A) 5 km West
B) 7 km West
C) 7 km East
D) 5 km North-East
Correct Answer: D
Solution :
Clearly, Kunal moves from A 10 km Northwards upto B, then moves 6 km Southwards upto C, turns towards East and walks 3 km upto D. Then, \[AC=(AB-BC)=(10-6)+4\,km;\] \[CD=3\,\,km.\] So, Kunal's distance from starting point A \[=AD=\sqrt{A{{C}^{2}}+C{{D}^{2}}}=\sqrt{{{4}^{2}}+{{3}^{2}}}=5\,\,km\] Also D is to the North-East of A.You need to login to perform this action.
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