At what time between 10,text{O'clock} and 11, | vedclass

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Question

(B) Let $x$ be the number of minutes past $10\,	ext{O'clock}$.At $10\,	ext{O'clock}$,the hour hand is at $300^{\circ}$ from the $12\,	ext{O'clock}$

Vedclass · Accepted Answer

$10\,	ext{h } 9\,	ext{m } 14\,	ext{s}$

Vedclass · Answer

(B) Let $x$ be the number of minutes past $10\,	ext{O'clock}$.At $10\,	ext{O'clock}$,the hour hand is at $300^{\circ}$ from the $12\,	ext{O'clock}$ position.In $x$ minutes,the minute hand moves $6x^{\circ}$ from the $12\,	ext{O'clock}$ position.The hour hand moves $\frac{x}{2}^{\circ}$ in $x$ minutes,so its position is $(300 + \frac{x}{2})^{\circ}$ from $12\,	ext{O'clock}$.For the hands to be symmetric with respect to the vertical line (the $12-6$ line),the angle of the minute hand from $12\,	ext{O'clock}$ (clockwise) must equal the angle of the hour hand from $12\,	ext{O'clock}$ (counter-clockwise).The angle of the hour hand from $12\,	ext{O'clock}$ (counter-clockwise) is $360^{\circ} - (300 + \frac{x}{2})^{\circ} = (60 - \frac{x}{2})^{\circ}$.Setting these equal: $6x = 60 - \frac{x}{2}$.$12x = 120 - x$ $\Rightarrow 13x = 120$ $\Rightarrow x = \frac{120}{13} \approx 9.2307\,	ext{minutes}$.$0.2307 	imes 60 \approx 13.84\,	ext{seconds}$,which rounds to $14\,	ext{seconds}$.Thus,the time is $10\,	ext{h } 9\,	ext{m } 14\,	ext{s}$.

At what time between $10\,\text{O'clock}$ and $11\,\text{O'clock}$ are the two hands of a clock symmetric with respect to the vertical line (give the answer to the nearest second)?

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