If 27 March, 1995 was a Monday, then what day of the week was 1 November, 1994?

The year 1990 has 356 days, i.e., 1 odd day, year 1991 has 356 days, v 1 odd day, year 1992 has 366 days, i.e., 2 odd days. Likewise year 1993, 1994, 1995 have 1 odd days, so calculated from years 1990-95 = (1 + 1 + 2 + 1 + 1 + 1)
= 7 odd days
= 0 odd day
Hence, the year 1996 will have the same calendar as that of the year 1990.

2nd July, 1984 means (1983 years 6 months and 2 days) 1900 years have 1 odd day 83 years have 20 leap years and 63 ordinary years
= (40 + 63) odd days
= 103
= 5 odd days
6 months and 2 days
Jan 31
Feb 29
Mar 31
Apr 30
May 31
June 30
July 02
= 184 days = 2 odd days
Total number of days = (1 + 5 + 2)
= 8 odd days
= 1 odd day
Hence, it was Monday on 2nd July,1984.

The year 2004 is a leap year. It has 2 odd days. The day on 8th February, 2004 is 2 days before the day on 8th February, 2005.
Hence, this day is Sunday

26th January (Republic day) 2008
? Friday
Clary, it will be on Wednesday in 2012.

Time from 5 am of a particular day to 10 pm on the 4th day is 89 h. Now, the clock loses 16 min in 24 h or on other words, we can say that 23 h 44 min of this clock is equal to 24 h of the correct clock. or (23 + 44/60)
? 356 h of this clock = 24 h of the correct clock
? 89 h of this clock
= (24 x 15/356 x 89) h of correct clock
= 90 h of the correct clock

Total number of odd days from December 17, 1899 to December 22, 1901
14 + 356 + 356 = 735
or, 735/7 = 105
= 0 odd days
it was Saturday on December 17, 1899.
So, it will be Saturday on December 22, 1901.

Friday will fall on 3, 10, 17, 24, 31
So, it will be 5th Friday on 31 st

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours.
(Because between 5 and 7 they point in opposite direction at 6 'o clock only). So, in a day, the hands point in the opposite direction 22 times.

Lets find out 1^st April from Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 1 (since 1^st April)
Month m = 2 (As march = 1, April = 2)
D is the last two digit of year here D = 01 (As year is 2001)
C is 1^st two digit of century here C = 20 (As year is 2001)
f = 1 + [13 x 2 - 1 / 5] + 01 +[01/4] + [20/4] -2 x 20
f = 1 + [25/5] + 01 + [0.5] + [5] - 40.
f = 1 + 5 + 01 + 0 + 5 - 40 = -28
This - ve value of f can be made positive by adding multiple of 7
So f = - 28 + 28 = 0
So number of odd days is 0,
So 1^st April 2001 is Sunday,
So 1^st Friday is on 6^th April , so next Fridays is 13^th, 20^th, 27^th April.

In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year)
Number of days in a period of 100 years is 365 x 100 + 23 or 365 x 100 + 24.
If century year is not a leap year then number of days = 365 x 100 + 23, and number of weeks is 5217 and 4 odd and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.