Problems from Problems with clocks

Development:

When it is $6$ o'clock exactly, the hour hand is at the $6$ (that corresponds to $45$ minutes) and the minute hand is at $12$ (that corresponds $0$ minutes).

We are considering $x$ the arch of the hour hand. Let's think then, what arch the minute hand will make.

This hand is at $0$ minutes which is why it will have to cover $45$ minutes plus what the hour hand has covered, which is what we have called $x$. So, it must cover $45+x$ to reach the hour hand.

Since we know that it is always $12$ times what the hour hand covers, we can raise the following equation:

$45+x=12x$

With reference to the unknown, we find that:

$x={\displaystyle \frac{45}{11}}\text{minutes}.$

Therefore, minutes will cross at $6$h $45+x$ minutes.

Let's see all the minutes and seconds correspond to $x={\displaystyle \frac{45}{11}}\text{minutes}$ to write this in a better way.

We know that

${\displaystyle \frac{45}{11}}\text{minutes}={\displaystyle \frac{44}{11}}+{\displaystyle \frac{1}{11}}=4\text{minutes}+{\displaystyle \frac{1}{11}}\text{minutes}$

where:

${\displaystyle \frac{1}{11}}\text{minutes}={\displaystyle \frac{1}{11}}\text{minutes}\cdot {\displaystyle \frac{60\text{seconds}}{1\text{minute}}}={\displaystyle \frac{60}{11}}\text{seconds}=5\text{seconds}$

Therefore, the hands will cross when the hands indicate $6h45+4^{\prime }{5}^{″}$, that is to say at $6h{49}^{\prime }{5}^{″}$.

Solution:

$6h{49}^{\prime }{5}^{″}$.

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