Calendar - Aptitude MCQ Questions and Solutions with Explanations

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year	Odd Day
2007	1
2008	2
2009	1
2010	1
2011	1
2012	2
2013	1
2014	1
2015	1
2016	2
2017	1

Sum = 14 odd days ≡ 0 odd days.
∴ Calendar for the year 2018 will be the same as for the year 2007.

The century divisible by 400 is a leap year.
∴ The year 700 is not a leap year.

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8^th Dec, 2007 will be 1 day beyond the day on 8^th Dec, 2006.
But, 8^th Dec, 2007 is Saturday.
∴ 8^th Dec, 2006 is Friday.

The year 2008 is a leap year. So, it has 2 odd days.
1^st day of the year 2008 is Tuesday (Given)
So, 1^st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.

The year 2007 is an ordinary year. So, it has 1 odd day.
1^st day of the year 2007 was Monday.
1^st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.

There is exactly 1 year, (365 days) between two dates.
2006 is an ordinary year. It has one odd day.
The day of the week on January 12^th 2007 is one day beyond Thursday ⇒ Friday

22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2200
= Number of odd days in 200 years
= 5 x 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday

350 days, $$\frac{{350}}{7}$$ = 50, no odd days, so it will be a Monday.

Day Codes
Saturday	0
Sunday	1
Monday	2
Tuesday	3
Wednesday	4
Thursday	5
Friday	6

Month Codes
January	1
February	4
March	4
April	0
May	2
June	5
July	0
August	3
September	6
October	1
November	4
December	6

Century Codes
1500 - 1599	0
1600 - 1699	6
1700 - 1799	4
1800 - 1899	2
1900 - 1999	0
2000 - 2099	6
2100 - 2199	4
2200 - 2299	2
2300 - 2399	0

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
$$\eqalign{ & = \frac{{24 + 0 + 11 + 2 + 6}}{7} \cr & = \frac{{43}}{7} \cr & = 1 \cr & = {\text{Sunday}} \cr} $$

Given that first day of a normal year was Friday
Odd days of the mentioned year = 1 (Since it is an ordinary year)
Hence First day of the next year = (Friday + 1 Odd day) = Saturday
Therefore, last day of the mentioned year = Friday