Calendar Questions and Answers

Answer: C

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.

Answer: D

The month has 30 days and begins on a Saturday, so:

\[ \text{Saturdays: } 1, 8, 15, 22, 29 \]

\[ \text{Sundays: } 2, 9, 16, 23, 30 \]

All five Sundays are holidays. "Every second Saturday" means alternate Saturdays - the 8th and the 22nd:

\[ \text{holidays} = 5 + 2 = 7 \]

\[ \text{working days} = 30 - 7 = 23 \]

Treating only the single "second Saturday" as a holiday would leave 24 days.

Answer: A

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).

Answer: D

Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.

Answer: D

Find the day for 1 April 2001, then step in sevens.

Odd days to the end of 2000: a multiple of 400 years leaves 0. For 1 January to 1 April 2001, an ordinary year:

\[ 31 + 28 + 31 + 1 = 91 \equiv 0 \pmod 7 \]

So 1 April 2001 was a Sunday, which puts the first Wednesday on the 4th. Adding weeks gives

\[ 4,\; 11,\; 18,\; 25 \]

April has 30 days, so a fifth Wednesday would need the 32nd and does not exist - which is why this list has four dates where the other options have five.

Answer: C

The year being crossed here contains a 29 February, so the shift is two days rather than one.

From 8 February 2004 to 8 February 2005 the span includes 29 February 2004, making it 366 days:

\[ 366 \equiv 2 \pmod 7 \]

Stepping back from the given Tuesday:

\[ \text{Tuesday} - 2 = \textbf{Sunday} \]

Subtracting only one day, the reflex from ordinary years, gives Monday - the trap this question is built around.

Answer: A

The Gregorian calendar repeats exactly every 400 years, because 400 years contain a whole number of weeks:

\[ 400 \text{ years} \rightarrow 0 \text{ odd days} \]

\[ 2016 - 1616 = 400 \]

So 9 August 1616 fell on the same weekday, Saturday. The 400-year block contains 97 leap years, and \( 400 \times 365 + 97 = 146{,}097 \) days, which is exactly \( 20{,}871 \) weeks.

Answer: B

Given Today is 20th January 2017, Sunday

In january, we have 31 days

February - 28 days (Non leap year)

March - 31 days

April - 30 days

=> Remaining days => 31 - 20 = 11 in Jan

+ 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days

More 20 days to complete 90 days => upto 20th April

Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.

Now, the day of the week will be 

90/7 => Remainder '6'

As the day starts with '0' on sunday 

6 => Saturday.

Required day is 20th April, Saturday.

Answer: A

Count the days from 9 July 2013 to 7 January 2014:

\[ 182 = 7 \times 26 \quad\Longrightarrow\quad 182 \equiv 0 \pmod 7 \]

The interval is a whole number of weeks, so the day does not shift at all: 7 January 2014 falls on the same day as 9 July 2013, namely Sunday.

Answer: D

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.