Calendar Questions and Answers

Answer: D

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year :       2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

Odd day : 1         2      1         1      1         2     1       1       1       2       1

Sum = 14 odd days  0 odd days.

 Calendar for the year 2018 will be the same as for the year 2007.

Answer: A

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

• \( 800 \div 400 = 2 \) ✓
• \( 1200 \div 400 = 3 \) ✓
• \( 2000 \div 400 = 5 \) ✓
• \( 700 \div 400 = 1.75 \) ✗

So 700 is not a leap year, even though it is divisible by 4. The 400-year rule exists because the tropical year is a little under \( 365.25 \) days.

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.

Answer: C

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

2008 is divisible by 4 and is not a century year, so it is a leap year with 366 days.

\[ 366 = 52 \times 7 + 2 \quad\Longrightarrow\quad 2 \text{ odd days} \]

\[ \text{Tuesday} + 2 = \textbf{Thursday} \]

An ordinary year advances the date by one weekday; a leap year by two.

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

15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.
=> Total number of odd days in 400 years = 0
Hence, total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)
Odd in days in the period 2001-2009:
7 normal years + 2 leap yeas
=> (7*1) + (2*2) = 11
=> Odd days will be 11- (7*1) = 4
Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15
= 227 days.
= 32 weeks and 3 days, this gives additional 3 odd days.
=> Total odd days= 3+4=7
=> 7 odd days=1 week= 0 odd days
=> 0 odd days= Sunday
Thus, 15th August 2010 was a Sunday.

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

To begin with we discover the day on 1.3.2005.
1.3.2005= (2004 year+ period from 1.1.2005 to 1.3.2005
Odd days in 1600 years = 0
Odd days in 400 years = 0
4 years = (1 jump year+ 3 normal years)
= (1*2+3*1) odd days = 5 odd days. 
Jan + Feb + March
31 + 28 + 1 = 60 days = (8 weeks +4 days)=4 odd days.
∴ Total number of odd days = (0+0+5+4) =9=2 odd days.
∴ 1.3.2005 was Tuesday. So, Friday lies on 4.3.2005.
Thus, Friday lies on 4th, 11th, 18th and 25th of March, 2005.

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

The year 2008 is a jump year.
So, it has 2 odd days.
First day of the year 2008 is Tuesday (Given).

In this way, first day of the year 2009 is
2 days past Tuesday.
Subsequently, it will be Thursday.