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Gaurav starts from Point A and drives 22 km towards the north. He then takes a right turn, drives 11 km, turns right and drives 25 km. He then takes a right turn and drives 17 km. He takes a final right turn, drives 3 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
6 km to the east
6 km to the west
5 km to the east Question ID : 4410091567496 Option 1 ID : 4410096191299 Option 2 ID : 4410096191296 Option 3 ID : 4410096191297 Option 4 ID : 4410096191298 Status : Answered Chosen Option : B
6 km to the east
Track the net displacement by assigning signs: North/East as positive and South/West as negative. Sum the movements on the X-axis and Y-axis separately.
Track the net displacement by assigning signs: North/East as positive and South/West as negative. Sum the movements on the X-axis and Y-axis separately.
Define Cartesian movements from Point A
Let Point A be (0, 0). 1: North 22 km = (0, +22). 2: Right turn (East) 11 km = (+11, 0). 3: Right turn (South) 25 km = (0, -25). 4: Right turn (West) 17 km = (-17, 0). 5: Right turn (North) 3 km = (0, +3).
Calculate net displacement
Net X = 11 - 17 = -6 km (6 km West). Net Y = 22 - 25 + 3 = 0 km. The final position relative to A is 6 km West. To return to A, Gaurav must move in the opposite direction: 6 km East.
A: 5 km to the west is incorrect as it fails to balance the X-axis displacement. C: 6 km to the west is the current position relative to A, not the required travel direction. D: 5 km to the east is incorrect as it does not compensate for the -6 km net X-displacement.
B is correct because the final position is 6 km to the west of Point A, meaning one must travel 6 km to the east to return to the starting point.
In direction problems, always treat the path as a series of vectors to avoid confusion with complex turns; summing coordinates is faster and more accurate than drawing diagrams.