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ATQ, Let the length of the platform be '10a' metres So, length of the train = 10a × 0.6 = '6a' So, speed of the train = {(6a + 10a)/20} = '0.8a' m/s So, new speed of the train = 0.8a × 1.25 = 'a' m/s So, required time = 6a ÷ a = 6 seconds
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