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) 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.
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
Question1: 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.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
Question2: 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
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.
Question3 : 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.
Question4 : 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
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Question 1 of 13
1. Question
1 pointsWhat was the day on 26 Jan 1948?
Correct
26 Jan, 1948 = (1947 years + Period from 1.1.1948 to 26.1.1948)
Counting of odd days:
Number of odd days in 1600 years = 0.
Number of odd days in 300 years = 15 = 1
(Because when 15 divided by 7, we get remainder 1)47 years = 11 leap years + 36 ordinary years
= (11 × 2 + 36 × 1)odd days = 58 odd days
= (8 weeks + 2 days) = 2 odd days
∴ 1947 years have (0 + 1 + 2) = 3 odd days
Jan
26 = 26 days26 days = (3 weeks + 5 days) = 5 odd days
∴ Total number of odd days = (5 + 3) = 8 = 1
Hence, the required day is Monday.
Incorrect
26 Jan, 1948 = (1947 years + Period from 1.1.1948 to 26.1.1948)
Counting of odd days:
Number of odd days in 1600 years = 0.
Number of odd days in 300 years = 15 = 1
(Because when 15 divided by 7, we get remainder 1)47 years = 11 leap years + 36 ordinary years
= (11 × 2 + 36 × 1)odd days = 58 odd days
= (8 weeks + 2 days) = 2 odd days
∴ 1947 years have (0 + 1 + 2) = 3 odd days
Jan
26 = 26 days26 days = (3 weeks + 5 days) = 5 odd days
∴ Total number of odd days = (5 + 3) = 8 = 1
Hence, the required day is Monday.

Question 2 of 13
2. Question
1 pointsWhat was the day on 6th Feb 1994?
Correct
6th Feb, 1994 = (1993 years + Period from 1.01.1993 to 6.02.1994)
Counting of odd days:
Number of odd days in 2000 years = 0 odd day
Number of odd days in 300 years = (5 × 3) = 15 = 1
(When 15 is divided by 7, it gives 1 as the remainder.)93 years = (23 leap year + 70 ordinary year)
⇒ (23 × 2 + 70 × 1) ⇒ (46 + 70) = 116 days.
⇒ 116 days = (16 Weeks + 4 days) = 4 odd days.
∴ 1993 years have = (0 + 1 + 4) = 5 odd days.
Jan Feb
31 + 6 = 37 odd days37 days = (5 weeks + 2 days) = 2 odd days
∴ Total numbers of odd days = (5 + 2) = 7.
Hence, the required day is Sunday.
Incorrect
6th Feb, 1994 = (1993 years + Period from 1.01.1993 to 6.02.1994)
Counting of odd days:
Number of odd days in 2000 years = 0 odd day
Number of odd days in 300 years = (5 × 3) = 15 = 1
(When 15 is divided by 7, it gives 1 as the remainder.)93 years = (23 leap year + 70 ordinary year)
⇒ (23 × 2 + 70 × 1) ⇒ (46 + 70) = 116 days.
⇒ 116 days = (16 Weeks + 4 days) = 4 odd days.
∴ 1993 years have = (0 + 1 + 4) = 5 odd days.
Jan Feb
31 + 6 = 37 odd days37 days = (5 weeks + 2 days) = 2 odd days
∴ Total numbers of odd days = (5 + 2) = 7.
Hence, the required day is Sunday.

