Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used. The article **Calendar Quiz 5** lists important **Calendar** practice questions for competitive exams like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams** and etc.

Examples:

- Years like 1988, 2008 are leap year (divisible by 4).

Centuries like 2000, 2400 are leap year ( divisible by 400).

Years like 1999, 2003 are not leap year (not divisible by 4).

Centuries like 1700, 1800 are not leap year ( not divisible by 400).

In a century, there is 76 ordinary year and 24 leap year.

An ordinary year has one odd day.

1st day of the year 2007 was Monday.

1st day of the year 2008 will be 1 day beyond Monday.

Hence, it will be Tuesday.

Year |
2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 |

Odd day |
1 | 2 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 2 | 1 |

Sum = 14 odd days = 0 odd day.

i.e, Calendar for the year 2018 will be the same as for the year 2007.

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

Counting of odd days:

2000 years have 0 odd day.

2001 to 2006 has 5 ordinary years and 1 leap year

i.e, 5 + 2 = 7 odd day = 0 odd day

Jan | Feb | Mar | Apr | May | June | July | Aug |

31 | 28 | 31 | 31 | 30 | 30 | 31 | 29 |

= 241 days = 34 weeks 3 days = 3 odd days.

Total No. of odd days = 0 + 0 + 3 = 3 odd days

August 29th 2007 falls on Wednesday

1st day of the year 2008 is Tuesday (Given)

So, 1st day of the year 2009 is 2 days beyond Tuesday.

Hence, it will be Thursday.

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