RRB RPF RPSF SI and CONS Model Papers 9 with solutions

(4) Let the length of train = x metres
∴ According to question,
Speed of the train $=\frac{\mathit{x}}{10}$m / sec.
And also the speed of the train $=\left(\frac{\mathit{x}+50}{14}\right)$m / sec.
Both the speeds should be equal, i.e.,
$\frac{\mathit{x}}{10}=\frac{\mathit{x}+50}{14}$
or, $14\mathit{x}=10\mathit{x}+500$
or, $14\mathit{x}-10\mathit{x}=500$
or, $4\mathit{x}=500$
$\therefore \mathit{x}=125$ metres
∴ Speed $=\frac{125}{10}$ = 12.5m/ sec. $=\frac{12.5\times 18}{5}$km / hr. = 45 km/hr.