Calendar Questions

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2005, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 2005 + 6) = 2011

Therfore, The calendar for the year 2005 will be same for the year 2011

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

We cannot find out the answer because the number of days of the current month is not given.

Let us find the day on 1st July, 2004.

2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1

= 183 days = (26 weeks + 1 day) = 1 t .

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '

:. 1st July 2004 was 'Thursday',-,-

Thus, 1st Monday in July 2004 _as on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .

First,we count the number of odd days for the left over days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result.