LIC AAO Model Paper 9

Let the length of train A be x metre.
∴ Length of train B = $\frac{\mathit{x}}{2}$ metre
Speed of train A = 126 kmph
= (126×5/{18}) m/sec
= 35 m/sec
Speed of train B = 90 kmph.
= $\left(90\times \frac{5}{18}\right)$ m/secc
= 25 m/sec
∴ Relative speed of A = $\frac{\mathrm{Length}\mathrm{of}\mathrm{trains}\mathrm{A}\mathrm{and}\mathrm{B}}{\mathrm{Time}\mathrm{taken}\mathrm{in}\mathrm{crossing}}$
⇒ 35 + 25 = $\frac{\mathit{x}+\frac{\mathit{x}}{2}}{9}$
⇒ 60 × 9 = $\frac{3\mathit{x}}{2}$
⇒ 3x = 90 × 12
⇒ x = $\frac{90\times 12}{3}$ = 360 metre
∴ Time taken in crossing the platform
$\frac{\mathrm{Length}\mathrm{of}\mathrm{train}\mathrm{A}\mathrm{and}\mathrm{platform}}{\mathrm{Speed}\mathrm{of}\mathrm{train}\mathrm{A}}$
= $\left(\frac{360+690}{35}\right)$ seconds
= $\frac{1050}{35}$ seconds
= 30 seconds