SSC CHSL 15 Nov 2015 Shift 1 Paper

Let length of train =$\mathit{l}$m and speed = $\mathit{x}$ m/s
Time taken to pass the first bridge (length = 500 m) = $100=\frac{\mathit{l}+500}{\mathit{x}}$
⇒ $\mathit{l}+500=100\mathit{x}$ --------(i)
Similarly, $60=\frac{\mathit{l}+250}{\mathit{x}}$
⇒ $\mathit{l}+250=60\mathit{x}$ --------(ii)
Dividing equation (i)by (ii), we get :
⇒ $\frac{\mathit{l}+500}{\mathit{l}+500}=\frac{100\mathit{x}}{60\mathit{x}}$
⇒ $\frac{\mathit{l}+500}{\mathit{l}+500}=\frac{5}{3}$
⇒ $3\mathit{l}+1500+5\mathit{l}+1250$
⇒ $5\mathit{l}-3\mathit{l}=1500-1250$
⇒ $2\mathit{l}=250$
⇒ $\mathit{l}=\frac{250}{2}=125$m