Subject: General Aptitude · Chapter: Problems on Trains · Exam: 2008 · Marks: · Difficulty:
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 $7 \frac{1}{2}$ seconds. Subsequently, it overtook a second person and passed him in $6 \frac{3}{4}$ seconds. At what rate was the second person travelling?
(a)1 km/hr
(b)2 km/hr
(c)4 km/hr
(d)5 km/hr
Answer
Answer (as printed): B
Explanation
Speed of the train relative to first man $$\begin{aligned} & =\left(\frac{75}{7.5}\right) \mathrm{m} / \mathrm{sec}=10 \mathrm{m} / \mathrm{sec} \\ & =\left(10 \times \frac{18}{5}\right) \mathrm{km} / \mathrm{hr}=36 \mathrm{km} / \mathrm{hr} . \end{aligned}$$ Let the speed of the train be $x \mathrm{km} / \mathrm{hr}$. Then, relative speed $=(x-6) \mathrm{km} / \mathrm{hr}$. $$\therefore x-6=36 \Leftrightarrow x=42 \mathrm{km} / \mathrm{hr} .$$ Speed of the train relative to second man $$\begin{aligned} & =\left(\frac{75}{6 \frac{3}{4}}\right) \mathrm{m} / \mathrm{sec}=\left(75 \times \frac{4}{27}\right) \mathrm{m} / \mathrm{sec} \\ & =\left(\frac{100}{9}\right) \mathrm{m} / \mathrm{sec}=\left(\frac{100}{9} \times \frac{18}{5}\right) \mathrm{km}=40 \mathrm{km} / \mathrm{hr} . \end{aligned}$$ Let the speed of the second man be $y \mathrm{kmph}$. Then, relative speed $=(42-y) \mathrm{kmph}$. $$\therefore 42-y=40 \Leftrightarrow y=2 \mathrm{km} / \mathrm{hr} .$$