Calendar - Free MCQ Practice Test with solutions, CLAT Quant Techniques

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴ After 61 days, it will be Saturday.

1992 being a leap year it had two odd days show the first day of the year 1993 was today beyond so Wednesday i.e it was Friday.

On 31^st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31^st December 2009, it was Thursday.
Thus, on 1^st Jan, 2010 it is Friday.

28 May, 2006 i.e. 2005 years and Period from 1.1.2006 to 28.5.2006

Odd days in 1600 years = 0
Odd days in 400 years = 0

5 years = 4 ordinary years + 1 leap year = (4 x 1) + (1 x 2) = 6 odd days

Jan (31) + Feb (28) + March (31) + April (30) + May (28) = 148 days
∴ 148 days = 21 weeks + 1 day = 1 odd day
Total number of odd days = 0 + 0 + 6 + 1 = 7 = 0 odd day.

Thus, the given day is Sunday.

17^th June, 1998 i.e.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 = 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan (31) + Feb (28) + March (31) + April (30) + May (31) + June (17) = 168 days
Therefore 168 days = 24 weeks = 0 odd day.

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

Thus, the given day is Wednesday.

15^th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.

Jan.     Feb.   March   April     Mayb   June     July       Aug.
(31   +   28   +   31   +   30   +   31   +   30   +   31   +   15) = 227 days

∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.
Given day is Sunday.

The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.
Given that, 6^th March, 2005 is Monday.
∴ 6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

We shall find the day on 1^st April, 2001.
1^st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1)     = 91 days ≡ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1^st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4^th, 11^th, 18^th and 25^th.

x weeks x days = (7x + x) days = 8x days.

The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8^th Feb, 2004 is 2 days before the day on 8^th Feb, 2005.
Hence, this day is Sunday.

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

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 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th 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.

22 Apr 2222 i.e. 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 = Monday

Hence 22 Apr 2222 is Monday

Each day of the week is repeated after 7 days. So, after 63 days, it will be on Monday. After 61 days, it will be on Saturday.

350 days, 350/7 = 50, no odd days, so it will be a Monday.

number of odd days in 338 days = 338 / 7 = 248 complete weeks + 2odd days. 2nd day after Saturday is Monday.

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
= 43/7 =1
Thus, the day of the week on 24th July 2011will be Sunday.