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# How To Solve Calendar Question Quickly

## How To Solve Calendar Problems Quickly in Aptitude

**For the calendars of two different years to be the same, the following conditions must be satisfied. Firstly, ****Both years must be of the same type. i.e., both years must be ordinary years or both years must be leap years. And Secondly , ****1 ^{st}January of both the years must be the same day of the week.**

### How to Quickly Solve Calendar Questions in Aptitude

**Solve Quickly Calendar Problems with Definitions**

The days in a year are organized through a system of calendar The year is classified into two types :

**Ordinary Years**(365 days)**Leap year**(366 days)

The difference between leap and ordinary year and leap year is of 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 1/4th 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

**Type 1: How to Solve Calendars Quickly ( To find the day of given date)**

**Question 1. What was the day on 20 June 1776?**

**Options:**

**A. Thursday**

**B. Monday**

**C. Wednesday**

**D. Tuesday**

**Solution: **20 June 1776 = (1775 years + Period from 1.1.1776 to 20.6.1776)

To calculate number of odd days till 1775, we need

Number of odd days in 1600 years = 0

Number of odd days in 100 years = 5

Number of odd days in 75 years = 18 leap years + 57 ordinary years = 2 odd days

Therefore, 1775 years had 0 + 5 + 2 = 7 odd days = 1 week = 0 odd days

Now, for calculating odd days for 1.1.1776 to 20.6.1776,

January (31 days) + February (29 days because 1776 is a leap year) + March (31 days) + (April 30 days) + May (31days) + June (20) = 172 days

Total number of odd days in 172 days = (172/7) = 24 weeks + 4 odd days

Total Number of odd days corresponding to the given date = 0 + 4 = 4 odd days

As per the table, on 20 June 1776 the day was Thursday.

Days | Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |

Number of odd days | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

**Correct option: A**

** Question 2. Which of the following is not a leap year?**

**Options:**

**A. 2400**

**B.1896**

**C.1656**

**D. 1900**

** Solution: **The year which is divisible by 4, 100 and 400 is a leap year.

**Correct option: D**

**Question 3 What was the day on 15th August 1947?**

**Options:**

**A. Monday**

**B. Sunday**

**C. Friday**

**D. Tuesday**

** Solution: **15th August 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)

To calculate number of odd days till 1946, we need

Number of odd days in 1600 years = 0

Number of odd days in 300 years = 1

Number of odd days in 46 years = 11 leap years + 35 ordinary years = 1 odd days

Therefore, 1946 years had 0 + 1 + 1 = 2 odd days

Now, for calculating odd days for 1.1.1947 to 15.8.1947,

January (31 days) + February (28 days because 1947 is ordinary year) + March (31 days) + (April 30 days) + May (31days) + June (30 days) + July (31 days) + August (15 days) = 227 days

Total number of odd days in 227 days = ((227/7) = (32 weeks + 3 days) = 3 odd days

Total Number of odd days corresponding to the given date = 2 + 3 = 5 odd days

As per the table, on 15^{th} August 1947 the day was Friday.

Days | Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |

Number of odd days | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

**Correct option: C**

**Type 2: To find the day of the week for the given date when some other date or day is specified.**

**Question 1. It was a Wednesday on 1st March. Which month will have Wednesday as the 1st day in the same year?**

**Options:**

**A. March**

**B. November**

**C. April**

**D. October**

**Solution: **By the following table , we get the solution ,

Month | March | April | May | June | July | August | September | October |

Odd Days | 3 | 2 | 3 | 2 | 3 | 3 | 2 | 3 |

Total number of days =3+2+3+2+3+3+2+3 = 21 days

21/7= 0 odd days, Therefore, 1^{st} November will be Wednesday.

**Correct option: B**

**Question 2. Taimur celebrated his 23rd birthday on 28th December 2013. On what day will he celebrate his 25th birthday?**

**Options:**

**A. Sunday**

**B. Wednesday**

**C.Saturday**

**D. Monday**

**Solution: **Number of odd days in the given period,

28th December 2013 = (2000 years + 13years)

Number of odd days in 2000 years = 0

Number of odd days in 13 years = (10 ordinary years + 3 leap years) = 8

Therefore, 13 years has 8 odd days = (8/7) = 1 odd days

This means that 31^{st} December was Monday (2 odd days).

So, 28^{th} December would have been Friday.

Taimur celebrated his 23^{th} birthday on 28^{th} December 2015.

So, number of odd days between her 23^{rd} and 25^{th} birthday = 2 odd days (two ordinary years)

Therefore, he will celebrate his 25^{th} birthday on Sunday.

**Correct option: A**

**Question 3. 2 January, 2007 was Tuesday. What day of the week was on 2 January, 2008?**

**Options:**

**A. Wednesday**

**B. Thursday**

**C. Tuesday**

**D.Monday**

** Solution: **The year 2007 was an ordinary year.

We know that one ordinary year has 1 odd day.

If 2 January, 2007 was Tuesday, then 2 January, 2008 will be 1 day beyond Tuesday.

Therefore, 2 January, 2008 was Wednesday.

**Correct option: A**

**Type 3: Solve Quickly Calendar’s Ques. (Identify the year having same calendar.)**

**Question 1. Which year will have the same calendar as that of the year 1998?**

**Options:**

**A. 2004**

**B. 2000**

**C. 2005**

**D. 2003**

**Solution: **Count the number of odd days from the year 1998 onwards to get the sum equal to 0 odd day.

Year | 1998 | 1999 | 2000 | 2001 | 2002 | 2003 |

Number of Odd days | 1 | 1 | 2 | 1 | 1 | 1 |

Sum = 7 odd days = 0 odd days

Therefore, calendar of the year 2004 will be the same as that of the year 1998.

**Correct option: A**

**Question 2. Which year will have the same calendar as that of the year 1700?**

**Options:**

**A. 1705**

**B. 1707**

**C. 1800**

**D. 1706**

** Solution:** Count the number of odd days from the year 1700 onwards to get the sum equal to 0 odd day.

Year | 1700 | 1701 | 1702 | 1703 | 1704 | 1705 |

Number of Odd days | 1 | 1 | 1 | 1 | 2 | 1 |

Sum = 7 odd days = 0 odd days

Therefore, calendar of the year 1700 will have the same calendar as that of the year 1706

**Correct option: D**

**Question 3. Which year will have the same calendar as that of the year 2002?**

**Options:**

**A. 2008**

**B. 2013**

**C. 2012**

**D. 2003**

**Solution: **Count the number of odd days from the year 2002 onwards to get the sum equal to 0 odd day.

Year | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |

Number of Odd days | 1 | 1 | 2 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 2 |

Sum = 14 odd days = 0 odd days

Therefore, calendar of the year 2013 will have the same calendar as that of the year 2002.

Note: Number of odd day from 2002 – 2007 = 0 odd days. But 2008, cannot be same as 2002 because 2008 is a leap year.

**Correct option: B**

**Read also** – **Formulas to solve calendar questions**

**Read also** – **Tips and Shortcuts for Calendar problems**

**Read also** – **Practice question on Calendar**

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