anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lalaly
  • lalaly
This looks better XD
anonymous
  • anonymous
5:32:30?
anonymous
  • anonymous
Post solution, not answer :)

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More answers

anonymous
  • anonymous
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
anonymous
  • anonymous
Nope that'snot the right answer, it's not 30 seconds.
anonymous
  • anonymous
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...
anonymous
  • anonymous
Wouldn't the angle between hands have to be equal if the hands exchanged places?
anonymous
  • anonymous
Yes.
dumbcow
  • dumbcow
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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