Calendar Questions and Answers

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 = 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: D

Count odd days in stages.

2000 years leave 0. 2001 to 2005 holds one leap year, 2004:

\[ 4(1) + 1(2) = 6 \]

1 January to 28 May 2006, an ordinary year:

\[ 31 + 28 + 31 + 30 + 28 = 148 \equiv 148 - 147 = 1 \pmod 7 \]

\[ \text{total} = 0 + 6 + 1 = 7 \equiv 0 \pmod 7 \]

Zero odd days is Sunday.

Answer: B

The year 2004 is a jump year.
In this way, it has 2 odd days.
∴ The day on sixth March, 2005 will be 2 days.

past the day on sixth Walk, 2004.
Yet, sixth March, 2005 is Monday.
∴ sixth March, 2004 is Saturday.

Answer: C

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days

167 days = (23 weeks + 6 days) =6 odd days.

Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.

Given day is Wednesday.

Answer: A

Count the odd days up to the date.

2000 years. Complete centuries in multiples of 400 leave no odd days, so 2000 years contribute 0.

2001 to 2009. Nine years of which 2004 and 2008 are leap:

\[ 7(1) + 2(2) = 11 \equiv 4 \pmod 7 \]

1 January to 15 August 2010. 2010 is ordinary:

\[ 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 \equiv 227 - 224 = 3 \pmod 7 \]

\[ \text{total odd days} = 0 + 4 + 3 = 7 \equiv 0 \pmod 7 \]

Zero odd days corresponds to Sunday.

Answer: A

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

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

Given day is Wednesday.

Answer: D

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

Step back one year from 8 February 2005 to 8 February 2004. That interval contains 29 February 2004, so it is 366 days long.

\[ 366 \rightarrow 2 \text{ odd days} \]

\[ \text{Tuesday} - 2 = \textbf{Sunday} \]

Going backwards the odd days are subtracted, and the leap day must be counted because 29 February 2004 lies inside the interval.

Answer: A

Days in 4 years => 

Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.

4 years => 365 + 365 + 365 + 366(Leap year)

4 years => 730 + 731 = 1461

Therefore, Number of Days in 4 Years = 1461 Days.

Answer: B

Multiples of 4 in a century:

\[ \frac{100}{4} = 25 \]

But the century year at the end is a leap year only if divisible by 400, and in three centuries out of four it is not:

\[ 25 - 1 = 24 \]

So an ordinary century has 24 leap years - the standard figure behind the 5 odd days in 100 years.

Answer: B

We know that,

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds.

Then, 1 year = 365 x 24 x 60 x 60 seconds.

= 8760 x 3600

1 year = 31536000 seconds.

Hence, 10 years = 31536000 x 10 = 315360000 seconds.