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 speeds by Charlie)

Hi!

My numbers don't agree if the first two comments.

On the second trip the spider covers 80 minutes tick marks of the clock face in the time the minute hand covers twenty munites.  So the spider is traveling 4 times faster.

On the first trip, together they cover all 60 tick marks of the clock face once.  To do this at that speed means the minute hand covered 12 ticks and the spider covered 48 (4 times).  For the first trip to complete in 6 minutes the spider would have to cover the remaining 54 ticks or be 9 times faster.

So the first trip was done at 6:12, the second trip completed at 6:32. 

Or is somthing wrong with my math?

 

Posted by Patrick on 2006-05-05 12:05:20