Modern Mathematics: Concept of Calendars LR Notes | EduRev

UPSC CSAT Preparation

CAT : Modern Mathematics: Concept of Calendars LR Notes | EduRev

The document Modern Mathematics: Concept of Calendars LR Notes | EduRev is a part of the CAT Course UPSC CSAT Preparation.
All you need of CAT at this link: CAT

What is Calendar?

A Calendar is a chart or series of pages showing the days, weeks, and months of a particular year, or giving particular seasonal information. 

Modern Mathematics: Concept of Calendars LR Notes | EduRev

Basic Structure of a Calendar

  1. Ordinary year: Any year which 365 days is called an ordinary year.Ex: 1879, 2009, 2019, etc.
  2. Leap year: Any year which has 366 days is called a leap year.Ex: 2012, 2016 2020 etc.
  3. The division of the number 365 by 7 gives the quotient 52 and remainder 1 which indicates that an ordinary year has 52 weeks and one extra day. This extra day is referred to as an “odd day” throughout the calendar topics.
  4. A leap year has 366 days, the division of the number 366 by 7 gives the quotient 52 and remainder 2. This indicates that a leap year has 52 weeks and 2 extra days. These two extra days are also referred to as “odd days”.

Concept of Odd Days

  • In an ordinary year, there are 365 days, which means 52 x 7 + 1, or 52 weeks and one day. This additional day is called an odd day. The concept of odd days is very important in calendars.
  • In a century - i.e. 100 years, there will be 24 leap years and 76 non-leap years. This means that there will be 24 x 2 + 76 x 1 = 124 odd days. Since, 7 odd days make a week, to find out the net odd days, divide 124 by 7. The remainder is 5. This is the number of odd days in a century.
  • You may memorize the following points related to the concepts of calendars to save time during the paper:
    (i) 100 years give us 5 odd days as calculated above.
    (ii) 200 years give us 5 x 2 = 10-7 (one week) => 3 odd days.
    (iii) 300 years give us 5 x 3 = 15-14 (two weeks) => 1 odd day.
    (iv) 400 years give us {5 x 4 + 1 (leap century)}-21} => 0 odd days.
  • Now, if we start from 1st January 0001 AD; for 0 odd days, the day will be Sunday; for 1 odd day, the day will be Monday; for 2 odd days, it will be Tuesday; for 3 odd days, it will be Wednesday and so on.

Concept of Leap Year

  • Every 100th year starting from 1st AD is a non-leap year, but every 4th century year is a leap year.
  • Any year divisible by 400 will be a leap year e.g.: 1200, 1600 and 2000. And the years 1800, 1900 will be non-leap years as they are divisible by 100, but not 400.

Calendar Calculations

The year consists of 365 days, 5 hours, 48 minutes (52 weeks and 1 odd day). An extra day is added once in every fourth year which was called the leap year, which has 366 days (52 weeks and 2 odd days). 

To find the day of any given date of the year, you need to understand the calendar calculations:

  • 1st January 1 AD was Monday therefore; we must count days from Sunday. This means the 0th day was Sunday, so the 7th day was Sunday again, and so on and so forth.
  • The day gets repeated after every seventh day (concept of a week), if today is Monday, then 28th day from now will also be Monday as it is a multiple of 7 (28/7 = 4, so four weeks). Here the 30 day will be calculated by 30/7, which is 4 weeks and 2 days, these two days are called odd days. With starting day as Monday and two odd days, the day will be Wednesday; this point is the most critical in calendars. The other of looking at it is since the 28th day is Monday, so the 30th day will be Wednesday. But you have to understand and use the concept of odd days as the question may be about thousands of years.
  • In a normal year, there are 365 days so 52 weeks and 1 odd day, in a leap year there are 366 days so 52 weeks and 2 odd days.
  • In 100 years there are 24 leap years and 76 normal years, so the number of odd days are 24(2) + 76 = 124, which is 17 weeks + 5 odd days, so 100 years have 5 odd days.
  • In 200 years the number of odd days is twice the number in 100 years which is 10, which is one week and 3 odd days, so 200 years have 3 odd days. In 300 years, the number of odd days is 15, which is two weeks and 1 odd day, so 300 years have one odd day.
  • 400 year is a leap year; similarly, the multiples of 400 also leap years.
  • In 400 years, the number of odd days become 20 + 1(from the leap year), so total days are 21, which is three weeks and 0 odd days. In 400 years there are 0 odd days
Try yourself:If 15 March 1816 was Friday, what day of the week would 15th April 1916 be?
View Solution


Important Formulas

Odd Days:

We are supposed to find the day of the week on a given date. For this, we use the concept of 'odd days'. In a given period, the number of days more than the complete weeks are called odd days.

Leap Year:

i. Every year divisible by 4 is a leap year if it is not a century.
ii. Every 4th century is a leap year and no other century is a leap year.
Note: A leap year has 366 days.

Examples: 

1) Each of the years 1948, 2004, 1676 etc. is a leap year.
2) Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.
3) None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year. 

Ordinary Year:

The year which is not a leap year is called an ordinary year. An ordinary year has 365 days.

