SSC CHSL Model Paper 8

Relative speed = (45 + 30) km/hr
$=75\times \frac{5}{18}$ m/sec $=\frac{125}{6}$ m/sec
We have to find the time taken by the slower train to pass the driver of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, distance covered = 500 m
∴ Required time = $\left(500\times \frac{6}{125}\right)=24$ sec