Final Round of National Olympiad

There are $\frac{3}{5} \cdot \frac{1}{3} = \frac{1}{5}$ of pupils who both do sports and go to math club, whereas $\frac{1}{2} \cdot \frac{2}{15} = \frac{1}{15}$ of pupils are boys who both do sports and go to math club. Thus $\frac{1}{5} - \frac{1}{15} = \frac{2}{15}$ of pupils are girls who both do sports and go to math club. Girls who do sports constitute $\frac{3}{5}(1 - \frac{5}{9}) = \frac{4}{15}$ of all pupils. Hence girls who do sports but do not go to math club constitute $\frac{4}{15} - \frac{2}{15} = \frac{2}{15}$ of all pupils. Girls who neither do sports nor go to math club constitute $\frac{1}{2} \cdot \frac{2}{15} = \frac{1}{15}$ of all pupils. Thus girls not going to math club constitute $\frac{2}{15} + \frac{1}{15} = \frac{1}{5}$ of all pupils. Other girls who constitute $\frac{1}{2} - \frac{1}{5} = \frac{3}{10}$ of all pupils go to math club. They constitute $\frac{3}{10} : \frac{1}{2} = \frac{3}{5} = 60\%$ of all girls.