SSC CGL Tier II 21 Sep 2014 Paper

Question : 36 of 100

Marks: +1, -0

A man goes 24 m due west and then 10 m due north. Then the distance of him from the starting point is

Solution:

(2)

∠ABC = 90°
AB = 24 metre , BC = 10metre

∴ A C=√{A B^{2}+B C^{2}}

=√{24^{2}+10^{2}}

=√{576+100}=√{676}

=26

metre

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