Question 3 of 13
3. Question
1 pointsWhat was the day 21st August 1992?
Correct
21st august 1992 = (1991 years + Period from 1.01.1992 to 21.08.1992)
Counting of odd numbers:
Numbers of odd days in 1600 = 0
Numbers of odd days in 300 years = 5 × 3 = 15 = 1
(When 15 is divided by 7, it gives 1 as the remainder.)91 years = (22 leap yeas + 69 ordinary years)
⇒ (22 × 2 + 69 × 1) = (44 + 69) = 113 days
⇒ 113 days = (16 weeks + 1 day) = 1 day
∴ 1991 years have = (0 + 1 + 1) = 2 odd day
Jan Feb Mar Apr May June July Aug
31 + 29 + 31 + 30 + 31 + 30 + 31 + 21 = 234 days.234 days = (33 weeks + 3 odd days) = 3 odd days
∴ Total odd days in numbers = (3 + 2) = 5.
So the day is Friday.
Incorrect
21st august 1992 = (1991 years + Period from 1.01.1992 to 21.08.1992)
Counting of odd numbers:
Numbers of odd days in 1600 = 0
Numbers of odd days in 300 years = 5 × 3 = 15 = 1
(When 15 is divided by 7, it gives 1 as the remainder.)91 years = (22 leap yeas + 69 ordinary years)
⇒ (22 × 2 + 69 × 1) = (44 + 69) = 113 days
⇒ 113 days = (16 weeks + 1 day) = 1 day
∴ 1991 years have = (0 + 1 + 1) = 2 odd day
Jan Feb Mar Apr May June July Aug
31 + 29 + 31 + 30 + 31 + 30 + 31 + 21 = 234 days.234 days = (33 weeks + 3 odd days) = 3 odd days
∴ Total odd days in numbers = (3 + 2) = 5.
So the day is Friday.

Question 4 of 13
4. Question
1 pointsWhat day was it on 1st Jan 2004?
Correct
1st Jan, 2004 = (2003 years + 1.01.2004)
Counting of odd days:
Numbers of odd days in 2000 = 0
Numbers of odd days in 3 ordinary years = 3.
Total odd day in year 2003 = (0 + 3) = 3.
Jan
1 = 1 odd day .Total odd days = (3 + 1) = 4
Hence, the required day is Thursday.
Incorrect
1st Jan, 2004 = (2003 years + 1.01.2004)
Counting of odd days:
Numbers of odd days in 2000 = 0
Numbers of odd days in 3 ordinary years = 3.
Total odd day in year 2003 = (0 + 3) = 3.
Jan
1 = 1 odd day .Total odd days = (3 + 1) = 4
Hence, the required day is Thursday.

Question 5 of 13
5. Question
1 pointsWhat day was it on 2nd may 2010?
Correct
2nd may 2010 = (2009 years + period from 1.01.2010 to 2.05.2010)
Counting odd days:
Numbers of odd days in 2000 = 0
Numbers of odd days 9 years = (2 leap years + 7 ordinary years) = (2 × 2 + 7 × 1) = 11 odd days = 4 odd days.
Total Odd day in 2009 year = (0 + 4) = 4 odd day
Jan Feb Mar Apr May
31 + 28 + 31 + 30 + 2 = 122 days⇒ 122 days = (17 weeks + 3 day) = 3 odd day
∴ Total number of days = (3 + 4) = 7.
Hence, the required day is Sunday.
Incorrect
2nd may 2010 = (2009 years + period from 1.01.2010 to 2.05.2010)
Counting odd days:
Numbers of odd days in 2000 = 0
Numbers of odd days 9 years = (2 leap years + 7 ordinary years) = (2 × 2 + 7 × 1) = 11 odd days = 4 odd days.
Total Odd day in 2009 year = (0 + 4) = 4 odd day
Jan Feb Mar Apr May
31 + 28 + 31 + 30 + 2 = 122 days⇒ 122 days = (17 weeks + 3 day) = 3 odd day
∴ Total number of days = (3 + 4) = 7.
Hence, the required day is Sunday.

