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Pranay starts from Point A and drives 45 km towards North. He then takes a right turn, drives 35 km, turns right and drives 75 km. He then takes a right turn and drives 25 km. He takes a final right turn, drives 30 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 otherwise specified.)
10 km towards North
10 km towards South
10 km towards East
10 km towards West
10 km towards West
Track net displacement on X (East-West) and Y (North-South) axes separately. Total Y = (+45 + 75 - 30) = +90 (North), Total X = (+35 - 25) = +10 (East).
Track net displacement on X (East-West) and Y (North-South) axes separately. Total Y = (+45 + 75 - 30) = +90 (North), Total X = (+35 - 25) = +10 (East).
Establish Coordinates
Starting at Point A (0,0). North is +Y, South is -Y, East is +X, West is -X.
Calculate Sequential Moves
1. 45km North: (0, +45). 2. Turn Right (East) 35km: (+35, +45). 3. Turn Right (South) 75km: (+35, 45-75 = -30). 4. Turn Right (West) 25km: (35-25 = +10, -30). 5. Turn Right (North) 30km: (+10, -30+30 = 0). Final position P is at (+10, 0).
Calculate Return Vector
Point P is at (+10, 0) relative to A (0,0). To reach A from P, move -10 units on X-axis (West) and 0 on Y-axis. Therefore, 10 km towards the West.
A: North is incorrect, as the horizontal displacement must be nullified. B: South is incorrect, as vertical displacement is already net zero. C: East would move further away from A.
D is correct because point P is 10 km to the East of point A, meaning Pranay must drive 10 km towards the West to return to point A.
In long sequence puzzles, always maintain a running (X, Y) tally to avoid missing turns or miscounting net distances.
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