The year next to 1990 will have the same calender 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

At 3 O'clock, then minute hand is 15 min away from the hour hand. To coincide, it must gain 15 min as 55 min are gained in 60 min.
15 min are gained in (60/55) x 15 min = 16⁴/₁₁ min
Hence, the hands are coincide at 16⁴/₁₁ min past 3.

Here, 27th March 1995 was Monday. Now for calculating total number of odd days. First, we calculate total number of days till 1 Number 1994
? Number of days in March 1995 = 27
Number of days in February 1995 = 28
Number of days in January 1995 = 31
Number of days in December 1994 = 31
Number of days in November 1994 = 29/146
? Number of odd days = 146/7 = 20⁶/₇
So, 6 odd days.
? On November 1, 1994 = Monday - 6
= Tuesday

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.