A clock was 2 mins late at 8 o' clock in the morning and 1 min fast at 6 o'clock in the evening. If the clock was running fast uniformly at what time did the clock show the right time?
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I think at 1 o'clock
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Answer: (600+3/600) *t=t+1
=>t=200 min
Convert to hours
3 hours 20 min
Add to 8 am
It will 11:20 am, hence the ans
Step-by-step explanation:
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