A person walks 12 km due north, then 15 km due east, after that 19 km due west and then 15 km due south. How far is he from the starting point ?
- (a) 5 km ✓ UPSC's answer
- (b) 9 km
- (c) 37 km
- (d) 61 km
Why the answer is (a)
• North–south: 12 km north, then 15 km south, leaves him 3 km south of the start.
• East–west: 15 km east, then 19 km west, leaves him 4 km west of the start.
• Straight-line distance: √(3² + 4²) = √25 = 5 km.
• Hence 5 km, option (a).
Why the other options are wrong
- (b) 9 km
- 9 km is not the distance because the net displacement is a 3 km by 4 km right triangle, not a 9 km straight-line distance.
- (c) 37 km
- 37 km is the sum of the four walked distances (12 + 15 + 19 + 15), not the straight-line distance from the starting point.
- (d) 61 km
- 61 km is not obtained from the net 3 km south and 4 km west displacement, which gives 5 km.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2016, held on 7 August 2016. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.