Counting of Odd Days:

  • 1 ordinary year = 365 days = (52 weeks + 1 day.)
    1 ordinary year has 1 odd day
  • 1 leap year = 366 days = (52 weeks + 2 days)
    1 leap year has 2 odd days
  • 100 years = 76 ordinary years + 24 leap year
    = (76 x 1 + 24 x 2) odd days = 124 odd days
    = (17 weeks + days) 5 odd days.
  • A number of odd days in 100 years = 5
  • A number of odd days in 200 years = (5 x 2) 3 odd days.
  • A number of odd days in 300 years = (5 x 3) 1 odd day.
  • A number of odd days in 400 years = (5 x 4 + 1) 0 odd day.
  • Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.
Modern Mathematics: Concept of Calendars LR Notes | EduRev

Solved Examples

Example 1: What was the day on 9th February 1979?

  • You know that in 1600 years, there will be 0 odd days. And in the next 300 years, there will be 1 odd day. 
  • From 1901 to 1978 we have 19 leap years and 59 non-leap years. So, the total number of odd days up to 31st Dec. 
  • 1978 is 19 x 2 + 59 = 97. On dividing 97 by 7 we get 6 as the remainder, which is the total number of odd days in these years. 
  • So, till 31st Dec. 1978, we have 1 + 6 = 7 odd days, which forms one complete week. 
  • Now, in 1979, we have 3 odd days in January, and 2 odd days in the month of February (up to 9th Feb). So, the total odd days are 3 + 2 = 5. 
  • Hence, 9th February 1979 was a Friday.

Try yourself:What was the day on 15th august 1947?
View Solution


Example 2: If May 10, 1997, was a Monday, what will be the day on Oct 10, 2001?

  • In this question, the reference point is May 10, 1997, and you have to find the number of odd days from May 10, 1997, up to Oct 10, 2001. 
  • Now, from May 11, 1997 - May 10, 1998 = 1 odd day
    May 11, 1998 - May 10, 1999 = 1 odd day
    May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
    May 11, 2000 - May 10, 2001 = 1 odd day
  • Thus, the total number of odd days up to May 10, 2001 = 5.
  • Now, the remaining 21 days of May will give 0 odd days. In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in September, 2 odd days and up to 10th October, we have 3 odd days. 
  • Hence, total number of odd days = 18 i.e. 4 odd days. Since, May 10, 1997 was a Monday, then 4 days after Monday will be Friday. So, Oct 10, 2001, would be a Friday.


Example 3: If 11th April 1911 was a Tuesday, what would be the day on 17th September 1915?

  • Firstly in terms of years, the year 1911 to 1912 would give us 2 odd days and 1913, 1914, 1915 would give 1, 1 and 1 odd day respectively.
  • Now shift the focus on months. If you move one month ahead i.e. from 11th April to 11th May, the month ending in between is April, which gives you 2 days. Now after that, the month of May, June, July, and August gives you 3, 2, 3, and 3 odd days respectively.
  • With this, you reach on 11th September 1915. After this, there are 6 more September days (from 11th to 17th September).
  • The total number of odd days is 2 + 1 + 1 + 1 + 2 + 3 + 2 + 3 + 3 + 6 = 24.
  • Subtracting 21 (3 full weeks) from this the odd number of days left is 3. Adding three days to the day given i.e. Tuesday, the answer becomes Friday.

Example 4: If 15 March 1816 was Friday, what day of the week would 15th April 1916 be? 

  • We are given that 15th March 1816 was a Friday.
  • Now we know that 100 years have 5 odd days. So till 15th March 1916, we will be having 5 odd days. So if we move from 15th March 1816 to 15th March 1916, we will encounter 5 odd days.
  • Now from 15th March 1916 to 15th April 1916 there would be 3 odd days.
  • So total number of odd days = 5+3 =8
    8 mod 7 = 1
    So, 15th April 1916 would be Friday + 1= Saturday

Example 5: The leap year 1895 is having the same calendar as that of the year X. Which of the following is a possible value of X.

  • 1895 is not a leap year. 
  • So, it will have 1 odd day.
  • Since, 1896 is a leap year, it will add 2 odd days.
  • Similarly, 1987, 1898, 1899, 1900 will add 1,1,1,1 odd days.Now the total number of odd days add up to 7.
  • So, the next year 1901 will have the same calendar as 1895.

Modern Math: Key Learnings

  • Whenever the reference date is given, you should take it to be blindly true. While moving from reference, take only the odd number of days in terms of years, months, and days to keep the calculative part on the minimal side.
  • It can be concluded here that the questions based on modern math are not difficult to handle. It is just a matter of learning the basics well and then practising it sufficiently to get the confidence of doing it in the exam.
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Free

,

Extra Questions

,

shortcuts and tricks

,

Semester Notes

,

MCQs

,

Modern Mathematics: Concept of Calendars LR Notes | EduRev

,

Sample Paper

,

past year papers

,

video lectures

,

study material

,

pdf

,

Previous Year Questions with Solutions

,

Summary

,

practice quizzes

,

ppt

,

Viva Questions

,

Modern Mathematics: Concept of Calendars LR Notes | EduRev

,

Objective type Questions

,

Modern Mathematics: Concept of Calendars LR Notes | EduRev

,

Exam

,

Important questions

,

mock tests for examination

;