UPSC Prelims 2016·CSAT·Quantitative Aptitude·Geometry and Mensuration

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'?

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Last updated 23 May 2026, 3:31 pm IST
  1. A7 km
  2. B13 kmCorrect
  3. C17 km
  4. D11 km

Explanation

The problem describes a journey that forms a right angled triangle. First, the person moves from point O to point A in a north east direction for 5 km. From point A, the person moves to point B in a north west direction for 12 km. In navigation, north east and north west directions are perpendicular to each other. This means the angle at point A is 90 degrees. Therefore, triangle OAB is a right angled triangle where OA and AB are the two sides and OB is the hypotenuse. To find the distance from the starting point O to the final point B, we use the Pythagoras theorem: OB squared equals OA squared plus AB squared. OB squared equals 5 squared plus 12 squared. OB squared equals 25 plus 144. OB squared equals 169. OB equals the square root of 169, which is 13. The person is 13 km away from the starting point. Thus, option B is correct.
Quantitative Aptitude: A person climbs a hill in a straight path from point 'O' on the ground in the direction of north-east and reaches a poin
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