Two trains are moving in opposite direction @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in second is

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Solution

The correct option is B 48
Relative speed $= (60 + 90) k m / h r = 150 k m / h r = 150 \times \frac{5}{18} m / s e c = \frac{125}{3} m / s e c$
Distance covered by the slower train to cross the faster train $= (1.1 + 0.9) k m = 2 k m = 2000 m$
The time taken by the slower train to cross the faster train in second $= \frac{d i s t a n c e}{s p e e d} = \frac{2000}{\frac{125}{3}} = 16 \times 3 = 48$ $s e c o n d s$

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Two trains are moving in opposite direction @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in second is

The correct option is B 48Relative speed =(60+90)km/hr=150km/hr=150×518m/sec=1253m/secDistance covered by the slower train to cross the faster train =(1.1+0.9)km=2km=2000mThe time taken by the slower train to cross the faster train in second =distancespeed=20001253=16×3=48 seconds