Aptitude - Calendar - Discussion

Yes, Thursday is the right Answer.

09-last two digit of the given year.
09/4=2[quoteint].
1-date given.
0-number of month (JFM=033, AMJ=614, JAS =625 and OND= 035).
6-no.of year (1600-1699=6, 1700-1799=4, 1800-1899=2,1900-1999=0, and 2000-2099=6).
Sum of all=18.
Sum/7= 4(remainder).
4 = Thursday (0=Sunday......,.6 = saturday).

the answer should be friday according to me because 2008 odd days are 2 and 1 of 1st jan 2009 so it should be 2+1=3 tues +3 i.e friday

@All.

The question in 1 Jan 2008, we know that 2008 is a leap year, assume that if 2008 was an non-leap year then it.

Will have 365 days, so when you divide 365/7 = 1 odd day all right, the day will we Wednesday but as it is a leap year it has 366 days now 366/7 = 2 odd days, now add to the date of 1 Jan 2008 = Tuesday + 2 Thursday.

If the question was like 1 Jan 2008 in Tuesday then what is 1 Jan 2007, the answer would be Monday, as 2007 is the simple year.

Right @Divya Agrawal.

I also got Wednesday as answer.

Answer will be Wedesday.

@Priyanka

Don't add 1st Jan 2008 in your calculation. (2nd Jan 2008 to 1st Jan 2009) in total 366 days, hence 2 odd days.

I had a doubt. From 1st Jan 2008 to 31st December 2008 is 366 days. So checking divisibility by 7, we get 2 odd days. So we can say that the last 2 days of the year (i.e. 30th and 31st Dec 2008) will be Wednesday and Thursday.

So from the question, actually 1st-Jan-2009 should be the next day i.e. 'Friday' right?

Please correct me if I am wrong!

Simply,

Divide this year by 4 i.e. 2008/4. It is divisible so we got to know that it is a leap year and leap year has 2 odd days as 366/4=52 weeks and 2 days. So the day will be 2 days ahead. So Tuesday + 2 = Thursday.

If 1st february, 2012 is wednesday, what was the day of the week on 1st february, 2011? Please explain the answer.