# Calendar shortcuts and important formulas for all competitive exam.

CALENDAR : calendar is used to find many problems related to odd days, leap year, and counting of odd days and many. Condition for Calender :
1. Odd Days:
We are supposed to find the day of the week on a given date. For this, we use the concept of 'odd days'.
or
In a given period, the number of days more than the complete weeks are called odd days.

2. Leap Year:
(a). Every year divisible by 4 is a leap year, if it is not a century.
(b). Every 4th century is a leap year and no other century is a leap year.

Note: A leap year has 366 days.

Examples:
(a) Each of the years 1948, 2004, 1676 etc. is a leap year.
(b) Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.
(c) None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3.Ordinary Year:

The year which is not a leap year is called an ordinary years. An ordinary year has 365 days

Counting of Odd Days:

1. 1 ordinary year = 365 days = (52 weeks + 1 day.) 1 ordinary year has 1 odd day.

2. 1 leap year = 366 days = (52 weeks + 2 days) $\therefore$ 1 leap year has 2 odd days.

3 . 100 years = 76 ordinary years + 24 leap years

= (76 x 1 + 24 x 2) odd days = 124 odd days.

= (17 weeks + days) = 5 odd days.

$\therefore$ Number of odd days in 100 years = 5

Number of odd days in 200 years = (5 x 2) 3 odd days.
Number of odd days in 300 years = (5 x 3) 1 odd day.
Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days

## Calendar Important Examples and Formulas

Question-1: It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Solution: On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

$\therefore$ On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

Question-2: What was the day of the week on 28th May, 2006?
Solution : 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March. April. May
(31 + 28 + 31 + 30 + 28 ) = 148 days
$\therefore$ 148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day. Given day is Sunday.

Question-3 : On what dates of April, 2001 did Wednesday fall?

Solution : We shall find the day on 1st April, 2001.

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.

Question-4 : Today is Monday. After 61 days, it will be:

Solution : Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday. After 61 days, it will be Saturday

## Calendar Important Questions from Previous Year Exams

Calender Aptitude

## Calender Important Video

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