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    A train travelling at 48 kmph completely crosses another train having half of its length and travelling in opposite direction at 42 kmph , in 12 secon
    Question

    A train travelling at 48 kmph completely crosses another train having half of its length and travelling in opposite direction at 42 kmph , in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is :

    A.

    400 mt .

    B.

    450 mt .

    C.

    560 mt .

    D.

    600 mt .

    Correct option is A

    Given:

    1. The speed of the first train = 48 km/h.
    2. The speed of the second train = 42 km/h.
    3. The length of the second train = half the length of the first train.
    4. The first train completely crosses the second train in 12 seconds.
    5. The first train passes a platform in 45 seconds.
    6. We need to find the length of the platform.

    Formula Used:

    1. Speed Conversion:

    Speed (in m/s)=Speed (in km/h)×518\text{Speed (in m/s)} = \text{Speed (in km/h)} \times \frac{5}{18}​​

    2. Time Relation:

    Relative Distance=Relative Speed×Time\text{Relative Distance} = \text{Relative Speed} \times \text{Time}​​

    3. Passing a Platform:

    Length of Train+Length of Platform=Speed of Train×Time\text{Length of Train} + \text{Length of Platform} = \text{Speed of Train} \times \text{Time}​​

    Solution:

    1. Convert speeds to m/s:
    - Speed of the first train:

    Speed of Train 1=48×518=13.33 m/s\text{Speed of Train 1} = 48 \times \frac{5}{18} = 13.33 \, \text{m/s}​​

    - Speed of the second train:

    Speed of Train 2=42×518=11.67 m/s\text{Speed of Train 2} = 42 \times \frac{5}{18} = 11.67 \, \text{m/s}​​

    2. Relative speed of the two trains (opposite directions):

    Relative Speed=13.33+11.67=25 m/s\text{Relative Speed} = 13.33 + 11.67 = 25 \, \text{m/s}​​

    3. Relative distance when crossing each other:
    Let the length of the first train be L meters. Then, the length of the second train is L2\frac{L}{2}​ .
    Total distance = L+L2=3L2L + \frac{L}{2} = \frac{3L}{2} ​​

    Using the formula:

    3L2=25×12\frac{3L}{2} = 25 \times 12​​

    Simplify:

    3L2=300 L=300×23=200 m\frac{3L}{2} = 300 \implies L = \frac{300 \times 2}{3} = 200 \, \text{m}​​

    Thus, the length of the first train is 200 m

    4. Passing the platform:
    The total distance covered when passing the platform = Length of Train + Length of Platform.
    Using the formula:

    L+Platform Length=Speed of Train 1×TimeL + \text{Platform Length} = \text{Speed of Train 1} \times \text{Time}​​

    Substitute values:

    200+Platform Length=13.33×45200 + \text{Platform Length} = 13.33 \times 45​​

    Simplify:

    200+Platform Length=600200 + \text{Platform Length} = 600​​

    Platform Length=600200=400 m\text{Platform Length} = 600 - 200 = 400 \, \text{m}​​

    Final Answer:

    The length of the platform is 400 meters
    **Option A: 400 m**

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