Calendar Questions and Answers

Answer: B

Correct answer: 1800

A century year is a leap year only when it is divisible by 400. 1600, 2000 and 2400 all satisfy that test.

1800 is divisible by 4 and by 100 but not by 400, so it is an ordinary year of 365 days.

Answer: A

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.

Given day is Sunday

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: 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: C

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3)  1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days

 168 days = 24 weeks = 0 odd day.

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

Given day is Wednesday.

Answer: B

On 31st December, 2005 it was Saturday. 

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

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

Answer: D

17th June, 1998= (1997 year Period structure 1.1.1998 to 17.6.1998)
Odd days in 1600 year=0.
Odd days in 300 year = (5*3) =1.
97 year has 24 jump year +73 standard years.
= 24*2 + 73*1
= 48 + 73
= 121 odd days
= 17 weeks + 2 odd days = 2 odd days
= 2 odd days.
∴ 1998 years have (0+1+2) = 3 odd days

Jan to June = (31+29+31+30+31)
= 152 days

Add 17 days of June.
= 152 + 17
= 169 days
= 24 weeks + 1 days
= 1 odd day.

∴ Total number of odd day = 3 + 1 = 4 odd days.
Hence 17.06.1998 was Thursday.

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: A

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.