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?
The solution to this puzzle is as follows.
By conditions of the problem, the distance traversed by the spider in the time taken by the minute hand to travel 20 minute marks is equal to 20+60 = 80 minute marks, giving the ratio between the speed of the minute hand and the spider as 20:80=1:4.
Let us assume that the minute hand traversed x minute marks (say) before the spider reached the minute hand for the first time. Accordingly, the spider must have traveled (30-x) minute marks at the same time in terms of the foregoing motion.
Consequently, x/(30-x) = ¼; giving, x=6, so that the second meeting occurred precisely 20+6 = 26 minutes following the commencement of the walk by the spider.
In other words, the second meeting occurred precisely at 26 minutes past 6 o'clock.
Edited on May 5, 2006, 8:53 am
Edited on May 5, 2006, 9:44 am
Edited on May 5, 2006, 1:54 pm