## Solution to: Confusing Clock

Suppose that a second pair of hands turns together with the wrong pair of hands, but then in the correct way.
When the wrong pair is in the same position as the correct pair, this means that the time is shown in the right way.
First, look at the minute hands that are at the twelve.
The "wrong" hand turns twelve times slower than the "correct" hand.
Let *x* be the distance (in minutes) that the "wrong" hand has progressed when the two minute hands are in the same position again.
The "correct" hand then has progressed 60+*x* minutes (one complete round more).
So it then holds that 12*x* = 60+*x*.
This means that *x* = 5 ^{5}/_{11} minutes.

For the hour hands that start at six holds the same.
The confused clock therefore shows the correct time again after 60 + 5 ^{5}/_{11} minutes,
so at 5 ^{5}/_{11} minutes past 7.

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