A person walks 10 m in front and 10 m to the right. Then every time turning to…

Direction-and-distance questions are solved by converting each leg of the walk into a displacement vector on a compass grid (North = +y, East = +x, South = −y, West = −x), tracking the walker's facing direction through every turn, and then finding the net displacement's magnitude using the Pythagorean theorem: distance = √(Δx2 + Δy2).

1. Fix the axes and the starting facing: let North = +y and East = +x. The person starts at the origin (0, 0) facing North (“in front”).

2. First leg — walks 10 m in front (North): position becomes (0, 10).

3. Second leg — walks 10 m “to the right”: he turns right (North → East) and walks 10 m, reaching (10, 10); he is now facing East.

4. Third leg — turning left (East → North) and walking 5 m: position becomes (10, 15).

5. Fourth leg — turning left again (North → West) and walking 15 m: position becomes (10 − 15, 15) = (−5, 15).

6. Fifth leg — turning left again (West → South) and walking 15 m: position becomes (−5, 15 − 15) = (−5, 0).

7. The final position (−5, 0) is measured against the starting point (0, 0): Δx = −5, Δy = 0.

Distance = √((−5)2 + 02) = √25 = 5 m.

Cross-check by summing the North–South and East–West legs independently, instead of tracing the path point by point:

• North–South legs: +10 m (leg 1) + 5 m (leg 3) − 15 m (leg 5) = 0 m net.

• East–West legs: +10 m (leg 2) − 15 m (leg 4) = −5 m net.

• Resultant magnitude: √(02 + (−5)2) = 5 m, matching the step-by-step trace and confirming the result.

So the person is 5 m from his starting point.