John noticed that the angle formed by the minute hand and hour hand on a standard 12-hour clock was 110 degrees when he left home after 6 p.m.; it was also 110 degrees when he returned before 7 p.m. that same night. If he left home for more than five minutes, for how many minutes was he away?
John noticed that the angle formed by the minute hand and hour hand on a standard 12-hour clock
was 110 degrees when he left home after 6 p.m.;
it was also 110 degrees when he returned before 7 p.m. that same night.
If he left home for more than five minutes, for how many minutes was he away?
\(\text{Let the angle formed by the minute hand and hour hand in degrees $\mathbf{ \Delta \alpha }$ } \\ \text{Let the time in hours $\mathbf{ t }$ }\)
The formula between the two values \(\Delta \alpha \) and \(t\) is:
\(\boxed{~ \Delta \alpha +360^\circ \cdot n = 330 \cdot t ,\ n\in \mathbb{Z} \\or\\ (360^\circ-\Delta \alpha) +360^\circ \cdot m = 330 \cdot t ,\ m\in \mathbb{Z}~ } \)
\(\begin{array}{|rl|rl|} \hline n & t = \dfrac{110^\circ +360^\circ \cdot n} {330} & m & t = \dfrac{(360^\circ-110^\circ) +360^\circ \cdot m} {330} \\ \hline \ldots&&\ldots \\ 5 & 5^h\ 47^m\ 16^s.\overline{36} & 4 & 5^h\ 7^m\ 16^s.\overline{36} \\ 6 & \mathbf{6^h\ 52^m\ 43^s.\overline{63}} & 5 & \mathbf{6^h\ 12^m\ 43^s.\overline{63}} \\ 7 & 7^h\ 58^m\ 10^s.\overline{90} & 6 & 7^h\ 18^m\ 10^s.\overline{90} \\ \ldots&&\ldots \\ \hline \end{array} \)
He was between \(6^h\ 12^m\ 43^s.\overline{63}\) and \(6^h\ 52^m\ 43^s.\overline{63}\) away.
So he left home for exact \(\mathbf{40}\) minutes