A clock takes 9 seconds to strike 4 times. In order to strike 12 times at the same rate, the time taken is:

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° .

According to question , we know that
Starting with 2000, count for number of odd days in successive years till the sum is divisible by 7.
2000 + 2001 + 2002 + 2003 + 2004 = 2 + 1 + 1 +1 + 2 = 7 odd days
? Number of odd days up to 2004 = 0 odd day
Hence , Calendar for 2000 will serve for 2005 also.

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 .