Maths Aptitude Test on Calendar

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

∴ On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

28 May, 2006 = (2005 years + 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. Feb. March April May (31 + 28 + 31 + 30 + 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.

Given day is Sunday

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.

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 May 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.

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.

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 6th 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,sup>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.

100 years contain 5 odd days.

∴ Last day of 1^st century is Friday.

200 years contain (5 x 2) 3 odd days.

∴ Last day of 2^nd century is Wednesday.

300 years contain (5 x 3) = 15 1 odd day.

∴ Last day of 3^rd century is Monday.

400 years contain 0 odd day.

∴ Last day of 4^th century is Sunday.

This cycle is repeated.

∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

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.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd day : 1 2 1 1 1 2 1 1 1 2 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.