Ans: $416$
Let $x$ and $y$ be the numbers of male and female applicants respectively.
Let $x$ and $y$ be the numbers of male and female applicants respectively.
$\left\{\begin{array}{ll}
x = y(1+28\%) & \ldots \unicode{x2460}\\
x – y = 91 & \ldots \unicode{x2461}
\end{array}\right.$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
y(1+28\%) – y & = & 91 \\
0.28 y & = & 91 \\
y & = & 325
\end{array}$
Sub. $y = 325$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
x & = & 325(1 + 28\%) \\
& = & 416
\end{array}$
Therefore, the number of male applicants in the recruitment exercise is $416$.
