Calendar Questions

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.

As we know that a week has 7 days,

k weeks k days = (7k + k) days = 8k days.

Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.

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.

Calculating Age has 2 conditions. Let your Birthday is on January 1st.

1. If the month in which you are born is completed in the present year i.e, your birthday, then

Your Age = Present year - Year you are born 

As of now, present year = 2018

i.e, Age = 2018 - 1995 = 23 years.

2. If the month in which you are born is not completed in the present year i.e, your birthday, then

Your Age = Last year - Year you are born

As of now, present year = 2018

i.e, Age = 2017 - 1995 = 22 years.

NOTE: For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year.

So, 1600

31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)

Odd days in 2000 years = 0

10 years has 2 leap years + 8 ordinary years.

Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.

31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.

Total number of odd days = (0 + 5 + 1) = 6.

Given day is saturday.