SSC CHSL 10 Jan 2017 Shift 2 Paper

A travels 12 km towards north and then takes a left turn and covers another 5 km towards west direction. From there, he turns 180° anticlockwise, i.e. will revert his direction and travels 10 km further towards east.
Thus, the minimum distance between his initial and final position = $\sqrt{{\left(12\right)}^2+{\left(5\right)}^2}$
= $\sqrt{144+25}=\sqrt{169}=13$km