UPSC Prelims 2016 · CSAT
Question 49: Basic Numeracy
A person climbs a hill in a straight path from point ‘O’ on the ground in the direction of north-east and reaches a point ‘A’ after travelling a distance of 5 km. Then, from the point ‘A’ he moves to point ‘B’ in the direction of north-west. Let the distance AB be 12 km. Now, how far is the person away from the starting point ‘O’ ?
AnswerUPSC’s answer is (b).
North-east and north-west directions are at right angles (45 degrees either side of north), so O, A and B form a right triangle with the right angle at A. OA = 5 and AB = 12, so OB = sqrt(25 + 144) = sqrt(169) = 13 km. This is the 5-12-13 triple, worth remembering.
Why each option is right or wrong
- (a) 7 km is 12 - 5, treating the two paths as along one line.
- (b) Right: sqrt(5^2 + 12^2) = 13.
- (c) 17 km is 5 + 12, adding the paths in a line.
- (d) 11 km does not follow from any right-angled combination of 5 and 12.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2016, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.