RRB Non Technical Model Papers 9 with solutions

(4) Let the speed of coming train from opposite direction= x km/h.
∴From question,
speed of 1st train
=$\frac{150}{15}$m/s = 10 m/s.
$=10\times \frac{18}{5}$km/h
= 36 km/h.
∴From question,
$\frac{\left(150+150\right)\mathit{m}}{12\mathit{s}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{d}}$
= (36 + x) km/h.
⇒$\frac{300}{12}$m/s = (36+x) km/h.
$\Rightarrow \frac{300}{12}\times \frac{18}{5}$ km/h.
= (36+x) km/h.
⇒ 90 km (36+x) km/h.
∴ x = 54 km/h.