Aptitude - Calendar - Discussion

Hello everybody ,
You must know some codes first.

Years, Days, Century code(for every 400 years ).
Jan-0, sun-0, 0:99-6.
Feb-3, mon-1, 1:199-4.
Mar-3, tue-2, 2:299-2.
Apr-6, wed-3, 3:399-0.
may-1, thr-4,
Jun-4, fri-5,
Jul-6, sat-6,
Aug-2.
Sep-5.
Oct-0.
Nov-3.
Dec-5.

Now formula is ,

Day = (Date+month code+no.of years+no.of leap years + century code)/(7).

In the given question 15th Aug 2010.

Day = (15+2+10+2+6)/7 = 7/7 = 0. Which is 'Sunday'. :).

How for march the value of month is 6? I don't understand the month value.

Hey, this type of questions can be easily done by following formula
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f= K+[(13m-1)/5]+ d+[d/4]+[c/4]-2c

Here k= date of which you have to calculate(15 in above que.)
m= month (for march, m=1 and so on..for jan m=11)
d= last two digit of given year(10 in above que.)
c= first two digit of given year(20 in above que.)
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note: take only integer value after division for ex. 13/5=2
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For given question

15th August, 2010?

f=15+[(13*6-1)/5]+10+[10/4]+[20/4]-2*20
=15+15+ 10 + 2 + 5 -40
=7

Now 7%7==0

So its SUNDAY.

@Rupali: First you go through the Important formluas which is given in the 1st page .

Please can you tell me how you write the.

Total number of odd days = (0 + 0 + 4 + 3) = 7 odd days.

I can't understand this step.

How can I know 9years has (2leap + 7 nonleap)?

@John

You can find the answer for your question in this page.

http://www.indiabix.com/aptitude/calendar/discussion-638

Hope this help you. Have a nice day!

Why we are taking the 2000 years as 1600 and 400 years ?

Thanks boney.

If we divide 1600/400 we will get zero as remainder so it has zero odd days.
so in 2000 yrs we have 0 odd days.
In 9 yrs there r 2 leap years
leap yrs have 2 odd days(366/7 2 as remainder) and
non leapyr have 1 odd day totaling we have 2*2+7*1=11 odd days
Now cal no: of days in 2010 upto aug 15 then dividing by 7
will give no of odd days 2010 which is 3.