Answer: Option B. -> 50 km/hr
Speed of the train relative to man
$$\eqalign{
& = \frac{{125}}{{10}}{\text{m/sec}} \cr
& = \frac{{25}}{2}{\text{m/sec}} \cr
& = \left( {\frac{{25}}{2} \times \frac{{18}}{5}} \right){\text{m/sec}} \cr
& = 45\,{\text{km/hr}} \cr} $$
Let the speed of the train be x kmph.
Then, relative speed = (x - 5) kmph
∴ x - 5 = 45 or
x = 50 km/hr

Answer: Option D. -> 64 seconds
Let the lengths of the train and the bridge be x meters and y meters respectively.
Speed of the first train
= 90 km/hr
= $$\left( {90 \times \frac{5}{{18}}} \right)$$  m/sec
= 25 m/sec
Speed of the second train
= 45 km/hr
= $$\left( {45 \times \frac{5}{{18}}} \right)$$  m/sec
= $$\frac{{25}}{2}$$ m/sec
Then, $$\frac{{{\text{x}} + {\text{y}}}}{{36}}$$ = 25
⇒ x + y = 900
∴ Required time
$$\eqalign{
& = \left[ {\frac{{\left( {{\text{x}} - 100} \right) + {\text{y}}}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr
& = \left[ {\frac{{\left( {{\text{x}} + {\text{y}}} \right) - 100}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr
& = \left( {800 \times \frac{2}{{25}}} \right){\text{sec}} \cr
& = 64\,{\text{sec}} \cr} $$