SSC CGL 23 Aug 2021 Shift 1 Paper

Let distance is a.
According to the question,
Speed of boat is $\mathit{x}$ and stream $=\mathit{y}$
Speed
So, $\frac{\mathit{a}}{\mathit{x}+\mathit{y}}+\frac{\mathit{a}}{\mathit{x}-\mathit{y}}=\mathit{T}$ . . . (i)
If speed is 2 times, time will be $\frac{80}{161}\mathit{T}$
$\frac{\mathit{a}}{2\mathit{x}+\mathit{y}}+\frac{\mathit{a}}{2\mathit{x}-\mathit{y}}=\frac{80}{161}\mathit{T}$ . . . (ii)
From Eqs. (i) and (ii),
$\Rightarrow \frac{\frac{\mathit{a}}{\left(\mathit{x}+\mathit{y}\right)}+\frac{\mathit{a}}{\left(\mathit{x}-\mathit{y}\right)}}{\frac{\mathit{a}}{\left(2\mathit{x}+\mathit{y}\right)}+\frac{\mathit{a}}{\left(2\mathit{x}-\mathit{y}\right)}}=\frac{\mathit{T}}{\frac{80}{161}\mathit{T}}$
$\Rightarrow \frac{\frac{\left(\mathit{x}-\mathit{y}+\mathit{x}+\mathit{y}\right)}{\left(\mathit{x}+\mathit{y}\right)\left(\mathit{x}-\mathit{y}\right)}}{\frac{\left(2\mathit{x}-\mathit{y}+2\mathit{x}+\mathit{y}\right)}{\left(2\mathit{x}+\mathit{y}\right)\left(2\mathit{x}-\mathit{y}\right)}}=\frac{161}{80}$
$\Rightarrow \frac{2\mathit{x}}{\left(\mathit{x}+\mathit{y}\right)\left(\mathit{x}-\mathit{y}\right)}\times \frac{\left(2\mathit{x}+\mathit{y}\right)\left(2\mathit{x}-\mathit{y}\right)}{4\mathit{x}}$ $=\frac{161}{80}$
$\Rightarrow 40\left(4{\mathit{x}}^2-{\mathit{y}}^2\right)=161\left({\mathit{x}}^2-{\mathit{y}}^2\right)$
$\Rightarrow 160{\mathit{x}}^2-40{\mathit{y}}^2=161{\mathit{x}}^2-161{\mathit{y}}^2$
$\Rightarrow {\mathit{x}}^2=121{\mathit{y}}^2$
$\frac{{\mathit{x}}^2}{{\mathit{y}}^2}=\frac{121}{1}$
$\Rightarrow \frac{\mathit{x}}{\mathit{y}}=\sqrt{\frac{121}{1}}=\frac{11}{1}$
The ratio of speed in still water and speed of river
$\Rightarrow \mathit{x}:\mathit{y}=11:1$