SSC CHSL 10 Jan 2017 Shift 1 Paper

Let the boy starts from point A and travels 4 km towards south to reach B and then takes a left turn and travels another 3 km east wards to finally reach point C.
Thus, ${\left(\mathit{A}\mathit{C}\right)}^2={\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{B}\mathit{C}\right)}^2$
⇒ ${\left(\mathit{A}\mathit{C}\right)}^2={\left(4\right)}^2+{\left(3\right)}^2$
⇒ ${\left(\mathit{A}\mathit{C}\right)}^2=16+9=25$
⇒ $\mathit{A}\mathit{C}=\sqrt{25}=5$ km
∴ He is 5 km in the south-east direction from his original position.