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# Calendar Quant Notes | EduRev

## Quantitative Techniques for CLAT

Created by: Gyanm Institute

## Quant : Calendar Quant Notes | EduRev

The document Calendar Quant Notes | EduRev is a part of the Quant Course Quantitative Techniques for CLAT.
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Chapter - 20

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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