How many Sundays will be in a period of 100 years.

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.

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.

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

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

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.

In a year we have 53 Sundays only when year start with Sunday (For non leap year) and either with Saturday of Sunday (For leap year).
From zeller's formula 1^st January 2001 was Monday
From number of odd days 1^st January 2006 will start with Sunday so it had 53 Sunday
Similar 1^stJanuary 2009 will start with Thursday.

Let us take two cases
Case (i) : When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays
Case (ii) : When year start with Sunday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays.

If month ends with Thursday then next month will start with Friday and it may have 5 Friday otherwise it may have 4 Fridays.