At exactly six o'clock, a spider started to walk at a constant speed from the hour hand anticlockwise round the edge of the clock face. When it reached the minute hand, it turned round (assume the turn was instantaneous) and walked in the opposite direction at the same constant speed. It reached the minute hand again after 20 minutes. At what time was this second meeting?

This was a nice easy math puzzle.  Knowing the spider travelled from the minute hand to the minute hand in a clockwise direction and that the minute hand had moved twenty minute marks during the interval, the spider travelled a total of 80 minute marks (60 + 20).

At 80 minute-marks/20 minutes, the spider travels at a constant rate of 4 minute-marks/minute. Thus the time in which the spider first encountered the minute hand was at 6:06 [30 minute-marks - (4 minute-marks/minute * 6 minutes)].  Adding the additional 20 minutes that elapsed between then and his second encounter with the minute hand, the time calculated is 6:26 at the second meeting.  

Posted by Dej Mar on 2006-05-05 18:38:08