UPPSC PCS - 2016
Krishna walks 1 km East of his house then 1 km North, again 2 km East, 3 km South and 3 km West. To reach his house he must walk
Year Asked: 2016
Correct Answer
D. 2 km North
Detailed Explanation
Track Krishna's position with coordinates, taking the house as origin (0, 0), East as +x and North as +y. Step 1: 1 km East gives (1, 0). Step 2: 1 km North gives (1, 1). Step 3: 2 km East gives (3, 1). Step 4: 3 km South gives (3, -2). Step 5: 3 km West gives (0, -2). So he is 2 km directly South of the house.
To get back, he must walk 2 km North, which is option D.
- A (1 km North): This would leave him 1 km short. It usually comes from misreading the final 3 km South leg.
- B (1 km South): Wrong direction and wrong distance. It would only work if the net southward displacement were 1 km, and it is not.
- C (2 km South): Right distance, wrong direction. Walking 2 km South takes him to (0, -4), further from home.
The core concept is net displacement. Only the final position relative to the start matters, not the total path length, and the return journey is the reverse of that net displacement. The same skill of tracking net position applies in map-reading and direction-sense questions across GS and reasoning papers.
Trap: Adding up the total distance walked (1+1+2+3+3 = 10 km) instead of tracking net displacement, and then choosing an option based on that sum.
This question appeared in UPPCS 2016 GS Paper 2.
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