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Kunal walks 10 km towards the North. From there he walks 6 km-towards South. Then, he walks 3 km towards East. How far and in which direction is he with reference to his starting point?

Answer:

Explanation:

Clearly, Kunal moves from A 10 km northwards up to B, then moves 6 km southwards up to C, turns towards East and walks 3 km up to D.

Then AC = (AB-BC)=(10-6)=4 km

CD=3 km

So, Kunal's distance from starting point A

$=AD = \sqrt{AC^{2}+CD^{2}}=\sqrt{4^{2}+3^{2}}=5 km$

D is the north east of A.