Introduction to Calendar Quant Notes | EduRev

Quantitative Techniques for CLAT

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Quant : Introduction to Calendar Quant Notes | EduRev

The document Introduction to Calendar Quant Notes | EduRev is a part of the Quant Course Quantitative Techniques for CLAT.
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CALENDAR

Ordinary year: An year having 365 days is called an 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 days 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 are 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, where as in a leap year Feb has one odd day.
1st 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 1: 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: If 10th January 1992 was Saturday, what day of the week was on 6th March, 1993.
Solution: Calculate number of odd days between 10th Jan 1992 and 6th Mar 1993 Jan 10, 1992 is Saturday
So from Jan 11 to Dec 31 of 1992 – days are 366 – 10 = 356 days
Jan 1993 = 31days
Feb 1993 = 28 days
6th March 1993 = 6 days
Total days = 421
= 60 weeks + 1 day
So No. of odd days = 1
Let us count one day after Saturday. The required day will be Sunday.

Example 2: on April 4, 1988 it was Monday. What day of the week was on 5th Nov. 1987.
Solutions: No. of days between 5th Nov. 1987 to 4th April 1988 6th Nov 1987 to 30 Nov  = 25 days
Dec 1987 = 31 days
Jan 1988 = 31 days
Feb 1988 = 29 days
March 1988 = 31 days
4th April 1988 = 4 days
Total     = 151 days
No. of odd days = 151 / 7 = 21 weeks – 4 days
So since 5th Nov. 1987 is prior to 4th April 1988
We are to cannot 4 days backwards from Monday
Requried 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 26th Jan. 1960.
Solution: No. of odd days upto 26th 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

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