Aptitude - Calendar - Discussion

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What was the day of the week on 17^th June, 1998?

[A].

Monday

[B].

Tuesday

[C].

Wednesday

[D].

Thursday

Answer: Option C

Explanation:

17^th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan.  Feb.   March    April    May      June 
(31 +  28  +  31   +   30   +   31   +   17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

Shruthi said: (Tue, Aug 17, 2010 03:16:40 PM)

how to calculate odd days for 300 years?

Bhupendra said: (Wed, Aug 18, 2010 12:26:15 PM)

499*2=899 = no of odd day in leap(for understand) 1996-499=1497 odd day in remaining year (for understand) odd day 899+1497= 2396 total odd day (for understand) 2396/7= 342+ 2/7=> 2odd day (actual odd day till 1996) 2+1= 3 odd day till 1997 (actual odd day till 1997) then 1 Jan 1998= thursday Jan 31 Feb 28 mar 31 april 30 may 31 june 17= wedenesday