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ATQ,
Let the original speed of the train = ‘y’ km/h
Then, decreased speed of the train = 0.75 × y = ‘0.75y’ km/h
According to the question,
(150/0.75y) – (150/y) = 1.5
Or, (600/3y) – (450/3y) = 1.5
Or, (150/3y) = (50/y) = 1.5
So, y = 50 ÷ 1.5 = 33.33 km/h
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