Clocks & Calendar Concept for CAT - Learn Quest


  • Learn Quest


    Clock is same as circular races. Like in circular races, two persons run on a circular track. Here also two persons are running on a circular track. But here, their name is Minute and hour ( hello ) and the track is called as Clock.

    So we know two persons are running we should know their speed as well. Lets find out their speed

    We know this circular track is of 360°
    So their speed will also be in degrees
    Now minute hand covers 360° in one hour
    So his speed 360°/60 => 6°/minute

    Similarly we know total distance is 360° is divided into 12 parts
    And so value of 1 part => 360/12 => 30°
    So basically hour hand covers 1 part (i.e 30°) in 1 hour
    So speed of hour hand => 30/60
    => (1/2)° / minutes
    and since both are moving in the same direction
    their relative speed => 6 - (1/2) => 5.5° /min

    You have observed the questions that usually come in circular races
    Two persons are running with some speed . when will they meet ?
    Similarly here When will the hour hand and minute hand meet for the first time ?
    So using the same concept discussed in the class
    First time they will meet is given by (total distance) /relative speed
    So here relative speed = 6 - (1/2) => (11/2)
    => 360/ (11/2)
    => (720/11)
    => 65 (5/11) minutes
    i.e in (12/11) hour or 65 (5/11) Two hands of a clock coincide

    How many times minute hand coincides with hour hand in 12 hours ?

    As we know they coincide first time after (12/11) hours
    So in 12 hours they will coincide 12/(12/11) times
    I.e 11 times

    Few Important Points

    In a period of 12 hours Hour hand and minute hand make an angle of
    0° with each other (i.e they coincide with each other)11 times
    180° with each other ( i.e they lie on the same straight line ) 11 times
    90° or any other angle with each other 22 times

    How many times the hands of a clock will be at 30° with each other ?

    As we know in 12 hours 30° --> 22 times
    So in 24 hours --> 44 times

    (i) statement 1 is sufficient to answer the question
    (ii) statement 2 is sufficient
    (iii) either statement 1 or 2 is sufficient
    (iv) both are required to answer the question
    (v) can not be answered even after using both
    What is the time shown in the clock?
    Ι. The minute hand is exactly on 9
    ΙΙ. The hour hand is exactly on 3

    The most common answer - statements Ι and ΙΙ together are sufficient.
    But the correct answer is statement ΙΙ alone is sufficient.
    Let us see why
    3 : 45 is not possible as at 3 : 45 hour hand will not be exactly on 3. It will be between 3 and 4.
    Then looking at the second statement, as the hour hand is exactly on 3, minute hand will be exactly on 12. So time is 3 o’clock.
    Hence statement ΙΙ alone is sufficient.

    Finding the angle between between the two hands at a given time

    Suppose we have to find angle between the two hands at 3:40 am/pm ?

    Let's see at 3:00
    Distance travelled by hour hand => 3 * 30 => 90°
    because one hour --> 30°
    Now since the clock is showing 3 : 40
    Obviously hour hand would have covered some distance extra 40 minutes
    So, 40 * (1/2) => 20°
    So 90 + 20 => 110°
    Now turn of minute hand
    So distance travelled by minute hand => 40 * 6° => 240°
    So angle between minute hand and hour hand => 240° - 110° => 130°

    Now you guys have read one formula to calculate angle between both the hands
    i.e 30H - 5.5M
    Let's see how to get that
    Suppose the clock is showing H : M
    Where H --> hour
    And M --> minutes
    So because of H
    Distance travelled by Hour hand --> 30H and because of M , distance travelled by hour hand
    M * (1/2) => 1/2 M
    So total distance travelled by hour hand => 30H + (1/2)M
    Now Distance travelled by Minute hand M * 6 => 6M
    So Angle between them
    => 30H + (1/2) M - 6M
    => 30H - 5.5M
    And here is a catch
    Angle can be 5.5M - 30H as well
    Depending on which value is bigger Or distance covered by which hand is bigger

    What is the angle between two hands of a clock at 7:35 ?

