A man who went out between 3 and 4 and return | vedclass

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(D) Let the man go out at $3:x$ and return at $8:y$,where $x$ and $y$ are minutes past the hour.When he goes out at $3:x$,the hour hand is at $(3 + x/

Vedclass · Accepted Answer

$18 \frac{6}{13}$ minutes past $8$

Vedclass · Answer

(D) Let the man go out at $3:x$ and return at $8:y$,where $x$ and $y$ are minutes past the hour.When he goes out at $3:x$,the hour hand is at $(3 + x/60)$ units and the minute hand is at $x$ units.When he returns at $8:y$,the hour hand is at $(8 + y/60)$ units and the minute hand is at $y$ units.Since the hands have interchanged positions:The position of the hour hand at return must be the position of the minute hand at departure: $(8 + y/60) = x/5$.The position of the minute hand at return must be the position of the hour hand at departure: $y = 5(3 + x/60) = 15 + x/12$.From the first equation: $x = 5(8 + y/60) = 40 + y/12$.Substitute $x$ into the second equation: $y = 15 + (40 + y/12)/12 = 15 + 40/12 + y/144 = 15 + 10/3 + y/144 = 55/3 + y/144$.$y - y/144 = 55/3 \implies 143y/144 = 55/3$.$y = (55 	imes 144) / (3 	imes 143) = (55 	imes 48) / 143 = (5 	imes 11 	imes 48) / (13 	imes 11) = 240 / 13 = 18 \frac{6}{13}$.Thus,he returned at $18 \frac{6}{13}$ minutes past $8$.

$A$ man who went out between $3$ and $4$ and returned between $8$ and $9$ found that the hands of the watch had exactly changed places. He returned at

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