Answer: Option B
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} \cr
& = \left( {45 + 30} \right)\,{\text{km/hr}} \cr
& = \left(75 \times \frac{5}{18} \right)\, \text{m/sec} \cr
& = {\frac{{125}}{6}} \,{\text{m/sec}} \cr
& {\text{We}}\,{\text{have}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{time}}\,{\text{taken}}\,{\text{by}}\,{\text{the}} \cr
& {\text{slower}}\,{\text{train}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{DRIVER}}\,{\text{of}}\, \cr
& {\text{The}}\,{\text{faster}}\,{\text{train}}\,{\text{and}}\,{\text{not}}\,{\text{the}}\,{\text{complete}}\,{\text{train}}{\text{.}} \cr
& \cr
& {\text{So,}}\,{\text{distance}}\,{\text{covered = Length}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}. \cr
& {\text{Therefore,}}\,{\text{Distance}}\,{\text{covered = 500}}\,{\text{m}}. \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& = {500 \times \frac{6}{{125}}} \cr
& = 24\,\sec \cr} $$