Calendar types of questions are frequently asked in major aptitude and government papers. These questions are pretty easy to solve if you understand the concepts. The first part is to know; what is a calendar and what does the calendar contain? The calendar contains years, months, weeks and days. In an ordinary year, there are 365 days, which means 52 x 7 + 1, or 52 weeks and one day. This additional day is called an odd day. The concept of odd days is very important in calendars. In a century – i.e. 100 years, there will be 24 leap years and 76 non-leap years.
Table of content
1) Facts about ordinary year
2) Facts about leap year
3) Steps to solve calendar questions
4) Find the day based on Dates questions
5) Questions Based on Odd Days
Important Notes on Ordinary year
1. In an ordinary year there are 12 months.
2. In an ordinary year there are 52 weeks and one odd day.
3. In an ordinary year there are 365 days.
Important Notes on Leap year
1. In a Leap year there are 12 months.
2. In a Leap year there are 52 weeks and 2 odd days.
3. In a Leap year there are 366 days.
Important facts about calendar
Important note:
1. If any year is exactly divisible by 4 then the year will be called Leap Year.
For Example: Is 1996 an exact divisible by 4? Yes, it is exactly divisible by 4. So, 1996 is a leap year.
For Example: Is 1993 an exact divisible by 4? No, it is not exactly divisible by 4. So,1993 is an ordinary year.
2. If any century is exactly divisible by 400 then the century is a leap year.
For Example: Is 1600 an exact divisible by 400? Yes, it is exactly divisible by 400. So, 1600 is a leap year.
For example: Is 1700 an exact divisible by 400? No, it is not exactly divisible by 400. So, 1700 is not a leap year.
There are some codes which is required to know before we dive into some questions?
Day , Months and Year important Codes to remember
| Days | Code |
|---|---|
| Sunday | 0 |
| Monday | 1 |
| Tuesday | 2 |
| Wednesday | 3 |
| Thursday | 4 |
| Friday | 5 |
| Saturday | 6 |
| Month | Code |
|---|---|
| January | 0 |
| February | 3 |
| March | 3 |
| April | 6 |
| May | 1 |
| June | 4 |
| July | 6 |
| August | 2 |
| September | 5 |
| October | 0 |
| November | 3 |
| December | 5 |
| Year | Code |
|---|---|
| 1600 – 1699 | 6 |
| 1700 – 1799 | 4 |
| 1800 – 1899 | 2 |
| 1900 – 1999 | 0 |
| 2000 – 2099 | 6 |
Steps to solve the calendar questions
This is the format on which you can solve a typical type of questions
1. Last two digit of the year.
2. Divide the last two digit of the year by 4 and take the quotient alone
3. Take the date
4. Take the code of the number of the month
5. Take the code of the number of the year on where it lies
This is the final step. Add it all and divide by 7 and match the remainder with the days code to figure what day it is.
Questions – Find the day on the basis of Date
Question 1) What was the day on 26th Jan 1947?
Solution: Now, let’s go as per the format:
1. Last two digit of the year = 47 (1947)
2. Divide the last two digit of the year by 4 and take the quotient alone = 11 (47/4 = 11 is the quotient)
3. Take the date = 26
4. Take the code of the number of the month = 0 (The code of the January is ‘0’)
5. Take the code of the number of the year where it lies = 0 (The value between 1900 – 1999 is ‘0’)
The addition to all this = 84.
![\[ \frac{84}{7}\]](https://padhle.com/wp-content/ql-cache/quicklatex.com-1d5e9f1d4a0a3a49c0f896c8b2dd935c_l3.png "Rendered by QuickLaTeX.com")
84 is an exact divisible of 7 and so the remainder is 0.
‘Sunday’ is coded as ‘0’.
So, the answer must have been sunday.
Answer: Sunday
Question 2) What was the day of the week on 5th October 2016?
Solution: Let’s proceed as per the format.
1. Take the last two digit of the year = 16 (2016)
2. Divide the last two digit of the year and take the quotient alone = 4 (16/4 = 4 (Quotient)
3. Take the date = 5
4. Take the code of the number of the month = 0 (The code for the October is ‘0’)
5. Take the code of the year on where it lies = 6 ( 2016 lies between 2000 – 2099)
The addition = 31
![\[ 31 = 7 * 4 = 28 + 3\]](https://padhle.com/wp-content/ql-cache/quicklatex.com-d2688a414a11a6691064b390f75605e8_l3.png "Rendered by QuickLaTeX.com")
The remainder is 3.
