Pat and Mat went on a trip. They set out in the morning after eight o'clock, when the large and small hands of Pat's clock lay on opposite rays. The hands of Pat's clock were on opposite rays again when both friends returned before noon. Mat measured the duration of the trip with a stopwatch. Determine with precision to the second how long the trip lasted. Assume that Pat's clock and Mat's stopwatch were both accurate.

(M. Volfová)
The speed of the small hand is $30^{\circ}$ in $60 \, \text{min}$, i.e., $0.5^{\circ}$ per $1 \, \text{min}$. The speed of the large hand is $360^{\circ}$ in $60 \, \text{min}$, i.e., $6^{\circ}$ per $1 \, \text{min}$. The position of the hand on the clock face that points to the number 12 is called the "basic position." At 8:00, the large hand is in the basic position, and the small hand at 8:00 points to the number 8, which is $240^{\circ}$ from the basic position. The time that has elapsed from 8:00 to the moment when the hands were on opposite rays and the trip began is denoted as $x$ min. At the desired moment, the large hand is rotated from the basic position by $(0 + 6x)^{\circ}$, and the small hand by $(240 + 0.5x)^{\circ}$. At that moment, the hands lie on opposite rays, so the rotation of the small hand from the basic position is $180^{\circ}$ greater than the rotation of the large hand. This leads to the equation, whose solution will determine $x$:

$$
\begin{aligned}
(0 + 6x) + 180 & = 240 + 0.5x, \\
5.5x & = 60, \\
x & = 10.\overline{90}.
\end{aligned}
$$

The value $10.\overline{90}$ min is approximately $10 \, \text{min} \, 54.5 \, \text{s}$; the trip therefore began at 8:10:54.5.

Similarly, we can set up an equation where $y$ represents the time in minutes that has elapsed from 11:00 to the moment when the trip ended:

$$
\begin{aligned}
(0 + 6y) + 180 & = 330 + 0.5y, \\
5.5y & = 150, \\
y & = 27.\overline{27}.
\end{aligned}
$$

The value $27.\overline{27}$ min is approximately 27 min 16.4 s; the trip therefore ended at 11:27:16.4.

By subtracting the two found times, we conclude that the trip lasted $3 \, \text{h} \, 16 \, \text{min} \, 22 \, \text{s}$.

Note. If all time values are expressed from the beginning in hours, the previous equation for the unknown $x$ looks like this:

$$
(0 + 360x) + 180 = 240 + 30x,
$$

and its solution is $x = \frac{2}{11} \, (\text{h})$. Similarly, the unknown $y$ is $\frac{5}{11} \, (\text{h})$ and the entire trip lasted

$$
\left(11 + \frac{5}{11}\right) - \left(8 + \frac{2}{11}\right) = 3 + \frac{3}{11} \, (\text{h}),
$$

i.e., approximately 3 h 16 min 22 s.

Alternative solution. During the half day, i.e., from 0:00 to 12:00, the hands lie on opposite rays a total of eleven times. The time intervals between such moments are always the same (both hands move at a constant speed). Therefore, the length of each interval is $\frac{12}{11} \, \text{h}$. The duration of the trip corresponds to three of these intervals; the correct result can also be obtained by this calculation:

$$
3 \cdot \frac{12}{11} = \frac{36}{11} = 3 + \frac{3}{11} \, (\text{h})
$$