The document Modern Mathematics: Concept of Calendars LR Notes | EduRev is a part of the LR Course Logical Reasoning (LR) and Data Interpretation (DI).

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Learn the methods, formulas and shortcuts to deal with the questions based on calendars.**Modern mathematics** includes the topics like clocks, calendars and stocks and shares. The questions from these topics appear off and on in all the major competitive exams. These areas are not covered in a detailed manner in the higher secondary classes and it is due to this reason that the students skip these topics usually. However, these topics, if understood properly, are easier to handle as compared to a few other topics of mathematics.

In this article, we are going to discuss the topic of calendars. To solve questions from this topic, you first need to understand the concept of 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.

â€¢ 100 years give us 5 odd days as calculated above.

â€¢ 200 years give us 5 x 2 = 10-7 (one week) => 3 odd days.

â€¢ 300 years give us 5 x 3 = 15-14 (two weeks) => 1 odd day.

â€¢ 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:**

â€¢ Every 100th year starting from 1st AD is a non-leap year, but every 4^{th} century year is a leap year.

â€¢ 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.

The following examples exhibit the application of the concepts explained above.**Solved Examples:****Example 1: **What was the day on 9^{th} February 1979?**Solution:** 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 31^{st} 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 9^{th} Feb). So, the total odd days are 3 + 2 = 5. Hence, 9^{th} February 1979 was a Friday.**Example 2:** If May 10, 1997 was a Monday, what will be the day on Oct 10, 2001?**Solution:** 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 10^{th} 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 17^{th} September 1915?**Solution: **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.**Modern Math: Key Learnings**

â€¢ 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.

â€¢ 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 practicing it sufficiently to get the confidence of doing it in the exam.

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