Calendar Questions and Answers

Answer: A

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

Answer: C

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

1600 years have 0 odd day. 300 years have 1 odd day.

47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 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,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

Answer: A

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

Answer: C

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

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result

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: C

Now, the given year is 2018

We know that 2016 is a Leap year.

2016  2017  2018  2019  2020

 LY    1st     2nd    3rd    LY

Here 2018 is at 2 nd position after the Leap year.

According to rule ( b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.

Answer: B

A calendar repeats when the accumulated odd days return to zero. 2024 is a leap year, and leap-year calendars repeat on a 28-year cycle:

\[ 2024 + 28 = 2052 \]

The cycle holds as long as no skipped century year intervenes, and none does between 2024 and 2052.

Answer: C

Count the days from one date to the other and reduce modulo 7.

From 18 March 1994 to 18 March 1995 is a full ordinary year, 365 days, so

\[ 365 \equiv 1 \pmod 7 \]

Going back from 18 March 1995 to 25 February 1995 is 21 days, which is exactly 3 weeks and so leaves no odd day. Net shift from the given Friday:

\[ +1 - 0 = +1 \quad\Longrightarrow\quad \text{Friday} + 1 = \textbf{Saturday} \]

1995 is not a leap year and February 1995 has 28 days, so no extra day creeps in.

Answer: C

1988 is a leap year, and leap-year calendars repeat every 28 years:

\[ 1988 + 28 = 2016 \]

Both years must be leap years starting on the same weekday for every date to line up, which the 28-year cycle guarantees here.

Answer: A

Count the days from 9 July 2013 to 7 January 2014:

\[ 182 = 7 \times 26 \quad\Longrightarrow\quad 182 \equiv 0 \pmod 7 \]

The interval is a whole number of weeks, so the day does not shift at all: 7 January 2014 falls on the same day as 9 July 2013, namely Sunday.