An ordinary year has 365 days. Week starting with
Friday will end in Thursday. Hence, the 364th day (end
of complete 52 weeks) will be Thursday. The 365th day
will be Friday. Thus, the first and last day of an ordinary year are same.
Mathematical Aptitude MCQ - Calendar Questions
Odd days in 2007 = 1 (2007 is an ordinary year and we
are doing calculation from January 1)
Odd days in 2008 = 2 (2008 is a leap year)
Odd days in 2009 = 1 (ordinary year)
Odd days in 2010 = 1 (ordinary year)
Thus, January 1, 2011, will be Monday plus 5 days, i.e., Saturday.
Odd days in 2008 = 2 (2008 is a leap year)
Odd days in 2009 = 1 (ordinary year)
Odd days in 2010 = 1 (ordinary year)
Thus, January 1, 2011, will be Monday plus 5 days, i.e., Saturday.
The earlier three Saturdays are on 15th, 8th and 1st.
If 15th is Saturday and hence, Thursday falls on 13th.
Thus, 13th is Thursday.
There are two conditions for two years having the
same calendar: both having same length in terms of
number of days and first day starting with same day of the week.
The year 1991 has 365 days, that is, 1 odd day, and year 1992 has 366 days, that is, 2 odd days, while year 1993 has 365 days, that is, 1 odd day. The years 1994,1995, and 1996 have 1 odd day each.
The sum of odd days so calculated from year 1991 to 1996.
(1 + 2 + 1 + 1 + 1 + 1) = 7 odd days.
Hence, the year 1997 will have the same calendar as that of the year 1991.
The year 1991 has 365 days, that is, 1 odd day, and year 1992 has 366 days, that is, 2 odd days, while year 1993 has 365 days, that is, 1 odd day. The years 1994,1995, and 1996 have 1 odd day each.
The sum of odd days so calculated from year 1991 to 1996.
(1 + 2 + 1 + 1 + 1 + 1) = 7 odd days.
Hence, the year 1997 will have the same calendar as that of the year 1991.
28th May, 2007 = (2000 years + 6 years + period from 1.1.2007 to 28.5.2007)
Odd days in 2000 years = 0
Odd days till 2006 = (5 ordinary years + 1 leap year) =(5 × 1 + 1 × 2) = 7 odd days
148 days = (21 weeks + 1 day) ⇒ 1 odd day
Total number of odd days = (2000 years + 6 years + period from 1.1.2007 to 28.5.2007)
= (0 + 7 + 1) = 8 odd days, i.e., again 1 odd day. Hence, Monday is the answer.
Odd days in 2000 years = 0
Odd days till 2006 = (5 ordinary years + 1 leap year) =(5 × 1 + 1 × 2) = 7 odd days
| Jan | Feb | March | April | May | Total |
| 31 | 28 | 31 | 30 | 28 | 148 |
Total number of odd days = (2000 years + 6 years + period from 1.1.2007 to 28.5.2007)
= (0 + 7 + 1) = 8 odd days, i.e., again 1 odd day. Hence, Monday is the answer.
