ADURO Quant Lessons - Clock Related Problems


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    The dial of the clock behaves like a circular track and where minute- hand is a fast runner and hour- hand is a slower one.

    For better understanding with clocks, assume 60 min on the dial as 60 points.

    The hour hand covers 5 point in 1 hour. The minute hand covers 60 point in 1 hour.

    The relative speed of two hands of clock is 55 points per hour. 1 point = 6 degree = > 60 point=360 degree

    Time taken to coincide minute hand with hour hand for 1st time=distance/ relative speed =60/55= > 12/11hours

    Now since in 12/11 hours they meet for first time so in 12 hours they will meet 12/(12/11) = 11 times

    For many times minute hand will coincide with hour hand in 24 hours?

    Number of times they will coincide
    =24/(12/11)
    =22 times

    Moving to applications of clocks - There are 4 types of problems on clocks:

    1. To calculate the angle between the two hands when time is given.

    2. To calculate the time when both the hands will be at some angle.

    3. Concept of slow and fast clocks.

    4. Overall gain/loss

    What is the angle between two hands at 3:10am?

    Now we will do it from 2 methods
    1- logical
    2- shortcut

    Method 1-

    At 3:00 The angle will be 90deg.
    Now as the minute hand will travel 10 points, to some extent the hour hand will also get displaced.
    So minute hand will cover 10×6=60deg in 10 min
    And hour hand will cover 10×1/2=5deg in 10 min

    So total angle= hour-minute
    = (90+5)-60
    =35

    Method 2-
    Formula for the angle between the hands = | 30H - (11/2)M |
    Where H → Hour reading & M → Minute reading

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

    | 30H - (11/2)M | = |210 - 385/2| = 17.5

    In 12 hours, How many times the two hands of clock will be opposite to each other ?

    So now solution-
    When they will be opposite they will make an angle of 180deg.
    180 deg = 30 points
    Now first they will meet when they will cover the distance of 30 points. But for next meet they will cover the distance of 60 and so on.
    So 30/55+60/55+60/55+...... = 12
    12 is the total hours
    6+12n=132
    N=10
    So total =11

    If between H and (H +1) o’clock, the two hands are together at an angle θ then required time = 2/11 [H× 30± θ] minutes

    Where H is reading of hour and theta is the angle between two hands.

    At what time between 4 and 5 o’clock are the hands of the clock together ?

    Required time = 2/11[H × 30 ± θ]
    Here θ = 0
    (Hands of clock are together) and H = 4
    ∴ Required time =
    2/11[4 × 30 ± 0] =240/11
    Therefore, the hands will be together at 240/11min past 4.

     

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