Aptitude - Calendar - Discussion

@Sundar.

How you take Jan 1, 2006 is Sunday?

Thank you all for explaining.

2006 - Sunday.
2007 - Monday.
2008(leap year) - wednesday.
2009 - Thursday.
2010 - Friday.

Only takes a remainder number of year.

As 2006-1
2007-1
2008-2
2009-1.
Total remainders number=5.
Then, Sunday from 5 days next Friday.

If 15 August 1947- Friday.

Then what will be 7 May 2019? Anyone solve this.

6th day of August 1989 falls on Thursday.

Then the number of Sunday in that month are? Can anyone explain to me?

@Soumya.

6th Aug is Thursday it means 2nd Aug is Sunday.
After every 7 days, the same day will fall.
2nd Aug, 9th Aug,16th Aug, 23th Aug, 30th Aug are Sunday. It means a total of 5 Sunday in Aug.

It is very simple remeber these 2 basic terms:

1)whenever we go from general year to general year or leap year then add 1.
2)whenever we go from leap year to general year then add 2.

Now come to question:

Given Jan 1 , 2006 was Sunday then;
Jan 1, 2010 = ?.

So here we go;

2006(general year) to 2007(general year)=1
2007(general year) to 2008(leap year)=1
2008(leap year) to 2009(general year)=2
2009(general year) to 2010(general year)=1

Adding them=1+1+2+1=5.
now Jan 1, 2010 = sunday+5= Friday.

Hope you guys got it. It's very simple method.

Hi,

Jan 1 2006 is sun.

Jan 1 2010 is ---date is same, year is different so if you are going general year to leap year add +1 days (leap to general add +2 days).

So 2007 is mon, 2008 is a leap year (divisible by 4) so add 2 days i.e. wed, 2009 is thus, 2010 is Friday.

If 3-March-2018 is Thursday.

Then 03-March-2008 which day will come?