1. Train A, travelling at 'S' m/sec, can cross a platform double its length in 21 sec. The same train, travelling at (S + 5) m/sec, can cross the same platform in 18 sec. What is the value of 'S'?






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  • By: anil on 05 May 2019 01.21 am
    Let length of train = $$x$$ m => Length of platform = $$2x$$ m Using, $$time = frac{distance}{speed}$$ While travelling at $$s$$ m/s, time taken => $$frac{x + 2x}{s} = 21$$ => $$x = 7s$$ --------------Eqn(1) Also, $$frac{x + 2x}{s + 5} = 18$$ => $$3x = 18 (s + 5)$$ Using eqn(1), we get : => $$3 imes 7s = 18s + 90$$ => $$21s - 18s = 3s = 90$$ => $$s = frac{90}{3} = 30$$ m/s
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