This type of problem can be solved with the familiar D=RxT equation, although in these problems D is measured in 'minutes measured from a certain hour (say 3:00am) on the face of a clock', rather than the more familiar 'miles on the road'. And since we are trying to find the time when they meet (coincide), we turn the equation around to T=D/R.
The minute hand travels at a rate of 60 minutes per hour (60m/h), while the hour hand travels at a rate of 5 minutes per hour (5m/h).
So for example, to find the time X when the minute hand and the hour hand first coincide, starting from 3:00am, the minute hand will travel 15+X minutes while the hour hand will travel X minutes. The equation can be written as
(15+X)/(60m/h)=(X)/(5m/h), or (15+X)(5m/h)=X(60m/h).
So we get 75+5X=60X, or 75=55X, or X= 75/55 = 15/11 = 1 4/11 minutes after 3:00am.
Similar equations can be used to find the time X when theminute and hour hands coincide after 4:00 and the equation becomes (20+x)5m/h)=60X and X will equal 20/11 =1 9/11 minutes after 4:00.
Continuing, the times can be easily computed for each hour and the progression will be 25/11 = 2 3/11 minutes after 5:00
30/11 = 2 8/11 minutes after 6:00,
35/11 = 3 2/11 minutes after 7:00,
40/11 = 3 7/11 minutes after 8:00,
45/11 = 4 1/11 minutes after 9:00
50/11 = 4 6/11 minutes after 10:00
55/11 = 5 minutes after 11:00, and this will be at exactly 12:00
The two hands therefore will not coincide during the 60 minutes from 12:00 to 1:00, since we have already counted the 12:00 time.
Continuing, the hands will coincide at 5/11 minutes after 1:00 and at 10/11 minutes after 2:00.