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?
(In reply to
re: speeds by Patrick)
It would be observed in terms of the puzzle under reference that the spider commenced its walk from the hour hand at precisely at 6 o'clock, giving the point of commencement of the spider as the 6 hour mark (or 30 minute mark), while the minute hand was situated at 12 hour mark. Since it has been conclusively
shown that spider's speed is precisely 4 times that of the minute hand and since on the first trip, together they cover all 30 minute marks( not 60 minute marks) of the clock face once, it follows that the spider met the minute hand for the first time at 6:06. Adding 20 minutes to this yields 6:26 as the required time, that is, the time when the spider met the minute hand for the second time.
Edited on May 5, 2006, 1:43 pm
Edited on May 5, 2006, 1:46 pm
Edited on May 5, 2006, 1:51 pm
Edited on May 5, 2006, 1:58 pm
re:speeds(2)
