A simple clock problem A man on his way to dinner shortly after $6.00$ pm observes that the hands of his watch form an angle of $110°$. 
Returning before $7.00$ pm he notices that again the hands of his watch form an angle of $110°$. 

Find the number of minutes that he has been away.

A clear explanation would be nice.
 A: Hint:  at exactly $6$ PM the hands make an angle of $180^\circ$.  The minute hand moves $360^\circ$ in one hour or $6^\circ$ per minute.  The hour hand moves $30^\circ$ in one hour, or $0.5^\circ$ per minute.  How many minutes elapse before the angle is reduced to $110^\circ$?  Then for the return, the minute hand must pass the hour hand and gain $110^\circ$ on the other side.
A: HINT:
As the hour hand takes $12$ hours $=12\cdot60$ minutes to rotate $360^\circ$
So, in $t$ minutes it rotates $\frac t2^\circ$
Similarly, the minute hand takes $60$ minutes  to rotate $360^\circ$
So, in $t$ minutes it rotates $6t^\circ$
So, in $t$ minutes the angle difference is $(6t-\frac t2)^\circ=\frac{11t}2^\circ$
At $6$ PM, the angle is $180^\circ$
So, we need $(180-110)^\circ=70^\circ$ more for the first case
and $\{360-(180-110)\}^\circ=290^\circ$ more for the second case
Now, the difference is $\frac{11t}2^\circ$ in $t$ minutes
The difference is $1^\circ$ in $\frac t{\frac{11t}2}=\frac2{11}$ minutes
The difference will be $70^\circ$ in $\frac2{11}\cdot70$ minutes $=\frac{140}{11}$ minutes $=12\frac8{11}$ minutes
The difference will be $290^\circ$ in $\frac2{11}\cdot290$ minutes $=\frac{580}{11}$ minutes $=52\frac8{11}$ minutes
A: At $12$ midnight the hands are together. $60$ minutes later at $1.00$am, the hands are $1/12$ of the dial, or $30^{\circ}$ apart. In one hour the minute hand has "caught up" $330^{\circ}$ at a rate of $\frac {330^{\circ}}{60}=5.5^{\circ}$ per minute.
In the question, the minute hand starts $110^{\circ}$ behind the hour hand and ends $110^{\circ}$ ahead of the hour hand. This is a [relative] difference of $220^{\circ}$ and at $5.5^{\circ}$ per minute this takes $40$ minutes.
You don't need to calculate the initial time or the final time to solve the problem.
Note for future reference - Because the minute hand travels twelve times as fast as the hour hand, the difference in angular speeds is $11$ times the (slower) speed of the hour hand. So the factor $11$ often comes up in problems involving the relative positions of minute and hour hands. Its presence in the question here (the $110$) is a give-away that there will be a nice answer.
