A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on…

$\mathrm{\Delta }QPA :\frac{h}{x+y}=\tan {30}^{°}$

$\Rightarrow \sqrt{3}h=x+y$ ......$\left(\mathrm{i}\right)$

$\mathrm{\Delta }QPB :\frac{h}{y}=\tan {60}^{°}$

$\Rightarrow h= \sqrt{3}y$   ......$\left(\mathrm{ii}\right)$

From equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right),$ we get

$3y=x+y$

$\Rightarrow y=\frac{x}{2}$

$\because$ Speed is uniform

Therefore, distance $x$ walked by him in $10 \min .$

and distance $\frac{x}{2}$ walked by him in $5 \min .$