RRB Non Technical Model Papers 8 with solutions

(4) Let the original speed of the train be x kmph ∴ Original time$=\frac{400}{\mathit{x}}$hr New speed of train
= (x + 10) kmph and new time
required =$\frac{400}{\mathit{x}+10}$hr According to
question, $\frac{400}{\mathit{x}}\mathit{–}\frac{400}{\mathit{x}}+10=2$
$\Rightarrow {\mathit{x}}^2+10\mathit{x}\mathit{–}2000=0\Rightarrow \mathit{x}=40$
and –50 Disregarding the negative value, required original speed of train = 40 kmph