A,B and C walking on equilateral triangle

(CAT 1993)

ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C.
1. Then D is
(a) 3 km east and 1 km north of A
(b) 3 km east and sqrt3 km north of A
(c) sqrt3 km east and 1 km south of A
(d) sqrt3 km west and 3 km north of A

2. The total distance walked by the person is
(a) 3 km
(b) 4 km
(c) 2 sqrt3 km
(d) 6 km

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