(c) 2 km/hr
Let the speed of the train be x m/s.
Speed of the 1st man = 6 km/hr = \(\big(6\times\frac{5}{18}\big)\) m/s = \(\frac53\) m/s
\(\frac{75}{(x-\frac53)} = \frac{15}{2}\)
⇒ 150 = 15 x \(\frac{(3x-5)}{3}\) = 15\(x\) - 25
⇒ 15\(x\) = 175 ⇒ \(x\) = \(\frac{175}{15}=\frac{35}{3}\) m/s
Let the speed of the second man be y m/s. Then
\(\frac{75}{\big(\frac{35}{3}-y\big)}=\frac{27}{4}\)
⇒ 300 = 27 x \(\big(\frac{35-3y}{3}\big) = 315 - 27y\)
⇒ 27y = 15 ⇒ y = \(\frac{15}{27}\) m/s = \(\big(\frac{15}{27}\times\frac{18}{5}\big)\) km/hr = 2 km/hr.