MBA
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ABC forms an equilateral triangle in which B is 2km east of C AND A is to the north of both B AND C .He starts walking from B in a direction parallel to AC and stops when he reaches a point D directly to the east of A.He,then,reverses direction and walks till he reaches a point E, directly to the south of C.Total distance walked by him is 1) 6 2) 5 3) 6root3 4) 7
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