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Important Calendar Aptitude formula

Odd days - Some question date and month will be given. You have to find the day regarding the given date. For these type of questions you can use odd day concept. Odd day concept is nothing but in the given period of time, number days more than the complete week.
Ordinary year - Only 365 days are present in the Ordinary year.
Leap year - A year which is divisible by 4, but it won't be century. But you have another option that every 4th century should be a leap year. It has 366 days.
1 ordinary year = 1 odd day; 1 leap year= 2 odd days; 100 years= 76 ordinary years + 24 leap years; total odd days= 76x1 + 24x2 = 124 odd days= 17 weeks + 5 days; in 100 yrs, there are 5 odd days; in 200 yrs 5x2= 3 odd days; in 300 yrs 5x3= 1 odd day; in 400 yrs (5x4)+1= 0 odd day; in 800 yrs (5x8)+2= 0 odd day
Example for Leap year - The years 2004, 1948 are leap years. 2001, 2007, 2005 are not a leap year.
Counting odd days -
Rule 1 - Ordinary year = 365 days. It has 52 weeks plus 1 day. So ordinary years has 1 odd day.
Rule 2 - Leap year = 366 days. Totally 52 weeks plus 2 days. Leap year has two odd days.
Rule 3 - 50 years = 38 ordinary years + 12 leap years
Totally = (38x1 odd day) + 12x2 odd days)
Therefore 62 odd days, then separate it in to weeks.
So 8 weeks+ 6 odd days.
For 50 years there will be a 6 odd days.
 Calendar aptitude solve examples :
It was Wednesday on February 1 2006, find what day will be on 2010 February 1?
Solutions -
On January 31st 2006 is a Tuesday.
No of odd day from year 2006 to 2009 = 5.
So count 5 days after Tuesday, there will be Wednesday, Thursday, Friday, Saturday, Sunday. The day which will come after Sunday, will be Monday this is the day of February 1 2010.

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