SSC CPO SI and ASI Model Paper 10

(2) Let the usual speed of the trainbe x km/h.
∴ Time required to reach destination= $\frac{100}{\mathit{x}}$h
Increased speed $=\left(\mathit{x}+5\right)$ kmph
Time to reach destination $=\frac{100}{\mathit{x}}+5\mathit{h}$
$\therefore \frac{100}{\mathit{x}}-\frac{100}{\mathit{x}}+5=\frac{1}{100}$
$\Rightarrow \frac{1}{\mathit{x}}-\frac{1}{\mathit{x}}+5=\frac{1}{100}$$\Rightarrow {\mathit{x}}^2+5\mathit{x}\mathit{–}500=0$$\Rightarrow \mathit{x}=20\mathit{o}\mathit{r}\mathit{–}25$
Disregarding the negative value, theusual speed of the train = 20 kmph