Calendar Questions and Answers

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.

Answer: C

On 31st December,2005 it was Saturday.
Number of odd days frame the year 2006 to the year 2009=(1+1+2+1)=5 days.

∴ On 31st December 2009 , it was Thursday.
In this way, on first jan, 2010 it is Friday.

Answer: C

\[ k \text{ weeks } + k \text{ days} = 7k + k = 8k \text{ days} \]

The week count is multiplied by 7 and the day count added, giving a single term in \( k \).

Answer: B

Leap years normally come four years apart, but a century year that is not divisible by 400 is skipped.

\[ 1896 \ \rightarrow \ \text{(1900 skipped)} \ \rightarrow \ 1904 \]

\[ 1904 - 1896 = 8 \text{ years} \]

So the maximum gap is 8 years. It happens around 1700, 1800, 1900, 2100 and so on - but not 2000, which is divisible by 400.

Answer: D

15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.
=> Total number of odd days in 400 years = 0
Hence, total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)
Odd in days in the period 2001-2009:
7 normal years + 2 leap yeas
=> (7*1) + (2*2) = 11
=> Odd days will be 11- (7*1) = 4
Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15
= 227 days.
= 32 weeks and 3 days, this gives additional 3 odd days.
=> Total odd days= 3+4=7
=> 7 odd days=1 week= 0 odd days
=> 0 odd days= Sunday
Thus, 15th August 2010 was a Sunday.