# Easy Solution Of Calendar Problems In Seconds

Calendar Problem like find the future date days or find out the day of past years are the most common and most likely important questions that are asked in Public Service Commission - Lok Sewa Aayog or any IQ - GK questions for selecting government or semi-government jobs.

Many people get confused with these problems. You can find a lot of solutions for getting answer these questions but the method of solving the calendar problems are different and may be hard to remember this procedure.

So, we are here to give the best easy ideas by which you can solve any kind of problems in few seconds. This method is very easy to understand and remember. It contains very few steps to solve any type of questions that mostly asked in job selecting exams.

So lets begin,

### First of all, here is 3 very easy tables that you have to remember:

**Table 1:**

Year | Odd Days |
---|---|

20 ** | 6 |

19 ** | 0 |

**Table 1:**Questions are in range of 1900 to 2999 only, so just remember if the year is on 20's, the odd days are 6 and for 19's year the odd day is 0.

**Table 2**

Odd Days | ||||||||||||

Jan | Feb | Mar | Apr | May | Jun | July | Aug | Sep | Oct | Nov | Dec | |

Normal Year | 1 | 4 | 4 | 0 | 2 | 5 | 0 | 3 | 6 | 1 | 4 | 6 |

Leap Year | 0 | 3 | 4 | 0 | 2 | 5 | 0 | 3 | 6 | 1 | 4 | 6 |

**Table 2:**For each months the odd days as mentioned above. Don't panic its very easy to remember, just remember in this pattern:

**144, 025, 036, 146**for Normal Year, And for Leap Years most of all are same only in Jan and Feb is different, so just remember in leap year Jan 0 and Feb 3, then all are same as normal year.

**Table 3**

Day | Sun | Mon | Tue | Web | Thu | Fri | Sat |

Odd Days | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

**Table 3:**This table is also very easy, just start 1 for Sunday and continue for other days increasing numbers.

OK,Great Work, Now the do these easy steps for finding days for required date.

### Method for finding out day of given date:

**For example: Find out the day of 2030 Jan 1?**

- First of all check the given year is leap year or not. For this just divide the given year by 4. (2030/4=507.5, i.e. it is not Leap Yea, it is Normal Year)
- Separate given year in two parts of two digits. (2030=> 20 | 30)
- Now lets calculate the total odd days of given date 2030 Jan 1:
- Take the day same as it is. = 1 ( 1 )
- Take the odd value of month. = 1 (January of Normal Year: 1 from table 2)
- Take last two digit of given year as it it. = 30 (30 last two digit of year as separated in step 2)
- Divide last 2 digit of year by 4 and take only quotient. = 7 (30/4= quotient:7, remainder:2)
- Take the odd number of first two digit of given year. 6 (20 | 00 => 6 from table 1)
- Now sum all of above odd days. =1+1+30+7+6 = 45
- Now take remainder by dividing that total sum by 7. = 3 (45/7 = quotient: 6 remainder: 3)
- We have the final odd day = 3, Now check this number on Table 3 and find the answer Day.

**Answer: Tuesday (From table 3)**

**Example 2: If 1983 June 13 is Sunday, Then what will be the day of 2018 Sep 30.**

For this type of question, first of all we will find the odd days of both of the given dates.

**Date 1: 1983 June 13**

Checking Leap Year: 1983/4 = 495.75, so its normal year.

**Now, by applying above method steps for calculating Odd days:**

13 (given day)

5 (odd value of month June by table 2)

83 (last two digit of given year)

20 (quotient of last two digit of year dividing by 4)

0 (odd days of first two digit given year 19 | ** by table 1)

**Total:**13+5+83+20+0=121

Now remainder of this total while diving it by 7

121/7 = 2 (quotient: 17, remainder: 2)

**Therefore, Final Odd Number is: 2**

__As by Table 3:__

__The real day is__

__Monday__

Now lets find odd days of another year,

**Date 1: 2018 Sep 30**

Checking Leap Year: 2018/4 = 504.5, so its normal year.

**Now, by applying above method steps for calculating Odd days:**

30 (given day)

6 (odd value of month June by table 2)

18 (last two digit of given year)

4 (quotient of last two digit of year dividing by 4)

6 (odd days of first two digit given year 20 | ** by table 1)

**Total:**30+6+18+4+6 = 64

Now remainder of this total while diving it by 7

64/7 = 1 (quotient: 9, remainder: 1)

**Therefore, Final Odd Number is: 1**

__As by Table 3: The real day is Sunday__

**Now,**

As per question, 1983 June 13 is Sunday,

but as we calculated the real day is Monday.

**So the given day is just 1 day before of real day. ------- logic 1**

**So,**

according to this logic 1,

real day of required date 2018 Sep 30 is Sunday

Final answer = 1 day before of Sunday. i.e. Saturday

**Therefore, Our required Answer is: Saturday.**

Thus, by using common logic and using above mentioned easy steps you can solve any questions related to calendar problems. I hope you understand this easy method. If you have any queries you can contact me through comment below. Or can contact on

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