\section*{Problem 3 - 111223}

Klaus notices that the two hands of his pocket watch form an angle of \(110^{\circ}\) at exactly two times between 6 o'clock and 7 o'clock.

Determine the number of minutes that have passed from the first to the second of the mentioned times!
}

In one minute, the minute hand sweeps an angle of \(\frac{360^{\circ}}{60}=6^{\circ}\) and the hour hand sweeps an angle of \(\frac{360^{\circ}}{12 \cdot 60}=0.5^{\circ}\).

At x minutes past 6 o'clock, the minute hand forms an angle of \(6 x^{\circ}\) with the 12 o'clock position, and the hour hand forms an angle of \((180+0.5 x)^{\circ}\). For the two mentioned times, a minutes past 6 o'clock and \(\mathrm{b}\) minutes past 6 o'clock with \(a<b\), the following equations hold:

\[
180+0.5 a-6 a=110 \quad \text { and } \quad 6 b-(180+0.5 b)=110
\]

Adding these two equations yields \(\frac{11}{2} b-\frac{11}{2} a=220\) or \(b-a=40\). Therefore, exactly 40 minutes lie between the two times.