A man on his way to dinner shortly after $6.00$ p.m. observes that the hands of his watch form an angle of $110^{o}$ . Returning before $7.00$ p.m. he notices that again the hands of his watch form an angle of $110^{o}$ . Find the number of minutes that he has been away.

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Solution

The dinner is between $6 : 00$ and $7 : 00$ pm
Hour hand moves $= \frac{360^{o}}{12}$ hours
$\Rightarrow \frac{36^{o}}{12 \times 60} = 0.5 min$
Minute hand moves $\frac{360}{60}$ minutes $= 6 / m i n$
The relative rate difference $= 6^{o} - {0.5}^{o} = {5.5}^{0} / m i n$
In this case, minute hand movement in relation to hour hand was from:
$\Rightarrow 110^{o}$ (before dinner) + $110^{o}$ (after dinner) $= 220^{o}$
$∴ \frac{220^{o}}{5.5} = 40 m i n u t e s$

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A man on his way to dinner shortly after 6.00 p.m. observes that the hands of his watch form an angle of 110o. Returning before 7.00 p.m. he notices that again the hands of his watch form an angle of 110o. Find the number of minutes that he has been away.

A man on his way to dinner shortly after $6.00$ p.m. observes that the hands of his watch form an angle of $110^{o}$ . Returning before $7.00$ p.m. he notices that again the hands of his watch form an angle of $110^{o}$ . Find the number of minutes that he has been away.