Wednesday is coded as ‘3’.
Answer: Wednesday
Question 3) What was the day on the week of 26th September 1998?
Solution: Let’s proceed as per the format.
1. Take the last two digit of the year = 98 (1998)
2. Divide the last two digit of the year and take the quotient alone = 24 (98/4 = 24 (Quotient)
3. Take the date = 26
4. Take the code of the number of the month = 5 (September is coded as ‘5’)
5. Take the code of the year on where it lies = 0 (1998 lies between 1900-1999)
Addition = 153.
![\[ 153 = 7 * 21 + 6 \]](https://padhle.com/wp-content/ql-cache/quicklatex.com-1d6c15de5e566dc199bee61a2fb6f023_l3.png "Rendered by QuickLaTeX.com")
6 is the remainder
Saturday is coded as 6.
Answer: Saturday
Question 4) What is the day on the week of 24th March 2038?
Solution: Let’s proceed as per the format.
1. Take the last two digit of the year = 38 (2038)
2. Divide the last two digit of the year and take the quotient alone = 9 (38/4 = 9 (Quotient)
3. Take the date = 24
4. Take the code of the number of the month = 3 (March is coded as ‘3’)
5. Take the code of the year on where it lies = 6 (2038 lies between 20000-2099)
Addition = 80.
![\[ 153 = 7 * 11 + 3 \]](https://padhle.com/wp-content/ql-cache/quicklatex.com-d83999eaf29497f20d08c3b443675503_l3.png "Rendered by QuickLaTeX.com")
3 is the remainder
Wednesday is coded as 3
Answer: Wednesday
Important note:
When it is a leap year and if the month is January and February we are supposed to subtract 1 from the remainder.
Question 5) What was the day on 4th Feb 2004?
Solution: Is 2004 a leap year? Yes, it is a leap year. Is the month Jan or Feb? Yes, it is the month of Feb. This means we need to subtract 1 from the remainder.
Let’s proceed as per the format.
1. Take the last two digit of the year = 04 (2004)
2. Divide the last two digit of the year and take the quotient alone = 1 (04/4 = 1 (Quotient)
3. Take the date = 4 (4th Feb)
4. Take the code of the number of the month = 3 (February is coded as ‘3’)
5. Take the code of the year on where it lies = 6 (2038 lies between 20000-2099)
Addition = 18.
![\[ 18 = 7 * 2 + 4 \]](https://padhle.com/wp-content/ql-cache/quicklatex.com-b8e2a6880fe938f3963e93812521e0d9_l3.png "Rendered by QuickLaTeX.com")
4 is the remainder
But it is a leap year and the month is Feb. So, reduce 1 from the remainder.
Remainder = 4 -1 = 3
Wednesday is coded as 3
Answer: Wednesday
Questions Based on Odd days
Question 1 ) Today is Monday, after 30 days it will be?
Solution: It is given it is a Monday.
Divide 30 by 7 and find the remainder.
![\[ 30 = 7*4 + 2 \]](https://padhle.com/wp-content/ql-cache/quicklatex.com-2ca4bd778d96e5aff5a6540d3a1a0c30_l3.png "Rendered by QuickLaTeX.com")
Remainder = 2
So, Monday + 2 = Wednesday
Answer: Wednesday
Question 2 ) Jan 4, 2016 falls on Monday. What day of the week falls on Jan 4, 2017?
Solution: We are going to solve this sum with the help of odd days.
In an ordinary year, odd days = 1
In a leap year, odd days = 2
Since 2016 is a leap year. Add 2 to the Monday.
Jan 4, 2017 = Monday + 2 (Leap year odd days)
Jan 4, 2017 = Wednesday
Answer: Wednesday
Question 3 ) 1st March 2006 falls on Wednesday. What day does 1st March 2010 fall?
Solution: We are going to solve it in odd days.
From 2006 – 2010 there are 4 years namely 2007, 2008, 2009 and 2010.
Let’s find the odd days for these years namely
2007 = 1 odd day (Non leap year)
2008 = 2 odd day (Leap year)
2009 = 1 odd day (Non leap year)
2010 = 1 odd day (Non leap year)
Total = 5 days.
So, Wednesday + 5 days = Sundays.
Answer: Sundays
Question 4 ) Today is Monday. After 64 days it will be?
Solution: We are going to solve it in odd days.
