Calendar with Explanation

Answer- Option B

Explanation-

17 October 2005 = (2004 years + Period from 1.1.2005 to 17.10.2005)
Odd days in 2000 years = 0 odd days
4 years has 3 ordinary days + 1 leap year
Number of odd days in 4 years = (3 * 1 + 1 * 2) = 5 odd days

 Jan.   Feb.   March   April   May   June   July   Aug.   Sept.   Oct. 

 31   + 28   + 31    + 30    + 31  + 30   + 31   + 31   + 30    + 17  = 290 days 

290days = 3 odd days
Total number of odd days = 3 + 5 = 8
1 odd day
The day was Monday on 17^th October 2005

Answer- Option D

Explanation-

25^th March 1997 = (1996 years + Period from 1.1.1997 to 25.03.1997)
Odd days in 1600 years = 0 odd days
Odd days in 300 years = 1 odd days
96 years has 24 leap days + 72 ordinary years
Number of odd days in 96 years = (24 * 2 + 72 * 1) = 120 odd days = 1 odd day

 Jan.   Feb.   March 

 31   + 28   + 25 = 84 days 

84 days = 0 odd days
Total number of odd days = 1 + 1 + 0 = 2
2 odd days
The day was Tuesday on 25^th March 1997

Answer- Option A

Explanation-

30^th April 2008 = (2007 years + Period from 1.1.2008 to 30.04.2008)
Odd days in 2000 years = 0 odd days
7 years has 6 ordinary years + 1 leap year
Number of odd days in 7 years = (6 * 1 + 1 * 2) = 8 odd days = 1 odd day

 Jan.   Feb.   March   April 

 31   + 29   + 31    + 30 = 121 days 

121 days = 2 odd days
Total number of odd days = 1 + 2 = 3
3 odd day
The day was Wednesday on 30^th April 2008

Answer- Option A

Explanation-

First of all, we will determine the day on 1^st June 2002
1^st 2002 = (2001 years + period from 1.1.2002 to 1.06.2002)
Odd days in 2000 years = 0 odd days
1 year has 1 ordinary day
Number of odd days in 1 year = 1 odd day

 Jan.   Feb.   March   April   May   June 

 31   + 28   + 31    + 30    + 31  + 1  = 152 days 

152 days = 5 odd days
Total number of odd days = 1 + 5 = 6
On 1^st June 2002, it was Saturday
So, in June 2002 Thursday falls on
6^th, 13^th, 20^th, 27^th

Answer- Option C

Explanation-

To show this problem, we have to count the number of odd days from 2009 onwords to obtain the sum which is equal to 0 odd day

 Year   :   2009   2010   2011   2012   2013   2014 

 Odd day:   1     1       1      2      1      1 

Sum = 7 odd days = 0 odd day
Year 2015 will be same as of year 2009

Answer- Option B

Explanation-

There is 1 odd day in the year 2009
So, the day on 15^th September 2010 will be one day beyond the day on 15^th September 2009
But, 15^th September 2010 is Wednesday
On 15^th September 2009, it was Tuesday

Answer- Option C

Explanation-

There are 2 odd days in the year 2004
So, the day on 21^th January 2005 will be two days beyond the day on 21^th January 2004
On 21^th January 2004, it was Wednesday