## Concept of Calender, Date, Odd Days, Leap Year, Ordinary Year, Finding Day of given dates etc

Calender
A calendar is chart or series of pages showing the days, weeks and months of a particular year. A calendar consist of 365 days divided into 12 months.

## Ordinary Year

A year having 365 days is called an ordinary year. An ordinary year contains 52 complete weeks + 1 extra days = 365 days

## Leap Year

A leap year has 366 days ( the extra day is 29th of February) ( 52 complete weeks +2 extra days = 366 days).
A leap year is divisible by 4 except for a century. For a century to be a leap year, it must be divisble by 400 . e.g.
·         Years like 1988, 2008 are leap year (divisble by 4).
·         Centuries like 2000, 2400 are leap year ( divisble by 400).
·         Years like 1999, 2003 are not leap year (not divisble by 4).
·         Centuries like 1700, 1800 are not leap year ( not divisble by 400).
·         In a century, there is 76 ordinary year and 24 leap year.

## Odd Days

Extra days, apart from the complete weeks in a given periods are called odd days. An ordinary year has 1 odd day while a leap year has 2 odd days.
To find the numbers of odd days
·         Number of days in an ordinary year = 365 = (52 x 7) + 1 = 52 weeks + 1 odd day. Thus, an ordinary year has 1 odd day.
·         Number of days in a leap year = 366 = (52 x7) + 2 = 52 weeks + 2 days. Thus, a leap year has 2 odd days.
·         Number of days in a century (100 years) = 76 ordinary years + 24 leap years = ( 76 x 1 + 24 x 2) = 124 = 17 x 7 +5 = 17 weeks + 5 odd days. Thus, 100 years has 5 odd days.
·         Number of odd days in 200 years = 5 x 2 = 10 days = 1 week + 3 days = 3 odd days
·         Number of odd days in 300 years =  5 x 3 = 15 days = 2 weeks + 1 days = 1 odd day
·         Number of odd days in 400 years = (5 x 4 +1) days = 21 days= 3 weeks = 0 odd days . Here please note that as 400th is a leap year, therefore 1 more day has been taken.
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years, 2400 years etc. has no odd days. Remember the adjacent table for the number of odd days in different  months of an year.
 ·         In an ordinary year, February has no odd days, but in a leap year, February has one odd day. ·         The 1st day of a century must be Tuesday, Thursday or Saturday. ·         The last day of a century cannot be Tuesday, Thursday or Saturday.

 Months Odd days January 3 February 0/1  (Orinary/ leap) March 3 April 2 May 3 June 2 July 3 August 3 September 2 October 3 November 2 December 3

# Day Gain/ Loss

## Ordinary year ( +/- 1 day)

·         When we proceed forward by 1 year, then 1 day is gained. For example 9th August 2014 has to be Friday + 1 = Saturday.
·         When we move backward by 1 year, then 1 day is lost. For example – 24th December 2013 is Tuesday, then 24th December 2012 has to be Tuesday – 1 = Monday.

## Leap Year ( +/-  2 days)

·         When we proceed forward by 1 leap year, then 2 days are gained. For example , If it is Wednesday on 25th December 2011, then it would be Friday on 25th December 2012 ( Wednesday +2) because 2012 is a leap year.
·         When we move backward by 1 leap year, then 2 days are lost. For example, If it is Wednesday on 18th December 2012, then it would be Monday on 18th December 2011. [Wednesday – 2] because 2012 is a leap year.

# Exceptions

·         The day must have crossed 29th february for adding 2 days otherwise 1 day. For Example, If 26thJanuary 2011 is Wednesday, 26th January 2012 would be Wednesday +1 = Thursday ( even if 2012 Is a leap year , we added +1 day because 29th February is not crossed). If 23rd March 2011 is Wednesday, then 23rd March 2012 would be Wednesday +2 = Friday ( +2 days because 29thFebruary is crossed)

# To Find a Particular Day on the Basis of Given Day and Date

Following steps are taken into consideration to solve such questions.
Step – 1 : Firstly, you have to find the number of odd days between the given date and the date for which the day is to be determined.
Step-2: The day ( for a particular date) to be determined, will be that day of the week which is equal to the total number of odd days and this number is counted forward from the given day, in case the given day comes before the day to be determined. But, if the given day comes after the day to be determined, then the same counting is done backward from the given day.
Ex- If 5th January 1991 was staurday, what day of the week was it on 4th march 1992?
Sol – Number of days between 5th January 1991 and 4th March 1992 = (265 -5) days of the year 1991 +  31 days of January 1992 + 29 days of February 1992 + 1 day of March 1992 (as 1992 is completely divisible by 4, hence it is a leap year and that’s why February has 29 days)
= 360 + 31 + 29 + 4 = 424
= 60 weeks + 4 days
Hence, number of odd days = 4
4th March 1992 will be the 4th day beyond Saturday. So, the required day will be Wednesday.

# To Find a Particular Day without Given Date and Day

Following steps are taken into consieration to solve such questions
Step-1 : Firstly, you have to find the number of odd days upto the date for which the day is to be determined.
Step -2: You are required day will be according to the following conditions:
1.       If the number of odd days = 0, then required day is Sunday.
2.       If the number of odd days = 1, then requried day is Monday.
3.       If the number of odd days =2, then required day is Tuesday.
4.       If the number of odd days =3, then required day is Wednesday.
5.       If the number of odd days =4, then required day is Thursday.
6.       If the number of odd days = 5, then required day is Friday.
7.       If the number of odd days =6, then required day is Saturday.
Ex- Find the day of the week on 26th January 1950.
Sol –Number of odd days upto 26th January 1950 = Odd days for 1600 years + Odd days for 300 years + Odd days for 49 years + Od days of 26 days of January 1950
= 0 +1 + (12 x 2 + 37 ) + 5
=0 + 1 + 61 + 5
= 67 days
=9 weeks + 4 days
= 4 odd days
Thus, it was Thursday on 26th January 1950.
Note- 49 years has 12 leap year and 37 ordinary year.

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