A) 5 km West
B) 5 km North-East
C) 7 km East
D) 7 km Wast
Correct Answer: B
Solution :
The movements of Kunal are as shown in adjacent figure. (A to B, B to C and C to D). \[AC=(AB-BC)=(10-6)km=4\,\,km\] Clearly, D is to the North-east of A. \[\therefore \] Kunal's distance from starting point A \[=AD=\sqrt{A{{C}^{2}}+C{{D}^{2}}}\] \[=\sqrt{{{4}^{2}}+{{3}^{2}}}=\sqrt{25}=5\,\,km\] So, Kunal is 5 km to the North-east, of his starting point.You need to login to perform this action.
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