Explanation

(c) The reflection sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity.
$v^{'} = \frac{v\left( v + v_{o} \right)}{\left( v - v_{s} \right)}$
$= v\left( \frac{v + v_{s}}{v - v_{s}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because v_{o} = v_{s} \right)$
$\Longrightarrow v^{'} = 1.2\left( \frac{350 + 50}{350 - 50} \right)$
$= \frac{1.2 \times 400}{300} = 1.6\ kHz$