How to Solve Calendar Based Questions in CLAT 2020?
Concept and approach
A calendar is an instrument which is used to keep, indicate and organize the days in a year. We follow the Gregorian calendar which is also known as the Christian calendar or the Western calendar.
The Western calendar is widely used internationally as the civil calendar. The Gregorian calendar starts on 1st January and ends on 31st December. The Gregorian calendar is named after Pope Gregorian XIII, who introduced the calendar in October 1582.
Based on the given information in the questions, if you are asked to find out what a day of the week was September 15 2008, you can easily calculate it on your fingertips but if a weekday on January 28th 2008, is to be found then it may be a tough job.
The question on the topics are very common in various competitive exams the method of solving such question lies in the concept of obtaining the number of odd days before jumping on the topic Let’s start with some of the basic concepts.
What is a leap year and an ordinary year?
Consider the year 1997 is the number 1997 is exactly divisible by 4 the answer is no it is not let us take the year 1996 if the number 19 and 6 exactly divisible by 4 the answer is yes it is.
Now extending the given concept on to the following year such as 1980 1919 87 1964 1600 1951,1300,1988, etc we can conclude :
- Whenever the number of here is exactly divisible by 4 except the century year then it is the leap year
- Whenever the number of the year is not divisible by 4 then it is an ordinary year.
- In the case of century year if the number of here is exactly divisible by 400 then it is a leap year
- Whenever the number of here is not divisible by 400 then it is an ordinary year.
An ordinary year can be defined as the year having 365 days which is equal to 52 weeks and an extra day
A year is a century year if it is divisible by 100.
A year is a non-century year if it is not a century year or not divisible by a hundred.
A year is a leap year if it is a non-century year that is divisible by 4 or a century your that is divisible by 400
How to find out the number of odd days?
The total number of days for a specific period of time when divided / 7 gives a reminder that reminder is termed as the odd day(s).
Counting of odd days
- 1 ordinary year=365 days = 52 weeks + 1 odd day. An ordinary year has one odd day.
- 1 leap year = 366 days= 52 weeks +2 odd days.
- 1 Century year = 100 years = 76 ordinary year + 24 leap years = 76 +2 × 24= 124 odd days = 5 odd days.
Now based on the above fact we can conclude that:
- Number of odd days in 100 years =5
- Number of odd days in 200 year = 10 days i.e 1 week + 3odd days =3
- Number of odd days in 300 years = 15 days = 2 weeks + 1 odd day = 1
- Number of odd days in 400 years = (20 + 1)= 3 weeks = 0
- 1st January 1 A.D was a Monday
- In an ordinary year, the calendar of the month of January is the same as the calendar of the month of October.In short in an ordinary year January = October.
- In a leap year, the calendar of the month of January is the same for the month of July in short,
In a leap year January = July
- January 1st 1992 was a Wednesday. what day of the week was January 1, 1993?
Solution: 1992 being a leap year it had two odd days show the first day of the year 1993 was today beyond so Wednesday i.e it was Friday.
- January 12, 1980, was a Saturday the day of the week on January 12, 1979, was?
Solution: Dahiya 1979 being an ordinary year had one or day. 12th January 1980 was a Saturday. But it was one day beyond 12th January 1979. Therefore, 12 January 1979 was Friday.
- India got independence on 15th August 1947 what was the day of the week on that day?
Solution: 15th aug, 1947 = (1600+300+46) years + 227 (1 January to 15 August of 1947).
Number of odd days = 0+1+1+3 = 5 odd days (Friday)
- If today is Saturday, then what day of the week will be on the 338 days from today?
Solution: number of odd days in 338 days = 338 / 7 = 248 complete weeks + 2odd days. 2nd day after Saturday is Monday.