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

If there is no second′s hand on  a clock and the minute and hour  hand move in continuous fashion,  then exactly at what time between  02:10  and 02:15 does the position  of the two hands exactly coincide?

$${If}\:{there}\:{is}\:{no}\:{second}'{s}\:{hand}\:{on} \\ $$$${a}\:{clock}\:{and}\:{the}\:{minute}\:{and}\:{hour} \\ $$$${hand}\:{move}\:{in}\:{continuous}\:{fashion}, \\ $$$${then}\:{exactly}\:{at}\:{what}\:{time}\:{between} \\ $$$$\mathrm{02}:\mathrm{10}\:\:{and}\:\mathrm{02}:\mathrm{15}\:{does}\:{the}\:{position} \\ $$$${of}\:{the}\:{two}\:{hands}\:{exactly}\:{coincide}? \\ $$

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

say the time is x minutes after  2 o′clock.  position of minute hand: x  position of hour hand: 10+((5x)/(60))  when both hands coincide:  x=10+((5x)/(60))  ⇒x=((120)/(11))=10^m 54^s

$${say}\:{the}\:{time}\:{is}\:{x}\:{minutes}\:{after} \\ $$$$\mathrm{2}\:{o}'{clock}. \\ $$$${position}\:{of}\:{minute}\:{hand}:\:{x} \\ $$$${position}\:{of}\:{hour}\:{hand}:\:\mathrm{10}+\frac{\mathrm{5}{x}}{\mathrm{60}} \\ $$$${when}\:{both}\:{hands}\:{coincide}: \\ $$$${x}=\mathrm{10}+\frac{\mathrm{5}{x}}{\mathrm{60}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{120}}{\mathrm{11}}=\mathrm{10}^{{m}} \mathrm{54}^{{s}} \\ $$

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

generally: y o′clock x minutes  x=5y+((5x)/(60))  ⇒x=((60y)/(11))  y=0^h : x=0  y=1^h : x=((60)/(11))=5^m 27^s   y=2^h : x=((120)/(11))=10^m 54^s   y=3^h : x=((180)/(11))=16^m 21^s   y=4^h : x=((240)/(11))=21^m 49^s   y=5^h : x=((300)/(11))=27^m 16^s   y=6^h : x=((360)/(11))=32^m 43^s   ......

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

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

let after 2 o′clock hour hand  moves by angle α.  12α=60°+α  α=((60°)/(11))  6° of min hand=1min  60°+((60°)/(11))=(10+((10)/(11)))min  =((120)/(11))min  2h10min+54(6/(11))sec

$${let}\:{after}\:\mathrm{2}\:{o}'{clock}\:{hour}\:{hand} \\ $$$${moves}\:{by}\:{angle}\:\alpha. \\ $$$$\mathrm{12}\alpha=\mathrm{60}°+\alpha \\ $$$$\alpha=\frac{\mathrm{60}°}{\mathrm{11}} \\ $$$$\mathrm{6}°\:{of}\:{min}\:{hand}=\mathrm{1}{min} \\ $$$$\mathrm{60}°+\frac{\mathrm{60}°}{\mathrm{11}}=\left(\mathrm{10}+\frac{\mathrm{10}}{\mathrm{11}}\right){min} \\ $$$$=\frac{\mathrm{120}}{\mathrm{11}}{min} \\ $$$$\mathrm{2}{h}\mathrm{10}{min}+\mathrm{54}\frac{\mathrm{6}}{\mathrm{11}}{sec} \\ $$

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