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anonymous

  • 4 years ago

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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  1. lalaly
    • 4 years ago
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    This looks better XD

  2. anonymous
    • 4 years ago
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    5:32:30?

  3. anonymous
    • 4 years ago
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    Post solution, not answer :)

  4. anonymous
    • 4 years ago
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    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

  5. anonymous
    • 4 years ago
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    Nope that'snot the right answer, it's not 30 seconds.

  6. anonymous
    • 4 years ago
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    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...

  7. anonymous
    • 4 years ago
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    Wouldn't the angle between hands have to be equal if the hands exchanged places?

  8. anonymous
    • 4 years ago
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    Yes.

  9. dumbcow
    • 4 years ago
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    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<x<35 and 25<y<30 (x-30)/5 will give you the percentage of how far x is in the interval 30-35. Now if we assume x is the hour hand, then to find where the minute hand is we just multiply by 60. Example: If x=32.5, hour hand is right in middle, then minute hand must be at 30. Set up 2 equations: y = (x-30)/5 *60 --> 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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