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The hour needle of a clock is malfunctioning and travels in the anti-clockwise direction, i.e., opposite to the usual direction, at the same speed it would have if it was working correctly. The minute needle is working correctly. Suppose the two needles show the correct time at $12$ noon, thus both needles are together at the $12$ mark. After how much time do the two needles meet again?

  1. $\dfrac{10}{11}$ hour
  2. $\dfrac{11}{12}$ hour
  3. $\dfrac{12}{13}$ hour
  4. $\dfrac{19}{22}$ hour
  5. One hour
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Since, minute needle takes $60$ minutes to cover $360^{\circ}$, So, in $1$ minute, it will cover $6^{\circ}.$ Similarly, hour needle takes $12 \times 60 = 720$ minutes to cover $360^{\circ}$, So, in $1$ minute, it will cover $(\frac{1}{2})^{\circ}.$

When both needles are at $12$ and starts traveling in their direction. Suppose, after $t$ minutes, minute needle covers $x^{\circ}$ with the speed of $6^{\circ}/min$ and in the same time, hour needle covers $(360-x)^{\circ}$ with the speed of $(\frac{1}{2})^{\circ}/min$.

When they meet, then $t = \frac{x^{\circ}}{6^{\circ}/min} = \frac{(360-x)^{\circ}}{(\frac{1}{2})^{\circ}/min}$

$\Rightarrow x = (\frac{12*360}{13})^{\circ}$

Since, $t = \frac{x^{\circ}}{6^{\circ}/min}$

So, $t = \frac{(\frac{12*360}{13})^{\circ}}{6^{\circ}}\; min$

$t = \frac{12}{13} \times 60\; min$

$t = \frac{12}{13} Hour$
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