In a month of 31 days, third Thursday falls on 16th. What will be the last day of the month?

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.

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

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

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.

Calendar of 2 years is similar if number of odd days between these two years is zero. Consider option one by one
(a) Between 1940 to 1946 we have two leap years 1940 and 1944 so number of odd days is 2 + 1 + 1 + 1 + 2 + 1 = 8 or 1 hence calendar of these two years is not similar.
(b) Between 1977 to 1982 we have one leap years 1980 so number of odd days is 1 + 1 + 1 + 1 + 2 + 1 = 6 calendar of these two years is not similar.
(c) Between 1912 to 1916 we have one leap years 1912 so number of odd days is 2 + 1 + 1 + 1 = 5 hence calendar of these two years is not similar.

Number of odd days between 2013 and 2015 is 2, so we have to find the year which will have 2 odd days between 1977 and required year.
Consider options one by one -
(a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days
(b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days
(c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.