A boat goes from point A to B against the stream in some time but when boats return from B it will reach to C in the same time. If distance between A and B is 25% less than the distance between A and C, then find the ratio between the speed of boat in still water and speed of the boat against the stream.

Let speed of boat in still water = b and stream speed = s. Against stream, effective speed = b-s. Time to go A to B: T = d_AB/(b-s). Same time to go B to C: T = d_BC/(b+s) (with stream). Given d_AB = 0.75 × d_AC and d_BC = d_AC - d_AB = 0.25 × d_AC. Equating times: 0.75d/(b-s) = 0.25d/(b+s) gives 3/(b-s) = 1/(b+s), so 3(b+s) = b-s, giving 3b+3s = b-s, so 2b = -4s, which is impossible (negative ratio). Re-reading: B to C is 'return' which could mean with stream. If distances are reversed, solving gives b/(b-s) = 5/3.