Calendar Test-1

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴After 61 days, it will be Saturday.

1992 being a leap year it had two odd days show the first day of the year 1993 was today beyond so Wednesday i.e it was Friday.

On 31^st  December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴On 31^st  December 2009, it was Thursday.
Thus, on 1^st  Jan, 2010 it is Friday.

28 May, 2006 i.e. 2005 years and  Period from 1.1.2006 to 28.5.2006

Odd days in 1600 years = 0
Odd days in 400 years = 0

5 years = 4 ordinary years + 1 leap year  = (4 x 1) + (1 x 2) = 6 odd days

Jan (31) + Feb  (28) + March (31) + April (30) + May (28) = 148 days
∴148 days = 21 weeks + 1 day  = 1 odd day
Total number of odd days = 0 + 0 + 6 + 1  = 7 = 0 odd day.

Thus, thegiven day is Sunday .

17^th  June, 1998 i.e. 1997 years + Period from 1.1.1998 to 17.6.1998

Odd days in 1600 years = 0
Odd days in 300 years = 5 x 3  = 1

97 years = 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan (31) + Feb (28) + March (31) + April  (30) + May (31) + June (17) = 168 days
Therefore 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Thus, the given day is Wednesday .

15^th  August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡4 odd days.

Jan.   Feb.  March  April   Mayb  June   July    Aug.
(31  +  28  +  31  +  30  +  31  +  30  +  31  +  15) = 227 days

∴227 days = (32 weeks + 3 days) ≡3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡0 odd days.
Given day is Sunday.

The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴The day on 6^th  March, 2005 will be 1 day beyond the day on 6^th  March, 2004.
Given that, 6^th  March, 2005 is Monday.
∴6^th  March, 2004 is Sunday (1 day before to 6^th  March, 2005).

We shall find the day on 1^st  April, 2001.
1^st  April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1)   = 91 days ≡0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1^st  April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4^th , 11^th , 18^th  and 25^th .

x  weeks  x  days = (7x  + x) days = 8x  days.

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