# Problems from Problems with clocks

At what time between $$6$$ and $$7$$ o'clock will the clock hands overlap?

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### 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:

$$x=\dfrac{45}{11} \ \mbox{minutes}.$$

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

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

We know that

$$\dfrac{45}{11} \ \mbox{minutes}=\dfrac{44}{11}+\dfrac{1}{11}=4 \ \mbox{minutes} + \dfrac{1}{11} \ \mbox{minutes}$$

where:

$$\dfrac{1}{11} \ \mbox{minutes}=\dfrac{1}{11} \ \mbox{minutes} \cdot \dfrac{60 \ \mbox{seconds}}{1 \ \mbox{minute}} = \dfrac{60}{11} \ \mbox{seconds} = 5 \ \mbox{seconds}$$

Therefore, the hands will cross when the hands indicate $$6h \ 45+4' \ 5''$$, that is to say at $$6h \ 49' \ 5''$$.

### Solution:

$$6h \ 49' \ 5''$$.

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