SSC CHSL 11 Aug 2021 Shift 2 Paper

Given, Distance covered by Ajit Singh from $9:30$ am to 1: $30\mathit{p}\mathit{m}$.
(in $4\mathit{h})=44\times 4=176\mathit{k}\mathit{m}$
$\therefore$ Remaining distance
$=416-176=240\mathit{k}\mathit{m}$
Now, David Raj left for $\mathit{P}$.
$\therefore$ They will meet after $\frac{240}{\left(52+44\right)}\mathit{h}$
$=\frac{240}{96}=2\frac{1}{2}\mathit{h}$
$\therefore$ Time when they meet each other
$=1:30+2:30=4\mathit{p}\mathit{m}$