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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Similar to the previous question, we should try to avoid the hypotenuse
The total path distance would be
300(CD) + 400(DE) + 300(EF) + 200(FK) + 400(KN) + 300(NO) + 150(OP) + 400(PI) + 150(IJ) + 200(JG) + 400(GB) + 300(BC)
Adding up to 3,500 meters
Therefore, 3500 is the correct answer.