Calendar Questions and Answers

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:

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

Answer: A

17th 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

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday

Answer: C

Correct answer: 2018

A calendar repeats when the odd days since the starting year sum to a multiple of 7. Counting from 2007 onward, ordinary years contribute 1 each and the leap years 2008, 2012 and 2016 contribute 2 each.

\[1+2+1+1+1+2+1+1+1+2+1 = 14\]

Fourteen is a multiple of 7, and that total is reached at the end of 2017, so 2018 repeats the calendar of 2007.

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.

\[ 2000 \div 400 = 5 \ \checkmark, \qquad 2008 \div 4 = 502 \ \checkmark, \qquad 2012 \div 4 = 503 \ \checkmark \]

\[ 2009 \div 4 = 502.25 \ \times \]

So 2009 is not a leap year.