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?

Guest Jul 19, 2019

#1**+3 **

**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

heureka Jul 19, 2019