2020-I-05

Ans: $416$
Let $x$ and $y$ be the numbers of male and female applicants respectively.

$\{\begin{array}{ll}x=y(1+28\mathrm{\%})& \dots \text{①}\\ x–y=91& \dots \text{②}\end{array}$

Sub. $\text{①}$ into $\text{②}$, we have

$\begin{array}{rcl}y(1+28\mathrm{\%})–y& =& 91\\ 0.28y& =& 91\\ y& =& 325\end{array}$

Sub. $y=325$ into $\text{①}$, we have

$\begin{array}{rcl}x& =& 325(1+28\mathrm{\%})\\ & =& 416\end{array}$

Therefore, the number of male applicants in the recruitment exercise is $416$.