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?

First, we set up the equation for the time (t) that the spider meets the minute hand the first time. The minute hand moves at 6 degrees per minute, and the spider moves at x degrees per minute. Since the spider starts at the 6 or at 180 degrees away from the minute hand,

6(t) = 180 - x(t)

Now if we say that the spider and minute hand are at the same spot, and start moving in the same direction, and the spider laps the minute hand (360 degrees plus the distance traveled by the minute hand, or 360 + 6(20) or 480 total degrees) then the distance traveled by the spider in 20 minutes must be 480 degrees. The spider, therefore, travels 24 degrees per minute.

Plug 24 degrees per minute into the equation for x, and solve:

6(t) = 180 -24(t)
6(t) + 24(t) = 180
30(t) = 180
t = 6

So the first meeting is at 6 minutes after 6, or 6:06. 20 minutes later at the second meeting is 6:26.

As desired.

Posted by Andy on 2006-05-18 19:18:48