Aptitude - Calendar - Discussion

The question is to find the one that not end with a century then we need to find which days will end with a century.

100/7 = (remainder 2 therefore Monday)
200/7 = (rem 4 therefore wed)
300/7(rem 6 therefore Friday)
400/7 = (rem 0 so Saturday)

Tuesday the only one day given in the options which don't end in a century.

I hope it was easy to understand.

No, this is wrong: because the day on 31st December 1799 was a Tuesday (So last day of century can be a Tuesday as of 17th century last day was Tuesday).

No. Off odd days in 100 years = 5;
But only from 1601 to 1700 (as 1700 is divisible by 4 but not by 400 hence not leap year) BUT No. Off odd days in 100 years = 6 from 1600 to 1699 (as 1600 is a leap year so 25 leap years and 75 normal so 6 odd days).

Hence: The last days are Friday + 6 odd days = Thursday (6odd days because the here year 1400, 1600, 2000 is considered).

Thursday + 5 odd days = Tuesday.

Tuesday + 5 odd days = Sunday.

Sunday + 5 odd days = Friday.

For confirmation please check actual days: the day on 31st December 1599 is Friday; a day on 31st December 1699 is Thursday; a day on 31st December 1799 is Tuesday; day on 31st December 1899 is Sunday; a day on 31st December 1999 is Friday.

@All.

Remember that the last day of the century can't be Tuesday, Thursday and Saturday.

100 years = 24 leap years + 76 ordinary years.
= (24*2) +(76*1).
= 124 odd days.
= 124÷7.
=(17 weeks) +5odd days
= 5(Friday).
200 years =5*2=10
=1 week + 3odd days
=3(Wednesday).

300 years=5*3=15.
=2 weeks + 1 odd day.
=1 (Monday).

I'm taking the solution as:

100/7=14 remainder 2. ie Tuesday.

Every century has this one
100 = 5odd days=>friday.
200 = 3odd days=>wednesday.
300 = 1odd days=>monday.
400 = 0 odd days=>sunday.

Remainig=>(tuesday,thusday,saturday)=>tuesday only is their so correct answer is Tuesday.

First day of any century eg 21 century is 01/01/2001 and not 01/01/2000 and first day of century can start either Monday Tuesday Thursday and Saturday.

Thus last day of century can be Sunday Monday Wednesday and Friday as explained by many which I agree.

But first day of 1stcentury century 01/01/0001 or 1st day of AD1 is a Sunday is that an exception? And the one who need why 100? Is not a leap year here it goes?

Actually a year contains 365.24237 days thus to make up 365.25 first we have one day extra after every forth year to make up the average then to make it. 01 less we do not have multiple of 100 as leap year so average bounces to 365.24 now to make more precise or to make 365.2425 we have then every multiple of 4000 as a leap year.

We can even make it more precise by not having leap year on 4000 otherwise we end up with one day advance by 8000 years but that is not done cause earths revolution speed or say velocity is decreasing it is descending but with a very slow rate which will make greater effect by then.

Century (100,200,300) has no leap years. So we did not get 2 as odd day i.e., Tuesday.

100 is divided by 4we will get 25. But 100 is not a leap year. So we have only 24 leap years and 76 ordinary years.

As per the given solution, if we calculate, then we get:

100 - 5 odd days - friday
200 - 3 odd days - wednesday
300 - 1 odd day - monday
400 - 0 odd day - sunday
500 - 4 odd days - Thursday
600 - 2 odd days - Tuesday
700 - 0 odd day - Sunday
800 - 0 odd day - Sunday.

I guess ideally the answer here should be none of these. Am ai right? Please, anyone explain me.