Calendar Questions and Answers

Answer: C

Given Year is 2005, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 2005 + 6) = 2011

Therfore, The calendar for the year 2005 will be same for the year 2011

Answer: C

1939 is an ordinary year. Counting odd days forward until they total a multiple of seven:

\[ 1939:1,\ 1940:2,\ 1941:1,\ 1942:1,\ 1943:1,\ 1944:2,\ 1945:1,\ 1946:1,\ 1947:1,\ 1948:2,\ 1949:1 \]

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

So 1939's calendar repeats in 1950 - the usual 11-year gap for an ordinary year with three leap years in between.

Answer: C

A year is a leap year if it is divisible by 4, except that a century year must also be divisible by 400.

\[ 1996 \div 4 = 499 \ \checkmark, \qquad 1984 \div 4 = 496 \ \checkmark, \qquad 2004 \div 4 = 501 \ \checkmark \]

\[ 1863 \div 4 = 465.75 \ \times \]

So 1863 is not a leap year. None of these is a century year, so the 400-rule never comes into play.

Answer: C

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: B

Only the remainder on division by 7 matters - the odd days.

2007 is not divisible by 4, so it is an ordinary year of 365 days.

\[ 365 = 52 \times 7 + 1 \quad\Longrightarrow\quad 1 \text{ odd day} \]

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

Answer: D

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.

Answer: C

Correct answer: 5

In 100 years there are 24 leap years and 76 ordinary years, since 1900-style century years are not leap.

\[\text{Odd days} = 76 \times 1 + 24 \times 2 = 76 + 48 = 124\]

\[124 = 17 \times 7 + 5\]

So 100 years carry 5 odd days, which is why 200 years carry 3, 300 carry 1 and 400 carry none.

Answer: A

Given that the month begins on a Friday and has 31 days

Sundays = 3rd, 10th, 17th, 24th, 31st
=> Total Sundays = 5

Every second & fourth Saturday is holiday.
2nd & 4th Saturday in every month = 2

Total days in the month = 31

Total working days = 31 - (5 + 2) = 24 days.

Answer: A

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.

Answer: D

Going back one year shifts the day by the number of odd days in the year crossed.

The span from 8 December 2006 to 8 December 2007 lies wholly inside ordinary territory - 2008 is the leap year, and its 29 February is not in this range - so the span is 365 days:

\[ 365 \equiv 1 \pmod 7 \]

The later date is therefore one day further on, so the earlier one is one day back:

\[ \text{Saturday} - 1 = \textbf{Friday} \]