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Anil starts from point A and drives 6 km towards the east. He then takes a left turn, drives 2 km, turns right, and drives 3 km. He then takes a right turn and drives 5 km. He takes a final right turn, drives 9 km, and stops at point P. How far (shortest distance) and towards which direction should he drive in order to reach point A again? (All turns are 90┬░ turns only, unless specified.)
3 km towards the west
6 km towards the east
6 km towards the south
3 km towards the north
3 km towards the north
Represent movements on a 2D Cartesian plane using +X (East), -X (West), +Y (North), -Y (South) to calculate net displacement at the end.
Represent movements on a 2D Cartesian plane using +X (East), -X (West), +Y (North), -Y (South) to calculate net displacement at the end.
Mapping the path
Start at (0,0). 1: East 6km (+6, 0). 2: Left turn (North) 2km (+6, +2). 3: Right turn (East) 3km (+9, +2). 4: Right turn (South) 5km (+9, -3). 5: Right turn (West) 9km (0, -3).
Calculating displacement
Final position is (0, -3) relative to point A (0,0). To reach A (0,0) from (0, -3), one must travel 3 units in the positive Y direction (North).
A: 3km West is incorrect as the current x-coordinate is already 0. B: 6km East would put the person at x=6, away from the origin. C: 6km South would increase the gap from the origin (0,0) to (0, -9).
D is correct because the net displacement from point A is 3 km south, meaning he must travel 3 km north to return to the starting point.
Always treat North/East as positive and South/West as negative; summing the coordinates is much faster than drawing a complex diagram.