Calendar Questions and Answers

Answer: A

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004.

Given that, 5th March, 2005 is Monday.

5th March, 2004 is Sunday (1 day before to 5th March, 2005).

Answer: A

19th April, 1986 = (1985 years + Period from 1.1.1986 to 19.4.1986)

Odd days in 1600 years = 0

Odd days in 300 years = 1

85 years has 21 leap years + 64 ordinary years.

Number of odd days in 85 years =( 21 x 2 + 64) = 1 odd day.

Jan. Feb. March April

(31 + 28 + 31 + 19) = 109 days

109 days = 15 weeks + 4 days= 4 odd day.

Total number of odd days = (1 + 0 + 1 + 4) = 6.

Given day is Saturday.

Answer: C

Correct answer: Friday

Count the odd days up to that date. 1600 years give 0, the next 300 give 1, and the 46 years from 1901 to 1946 give \(35 + 22 = 57\), that is 1 odd day.

From 1 January to 15 August 1947 is 227 days, and \(227 = 32 \times 7 + 3\), giving 3 more.

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

Counting 0 as Sunday, 5 odd days lands on Friday.

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: B

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

Count the odd days from 1 January 2006 to 1 January 2010.

\[ 2006:1, \quad 2007:1, \quad 2008:2 \text{ (leap)}, \quad 2009:1 \]

\[ \text{total} = 5 \text{ odd days} \]

\[ \text{Sunday} + 5 = \textbf{Friday} \]

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.