SSC CHSL Model Paper 18

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter
Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)
Distance = 2x + x = 3x
Relative speed = 48 + 42 = 90 kmph
$=90\times \frac{10}{36}$ m/s = 25 m/s
Time = 12 s$\frac{\mathrm{Distance}}{\mathrm{time}}=\mathrm{speed}$
$\Rightarrow \frac{3\mathit{x}}{12}=25$
⇒ $\mathit{x}=25\times \frac{12}{3}=100$ meter
Length of the first train = 2x = 200 meter
Time taken to cross the platform = 45 s
Speed of train1 = 48 kmph = $\frac{480}{36}=\frac{40}{3}$m/s
Distance = 200 + y where y is the length of the platform
⇒ $200+\mathit{y}=45\times \frac{4}{3}=600$
⇒ $\mathit{y}=400$ meter