Introduction to Calendar - Quantitative Techniques for CLAT

Calendar

A calendar is a system of organizing units of time for the purpose of reckoning time over extended periods. By convention, the day is the smallest calendrical unit of time; the measurement of fractions of a day is classified as timekeeping.

• Ordinary year: An year having 365 days is called ordinary years.
• Leap year: Year having 366 days is called a leap year. Every leap year is exactly divisible by 4 and ordinary years are not completely divisible by 4.
• Odd days: In a given period, the days apart from complete weeks are called odd days. An ordinary year has one odd day, i.e. 365/7 = 52 weeks + 1 day While the leap year has two odd days, i.e. 366/7 = 52 + 2 days.
  
  

To Find the Number of Odd Days

• In 100 years there are 24 leap years + 76 ordinary years
  = 24 x 52 weeks + 24 x 2 days + 76 x 52 weeks + 76 days
  = 6 days  + 6 days
  = 12 days = 1 week and 5 days
  So, in 100 years there are 5 odd days similarly in 200 years there are 3 odd days and in 300 years there is 1 odd day in 400 years there is 0 odd day similarly in 800 years, 1200 years and 1600 years there is 0 odd day.
• Odd days in Feb: In an ordinary year, Feb has no odd day, whereas in a leap year Feb has one odd day.
• 1^st day of the century must be Tuesday, Thursday or Saturday and Last day of a century cannot be Tuesday, Thursday or Saturday.

To Find a Particular Day When a Day and a Date is Given

Step I: Find out the number of odd days between the given date and the date for which the day is to be found one.

Step II: From the given day count the odd days in the forward direction to arrive at the day on the given date.

Example 1: If 10th January 1992 was Saturday, what day of the week was on 6^th March 1993.
Solution: 

To find out the day of the week for 6th March 1993, we need to count the number of days between 10th January 1992 and 6th March 1993.

We can start by counting the number of days in January 1992, which is 31. Then, we add the number of days in February 1992, which is 29 since it was a leap year. Next, we add the number of days in all the months from March 1992 to February 1993, which is 365. Finally, we add the number of days in March 1993 up to the 6th, which is 6.

Therefore, the total number of days between 10th January 1992 and 6th March 1993 is 31 + 29 + 365 + 6 = 431 days.

Now, we can divide 431 by 7 to get the number of weeks and days. The quotient is 61 and the remainder is 4. This means that 431 days is equivalent to 61 weeks and 4 days.

Since 10th January 1992 was a Saturday, we can count 4 days forward from Saturday to get the day of the week for 6th March 1993. This means that 6th March 1993 was a Saturday.

Example 2: On April 4, 1988, it was Monday. What day of the week was on 5^th Nov. 1987.
Solutions: No. of days between 5^th Nov. 1987 to 4^th April 1988 6^th Nov 1987 to 30 Nov  = 25 days
 Dec 1987 = 31 days
Jan 1988 = 31 days
Feb 1988 = 29 days
March 1988 = 31 days
4^th April 1988 = 4 days
Total = 151 days
No. of odd days = 151 / 7 = 21 weeks – 4 days
So, since 5^th Nov. 1987 is prior to 4^th April 1988
We are to count 4 days backwards from Monday, the required day is Thursday.

To Find the Day on a Particular Date if Day and Date is not Given

• The procedure can be understood from the given example.

Example 3: Find the day of the week on 26^th Jan. 1960.
Solution: No. of odd days upto 26^th Jan. 1960
= Odd days for 1600 years + odd days for 300 years + odd days for 59 years + odd days of 26 days of Jan 1960
= 0 + 1 + 59 + 14 + 5 = 79 days
= 79 / 7 = 11 weeks + 2 days = 2 odd days
The required day is Tuesday Zero odd day means Sunday. We are to consider one odd day as Monday 2 odd days as Tuesday and so on.