Two clocks are set right at 10 am. One gains 20 seconds and the other loses 40 seconds in 24 hours. What will be the true time when the first clock indicates 4 pm on the following day?

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.

We can say that ,
During the interval we have two leap years as 1992 and 1996 and it contains February of both these years.
? The interval has ( 5 + 2 ) = 7 days = 0 odd day.
Hence, January 7, 1997 was also Tuesday.

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.