Calendar Questions and Answers

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.

Answer: A

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

Answer: B

Count forward until the accumulated odd days come back to zero.

\[ 2003:1,\ 2004:2,\ 2005:1,\ 2006:1,\ 2007:1,\ 2008:2,\ 2009:1,\ 2010:1,\ 2011:1,\ 2012:2,\ 2013:1 \]

\[ \text{total} = 14 = 2 \times 7 \quad\Longrightarrow\quad 0 \text{ odd days} \]

So the calendar of 2003 repeats in 2014. Leap years contribute two odd days each, which is why the cycle is 11 years and not 7.

Answer: B

16 July, 1776
= (1775 years + period from 1.1.1776 to 16.7.1776.
Checking of odd days.
No. of odd days in 1600 years =0
No. of odd days in 100 years = 5 

75 years= 18 jump years+57 conventional years
= (18*2+57*1) odd days = 93 odd days
= (13 weeks +2 days) = 2 odd days.
∴ 1775 years have= (0+5+2) odd days= 7 days= 0 days

Jan + Feb. + March + April + May + June + July
31 +29 + 31 + 30 + 31 + 30 +16 = 198 days

198 days = (28 weeks+2 days) = 2 odd days
=Total no. of odd days = (0+2) = 2

Thus, the required day is Tuesday.

Answer: C

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.

Answer: C

Count forward from the given day and reduce modulo 7:

\[ 25 - 3 = 22 \text{ days} \]

\[ 22 = 3 \times 7 + 1 \quad\Longrightarrow\quad 22 \equiv 1 \pmod 7 \]

So the 25th is one day past the day of the 3rd:

\[ \text{Tuesday} + 1 = \textbf{Wednesday} \]

Listing them confirms it: the Tuesdays are the 3rd, 10th, 17th and 24th, so the 25th is the Wednesday that follows.