SSC CGL 19 Oct 2014 Shift 2 Paper

Let both trains meet after t hours since 7 a.m.
Distance between stations A and B = x Km.
∴ $\frac{\mathit{x}}{4}$ × t + $\frac{\mathit{x}}{\frac{7}{2}}$ × (t - 1) = x
⇒ $\frac{\mathit{t}}{4}$ + $\frac{2\left(\mathit{t}-1\right)}{7}$ = 1
⇒ $\frac{7\mathit{t}+8\mathit{t}-8}{28}$ = 1
⇒ 15t - 8 = 28
⇒ 15 t = 28 + 8 = 36
⇒ t = $\frac{36}{15}$ = $\frac{12}{5}$ hours
= 2 hours 24 minutes
∴ Required time = 9 :24 a.m.