Discussion

Explanation:

For the clock to show correct time again, the clock needs to lose 12 hours.

The slower clock loses 3 minutes every hour, hence it will lose 12 hours in $\frac{12\times 60}{3}\times 1$ = 240 hours = 10 days.

⇒ 10 days after 6 a.m. of 1^st January will be 6 a.m. of 11^th January.

Hence, option (d).

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