 # Calendars

Calendar Problems in Aptitude: Tricks for Calendar Aptitude, Calendar Aptitude Formulas, Calendar Aptitude Questions with Solutions and Shortcut for Problems

## Calendar Problems in Aptitude: Tricks for Calendar Aptitude, Calendar Aptitude Formulas, Calendar Aptitude Questions with Solutions and Shortcut for Problems # 3 TIPS on cracking Aptitude Questions on Calendars

Looking for Questions instead of tips? - You can directly jump to  Aptitude Test Questions on Calendars

## Tip #1: Understand the basics of the Gregorian calendar

The day of the week repeats after every seven days.

For any time period, the no. of days in excess over complete weeks are called odd days.

A normal year has 365 days (i.e.) 52 weeks plus 1 odd day. When we proceed forward by one year, then 1 day is gained and vice-versa.

The number of days per month in a normal year are: A year is considered to be a leap year if:

a)    The year can be evenly divided by 4;

b)    If the year can be evenly divided by 100, it is NOT a leap year, unless;

c)    The year is also evenly divisible by 400. Then it is a leap year

Leap years have 366 days. The extra day is added to February which has 29 days in a leap year.

When we proceed forward by one leap year, then 2 odd days are gained because leap years have 366 days

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## Tip #2: If you have a reference day, then:  Reqd. day= Ref. day ± N weeks ± M days

Question: What day of week was it on 5th November 1989 if it was Monday on 4th April 1988 ?

Solution:

4th April 1988 was a Monday.

1988 was a leap year, hence it has 2 odd days. However, since we are starting from April 4th, February has already been crossed and hence the extra odd day can be ignored.

=> 4th April 1989 was a Tuesday

No of days between 4th April 1989 and 5th November 1989

= 26 + 31+ 30 + 31 + 31 + 30 + 31 + 5 = 215 days (which has 5 odd days.)

5th November 1989 was a Sunday

Question: Find the year nearest to 2007 in future which is the same calendar year as it.

Solution: When no. of odd days is multiple of 7, the year will have same calendar year as 2017.

In 2018, no. of odd days becomes 14. Thus, the reqd. year is 2018.

Note: The questions on calendars can be broadly categorized into 2 types:

1.    Those that provide a reference day and need you to find the day of any particular date using the reference day.

2.    Problems which do not provide any reference data but ask you to find the day of a particular date.

Although these might be complicated, there are some points and shortcuts that are helpful in solving the problems in just a few seconds.

If you have a reference day, express the required day in the form of

Required day = Reference day ± N weeks ± M days

Also, remember the following points:

a)    The day of the week repeated after every seven days.

b)    When we proceed forward by one year, then 1 day is gained and vice-versa.

c)    When we proceed forward by one leap year, then 2 days are gained and vice-versa.

d)    Every year divisible by 4 is a leap year, if it is not a century.

e)    Every 4th century is a leap year and no other century is a leap year.

## Tip #3: If you do not have a reference day, follow these 5 steps

First of all, memorize the following tables. Then follow the steps below.   Step 1: Add the day digits to last two digits of the year.

Step 2: Divide the last two digits of the year by four and add it to the result in step 1.

Step 3: Add Month Code and year codes to the result obtain in step 3.

Step 4: Divide the result of step 4 by seven.

Step 5: Obtain the remainder and match with the day code.

Question: What was the day of the week on 17th June, 1998?

Solution:

Year Code: 0, Month Code: 4, Day digits = 17, Last 2 digits of the year = 98.

1.    17 + 98 = 115.

2.    Quotient of 98 / 4 = 24, 115 + 24 = 139.

3.    139 + 4 + 0 = 143.

4.    Remainder of 143 / 7 = 3.

5.    Day code 3 = Wednesday.

Thus, 18th June 1998 was a Wednesday.

Question: Which days of the week cannot be the last day of a century?

Solution:

Year code = n, Month Code = 5, Day digits = 31, Last 2 digits of the year = 00, where n = 0, 2, 4 or 6.

1.    31 + 00 = 31.

2.    Quotient of 00 / 4 = 0, 31 + 0 = 31.

3.    31 + 5 + n = 36 + n = 36, 38, 40 or 42.

4.    Remainder of: 36 / 7 = 1, 38 / 7 = 3, 40 / 7 = 5, 42 / 7 = 0.

5.    Day Code: 1 = Monday, 3 = Wednesday, 5 = Friday, 0 = Sunday.

Thus, the last day of a century cannot be Tuesday, Thursday or Saturday.

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Calendar Problems in Aptitude: Tricks for Calendar Aptitude, Calendar Aptitude Formulas, Calendar Aptitude Questions with Solutions and Shortcut for Problems

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