Calendar — Part 1 MCQs with Explanations

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 = (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 Therefore 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 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.

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 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 = (7 x + x ) days = 8 x 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 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

$$\eqalign{ & \frac{{10 + 4 + 96 + 24 + 0}}{7} \cr & = \frac{{134}}{7} \cr & = 1 \cr & = {\text{Sunday}} \cr} $$

20 th June, 1837 means 1836 complete years + first 5 months of the year 1837 + 20 days of June 1600 years give no odd days 200 years give 3 odd days 36 years give (36 + 9 ) or 3 odd days Thus 1836 years give 6 odd days From 1 st January to 20 th June there are 3 odd days Odd days: January 3, February 0, March 3, April 2, May 3, June 6 = 17 Thus the total number of odd days = 6 + 3 or 2 odd days This means that the 20th of June fell on 2nd day commencing from Monday. The required day was Tuesday.

Number of days from 26-Jan-1996 to 15-May-1996 (both days included) = 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May) = 111

94 days = (13 × 7) + 3 = 3 odd days. The required day is 3 days beyond Thursday i.e., Sunday

Counting the years 1600 + 300 + 74 In 1600 years, there are zero odd days In 300 years, there is one odd day In 74 years, there are 18 leap years and 56 normal years, so the odd days are: 18(2) + 56(1) = 36 + 56 = 92, Which is 13 weeks and 1 odd day In 25 days of January, 1975, there are 3 weeks and 4 odd days Total odd days = 0 + 1 + 1 + 4 six odd days, so it was a Saturday.

Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9 - 7 = 2, reduced perfect multiple of 7 from total days) Hence today = (Thursday + 2 odd days) = Saturday

1 - May - 2007 $$\eqalign{ & = \frac{{1 + 2 + 7 + 1 + 6}}{7} \cr & = \frac{{17}}{7} \cr & = 3 \cr & = {\text{Tuesday}} \cr} $$ = May 1 st → Tuesday + 5 days = Saturday = 5 th may 5 th may + 7 days = Saturday = 12 th may 12 th may + 7 days = Saturday = 19 th may 19 th may + 7 days = Saturday = 26 th may = Answer = 26 th may

Given that seventh day of a month is three days earlier than Friday ⇒ Seventh day is Tuesday ⇒ 14 th is Tuesday ⇒ 19 th is Sunday

This is a leap year. So, none of the next 3 years will be leap years. Each year will give one odd day so the day of the week will be 3 odd days beyond Monday i.e. it will be Thursday.

Given that 1st October is Sunday Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Hence 1 st November = (Sunday + 3 odd days) = Wednesday

Odd days ⇒ The number of days more than complete number of weeks in the given period are odd days. 123 = 7 × 17 + 4 ⇒ 4 odd days.

After every 400 years, the same day occurs. (Because a period of 400 years has 0 odd days) Thus, 18 th April 1603 is Thursday, After 400 years i.e., on 18 th April 2003 has to be Thursday.

$$\frac{{126}}{7}$$ = 0 Each day of the week is repeated after 7 days. So, after 126 days, it will be Friday. After 126 days, it will be Friday

Given that 25th August = Thursday Hence 29th August = Monday So 22nd,15th and 8th and 1st of August also will be Mondays Number of Mondays in August = 5

In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 11 or 4400 times. Thus the 29th day of the month occurs = 4400 + 97 = 4497 times.

Mentioned month begins on a Saturday and has 30 days Sundays = 2 nd , 9 th , 16 th , 23 rd , 30 th ⇒ Total Sundays = 5 Every second Saturday is holiday. 1 second Saturday in every month Total days in the month = 30 Total working days = 30 - (5 + 1) = 24

We know every year has 1 odd days We know leap year has 2 odd days Here, Number of Normal years= 22 and number of Leap years =8 So odd days =22+16 =38 i.e 3 odd days (remainder when 38 is divided by 7,i.e. 3) Hence, it was a Thursday.

1 st Jan 1901 = (1900 years + 1 st Jan 1901) We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400) Number of odd days in the period 1601 - 1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything) 1 st Jan 1901 = 1 odd day Total number of odd days = (0 + 1 + 1) = 2 2 odd days = Tuesday Hence 1 January 1901 is Tuesday.

This is a leap year. So, none of the next 3 years will be leap years. Each ordinary year has one odd day, so there are 3 odd days in next 3 years. So the day of the week will be 3 odd days beyond Wednesday i.e. it will be Saturday

Given that January 1, 2004 was Thursday. Odd days in 2004 = 2 (because 2004 is a leap year) (Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004) Hence January 1, 2005 = (Thursday + 2 odd days) = Saturday