Question 1

What will be the day of the week 15th August, 2010?

Accepted Answer

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.

Question 2

How many days are there in x weeks x days?

Accepted Answer

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

Question 3

What was the day of the week on 28th May, 2006?

Accepted Answer

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.

Question 4

Today is Monday. After 61 days, it will be:

Accepted Answer

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.

Question 5

The calendar for the year 2007 will be the same for the year:

Accepted Answer

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.

Question 6

The last day of a century cannot be

Accepted Answer

100 years contain 5 odd days. ∴ Last day of 1st century is Friday. 200 years contain (5 x 2) ≡ 3 odd days. ∴ Last day of 2nd century is Wednesday. 300 years contain (5 x 3) = 15 ≡ 1 odd day. ∴ Last day of 3rd century is Monday. 400 years contain 0 odd day. ∴ Last day of 4th century is Sunday. This cycle is repeated. ∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Question 7

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

Accepted Answer

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.

Question 8

If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?

Accepted Answer

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 6th March, 2005 will be 1 day beyond the day on 6th March, 2004. Given that, 6th March, 2005 is Monday. ∴ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Question 9

January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?

Accepted Answer

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

Question 10

On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?

Accepted Answer

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

Question 11

January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?

Accepted Answer

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

Question 12

On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?

Accepted Answer

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.

Question 13

Which of the following is not a leap year?

Accepted Answer

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

Question 14

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

Accepted Answer

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 31st December 2009, it was Thursday. Thus, on 1st Jan, 2010 it is Friday.

Question 15

On what dates of April, 2001 did Wednesday fall?

Accepted Answer

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 1st April, 2001 it was Sunday. In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

Question 16

The year next to 1991 will have the same calender as that of the year 1991.

Accepted Answer

For two years having same calendar must have two common conditions: both having same length in terms of number of days and first day of the week. The year 1991 has 365 days,,that is 1 odd day , year 1992 has 366 days , i.e. 2 odd days , year 1993 has 365 days i.e. 1 odd day. the years 1994,1995,1996 have 1 odd day each. the sum of odd days so calculated from year 1990 to 1996 . (1+2+1+1+1+1)=7 odd days. 5 hence, the year 1997 will have same calendar as that of year 1990.

Question 17

Today is Friday. After 62 days it will be?

Accepted Answer

Each day of the week is repeated after 7 days ∴ After 63 days, it would be Friday So, After 62 days, it would be Thursday

Question 18

On July, 2, 1985, it was Wednesday the day of the week on July 2, 1984 was?

Accepted Answer

The year 1984 being a leap year, it has 2 odd days. So, the day on 2nd July, 1985 is two days beyond the day on 2nd July, 1984. But, 2nd July 1985 was Wednesday. ∴ 2nd July, 1984 was Monday

Question 19

How many days are three from 2nd January 1993 to 15th march 1993?

Accepted Answer

No of days = Jan. + Feb. + March 30 + 28 + 15 = 73 days

Question 20

The year next to 1990 having the same calender as 1998 is?

Accepted Answer

Starting with 1988, we go on counting thye number of odd days till the sum is divisible by 7 Year 1988 odd days = 2 Year 1989 odd days = 1 Year 1990 odd days = 1 Year 1991 odd days = 1 Year 1992 odd days = 2 So total odd days = 7; ∴ Calendar for 1993 is the same as that of 1988.

Calendar Questions