Faizal is standing at point A facing North. He walks 15 meters to his left and…

Concept: In a direction-sense problem, track movement along two perpendicular axes (East-West and North-South). Movements along the same axis in opposite directions must be subtracted, not added — the net displacement is the difference between the two distances, and the side a person ends up on is decided by whichever distance is larger.

Applying it here:

1. Faizal starts at point A facing North.

2. Turning to his left while facing North points him West, so the first 15 meter walk takes him 15 meters West of A.

3. The about-turn (a 180° turn) changes his facing from West to East.

4. Walking 30 meters East from a point 15 meters West of A is a movement along the same East-West line but in the opposite direction to the first leg, so the net displacement is 30 − 15 = 15 meters, on the East side of A, since the second, longer walk overtakes the first.

Cross-check: Taking West as the negative direction and East as the positive direction on a number line with A at 0: the first move is −15, the second move is +30, giving a net position of −15 + 30 = +15, i.e., 15 units in the positive (East) direction — consistent with the step-by-step result above.

Faizal ends up 15 meters to the East of the starting point A.