SSC CPO SI and ASI Model Paper 1

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