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Anurag starts from Point Y and drives 11 km towards North. He then takes a right turn, drives 25 km, turns right and drives 27 km. He then takes a right turn and drives 11 km. He takes a right turn, drives 41 km. He then turns left, drives 14 km and stops at Point Z. How far (shortest distance) and towards which direction should he drive in order to reach Point Y again? (All turns are 90 degrees turns only unless specified)
36 km towards North
14 km towards North
30 km towards South
25 km towards South
25 km towards South
Track net displacement on North-South and East-West axes; calculate final position relative to start point Y (0,0).
Track net displacement on North-South and East-West axes; calculate final position relative to start point Y (0,0).
Mapping the path on Cartesian plane (Y as origin 0,0)
Start at (0,0). 1. North 11 km: (0, 11). 2. Right (East) 25 km: (25, 11). 3. Right (South) 27 km: (25, 11-27) = (25, -16). 4. Right (West) 11 km: (25-11, -16) = (14, -16). 5. Right (North) 41 km: (14, -16+41) = (14, 25). 6. Left (West) 14 km: (14-14, 25) = (0, 25). Point Z is at (0, 25).
Determining distance and direction to Y
Point Z is at (0, 25) and Point Y is at (0, 0). To move from (0, 25) back to (0, 0), one must travel 25 km in the negative Y direction, which corresponds to the South.
A: 36 km North is incorrect as it increases the distance from Y. B: 14 km North is incorrect as it doesn't account for the vertical coordinate change. C: 30 km South is mathematically inconsistent with the net displacement calculation.
D is correct because Point Z is at 25 km North of Point Y, so moving 25 km South returns Anurag to Point Y.
For long direction problems, always use the +X, -X, +Y, -Y method to ensure you don't lose track of turns and distances.