Question 6 of 13
6. Question
1 pointsWhat was the day on 8th March 1995?
Correct
8th March, 1995 = (1994 years + Period form 1.01.1995 to 8.03.1995)
Counting of odd days:
Numbers of odd days in 1600 years = 0.
Numbers of odd days in 300 years = (5 × 3) = 15 = 1.
(When 15 is divided by 7, it gives 1 as the remainder.)94 years = (23 leap years + 71 ordinary years)
⇒ (23 × 2 + 71 × 1) = 117 odd days ⇒ (16 weeks + 5 days) = 5 odd days
1994 years have = (0 + 5 + 1) = 6 odd days.
Jan Feb Mar
31 + 28 + 8 = 67 days = (9 weeks + 4 odd days) = 4 odd daysTotal numbers of odd days = (6 + 4) = 10 = 3,
Hence, the required day is Wednesday.
Incorrect
8th March, 1995 = (1994 years + Period form 1.01.1995 to 8.03.1995)
Counting of odd days:
Numbers of odd days in 1600 years = 0.
Numbers of odd days in 300 years = (5 × 3) = 15 = 1.
(When 15 is divided by 7, it gives 1 as the remainder.)94 years = (23 leap years + 71 ordinary years)
⇒ (23 × 2 + 71 × 1) = 117 odd days ⇒ (16 weeks + 5 days) = 5 odd days
1994 years have = (0 + 5 + 1) = 6 odd days.
Jan Feb Mar
31 + 28 + 8 = 67 days = (9 weeks + 4 odd days) = 4 odd daysTotal numbers of odd days = (6 + 4) = 10 = 3,
Hence, the required day is Wednesday.

Question 7 of 13
7. Question
1 pointsWhat was the day on 13th April 1999?
Correct
13th April, 1999 = (1998 years + Period from 1.01.199 to 13.04.1999)
Counting of odd days:
Numbers of odd days in 1600 years = 0.
Numbers of odd days in 300 years = (5 × 3) = 15 = 1.
(When 15 is divided by 7, it gives 1 as the remainder.)98 years = (24 leap year + 74 ordinary years) ⇒ (24 × 2 + 74 × 1) = (48 + 74) = 122 days.
⇒ 122 days = (17 weeks + 3 day) = 3odd days
∴ 1998 years have = (0 + 1 + 3) = 4 odd days.
Jan Feb Mar April
31 + 28 + 31 + 13 = 103 days⇒ 103 days = (14 weeks + 5 days) = 5 odd days
∴ Total number of odd days = (5 + 4) = 9 = 2.
Hence, the required day is Tuesday.
Incorrect
13th April, 1999 = (1998 years + Period from 1.01.199 to 13.04.1999)
Counting of odd days:
Numbers of odd days in 1600 years = 0.
Numbers of odd days in 300 years = (5 × 3) = 15 = 1.
(When 15 is divided by 7, it gives 1 as the remainder.)98 years = (24 leap year + 74 ordinary years) ⇒ (24 × 2 + 74 × 1) = (48 + 74) = 122 days.
⇒ 122 days = (17 weeks + 3 day) = 3odd days
∴ 1998 years have = (0 + 1 + 3) = 4 odd days.
Jan Feb Mar April
31 + 28 + 31 + 13 = 103 days⇒ 103 days = (14 weeks + 5 days) = 5 odd days
∴ Total number of odd days = (5 + 4) = 9 = 2.
Hence, the required day is Tuesday.

Question 8 of 13
8. Question
1 pointsWhat was the day on 14th Feb 2010?
Correct
14th Feb, 2010 = (2009 years + Period from 1.01.2010 to 14.02.2010)
Number of odd days in 2000 = 0.
Number of odd days in 9 years = (2 leap years + 7 ordinary years) ⇒ (2 × 2 + 7 × 1) = 11 days = 4 odd days
(When 11 is divided by 7, it gives 4 as the remainder.)∴ 2009 years have = (0 + 4) = 4 odd days
Jan Feb
31 + 14 = 45 days ⇒ (6 weeks + 3 days) = 3 odd days.∴ Total numbers of odd days = (0 + 4 + 3) = 7 odd days.
It means the required day is Sunday.
Incorrect
14th Feb, 2010 = (2009 years + Period from 1.01.2010 to 14.02.2010)
Number of odd days in 2000 = 0.
Number of odd days in 9 years = (2 leap years + 7 ordinary years) ⇒ (2 × 2 + 7 × 1) = 11 days = 4 odd days
(When 11 is divided by 7, it gives 4 as the remainder.)∴ 2009 years have = (0 + 4) = 4 odd days
Jan Feb
31 + 14 = 45 days ⇒ (6 weeks + 3 days) = 3 odd days.∴ Total numbers of odd days = (0 + 4 + 3) = 7 odd days.
It means the required day is Sunday.

