A business man looks at his watch before leaving the office for lunch. When he returns, he finds that the hour and minute hands have exchanged places from the position they had when he left the office. 1- Find the time when he left and 2 - the time when he returned.

Thanks for any help.

Guest Feb 14, 2020

#1**0 **

Since he left for lunch, we regard the time as around noon....

If he left at 12:05 and returned at 1:00 the hands are in exchanged places. I think.

_{.}

Guest Feb 14, 2020

#2**0 **

1 - The ratio of the minute hand to that traveled by the hour hand is 12:1. Since the distance traveled around the dial by the minute hand is 12/13 of a revolution and the distance traveled by the hour hand is 1/13 of a revolution, then: 12/13 x 60 =**55 5/13 minutes before noon.**

2 - Since the minute hand travels 12 times faster than the hour hand, then the distance between them was 11/12 of the distance of the minute hand from 12:00 noon. Hence the time the business man left for lunch is 11/12 of the distance, which is 1/13 of a revolution. Therefore, the distance is:12/11 x 1/13 x 60=5 5/143 minutes past noon. It then follows that the time when he came back is:5 5/143 + 55 5/13 =60 60/143. That is: **60/143 of a minute after 1:00 pm.**

Guest Feb 14, 2020