Solution to: Confusing Clock

Suppose that a second pair of hands turns together with the incorrect pair, but in the correct way. When the incorrect pair is in the same position as the correct pair, this means that the time is displayed correctly. First, consider the minute hands that are at twelve. The 'incorrect' hand turns twelve times slower than the 'correct' hand. Let x be the distance (in minutes) that the 'incorrect' 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). Thus, we have the equation 12x = 60 + x. This means that x = 5 ⁵⁄₁₁ minutes.
The same holds true for the hour hands that start at six. Therefore, the confused clock shows the correct time again after 60 + 5 ⁵⁄₁₁ minutes, which is at ⁵⁄₁₁ minutes past 7.

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