FOR FREE YEAR SOLVED

CALENDAR

Calendar is a definite arrangement of days, weeks, and months. Whenever we are asked the day of a particular date, we search the nearest calendar and scan it. But we do not give ourselves some time to think and get the answer. After learning some interesting concepts and tricks from this portal, you will not only develop your understanding but also be able to find the day of Date of Births of your acquaintances. Very Interesting!! Isn’t it? 

Basic Terms

> Days: One-Seventh of a week

> Week: One-Fifty Secondth of a year

> Month: One-twelfth of a year

> Year: A year is an approximate time taken by earth to complete one revolution around the sun. One year has 365 days ordinarily. A leap year has 366 days, because of 29th February.

Objectives

Here we will learn three important concepts and tricks for three important cases: 

1) To find the day of a particular date when a reference date is given.

2) To find the day of a particular date when a reference date is not given.

3) To find a week-day when another week-day is given.

Concepts

Before learning the concepts we must know – 

WHAT ARE ODD DAYS? 

ODD DAYS are the number of extra days from a multiple of 7. If a number is divided by 7, then the remainder is ODD DAYS (O.D.). 

For e.g. – Suppose we are told to find the odd days for a month having 30 days, then we should divide it by 7 and the remainder 2 is the odd days of the month.

Now, a reference chart (Chart 1) is required with the actual concepts to solve the Calendar sums, which is as follows:

Chart 1

Chart 2

Chart 3 (Very Important)

Now let us grab the concepts: 

1) To find the day of a particular date when a reference date is given

A reference date will be given. We have to traverse from the initial point or the reference date towards the required date, keeping Chart 2 in our account.

For e.g. – A reference date is given as 15th April 2003, falls on Monday. Now we have to find the day of the following dates:

a) 15th April 2004 (Exactly one year after)

b) 15th April 2002 (Exactly one year before)

c) 15th April 2010 (Exactly more than one year after)

d) 10th August 2003 (Any day on the same year)

e) 8th September 2005 (Any day after the reference date)

f) 17th July 2001 (Any day before the reference date)

Solutions:

a) We will traverse from 15th April 2003 to 15th April 2004 crossing the Leap Year day, 29th Feb 2004. So, this is an exact one leap year forward traversal. Therefore, the required day must be 2 odd days after the reference day, i.e. Monday. Hence, the required day is WEDNESDAY.

b) We will traverse from 15th April 2003 to 15th April 2002. So, this is an exact one ordinary year backward-traversal. Therefore, the required day must be 1 odd day before the reference day, i.e. Monday. Hence, the required day is SUNDAY.

c) We will traverse from 15th April 2003 to 15th April 2010 crossing the Leap Year days, 29th Feb 2004, and 29th Feb 2008. So, this is an exact two leap years and five ordinary years forward-traversal. Referring to Chart 3, the required day must be [2 X 2 + 1 X 5 = 9, which is greater than 7. 

So, the Remainder of (9/7) is 2] 

2 odd days after the reference day, i.e. Monday. 

Hence, the required day is WEDNESDAY.

d) We will traverse from 15th April 2003 to 10th August 2003. So, this is a same-year traversal. Referring to Chart 2, 

So, Total Odd days = 1 + 3 + 2 + 3 + 3 = 12 

→ Remainder of (12/7) = 5

Therefore, the required day is 5 days after the reference day, i.e. Monday. Hence, the required day is SATURDAY.

We will traverse from 15th April 2003 to 8th September 2005. Referring to Chart 2 and 3, 
 

So, Total Odd days = 2 + 1 + 3 + 2 + 3 + 3 + 1 = 15 

→ Remainder of (15/7) = 1

Therefore, the required day is 1 day after the reference day, i.e. Monday. Hence, the required day is TUESDAY.

e) We will traverse from 15th April 2003 to 17th July 2001. Referring to Chart 2 and 3, 
 

So, Total Odd days = 1 + 1 – 2 – 2 – 3 – 2 = -7 

→ Remainder of (-7/7) = 0

Therefore, the required day is 0 days after the reference day, i.e. Monday. Hence, the required day is MONDAY.

2)  To find the day of a particular date when a reference date is not given

> Unlike the previous case, there will be no reference date given. But we have to assume the last 10’s-multiple-year as the reference date. 

> From that year, odd days are calculated till the particular date is reached. 

> Say, we are asked the day of 6th September 1993. The last 10’s-multiple-year is 1990, whose odd day is 1, because of its ordinary in nature. So the odd days of 1991 and 1992 are 1 and 2 respectively. 

> Now the odd days from January 1993 to August 1993 are calculated as follows –

3 (Jan) + 0 + 3 + 2 + 3 + 2 + 3 + 3 = 19 

→ Remainder of (19/7) = 5.

And the odd days till 6th September 1993 is 6. 

So, the total NUMBER OF ODD DAYS = 1 + 1 + 2 + 5 + 6 = 15 

→ Remainder of (15/7) = 1.

If we refer the chart 1, 1 is the odd day number of MONDAY. 

And here it is the required day! Very simple and interesting, isn’t it?

3) To find a week-day when a reference week-day is given

First, answer the following – 

a) Day after yesterday – 

b) Day after the 2 days before yesterday – 

c) Day before tomorrow – 

d) 2 Days before tomorrow –

If you can answer these questions, you are done with this part. 

Answers are – 

TODAY, DAY BEFORE YESTERDAY, TODAY, YESTERDAY respectively.