Solution to Problem:

The circle on the face of the clock is divided into 60 equal parts. The minute hand travels around the circle at a speed of 60 spaces per hour. The hour hand follows it at a speed of 5 spaces per hour. So the minute hand gets ahead at a speed of 55 spaces per hour. The two hands will be together again when the minute hand is a full lap, or 60 spaces, ahead. To get that far ahead will take 60/55 hours, which is 1 1/11 hours, or 1 hour 5 5/11 minutes. So the hands will be together at 5 5/11 minutes past 1 o'clock.

Another solution is to use a RATE-TIME-DISTANCE table.
Let y = the spaces travelled by the hour hand.
Then y + 60 = the spaces travelled by the minute hand.
Let x = hours travelled by both the hour and the minute hands.

	Rate	Time	Distance
Minute Hand	60 sp/hr	x hr	y + 60 sp
Hour Hand	5 sp/hr	x hr	y sp

This yields 2 equations:
60 x = y + 60
5 x = y

Solving simultaneously, x = 60/55 hours or
approximately 1:05:27 (hr:min:sec).