Answer & Solution
Answer: Option B
Solution:
Speed of the train relative to first man
$$\eqalign{
& = \frac{{75}}{{7.5}}{\text{m/sec}} = 10\,{\text{m/sec}} \cr
& = \left( {10 \times \frac{{18}}{5}} \right){\text{km/hr}} = 36\,{\text{km/hr}} \cr} $$
Let the speed of the train be x km/hr.
Then, relative speed = (x - 6) km/hr
∴ x - 6 = 36
⇒ x = 42 km/hr
Speed of the train relative to second man
$$\eqalign{
& {\text{ = }}\frac{{75}}{{6\frac{3}{4}}}\,{\text{m/sec}} \cr
& = \left( {75 \times \frac{4}{{27}}} \right){\text{m/sec}} \cr
& = \frac{{100}}{9}{\text{m/sec}} \cr
& = \left( {\frac{{100}}{9} \times \frac{{18}}{5}} \right){\text{km}} \cr
& = 40\,{\text{km/hr}} \cr} $$
Let the speed of the second man be y kmph.
Then, relative speed = (42 - y) kmph
∴ 42 - y = 40
⇒ y = 2 km/hr