SSC CHSL 10 Jan 2017 Afternoon Shift Question 20

Question 20

A travels 12 km towards north and then takes a left turn and covers another 5 km. From there, he turns 180° anticlockwise and travels 10 km further. What is the minimum distance between his initial and final position?

Solution

A travels 12 km towards north and then takes a left turn and covers another 5 km towards west direction. From there, he turns 180° anticlockwise, i.e. will revert his direction and travels 10 km further towards east. 

Thus, the minimum distance between his initial and final position = $$\sqrt{(12)^2+(5)^2}$$

= $$\sqrt{144+25}=\sqrt{169}=13$$ km

=> Ans - (B)



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