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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
The first thing to realise here is the lengths of the paths.
KN should be equal to LM, giving the length of KO using Pythagoras theorem as 500m
Similarly, the length of HJ = OP = 150m and length of GJ = EL = 200m, giving the length of HG as 250 m
The shortest path from L to B and then to P (or the other way around would involve) using these hypotenuses as much as possible instead of the two adjacent sides. The shortest can be visualised as shown below or multitude of others variations, as there are multiple ways that would make one travel the shortest distance)
The below figure is the simplest one for visualisation.
Other possible paths are L-E-F-C-.. and following the same path.
The shortest distance in each of these instance would be LE + ED + DC + CB + BG + GI + IP + PO + OK + KL
Which would be 200 + 400 + 300 + 300 + 400 + 250 + 400 + 150 + 500 + 300 = 3200