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 3** 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.

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan Feb Mar Apr May Jun Jul

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days)

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

Required day was 'Tuesday'.

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 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

1st 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 1st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.

But, 8th Dec, 2007 is Saturday

So, 8th Dec, 2006 is Friday.

The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.

Hence, this day is Sunday

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