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)

"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."

On the first trip, together they cover only 30 of the tick marks--not all 60.  The spider goes from #30 down to #6, while the minute hand goes from #0 to #6.  The hand has gone 6 tick marks, and the spider has gone 24, which is 4 times as many (not 4 times more).

BTW that brings up: the spider is going 4 times as fast as the minute hand--not 4 times faster, which would be 5 times as fast. (One time faster would be twice as fast, for example, and one times more tick marks would be 2 times as many tick marks.)

Edited on May 5, 2006, 1:44 pm

Posted by Charlie on 2006-05-05 13:39:21