Calendar Questions and Answers

Answer: B

The Tuesday is 22nd July, 1980.

From 24 July 1979 to 24 July 1980 is one year that includes 29 February 1980, so it has 366 days:

\[ 366 = 52 \times 7 + 2 \]

The weekday therefore moves forward by 2 days: 24 July 1980 falls on Thursday. Counting back, 23 July 1980 is Wednesday and 22 July 1980 is Tuesday.

24 July 1980 would be a Tuesday only if the year had exactly 52 weeks, ignoring the extra days. 23 July would follow from adding just one odd day, as for a non-leap year.

Answer: B

2nd March, 2023 was a Thursday.

Count the days from 30 January to 2 March 2023: 1 more day in January, 28 days in February (2023 is not a leap year) and 2 days in March, which is 31 days.

\[ 31 = 4 \times 7 + 3 \]

So the day moves forward 3 from Monday: Tuesday, Wednesday, Thursday. Wednesday comes from counting one day too few, for example by leaving out 31 January.

Answer: C

The fourth Tuesday falls on the 27th.

If the 16th is the third Friday, the Fridays are the 2nd, 9th, 16th, 23rd and 30th. Tuesday comes 3 days before Friday, so the Tuesdays are the 6th, 13th, 20th and 27th. The fourth Tuesday is the 27th.

26 and 28 are Monday and Wednesday of that week, and 25 is the Sunday. Those options come from counting the offset from Friday wrongly.

Answer: B

The minimum is 22 working days.

Every month has exactly two 'second and fourth' Saturdays off, so the working days are the days of the month minus 2 minus the number of Sundays.

• 28 days (February): 4 Sundays, so 28 - 4 - 2 = 22.
• 29 days with 5 Sundays: 29 - 5 - 2 = 22.
• 30 days with 5 Sundays: 23.
• 31 days with 5 Sundays: 24.

So no month can have fewer than 22 working days.

The only months that reach the minimum are those with 28 days, or 29 days with five Sundays (a February in a leap year that starts on a Sunday).

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.