SSC CGL 17 Aug 2017 Shift 1 Paper

Let the speed of Duronto train = x km/h
and the speed of Express train = y km /h
∴ Total distance = 432 km
According to the question,
$\frac{432}{\mathit{y}}-\frac{432}{\mathit{x}}=1$
⇒ $\frac{1}{\mathit{y}}-\frac{1}{\mathit{x}}=\frac{1}{432}$ ...(i)
$\mathrm{Again},\frac{432}{\mathit{x}}-\frac{432}{\frac{150\mathit{y}}{100}}=2$
⇒ $\frac{1}{\mathit{x}}-\frac{2}{3\mathit{y}}=\frac{2}{432}$ ...(ii)
Adding Eqs. (i) and (ii),
$\frac{1}{\mathit{y}}-\frac{2}{3\mathit{y}}=\mathrm{}\frac{3}{432}$ $\Rightarrow \frac{3-2}{3\mathit{y}}=\frac{1}{144}$
⇒ $\frac{1}{3\mathit{y}}=\frac{1}{144}\therefore \mathit{y}=\mathrm{}\frac{144}{3}\mathrm{km}∕\mathrm{h}$
Putting the value of y in Eq.(i),
$\frac{3}{144}-\frac{1}{\mathit{x}}=\frac{1}{432}$
⇒ $\frac{3}{144}-\frac{1}{432}=\frac{1}{\mathit{x}}$
⇒ $\frac{9-1}{432}=\frac{1}{\mathit{x}}$
⇒ $\frac{8}{432}=\frac{1}{\mathit{x}}$
⇒ $8\mathit{x}=432$
∴ $\mathit{x}=\frac{432}{8}=54\mathrm{km}∕\mathrm{h}$