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# How To Solve Calendar Question Quickly

## How To Solve Calendar Problems Quickly in Aptitude

Here, On this page you will get to know How To Solve Calendar Problems Quickly in Competitive Exam. Go through this page to find useful Insights on Calendar based Problem.

### 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 |
---|---|---|---|---|---|---|---|

No. 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 |
---|---|---|---|---|---|---|---|

No. of Odd Days | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

**Correct option: C**

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**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 ,

Months | March | April | May | June | July | August | September | October |
---|---|---|---|---|---|---|---|---|

Odd Days | 9 | 7 | 11 | 9 | 8 | 6 | 5 | 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 |
---|---|---|---|---|---|---|

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 |
---|---|---|---|---|---|---|

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. Starting with **2002 having No. Of Odd Days = 1.**

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

Odd Days | 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**

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