A person travelled from A to B and returned back. He has two stoppages (C and

Let each equal segment for the onward journey be x km, so the onward journey time is (x/20) + (x/25) + (x/10) hours, which simplifies to (23x/100) hours. On the return trip, the time from B to D is (x/20) hours, the time from D to C is ((x+10)/40) hours, and the time from C to A is (x/25) hours. The return journey time is (19x/100) + (1/4) hours, and setting this equal to the onward journey time minus 3 minutes (which is 1/20 of an hour) gives the equation (19x/100) + (1/4) = (23x/100) - (1/20). Solving this equation yields x = 4, so the total distance between A and B is 3x, which equals 12 km.

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