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'?
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 :
=> $$3 \times 7s = 18s + 90$$
=> $$21s - 18s = 3s = 90$$
=> $$s = \frac{90}{3} = 30$$ m/s
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