Tips on Solving Aptitude Questions Based On Calendars

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3 TIPS on cracking Aptitude Questions on
Calendars
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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
• The year can be evenly divided by 4
• If the year can be evenly divided by 100, it is
  NOT a leap year, unless
• 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

Jan Feb Mar Apr May Jun July Aug Sep Oct Nov Dec
31 28 31 30 31 30 31 31 30 31 30 31
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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. gt 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
Year 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018
Odd days 1 2 1 1 1 2 1 1 1 2 1 1
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.
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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.
Month(s) Code
Jan, Oct 0
Feb, Mar, Nov 3
Apr, Jul 6
May 1
Jun 4
Aug 2
Sep, Dec 5
Day Code
Sunday 0, 7
Monday 1
Tuesday 2
Wednesday 3
Thursday 4
Friday 5
Saturday 6
Year Code
1600 1699 6
1700 1799 4
1800 1899 2
1900 1999 0
2000 2999 6
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.
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Tip 3 If you do not have a reference day,
follow these 5 steps (continued..)

• 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.
• 17 98 115.
• Quotient of 98 / 4 24, 115 24 139.
• 139 4 0 143.
• Remainder of 143 / 7 3.
• 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.
• 31 00 31.
• Quotient of 00 / 4 0, 31 0 31.
• 31 5 n 36 n 36, 38, 40 or 42.

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