Problems with clocks

When it is $6$ o'clock exactly, the hour hand is at the $6$ (that corresponds to $30$ 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 $30$ minutes plus what the hour hand has covered, which is what we have called $x$. So, it must cover $30+x$ 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:

$$30+x=12x$$

With reference to the unknown, we find that:

$$x=\dfrac{30}{11} \ \mbox{minutes}.$$

Therefore, minutes will cross at $6$h $30+x$ minutes.

Let's see all the minutes and seconds correspond to $x=\dfrac{30}{11} \ \mbox{minutes}$ to write this in a better way.

We know that

$$\dfrac{30}{11} \ \mbox{minutes}=\dfrac{22}{11}+\dfrac{8}{11}=2 \ \mbox{minutes} + \dfrac{8}{11} \ \mbox{minutes} $$

where:

$$\dfrac{8}{11} \ \mbox{minutes}=\dfrac{8}{11} \ \mbox{minutes} \cdot \dfrac{60 \ \mbox{seconds}}{1 \ \mbox{minute}} = \dfrac{480}{11} \ \mbox{seconds} \approx 44 \ \mbox{seconds} $$

Therefore, the hands will cross when the hands indicate $6h \ 30+2' \ 44''$, that is to say at $6h \ 32' \ 44''$.