9}}\times\ 100$$

= $$\frac{400y}{500y}\times\ 100$$

= 80%