Solution: Let $n_m$, $n_f$ denote the number of male and female employees in the firm respectively. Further assume that $$\begin{array}{rl}{x}_m& =\text{ the average salary of male employees }\\ x_f& =\text{ the average salary of female employees }\\ x& =\text{ average salary of all the employees }.\end{array}$$ According to the problem, $$\begin{array}{r}{x}_m=520x_w=420\text{ and }x=500.\end{array}$$ We have $$\begin{array}{rl}x=\frac{n_m{x}_m+n_f{x}_f}{n_m+n_f}& ⟹500=\frac{520n_m+420n_f}{n_m+n_f}\\ & ⟹500(n_m+n_f)=520n_m+420n_f\\ & ⟹20n_m-80n_f=0\\ & ⟹\frac{n_m}{n_f}=\frac{4}{1}.\end{array}$$ Therefore, the percentage of male employees in the firm will be $$\frac{4}{4+1}\times 100=80\mathrm{\%}$$ and the percentage of male employees in the firm will be $$\frac{1}{4+1}\times 100=20\mathrm{\%}.$$