$$\frac{5}{9}$$ part of population in a village are males. If 30% of males are married, percentage of unmarried females in total population is :

Let total population = x
Then, number of males = $$\frac{5}{9}$$x
Married males :
= 30% of $$\frac{5}{9}$$x
= $$\frac{30}{100}$$ × $$\frac{5}{9}$$x
= $$\frac{x}{6}$$
Married females = $$\frac{x}{6}$$;
Number of females :
= x - $$\frac{5}{9}$$x
= $$\frac{4x}{9}$$
Unmarried females :
= $$\frac{4x}{9}$$ - $$\frac{x}{6}$$
= $$\frac{5x}{18}$$
∴ Required percentage :
= $$\frac{5x}{18}$$ × $$\frac{1}{x}$$ × 100%
= $$27\frac{7}{9}$$%