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Question Number 88014 by ajfour last updated on 07/Apr/20

$$Ifthereisno{second}^{\prime }shandon$$$$aclockandtheminuteandhour$$$$handmoveincontinuousfashion,$$$$thenexactlyatwhattimebetween$$$$02:10and02:15doestheposition$$$$ofthetwohandsexactlycoincide?$$

Answered by mr W last updated on 08/Apr/20

$$saythetimeisxminutesafter$$$$2o^{\prime }clock.$$$$positionofminutehand:x$$$$positionofhourhand:10+\frac{5x}{60}$$$$whenbothhandscoincide:$$$$x=10+\frac{5x}{60}$$$$\Rightarrow x=\frac{120}{11}={10}^m{54}^s$$

Commented by mr W last updated on 08/Apr/20

$$generally:y{o}^{\prime }clockxminutes$$$$x=5y+\frac{5x}{60}$$$$\Rightarrow x=\frac{60y}{11}$$$$y=0^h:x=0$$$$y=1^h:x=\frac{60}{11}=5^m{27}^s$$$$y=2^h:x=\frac{120}{11}={10}^m{54}^s$$$$y=3^h:x=\frac{180}{11}={16}^m{21}^s$$$$y=4^h:x=\frac{240}{11}={21}^m{49}^s$$$$y=5^h:x=\frac{300}{11}={27}^m{16}^s$$$$y=6^h:x=\frac{360}{11}={32}^m{43}^s$$$$......$$

Commented by ajfour last updated on 07/Apr/20

thank you for the general treatment mrW sir; i have presented roughly my approach too.

Answered by ajfour last updated on 07/Apr/20

$$letafter2o^{\prime }clockhourhand$$$$movesbyangle\alpha .$$$$12\alpha =60°+\alpha$$$$\alpha =\frac{60°}{11}$$$$6°ofminhand=1min$$$$60°+\frac{60°}{11}=(10+\frac{10}{11})min$$$$=\frac{120}{11}min$$$$2h10min+54\frac{6}{11}sec$$

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