A train 75 m long overtook a person who was walking at the rate of 6 km/hr in the same direction and passed him in 71/2 seconds. Subsequently, it overtook a second person and passed him in 63/4 seconds. At what rate was the second person travelling?

Answer: Option B

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