A takes 2 hours more than B to walk P km, but if A doubles his speed, then he can make it in one hour less than B. How much time does B require for walking P km?
4 hours
Let B take t hours to walk P km. A is slower, taking 2 hours more, so A normally takes (t + 2) hours.
Key idea: for a fixed distance, time is inversely proportional to speed. So if A doubles his speed, his time is halved, becoming (t + 2)/2 hours.
The problem says with doubled speed A finishes one hour less than B, i.e. in (t − 1) hours.
So: (t + 2)/2 = t − 1
Multiply both sides by 2: t + 2 = 2t − 2
Solve: 2 + 2 = 2t − t, so t = 4.
Check: A normally takes 6 h; doubled speed makes it 3 h, which is indeed 1 h less than B's 4 h. Correct.
So B needs 4 hours. The value 3 hours is A's doubled-speed time, not B's; 5 hours and 3.5 hours do not satisfy the halving-and-one-hour-less condition.
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