More clocks and angles

In one hour, the hour hand moves $1/12$ of a revolution ahead, and the minute hand moves one revolution. Thus the position angle of the minute hand relative to the hour hand increases $11/12$ revolution per hour. Thus, the angle between the hands increases at a rate of $\frac{\Delta \theta}{\Delta t} = \frac{330^\circ}{\text{hr}}.$

There are two possible scenarios:

• if the minute hand was initially behind on the hour hand, $\theta = -18^\circ, \theta' = +18^\circ, \Delta \theta = 36^\circ;$

• if the minute hand was initially ahead of the hour hand, $\theta = 18^\circ, \theta' = 360 -18^\circ = 342^\circ, \Delta \theta = 324^\circ.$

Clearly, the first scenario results in a smaller change in angle and therefore a shorter time interval. We have $\Delta t = \frac{\Delta \theta}{330^\circ/\text{hr}} = \frac{36^\circ}{330^\circ/\text{hr}} \times \frac{60\times 60\ \text{s}}{\text{hr}} \approx \boxed{393}\ \text{s}.$

Minimum amount of time would be taken if the minute hand is behind the hour hand. The two hands first will meet and then separate by $18^o$ .
That is twice the time taken for meeting.
In one minute, minute hand rotates $6^o$ and hour hand $6/12=1/2^o$ both cw.
So in one minute, the distance between two hands is changed by $6 - 1/2 = 11/5^o$ .
We want a change of 2 * 18= $36^o$ .
So minimum amount of time=36/(11/5) minutes=36/(11/5) *60 sec $\approx$ 393 sec.