Let $s$ and $d$ be the original speed of train and total distance. Therefore,

$\left ( 50/s \right ) + \left ( d-50 \right )/\left ( 3s/4 \right ) = \left ( d/s \right )+35/60$ ---(1)

$\left ( 74/s \right ) + \left ( d-74 \right )/\left ( 3s/4 \right ) = \left ( d/s \right )+25/60$ ---(2)

Solving Equation (1) and (2), we get $s$ $=$ $48 km/hr$.