Suresh is facing towards west and starts walking until he reaches a point 30 m…

Concept: In a direction-sense problem, track the walk on a coordinate grid — every East/West leg changes the horizontal (x) position and every North/South leg changes the vertical (y) position. After all the legs and turns are accounted for, add up the horizontal shifts and the vertical shifts separately to get the net displacement (Δx, Δy) from the start. The straight-line distance back to the start then follows the Pythagorean rule, distance = √(Δx2 + Δy2), with the sign of Δx and Δy telling you the direction.

Application: Take the starting point A as (0, 0), with Suresh initially facing West. Trace each leg of the walk in order, applying each turn instruction (right = clockwise 90°, left / anticlockwise = counter-clockwise 90°) to the current facing direction before moving:

1. Facing West, walks 30 m to reach B: position (-30, 0).

2. Turns right (West → North), walks 20 m to reach C: position (-30, 20).

3. Turns left (North → West), walks 10 m to reach D: position (-40, 20).

4. Turns 90° anticlockwise (West → South), walks 40 m to reach E: position (-40, -20).

5. Turns left (South → East), walks 30 m to reach F: position (-10, -20).

So the net displacement from A is Δx = -10 (10 m West) and Δy = -20 (20 m South). The distance is √(102 + 202) = √(100 + 400) = √500 = 10√5 m, and since F is west and south of A, the direction is South-West.

Cross-check: add the East/West legs and the North/South legs separately instead of tracking coordinates. Westward legs: 30 m (leg 1) + 10 m (leg 3) = 40 m West; eastward legs: 30 m (leg 5); net horizontal = 40 - 30 = 10 m West. Southward legs: 40 m (leg 4); northward legs: 20 m (leg 2); net vertical = 40 - 20 = 20 m South. This matches the coordinate trace exactly, confirming the result.

Result: Suresh finishes 10√5 m from the starting point, in the South-West direction.