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Mathcounts/AMC8

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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?

Jul 19, 2019

#1
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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

Jul 19, 2019