RRB RPF RPSF SI and CONS Model Papers 6 with solutions

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