A train departs a station daily at 6:00 am and follows a fixed schedule. One day, it traveled at a constant speed of 50 km per hour and arrived at its destination 6 hours later than its scheduled time. The next day, it traveled three-fourths of its route in one-half of its total scheduled travel time, and the remaining distance at 40 km per hour to reach just on time. The scheduled arrival time of the train is:

Let the total distance be D and the scheduled time be T hours. On the first day, D/50 equals T plus 6. On the second day, the train travels 3D/4 in T/2 hours, meaning its speed for that part is 3D/(2T), and it travels D/4 at 40 kmph in exactly T/2 hours to be on time. This gives D/4 equals 20T, so D equals 80T; substituting into the first equation gives 80T/50 equals T plus 6, meaning 30T equals 300 and T equals 10 hours. Adding 10 hours to the 6:00 am departure time gives a scheduled arrival time of 4:00 pm.

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