SSC CPO SI and ASI Model Paper 4

(4) Let the length of the train be x mand the speed be y m/s.
$\therefore \frac{\mathit{x}}{\mathit{y}}=5$
⇒ x = 5y
Again length of train and bridgetogether = (x + 200) m
$\therefore \frac{\mathit{x}+200}{\mathit{y}}=10$
⇒ x + 200= 10y
⇒ 5y + 200 = 10y
⇒ y = 200/5 = 40
∴ Length of the train = 5 × 40 m =200 m and speed = 40 m/s
∴ Total length of the train and thetunnel is (200 + 600) m = 800 m
∴ Time required by the train tocompletely cross the tunnel
$=\frac{\mathit{D}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}}{\mathit{S}\mathit{p}\mathit{e}\mathit{e}\mathit{d}}=\frac{800}{40}=20$seconds