Calendar Questions and Answers

Answer: A

Look backwards from 1989 until the odd days total a multiple of seven.

\[ 1988:2,\ 1987:1,\ 1986:1,\ 1985:1,\ 1984:2,\ 1983:1,\ 1982:1,\ 1981:1,\ 1980:2,\ 1979:1,\ 1978:1 \]

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

So 1989 repeats the calendar of 1978.

Answer: C

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

Track the weekday of 5 January year by year. An ordinary year adds one day; a leap year adds two, but only if 29 February falls within the interval - which it does for the year following a 5 January in a leap year.

\[ 2018 \text{ Fri} \to 2019 \text{ Sat} \to 2020 \text{ Sun} \to 2021 \text{ Tue} \]

\[ \to 2022 \text{ Wed} \to 2023 \text{ Thu} \to 2024 \ \textbf{Fri} \]

The jump from 2020 to 2021 skips Monday because 29 February 2020 lies between the two dates.

Answer: D

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

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

148 days = (21 weeks + 1 day) 1 odd day.

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

Given day is Sunday.

Answer: B

Each week is seven days, so

\[ x \text{ weeks } + x \text{ days} = 7x + x = 8x \text{ days} \]

The two \( x \)'s are added, not multiplied - \( 7x^{2} \) would be the result of multiplying them.

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.

\[ 1984 \div 4 = 496 \ \checkmark \text{ and not a century year} \]

1863 is not divisible by 4; 1900 and 2500 are century years not divisible by 400.

So only 1984 has 366 days.

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.