Calendar Concepts for CAT - Learn Quest


  • Learn Quest


    What is the day on 3 Jan 2017 ?

    When reference date not given, take it as 1 Jan 01 = Monday and start calculating.
    So Every 400 years have 0 odd days.
    Hence till 2000, 0 odd days.
    So on 1 Jan 2001, it’s still Monday
    Now 1 Jan 2001 to 1 Jan 2017 16 years where 2004, 2008, 2012, 2016 are leap yrs
    Hence 12 normal and 4 leap years, so no off odd days = 12 * 1 + 4 * 2 = 20 odd days
    Hence day on 1 Jan 2017 = 1 + 20 = 21 mod 7 = 0 That is Sunday
    So 3 Jan 2017 means Tuesday

    Dec 9, 2001 is Sunday then what was the day on Dec 9, 1971? [CAT 2001]

    Dec 9 1971 to Dec 9 2001 => 30 years.
    Leap years in between => 1972, 1976, .. 2000 = (100-72)/4 + 1 = 8 leap years so 22 normal.
    Odd days = 22 + 8 * 2 = 38
    38 mod 7 = 3
    hence Dec 9 1971 must be 4 then only 4+3 = 7 that is 7 mod 7 = 0
    Hence Dec 9 1971 was 4 = Thursday

    Mayank was born on Feb 29th of 2012 which happened to be a Wednesday. If he lives to be 110 years old, after 2012, how many birthdays would he celebrate on a Wednesday?

    29/02/2012 => Wednesday
    Next birthday on 29/02/2016
    Odd days till then = 2 + 1 + 1 + 1 = 5. It will not be Wednesday, until its multiple of 7
    So odd days in 2020 will be 5 + 5 = 10 then in 2024 it will be 15 and so on.
    Multiple of 7 when we get 5 * 7 = 35 odd days, that will be on 7th leap year, that is after 28 years.
    Hence Wednesday will fall on 29/02/2012
    2012 + 28 = 2040
    2040 + 28 = 2068
    2068 + 28 = 2096
    Now 2100 is not leap, so things will change here
    So next will be 2104 => 7 + 2 = +9 odd days after 2096.
    2108 => 9 + 5 = 14 odd days, multiple of 7, hence 29/02/2108 will also be Wednesday.
    Total => 2040, 2068, 2096, 2108 => 4 times

    Probability of Monday being last day of century = A.
    Probability of Tuesday/Thursday or Saturday being last day of century = B.
    Find A + B.

    100 years contain 5 odd days.
    Last day of 1st century is Friday.
    200 years contain (5 x 2) 3 odd days.
    Last day of 2nd century is Wednesday.
    300 years contain (5 x 3) 1 odd day.
    Last day of 3rd century is Monday.
    400 years contain 0 odd day.
    Last day of 4th century is Sunday.
    This cycle is repeated.
    Last day of a century cannot be Tuesday or Thursday or Saturday.
    So Probability of last day of any century being Monday = A = 1/4 and B = 0
    A + B = 1/4 + 0 = 1/4

    The calendar for the year 2007 can be used again next for which year?

    For the same calendar to be used again, it must be 1st jan of both years should be same day, that is both years should have 0 odd days difference(or multiple of 7).

    2008 => +1
    2009 => +2 so 3 (coz crossed leap yr)
    2010 => +1 so 4
    2011 => +1 so 5
    2012 => +1 so 6
    2013 => +2 so 8
    2014 => +1 so 9
    2015 => +1 so 10
    2016 => +1 so 11
    2017 => +2 so 13
    2018 => +1 so 14
    Hence Calendar of 2007 can be used again next in 2018.

    [Note: Some of you might be thinking, leap yr is 2008, 2012 but we are adding +2 in next year. why? Reason is we are finding days on 1 jan 2008, until then part of leap year which gives extra odd day that is 29 feb didn’t even come in picture. So this comes in picture when you try to find day on 1 jan 2009 or 1 jan 2013 that is next year.]

    Zeller's Calendar Rule

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    15th Aug 1947. Solve Using Zeller.

    15 /aug/1947
    K=15 M=6 C=19 and D=47
    15 + [13 * 6-1/5] + 47 + [47/4] + [19/4] - 2 * 19
    = 15 +15 + 47 + 11+ 4 - 38
    = 54
    so 54 mod 7 =5 = Friday

    Is Zeller formula tough to remember?? Will teach u something even more amazing. Watch the video for the trick!


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