Two trains are moving in the same direction at speed of 60 km/hr and 70 km/hr. The time taken by faster train to cross a man sitting in the slower train is 2 minutes 42 seconds. What will be the length (in metres) of the faster train?

Relative speed = (70 – 60) km/h $=10 mathrm{~km} / mathrm{h}=left(frac{10 imes 5}{18} ight) mathrm{m} . / mathrm{sec} .=left(frac{25}{9} ight) mathrm{m} . / mathrm{sec}$ Time = 2 minutes 42 seconds = (120 + 42) seconds = 162 seconds $herefore$ Length of faster train $=left(frac{25}{9} imes 162 ight)$ mitre $=450$ metre

Alternative :

Let speed of 1st and 2nd train be $mathrm{S}_{1} & mathrm{~S}_{2}$

respectively.

$mathrm{S}_{1}=60 mathrm{kmph}$

$mathrm{S}_{2}=70 mathrm{kmph}$

Relative speed = (70 – 60) kmph = 10 kmph

Length of faster train = Relative speed × Time

$=10 imes frac{5}{18} imes(2 imes 60+42)=450$ metre