An easy problem for MrBank, A man, who went out between five and six o'clock an returned between six and seven o'clock, found that the hands of the watch had exchanged places. when did he go out?

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An easy problem for MrBank, A man, who went out between five and six o'clock an returned between six and seven o'clock, found that the hands of the watch had exchanged places. when did he go out?

Mathematics
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This looks better XD
5:32:30?
Post solution, not answer :)

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Other answers:

The hour hand must move to where the minute hand was at an hour ago and vice versa. We know the hour already, 5:00. Let m = the minute after 5:00 \[5h+m=5(h+1)+(60-m)\] \[5(5)+m=5(6)+(60-m)\] \[2m=65\] \[m=32.5\] So he left at 5:32:30 and came back at 6:27:30
Nope that'snot the right answer, it's not 30 seconds.
The angle between the clock faces can be calculated as \[\left| 30h+m/2-6m \right|\] Plugging in 5 for h, 32.5 for m for the 5:00 hr and 6 for h and 27.5 for m for 6:00 shows the angle between the clock hands being equal...
Wouldn't the angle between hands have to be equal if the hands exchanged places?
Yes.
Another way to solve this is to look at ratios...the hour hand will go from 5 to 6 in the same time it takes minute hand to go round the clock, they also move proportionally. we know he went out sometime between 5:30 -5:35 and returned between 6:25-6:30 Let x,y be in minutes where: 30 y=12(x-30) x = (y-25)/5*60 --> x = 12(y-25) Solving this system: x = 32.3 or 32 min 18 sec y = 27.7 or 27 min 42 sec He left at 5:32:18 and returned at 6:27:42

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