Question 63

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'?

Solution

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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