Since minute hand gains 5 minutes in every 60 minutes
⇒ second hand gains 5 seconds in every 60 seconds
∴ In every 60 seconds true time, it moves 65 seconds
⇒ 65 × 6° = 390°.
Hence , required answer will be 390° .

As we can say that ,
There are 3 intervals when the clock strikes 4 Time taken in 3 intervals = 9 seconds
? Time taken for 1 interval = 3 seconds
In order to strike 12, there are 11 intervals, for which the time taken is 11 × 3 seconds = 33 seconds.
Therefore , Required time taken is 33 seconds .

As per the given above question , we know that
Year 2000 was a leap year.
Number of days remaining in 1999 = 365 - [31 days of January + 28 days of February + 5 days March] = 301 days = 43 weeks, i.e., 0 odd day.
Number of days passed in 2000 = January 31 days have 3 odd days.
February 29 days (being leap year) have 1 odd day
March 5 days have 5 odd days.
? Total number of odd days = 0 + 3 + 1 + 5 = 9 days = 1 week + 2 odd days
Therefore, March 5, 2000 would be two days beyond Friday, i.e., on Sunday.

According to question , we can say that
Total number of odd days = 1600 years have 0 odd day + 300 years have 1 odd day + 49 years (12 leap years + 37 ordinary years) have 5 odd days + 26 days of January have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days.
So, the day was Thursday.

Given that , The day after tomorrow is Sunday.
Therefore today is Friday.
Hence, the day on tomorrow's day before yesterday is given by:
= Friday ? 1 day = Thursday
Therefore , required day will be Thursday .

We know that ,
In every 30 minutes the time of watch increased by 3 minutes = 12 × 3 = 36 minutes
So the time after 6 hours = 5 am + 6 hours + 30 minutes = 11 : 36 am.
Therefore , required time will be 11 : 36 am.