SSC CGL 18 Aug 2021 Shift 2 Paper

Let usual speed is $\mathit{S}\mathit{k}\mathit{m}∕\mathit{h}$.
We know that,

According to the question,
$\Rightarrow \frac{150}{\mathit{S}-10}-\frac{150}{\mathit{S}}=\frac{1}{2}$
[Time $=30\mathit{m}\mathit{i}\mathit{n}=\frac{1}{2}\mathit{h}$ ]
$\Rightarrow \frac{150\mathit{S}-150\left(\mathit{S}-10\right)}{\left(\mathit{S}-10\right)\mathit{S}}=\frac{1}{2}$
$\Rightarrow \left(150\mathit{S}-150\mathit{S}+1500\right)2$
$=\left({\mathit{S}}^2-10\mathit{S}\right)$
$\Rightarrow {\mathit{S}}^2-10\mathit{S}-3000=0$
$\Rightarrow {\mathit{S}}^2-60\mathit{S}+50\mathit{S}-3000=0$
$\Rightarrow \mathit{S}\left(\mathit{S}-60\right)+50\left(\mathit{S}-60\right)=0$
$\Rightarrow \left(\mathit{S}+50\right)\left(\mathit{S}-60\right)=0$
$\mathit{S}=-50,\mathit{S}=60$
Speed can't be negative.
So, speed $=60\mathit{k}\mathit{m}∕\mathit{h}$
Now, if distance $=90\mathit{k}\mathit{m}$,
then drive he $15\mathit{k}\mathit{m}∕\mathit{h}$ slower $=60-15$

$\therefore$ Time $=\frac{90}{45}=2\mathit{h}$