The particle's motion is restricted to the wedge bounded by the lines y=2x and y=x∕2, and the particle could reach any grid point on the boundary on the wedge just by using a single repeated move. Since the point (25,75) is outside this wedge, we want the closest point on the boundary of the wedge to this point. We must minimise (x−25)2+(2x−75)2, which occurs when 2x−50+8x−4⋅75=0, that is when x=35. The distance from (35,70) to (25,75) is √102+52=5√5. The answer is (b).