Calendar Questions and Answers

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.

Answer: B

Given that 2.12.91 is the first Tuesday
Hence we can assume that 4.12.91 is the first Thursday

If we add 7 days to 4.12.91, we will get second Thursday
If we add 14 days to 4.12.91, we will get third Thursday
If we add 21 days to 4.12.91, we will get fourth Thursday

=> fourth Thursday = (4.12.91 + 21 days) = 25.12.91.

Answer: A

First,we count the number of odd days for the left over days in the given period.

Here,given period is 15.8.2012 to 11.6.2013

Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun

16   30   31  30   31   31   28  31   30   31   11(left days)

2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days

So,given day Thursday + 6 = Wednesday is the required result.

Answer: B

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

\[ 61 = 8 \times 7 + 5 \quad\Longrightarrow\quad 5 \text{ odd days} \]

Counting five days on from Monday:

\[ \text{Tue, Wed, Thu, Fri, } \mathbf{Sat} \]

The eight complete weeks return to Monday and contribute nothing.

Answer: C

Correct answer: Saturday

Only the remainder after dividing by 7 matters, since each complete week returns to the same day.

\[61 = 8 \times 7 + 5\]

Counting five days on from Monday gives Saturday.

Answer: B

1856 is a leap year, so its calendar returns after 28 years:

\[ 1856 + 28 = 1884 \]

Both must be leap years beginning on the same weekday, which the 28-year cycle ensures - and 1884 is indeed divisible by 4.