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**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 D****ay 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:Calculate number of odd days between 10^{th}Jan 1992 and 6^{th}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

6^{th}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 5^{th}Nov.1987.No. of days between 5

Solutions:^{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.

Try yourself:It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

View Solution

**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.**

= 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.

Try yourself:What was the day of the week on 28th May, 2006?

View Solution

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