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Akshita starts from Point A and drives 22 km towards North. He then takes a left turn, drives 17 km, turns left and drives 27 km. He then takes a left turn and drives 21 km. He takes a final left turn, drives 5 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-degree turns only, unless specified.)
5 km to the west
5 km to the east
4 km to the east
4 km to the west
4 km to the west
Track net movement by assigning signs: North/South as Y-axis (+/-) and East/West as X-axis (+/-). Net displacement from A is the final coordinate (x,y), then inverse it to find the return direction.
Track net movement by assigning signs: North/South as Y-axis (+/-) and East/West as X-axis (+/-). Net displacement from A is the final coordinate (x,y), then inverse it to find the return direction.
Mapping Movements (N/S, E/W)
Starting at A(0,0): 1) North 22 (+22Y), 2) Left (West)₁₇ (-17X), 3) Left (South)₂₇ (-27Y), 4) Left (East)₂₁ (+21X), 5) Left (North)₅ (+5Y).
Calculate Net Displacement
Net X: -17 + 21 = +4 (East). Net Y: 22 - 27 + 5 = 0. Final relative position: 4 km East from A.
Determine Return Path
Since Akshita is at 4 km East, he must travel 4 km to the West to return to Point A.
A: Incorrect, suggests 5km West which ignores the net horizontal calculation. B: Incorrect, 5km East moves further away from A. C: Incorrect, 4km East moves further away from the origin.
D is correct because the net displacement is 4 km East, requiring a movement of 4 km West to return to the starting point.
In multi-turn problems, always resolve N-S and E-W separately before combining them to prevent calculation errors.