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Get access to the detailed solutions to the previous year's questions asked in XAT exam.
Given, CE=0.5, BC = 1.3 and ED=2.5
Triangle CEB is a right-angled triangle => EB = 1.2
Triangles ECB is similar to triangle EDA
EB/EC = AE/ED => AE = 6
Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3.5 = 11.5km