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: 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.

Answer: B

16 July, 1776
= (1775 years + period from 1.1.1776 to 16.7.1776.
Checking of odd days.
No. of odd days in 1600 years =0
No. of odd days in 100 years = 5 

75 years= 18 jump years+57 conventional years
= (18*2+57*1) odd days = 93 odd days
= (13 weeks +2 days) = 2 odd days.
∴ 1775 years have= (0+5+2) odd days= 7 days= 0 days

Jan + Feb. + March + April + May + June + July
31 +29 + 31 + 30 + 31 + 30 +16 = 198 days

198 days = (28 weeks+2 days) = 2 odd days
=Total no. of odd days = (0+2) = 2

Thus, the required day is Tuesday.

Answer: C

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

 On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

Answer: C

Count forward from the given day and reduce modulo 7:

\[ 25 - 3 = 22 \text{ days} \]

\[ 22 = 3 \times 7 + 1 \quad\Longrightarrow\quad 22 \equiv 1 \pmod 7 \]

So the 25th is one day past the day of the 3rd:

\[ \text{Tuesday} + 1 = \textbf{Wednesday} \]

Listing them confirms it: the Tuesdays are the 3rd, 10th, 17th and 24th, so the 25th is the Wednesday that follows.

Answer: C

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: B

18 Feb, 2016 = (2015 years + Period from 1.1.2016 to 18.2.2016)

Odd days in 1600 years = 0

Odd days in 400 years = 0

15 years = (12 ordinary years + 3 leap year) = (12 x 1 + 3 x 2)= 4 odd days

Jan  Feb

(31 + 18 ) = 49 days

49 days = (7 weeks + 0 day)= 0 odd day.

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

Given day is Thursday

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: A

Count the odd days up to 15 August 1947.

\[ 1600 \text{ years} \rightarrow 0 \text{ odd days} \]

\[ 300 \text{ years} \rightarrow 1 \text{ odd day} \]

\[ 46 \text{ years} = 35 \text{ ordinary} + 11 \text{ leap} \rightarrow 35 + 22 = 57 \rightarrow 1 \text{ odd day} \]

From 1 January to 15 August 1947 (not a leap year):

\[ 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 = 32 \times 7 + 3 \]

\[ \text{total odd days} = 0 + 1 + 1 + 3 = 5 \]

Five odd days corresponds to Friday - India's Independence Day fell on a Friday.

Answer: D

Count the days from 6 March to 11 November, then add the part-day hours.

\[ \text{March (after the 6th)} = 25, \ \text{Apr } 30, \ \text{May } 31, \ \text{Jun } 30 \]

\[ \text{Jul } 31, \ \text{Aug } 31, \ \text{Sep } 30, \ \text{Oct } 31, \ \text{Nov } 11 \]

\[ \text{total} = 250 \text{ days} \]

From 9 a.m. to 9 a.m. is \( 250 \times 24 = 6000 \) hours, and from 9 a.m. to 8 p.m. is another 11:

\[ 6000 + 11 = 6011 \text{ hours} \]