64 / 7 = 1 Remainder
1 is the remainder.
Add 1 to the Monday = Tuesday
Answer: Tuesday
Question 5) 8th March 2006 is Wednesday, then what was the day of the week on 8th March 2005?
Solution: Find the odd day in 2005 and subtract it from Wednesday as shown below.
2005 = 1 odd day (Non-leap year)
8th March 2005 = Wednesday – 1 Tuesday
Answer: Tuesday
Question 6 ) Radha celebrated her wedding anniversary on Tuesday 30th September 1997. When will she celebrate her next wedding anniversary on the same day in 2003 ?
Solution: Same day will be repeated every 7 odd days.
So, add from 1997 so that odd days add up to 7.
1998 odd days = 1 (Non-Leap year)
1999 odd days = 1 (Non-Leap year)
2000 odd days = 2 (Leap year)
2001 odd days = 1 (Non-Leap year)
2002 odd days = 1 (Non-Leap year)
2003 odd days = 1 (Non-Leap year)
There are 6 years in order to get 7 odd days.
Now if we divide 7 odd days by 7 – remainder is 0 , hence the day will be the same.
or Tuesday + 7 = Tuesday only [ After 7 days the day repeats it self ]
So, in 2003 Radha will be celebrating her next wedding anniversary on the same day.
Answer: 2003
Important note:
1. There are 5 odd days in 1 century (100 years)
2. There are 3 odd days in 2 centuries (200 years)
3. There are 1 odd day in 3 centuries (300 years)
4. There are no odd days in 4 centuries (400 years). All the multiples of 400 years have no odd days. For example, 400, 800, 1200, 1600…. have no odd days.
Question 7 ) The last day of the century cannot be?
Solution: Let’s calculate by odd days.
1. 100 years = 5 odd days. Code for 5 = Friday
2. 200 years = 3 odd days. Code for 3 = Wednesday
3. 300 years = 1 odd day. Code for 1 = Monday
4. 400 years = 0 odd day. Code for 0 = Sunday
The cycle of Friday, Wednesday, Monday and Sunday will be repeated.
The days which can’t be the last day of the century are Tuesday, Thursday and Saturday
Answer: Tuesday, Thursday and Saturday
Odd days of Months
Divide the total days of month by 7 and left remainder = odd days or extra days
| Month | Total days | Divide by 7 | Remainder/Odd days |
|---|---|---|---|
| January | 31 | 31 /7 | 3 |
| February | 28 – ordinary year 29 – leap year | 28 /7 – ordinary year 29 /7 – leap year | 0 – ordinary year 1 – leap year |
| March | 31 | 31 /7 | 3 |
| April | 30 | 30 /7 | 2 |
| May | 31 | 31 /7 | 3 |
| June | 30 | 30 /7 | 2 |
| July | 31 | 31 /7 | 3 |
| August | 31 | 31 /7 | 3 |
| September | 30 | 30 /7 | 2 |
| October | 31 | 31 /7 | 3 |
| November | 30 | 30 /7 | 2 |
| December | 31 | 31 /7 | 3 |
Question 8 ) Aman birthday is on Monday 19th June. On what day of the week will be Ajay Birthday in the same year if Ajay was born on 17th November? [SSC CGL Question]
Solution – Just calculate the odd days again –
19th June -> Monday
Calculate all odd days from 19th June to 17th November
June days left – 30 – 19 = 11 days -> 11/7 = 4 remainder = 4 odd days
July days -> 31 -> 31 /7 = 3 remainder – 3 odd days
August days -> 31 -> 31/7 = 3 remainder – 3 odd days
September days -> 30 -> 30/7 = 2 remainder = 2 odd days
October – 31 days – > 31 /7 = 3 remainder = 3 odd days
November – 17 days [ given 17th november ] – 17/7 = 3 remainder = 3 odd days
Total odd days = 4 + 3 + 3 + 2 + 3 + 3 = 18 odd days
divide by 7 again [ as days repeats after 7 days ] = 18 /7 = 4 remainder = 4 odd days
19th june = Monday then add 4 to monday => Friday on 17th November
Another Method – Find all days from 19th june to 17th november and divide by 7
11 + 31 + 31 + 30 + 31 + 17 = 151 [ total 151 days between 19th June to 17th november ]
Divide 151 by 7 = 4 remainder = 4 odd days
simply add 4 to monday = Friday
Whichever methods suits you – follow it.