Question 9 of 13
9. Question
1 pointsWhat was the day on 5th may 2003?
Correct
5th May, 2003 = (2002 years + Periods from 1.01.2003 to 5.05.2003)
Number of odd days in year 2000 = 0
Number of odd days in 2 ordinary years = 2
∴ 2002 years have = (0 + 2) = 2 odd days
Jan Feb Mar April May
31 + 28 + 31 + 30 + 5 = 125 days⇒ 125 days = (17 weeks + 6 days) = 6 odd days
Total number of odd days = (2 + 6) = 8 = 1 odd day
It means the required day is Monday.
Incorrect
5th May, 2003 = (2002 years + Periods from 1.01.2003 to 5.05.2003)
Number of odd days in year 2000 = 0
Number of odd days in 2 ordinary years = 2
∴ 2002 years have = (0 + 2) = 2 odd days
Jan Feb Mar April May
31 + 28 + 31 + 30 + 5 = 125 days⇒ 125 days = (17 weeks + 6 days) = 6 odd days
Total number of odd days = (2 + 6) = 8 = 1 odd day
It means the required day is Monday.

Question 10 of 13
10. Question
1 pointsThe calendar for the year 2013 will be the same for the year:
Correct
The same calendar year will be the next year to the one in which we get the counting of odd days as 0 (In the multiple of 7).
Calculation of odd days:
1 + 1 + 1 + 2 + 1 + 1 = 7 = 0 odd day
Hence, the next same calendar year will be 2019.
Incorrect
The same calendar year will be the next year to the one in which we get the counting of odd days as 0 (In the multiple of 7).
Calculation of odd days:
1 + 1 + 1 + 2 + 1 + 1 = 7 = 0 odd day
Hence, the next same calendar year will be 2019.

Question 11 of 13
11. Question
1 pointsWhich of the following is a leap year?
Correct
1336 is a leap year, because it's completely divided by 4.
Incorrect
1336 is a leap year, because it's completely divided by 4.

Question 12 of 13
12. Question
1 pointsToday is Sunday, after 39 days, it will be
Correct
Each day of the week is repeated after 7 days
So, after 35 days it will be Sunday.
The 39th day, hence, will be 35 + 4 = Thursday.Incorrect
Each day of the week is repeated after 7 days
So, after 35 days it will be Sunday.
The 39th day, hence, will be 35 + 4 = Thursday. 
Question 13 of 13
13. Question
1 pointsOn what dates of April 2005 did Tuesday fall?
Correct
First we find the day on 1.04.2005
1.04.2005 = (2004 years + period From 1.01.2005 to 1.04.2005)
Odd days in 2000 years = 0
4 years = (1 leap year + 3 ordinary years)
= (1 × 2 + 3 × 1) odd days = 5 odd days
Jan Feb Mar Apr
31 + 28 + 31 + 1 = 91 days = (13 weeks) = 0 odd dayTotal number of odd days = (0 + 5) = 5 odd days
On 1.04.2005 was Friday, so Tuesday lies on 5.04.2005
Hence, Tuesday lies on 5th of April, 2005.
Incorrect
First we find the day on 1.04.2005
1.04.2005 = (2004 years + period From 1.01.2005 to 1.04.2005)
Odd days in 2000 years = 0
4 years = (1 leap year + 3 ordinary years)
= (1 × 2 + 3 × 1) odd days = 5 odd days
Jan Feb Mar Apr
31 + 28 + 31 + 1 = 91 days = (13 weeks) = 0 odd dayTotal number of odd days = (0 + 5) = 5 odd days
On 1.04.2005 was Friday, so Tuesday lies on 5.04.2005
Hence, Tuesday lies on 5th of April, 2005.
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