Calendar Questions and Answers

Answer: D

The month has 30 days and begins on a Saturday, so:

\[ \text{Saturdays: } 1, 8, 15, 22, 29 \]

\[ \text{Sundays: } 2, 9, 16, 23, 30 \]

All five Sundays are holidays. "Every second Saturday" means alternate Saturdays - the 8th and the 22nd:

\[ \text{holidays} = 5 + 2 = 7 \]

\[ \text{working days} = 30 - 7 = 23 \]

Treating only the single "second Saturday" as a holiday would leave 24 days.

Answer: B

Given Today is 20th January 2017, Sunday

In january, we have 31 days

February - 28 days (Non leap year)

March - 31 days

April - 30 days

=> Remaining days => 31 - 20 = 11 in Jan

+ 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days

More 20 days to complete 90 days => upto 20th April

Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.

Now, the day of the week will be 

90/7 => Remainder '6'

As the day starts with '0' on sunday 

6 => Saturday.

Required day is 20th April, Saturday.

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

The Gregorian calendar repeats exactly every 400 years, because 400 years contain a whole number of weeks:

\[ 400 \text{ years} \rightarrow 0 \text{ odd days} \]

\[ 2016 - 1616 = 400 \]

So 9 August 1616 fell on the same weekday, Saturday. The 400-year block contains 97 leap years, and \( 400 \times 365 + 97 = 146{,}097 \) days, which is exactly \( 20{,}871 \) weeks.

Answer: B

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

• 1900 ÷ 400 is not whole ✗
• 1700 ÷ 400 is not whole ✗
• 2500 ÷ 400 is not whole ✗
• \( 1200 \div 400 = 3 \) ✓

So 1200 has 366 days. All four are divisible by 4, which is exactly why the century rule is needed to separate them.

Answer: C

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

Step back one year from 17 March 1997 to 17 March 1996. Although 1996 is a leap year, 29 February 1996 falls before 17 March 1996, so it lies outside the interval.

\[ 365 \text{ days} \rightarrow 1 \text{ odd day} \]

\[ \text{Monday} - 1 = \textbf{Sunday} \]

Whether the leap day counts depends on where it falls relative to the two dates, not merely on whether the year is a leap year.

Answer: D

The 2nd is a Sunday, so the 1st is a Saturday. But the weekday on which the next month begins depends on how long the current month is.

• 28 days → next month starts Saturday
• 30 days → next month starts Monday
• 31 days → next month starts Tuesday

Since the current month is not identified, its length is unknown and the answer cannot be determined.

Answer: C

Given Year is 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

Answer: B

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.

Answer: B

Count forward until the accumulated odd days come back to zero.

\[ 2003:1,\ 2004:2,\ 2005:1,\ 2006:1,\ 2007:1,\ 2008:2,\ 2009:1,\ 2010:1,\ 2011:1,\ 2012:2,\ 2013:1 \]

\[ \text{total} = 14 = 2 \times 7 \quad\Longrightarrow\quad 0 \text{ odd days} \]

So the calendar of 2003 repeats in 2014. Leap years contribute two odd days each, which is why the cycle is 11 years and not 7.