(a)
Let numbers of boys are $x$ and numbers of girls are$y$.
$\therefore$ $53(x + y) = 55y + 50x$
$\Rightarrow$ $3x = 2y$
$\Rightarrow$ $x = \frac{2y}{3}$
$\therefore$ total number of students $= x + y = \frac{2y}{3} + y = \frac{5}{3}\ y$
Hence, required percentage
$= \frac{y}{\frac{5y}{3}} \times 100\% = \frac{3}{5} \times 100\% = 60\%$