UPSC Prelims 2016 CSAT Paper II · Q58 of 79 Basic Numeracy easy

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 ?

  1. (a) 5 km ✓ UPSC's answer
  2. (b) 9 km
  3. (c) 37 km
  4. (d) 61 km

Why the answer is (a)

• North-south displacement: 12 km north minus 15 km south = 3 km south.

• East-west displacement: 15 km east minus 19 km west = 4 km west.

• The final position is 3 km south and 4 km west of the starting point.

• Distance = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 km.

• Therefore the correct option is (a) 5 km.

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.

Reading the answer is not the same as getting it right under a clock. Practise this question with UPSC's negative marking, and anything you miss goes into an error notebook until you get it right twice.

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