SSC CGL 9 Aug 2017 Shift 1 Paper

Let the distance between Jaipur and Delhi be x km.
Train 1,
10 a.m. to 4 p.m. = 6 hours
∴ Speed $=\frac{\mathit{x}}{6}$ km /h
Train 2,
12 p.m. to 5 p.m. = 5 hours
∴ Speed $=\frac{\mathit{x}}{5}$ km /h
Let the two trains meet each other after t hours.
Distance covered by the first train $=\mathit{t}\times \frac{\mathit{x}}{6}$
Distance covered by the second train $=\left(\mathit{t}-2\right)\times \frac{\mathit{x}}{5}$
According to the question,
$\mathit{x}=\frac{\mathit{x}\mathit{t}}{6}+\frac{\mathit{x}\left(\mathit{t}-2\right)}{5}$
⇒ $5\mathit{t}+6\mathit{t}-12=30\Rightarrow 11\mathit{t}=42$
∴ $\mathit{t}=\frac{42}{11}=3\frac{8}{11}$
Required time $=10\mathrm{a}.\mathrm{m}.+3\frac{8}{11}=1:49$p.m
So, the trains will meet at 1:49 p.m.