    30H - 5.5M
    => 30 * 7 = 210
    5.5 * 35 = 192.5
    So 210 - 192.5
    => 17.5°

    Find the time between 2 and 3'o clock at which hour hand and minute hand makes an angle of 60° with each other

    Time --> 2 : M
    And angle is 30H - 5.5M
    Or 5.5M - 30H

    Case (I)
    30 * 2 - 5.5M = 60°
    => 5.5M = 0
    => M = 0°
    So at 2'o clock

    Case (ii)
    5.5M - 30H = 60°
    => 5.5M -30*2 = 60°
    => 5.5M = 120
    => M = 240/11
    So at 2 : (240/11)

    So there two times when they make 60° angle between 2'o and 3'o clock

    A clock strikes ones at 1 O’clock, twice at 2 O’clock and so on. What is the total number of striking in a day?

    The clock strikes once at 1 O’clock,
    twice at 2 O’clock,
    thrice at 3 O’clock
    .
    .
    .
    So in 12 hours ,
    the total number of strikes = 1 + 2 + 3 + 4 + ---- + 12
    12 * 13/2 => 78
    So in 24 hours => 78 * 2 => 156

    The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time. How much in a day does the clock gain or lose?

    Generally minute hand overtake hour hand after 65(5/11) minutes but here it is overtaking after 65 minutes
    Clearly clock is fast
    So it gains (5/11) in every 65 minutes
    So in 24 hours (1440 minutes )
    Gain of => 1440 * (5/11) / 65
    => (1440/143) minutes

    A watch set correctly at 8 a.m. on a Sunday shows 20 min more than the correct time at 4 p.m. on that day.
    (a) If the correct time is 10 p.m. on the same day, then what is the time shown by the watch?
    (b) If the watch shows 8:30 p.m. on the same day, what is the correct time?

    From 8 a.m. to 4 p.m. the time duration is 8 hrs.
    During these 8 hrs, the clock gains 20 min.
    It gains 20/8
    i.e., 2.5 min in 1 hr.

    a) From 8 a.m. to 10 p.m. time duration is 14 hrs.
    ∴ Total time gain = 14 × 2.5 = 35 min.
    Hence watch will show 35 min more.
    Time shown will be 10 : 35 p.m.

    b) As the clock gains, 2.5 min per hour, clock shows 62.5 min elapsed for every 60 min.
    From 8 a.m. to 8:30 p.m. time duration is 12.5 hours
    So 12.5 * 60 => 750 minutes
    Correct time Faulty time
    60 min 62.5
    ? 750
    So ? => 60 * 750 / (62.5) => 720min

    A clock is set right at 7:10 am on Thursday, which gains 12 minutes in a day. On Sunday if this watch is showing 3: 50 pm. What is the correct time?

    Total number of hours => Thursday 7.10 a.m. to the 3 : 50 p.m on Sunday
    => 24 x 3 + 8 hours 40 minutes
    => 80 hours + 40 minutes
    => 242/3 hours
    The clock gains 12 minutes in every 24 hours.
    i.e after 24 hours it will be showing 24 hours 12 minutes [ 121/5] hours
    Means

    Faulty time Correct time
    121/5 24
    242/3
    Correct time = (242 /3) * 24 / (121/5)
    => 80 hours
    So correct time
    => 7:10 + 80 hours
    => 3:10 PM
    => 12 hours
    So correct time
    8am + 12 => 8pm

    A clock was correct at 2 p.m, but then it began to lose 30 minutes each hour. It now shows 6 pm, but it stopped 3 hours ago. What is the correct time now?

    The clock loses 30 minutes per hour.
    i.e 30 minutes of this faulty clock = 60 minutes of the correct clock
    From 2 p.m to 6 p.m, total number of hours = 4 hours
    4 hours of this faulty clock => 4 x 60/30= 8 hours of original clock
    So, the correct time when the clock show 6 p.m = 6 p.m + 4 = 10 p.m
    But the clock stopped 3 hours ago ,
    So present time is 10 p.m + 3 hours = 1 a.m

    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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