HomeLogical ReasoningQuick Revision of Calendar for CLAT

Quick Revision of Calendar for CLAT

The Logical reasoning section of CLAT includes the topic of Calendar. The calendar is an important concept and is widely used in our daily lives. It is not a difficult task to score full marks in this part provided you are thorough with the basics of the subject. In this article, we will be mentioning the principles and formulas to solve the calendar questions.

Points to remember:

  1. It is important to consider which month it is i.e. a month with 30 days, 28 days or 31 days.
  2. An ordinary year has 365 days and a leap year has 366 days.
  3. An ordinary year has one odd day and a leap year has two odd days.
  4. A month with 31 days, 30 days, 29 days and 2 days has 3 odd days, 2 odd days, 1 odd day and 0 0dd days respectively.
  5. Every fourth ordinary year is a leap year.
  6. Odd days

100 years: 5 odd days

200 years: 3 odd days

300 years: 1 odd day

400 years: 0 odd days

  1. It was a Monday on 1 January 1 A.D.

Problem Solving

To solve all kind of questions, it is important to keep in mind the following formulas to get accurate and quick answers to the questions from Calendars.

  1. When you are asked to calculate the odd days, you have to divide the total number of days by 7 and the remainder is the number of odd days. When the number of days involves multiple years or centuries, you can break it up and use the references of no. of odd days as mentioned above.
    Therefore, the general formula of no. of odd days = Total no. of days/ 7
  2. The questions often ask which day it is on a particular date by giving a reference date. In such cases, you have to find out the no. of odd days and calculate the day.
  1. In some questions, you are indirectly asked which day it is without a reference date. So, you have to take the reference date as 1st January 1 A.D.
  2. To find whether a year is a leap year or not:
    For the non-century year, if it is divisible by 4, it is a leap year.
    For a century year, if it is divisible by 400, it is a leap year.


  1. How many odd days are there in 1100 years?

Solution:  1100 years = 400 years + 400 years + 300 years
= 0 odd days + 0 odd days +  1 odd day
= 1 odd day

  1. If today is Tuesday, then what day of the week will be on the 335 days from today?

Solution:  335 days = 47 weeks + 6 odd days. Therefore it is Monday.

  1. Was 2000 a leap year?

Solution: Since 2000 is a century year, we have to see if it is divisible by 400.
2000/400 = 5.
Hence 2000 is a leap year


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