SSC CGL 18 Aug 2021 Shift 3 Paper

We know, Speed =
According to the question,
$\Rightarrow \frac{\mathit{x}}{72}-\frac{\mathit{x}}{90}=\frac{15}{60}-\frac{5}{60}$
$\Rightarrow \frac{5\mathit{x}-4\mathit{x}}{360}=\frac{10}{60}$
$\Rightarrow \frac{\mathit{x}}{360}=\frac{1}{6}$
$\Rightarrow \mathit{x}=60\mathit{K}\mathit{m}$
Now, at the speed of $72\mathit{k}\mathit{m}∕\mathit{h}$.
$\Rightarrow \frac{60}{72}\times 60=50\mathit{m}\mathit{i}\mathit{n}$
Now, we know that when the train moves $72\mathit{k}\mathit{m}∕\mathit{h}$ then, the train will be late $=15\mathit{m}\mathit{i}\mathit{n}$
So, total time required to complete the journey $=50\mathit{m}\mathit{i}\mathit{n}-15\mathit{m}\mathit{i}\mathit{n}=35\mathit{m}\mathit{i}\mathit{n}$.