Calendar Questions and Answers

Answer: A

The correct statement is that 2nd February of that year is a Tuesday.

February can have a fifth Monday only in a leap year in which 1 February is a Monday. The Mondays then fall on the 1st, 8th, 15th, 22nd and 29th, and the fifth one is 29 February, the person's birthday. So 2 February is a Tuesday.

The other statements fail. With 1 February a Monday, the month starts on Monday, not Sunday, and its Sundays are the 7th, 14th, 21st and 28th, only four. The Mondays have dates 1, 8, 15, 22 and 29, mostly odd, so they do not all have even dates.

Answer: B

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

Answer: A

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.

Jan. Feb. Mar.  Apr.  May.  Jun.  Jul.  Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 =  0 odd days.

Given day is Sunday

Answer: C

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.

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.