## 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?

1. lalaly

This looks better XD

2. anonymous

5:32:30?

3. anonymous

Post solution, not answer :)

4. 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

5. anonymous

Nope that'snot the right answer, it's not 30 seconds.

6. 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...

7. anonymous

Wouldn't the angle between hands have to be equal if the hands exchanged places?

8. anonymous

Yes.

9. 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<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