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**SCHEDULE DAY& DATE**

__IMPORTANT POINTS TO KNOW:__

- In this type of questions a particular day/date in a particular year is given
- Either a day before or after the date is to found out
- Using this information we have to find out the day which falls on the given date in the question
- A normal year consists of 365 days
- Every fourth year consists of 366 days i.e., 1 day extra known as the leap year
- The extra day in the leap year is 29
^{th}of February - When the number of days are converted into number of weeks we may get some remainder days known as odd days
- Normal year = 365 days = 52 weeks + 1 day

o Here the remaining 1 day = odd day

- Similarly, Leap year = 366 days = 52 weeks + 2 days

o Here there are 2 odd days in a leap year

- A month consists of 30, 31, 28 or 29 days depending on name of the month
- 31 days months are = January, March, May, July, August, October, December
- 30 days months are = April, June, September, November
- February consists of 28 days in a normal year and 29 days in a leap year
- A year can be determined as a leap year if it is divisible by 4

o Leap year = [year/4]

- There are 7 days in a week
- The number of odd days are crucial to find the day of the date given in the question

** TABLE1:**IF THE DAY TO BE FOUND IS AFTER THE REFERENCE DATE GIVEN IN THE QUESTION:

ODD DAY |
DAY OF THE WEEK |

0 | Previous Day |

1 | Given/Same day |

2 | Same day + 1 |

3 | Same day + 2 |

4 | Same day + 3 |

5 | Same day + 4 |

6 | Same day + 5 |

7 | Same day + 6 |

** TABLE 2:**IF THE DAY TO BE FOUND IS BEFORE THE REFERENCE DATE GIVEN IN THE QUESTION:

ODD DAY |
DAY OF THE WEEK |

0 | Next Day |

1 | Given/Same day |

2 | Same day – 1 |

3 | Same day – 2 |

4 | Same day – 3 |

5 | Same day – 4 |

6 | Same day – 5 |

7 | Same day – 6 |

__QUESTION1:__

The 1^{st} day of the year 1998 was Thursday. If the birthday of Arun falls on 26^{th} June, then on which day in 1998 was his birthday?

__GIVEN:__

1^{st} January 1998 = Thursday

__SOLUTION:__

First we have to check whether 1998 is a normal or a leap year

1998 is not divisible by 4 i.e., it is a normal year with February = 28 days

Number of days to be calculated from 1^{st}January 1998 to 26^{th} June 1998

January + February + March + April + May + June

= 31 + 28 + 31 + 30 + 31 + 26

= 177 days

Now to find the number of odd days convert the days into weeks

[177/7] = 25 weeks + 2 days

Therefore odd days = 2

Hence from table 1, 2 odd days = same day + 1

àThursday + 1 = Friday

Therefore,**26 ^{th} June 1998 was a Friday**

__QUESTION2:__

The Independence Day was celebrated on Friday, 15^{th} August in 1996. What was the 1^{st}day of 1996?

__GIVEN:__

15^{th} August 1996 = Friday

__SOLUTION:__

Here the day prior to the reference date is to be calculated

1^{st} we have to find out whether 1996 is a normal/leap year

[1996/4] = 499

Hence, 1996 is a leap year = February = 29 days

Now the number of days from 1^{st}January 1996 to 15^{th} August 1996 is to calculated

January + February + March + April + May + June + July + 15^{th} August

= 31 + 29 + 31 + 30 + 31 + 30 + 31 + 15

= 228 days

To find the number of odd days

[228/7] = 32 weeks + 4 days

From Table 2,

4 odd days = Same day – 3 = Friday – 3 = Tuesday (3 days before Friday)

Therefore**1 ^{st}January 1996 was a Tuesday**