**Calendar Tricks**

प्रतियोगी परीक्षाओं में रीजनिंग सबसे स्कोरिंग सेक्शन होता है, इसलिए इसका मतलब है कि हम इस सेक्शन के माध्यम से सर्वश्रेष्ठ स्कोर कर सकते हैं लेकिन अगर हम इस रीजनिंग भाग के लिए तैयार नहीं हैं तो हमारा स्कोर कम रह सकता हैं। कुछ अभ्यर्थी जो पहले ही इन परीक्षाओं से गुजर चुके हैं, इसे अधिक भ्रमित करने वाला हिस्सा मानते हैं।

One of the most important topics is Calendar. Most of the candidates face difficulty in solving calendar reasoning questions. So here we are providing some simple calendar reasoning tricks which will help to solve the calendar questions in short time and in easiest way. नीचे दिए गए कैलेंडर रीजनिंग ट्रिक्स की मदद से सेकंड्स में प्रश्न हल करें।

**Calendar Basic Concepts**

**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**

That year (except century) which is divisible by 4 is called a leap year, whereas century is a leap year by itself when it is divisible by 400.

**Note:** A leap year has 366 days.

**Examples:**

- Each of the years 1948, 2004, 1676 etc. is a leap year.
- Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.
- 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 years. 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 years = (76 × 1 + 24 × 2) odd days = 124 odd days. = (17 weeks + 5 days) = 5 odd days.
- Number of odd days in 100 years = 5
- Number of odd days in 200 years = (5 × 2) = 3 odd days
- Number of odd days in 300 years = (5 × 3) = 1 odd day
- Number of odd days in 400 years = (5 × 4 + 1) = 0 odd day
- Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc has 0 odd days.

**Codes to Remember**

**Year Code**

1600-1699 | 6 |

1700-1799 | 4 |

1800-1899 | 2 |

1900-1999 | 0 |

2000-2099 | 6 |

**Month Code**

January / October | 0 |

May | 1 |

August | 2 |

February / March / November | 3 |

June | 4 |

September / December | 5 |

July / April | 6 |

**Day Code**

Sunday | 0 |

Monday | 1 |

Tuesday | 2 |

Wednesday | 3 |

Thursday | 4 |

Friday | 5 |

Saturday | 6 |

**Number of Odd Days in a Month**

Month | Number of odd days |

January | 3 |

February(ordinary/leap) | (0/1) |

March | 3 |

April | 2 |

May | 3 |

June | 2 |

July | 3 |

August | 3 |

September | 2 |

October | 3 |

November | 2 |

December | 3 |

** **Check Here – __How to Prepare Reasoning for Competitive Exams__

**Calendar Reasoning Tricks (****कैलेंडर ****रीज़निंग ****ट्रिक्स)**

Trick 1 |

** ****Steps to Solve Calendar Problems**

**Step 1:** Add the day digit to last two digit of the year.

**Step 2:** Divide the last two digits of the year by four.

**Step 3:** Add the Quotient value in step 3 to result obtain in step 1.

**Step 4:** Add Month Code and year codes to the result obtain in step 3.

**Step 5:** Divide the result of step 4 by seven.

**Step 6:** Obtain the remainder and match with the day code.

**(Note: If year completely divide by 4 and Month is January / February, then the answer is one day before the day obtained by step 6) **

**Example:** What was the day on 27th November 1989?

**Solution:**

Year Code: 0, Month Code: 3, Day digits = 27, Last 2 digits of the year = 89.

**Step 1:** 27 + 89 = 116

**Step 2: **89 / 4 = 22.25

**Step 3:** 116 + 22 = 138

**Step 4:** 138 + 3 + 0 = 141

**Step 5:** Remainder of 141 / 7 = 1

**Step 6:** Day code 1 = Monday

So, the answer is Monday.

**Example:** What was the day on 17th July 1776?

**Solution:**

Year Code: 4, Month Code: 6, Day digits = 17, Last 2 digits of the year = 76.

**Step 1:** 17 + 76 = 93

**Step 2: **76 / 4 = 19

**Step 3:** 93 + 19 = 112

**Step 4:** 112 + 6 + 4 = 122

**Step 5:** Remainder of 122 / 7 = 3

**Step 6:** Day code 3 = Wednesday

So, the answer is Wednesday.

**Example:** What was the day on 26th January 2020?

**Solution:**

Year Code: 6, Month Code: 0, Day digits = 26, Last 2 digits of the year = 20.

**Step 1:** 26 + 20 = 46

**Step 2: **20 / 4 = 5

**Step 3:** 46 + 5 = 51

**Step 4:** 51 + 6 + 0 = 57

**Step 5:** Remainder of 57 / 7 = 1

**Step 6:** Day code 1 = Monday

So, the answer is Sunday.

**Example:** What was the day on 29th February 2020?

**Solution:**

Year Code: 6, Month Code: 3, Day digits = 13, Last 2 digits of the year = 20.

**Step 1:** 29 + 20 = 49

**Step 2: **20 / 4 = 5

**Step 3:** 49 + 5 = 54

**Step 4:** 54 + 6 + 3 = 63

**Step 5:** Remainder of 63 / 7 = 0

**Step 6:** Day code 0 = Sunday

So, the answer is Saturday.

**Example:** What was the day on 24th March 2020?

**Solution:**

Year Code: 6, Month Code: 3, Day digits = 24, Last 2 digits of the year = 20.

**Step 1:** 24 + 20 = 44

**Step 2: **20 / 4 = 5

**Step 3:** 44 + 5 = 49

**Step 4:** 49 + 6 + 3 = 58

**Step 5:** Remainder of 58 / 7 = 2

**Step 6:** Day code 2 = Tuesday

So, the answer is Tuesday.

Also Check – __How to Prepare For Written Examination__

Trick 2 |

Working rule to find the day of the week on a particular date when reference day is given:

**Step 1:** Find the net number of odd days for the period between the reference date and the given date (Exclude the reference day but count the given date for counting the number of net odd days).

**Step 2:** The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).

**Example:** January 11, 1997 was a Sunday. What day of the week was on January 7, 2000?

**Solution:**

Total number of odd days between January 11, 1997 and January 7, 2000

= (365 -11) in 1997 + (365 days in 1998) + (365 days in 1999) + (7 days in 2000)

= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) + (52 weeks + 1 odd day) + (7 odd days)

= 13 days

= 1 week + 6 odd days

Hence, January 7, 2000 would be 6 days ahead of Sunday, i.e., it was on Saturday.

**Example:** If it is Friday on 10 March 2003, then which day would come on 25 June, 2003?

**Solution:**

Total number of odd days between 10 March 2003 and 25 June 2003

= 21 + 30 + 31 + 25 = 107 = 15 weeks + 2 odd days

Hence, 25 June, 2003 would be 2 days ahead of Friday, i.e., it was on Sunday.

Download PDf – __Calendar Reasoning Tricks (____कैलेंडर ____रीज़निंग ____ट्रिक्स)__

We hope that these Calendar Reasoning Tricks (कैलेंडर रीज़निंग ट्रिक्स) will play an important role in your preparation and save your time. If you have any query or problem related to Calendar Reasoning Tricks, you can share with us through below given comment box.

__Something That You Should Put An Eye On__

Surds And Indices Tricks | Simple Interest Tricks |

Compound Interest Tricks | Profit and Loss Tricks |