# How To Solve Calendar Ques Quickly

## How to Quickly Solve Calendar ques

## Solve Quickly Calendar & Definitions

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

**(i) Ordinary years** (365 days)

**(ii) 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

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

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

### Ques 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

### Ques 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

### Ques 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.**

### Ques 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:

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

### Ques 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

### Ques 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.)**

### Ques 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

### Ques 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

### Ques 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.

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