**In this post, we will discuss the concept of Calendar and Smart Tricks to solve Calendar Problems in SSC Exams**

Many SSC aspirants face difficulty when it comes to Calendar Problems. They tend to skip calendar problems due to complex calculations. Calendar problems can be easily solved by using smart tricks that save time. In this post, we will discuss the basic concept of calendar and tricks to solve calendar problems.

**What is a Calendar?**

A calendar is an instrument which is used to keep, indicate and organize the days in a year. We follow the Gregorian calendar which is also called as the Christian calendar or the Western calendar. The Western calendar is the widely used internationally as the civil calendar. The Gregorian calendar starts on 1

^{st}January and ends on 31

^{st}December. The Gregorian calendar is named after Pope Gregorian XIII, who introduced the calendar in October 1582.

A year is classified into two types –

**Ordinary year –**An ordinary year has 365 days.

**Leap year –**A leap year has 366 days.

The difference between an ordinary year and a leap year is that a leap year has one extra day, designated as 29

^{th}February. A leap year occurs every four years, to keep the calendar in alignment with the earth's revolution around the sun. Earth takes 365 ¼ days to complete one revolution. Hence, to compensate 1/4^{th}day or 6 hours in the calendar we have a leap year.Therefore, a year cannot be a leap year if it is not divisible by 4 or 400.**For example –**1700, 1800, 1900 are not a leap year because it is not divisible by 400.

1600, 2000, 2400 are leap years because they are divisible by 400

.

**Calendar Problems can be classified into two broad categories –**

1. Questions that ask you to find the day of the week for the given date when some other date or day is specified.

2. Questions that ask you to find the day of the week for the given date when no other date or day is specified.

Calendar Problems can be solved using the concept of odd days

**Odd Days:**In a given time period, the number of days that are more than the weekly cycle are called odd days.

For example: 10 days = 1 week + 3 days.

Then the extra 3 days become odd days.

Likewise,

15 days = 2 weeks + 1 day (1 odd day)

30 days = 4 weeks + 2 days (2 odd day)

Ordinary years = 365 days = 52 weeks + 1 day (1 odd day)

Leap year 366 days = 52 weeks + 2 days (2 odd day)

Therefore, an ordinary year has 1 odd day and a leap year has 2 odd days.

**Reference Chart:**

*Sunday is taken as the reference point.

**Number of odd days for longer Periods**

è Calculate the Number of Odd days for the 1

^{st}Century – 100 years:**Solution:****Step 1:**
In the first century, there were 24 leap years. (100/4 = 25. Since 100

^{th}year is not divisible by 400 we deduct 1 year)
Ordinary years = 76

**Step 2:**

Number of odd days = 76 x 1(Since ordinary years have 1 odd day) + 24 x 2 (Since leap years have 2 odd days)

= 76 + 48

= 124

**Step 3:**

124 = 17 weeks + 5 days

Therefore, there are 5 odd days

Hence, the last day of 1

^{st}century i.e. 31^{st}December is Friday (according to reference chart)
è Calculate the number of odd days for the first 200 years:

**Solution:****Step 1:**

Odd days = 5 x 2 (Since the second century has the same number of odd days)

= 10 odd days

**Step 2:**

10 odd days = 1 week + 3 days

Therefore there are 3 odd days in the first 200 century.

Hence, the last day of 200 years i.e. 31

^{st}December is Wednesday (according to reference chart)**Calendar Problem Solved using Smart Tricks**

**Problem:**Which day of the week was 15

^{th}August 1947?

**Solution:**

**Step 1:**

To simplify the calculation, break the years

1600 years + 300 years + 46 years + (January – July) + 15 days

**Step 2:**

Calculate the odd days for the years

1600 years = 0 odd days

300 years = 1 odd day

46 years = (35 ordinary years + 11 leap years)

= (35 odd days + 22 odd days) = 57 odd days = 1 odd day

= (35 odd days + 22 odd days) = 57 odd days = 1 odd day

January = 31 days = 3 odd days

February = 0 odd days

March = 31 days = 3 odd days

April = 30 days = 2 odd days

May = 31 days = 3 odd days

June = 30 days = 2 odd days

July = 31 days = 3 odd days

August = 15 days = 1 odd day

**Step 3:**

Adding the odd days = 0 + 1 + 1 + 3 + 0 + 3 + 2 + 3 + 2 + 3 + 1

= 19 odd days

= 2 weeks + 5 odd days

= 5 odd days

Therefore, 15

^{th}August 1947 was a Friday, since 5 odd days indicate Friday.**Reference Chart that indicates odd days for the given months**

**Watch our expert faculty explain Smart Tricks to solve Calendar problems**

Stay tuned for more Calendar Problems

Should give an example ....so that can have better understanding

ReplyDeleteHi Vishal.. The examples are given in the next blog in this series. Find it below:

ReplyDeletehttp://blog.talentsprint.com/2017/06/calendar-2-calendar-problems-for-ssc.html