Subject: General Aptitude · Chapter: Problems on Trains · Exam: 2005 · Marks: · Difficulty:
Two trains start simultaneously (with uniform speeds) from two stations 270 km apart, each to the opposite station; they reach their destinations in $6 \frac{1}{4}$ hours and 4 hours after they meet. The rate at which the slower train travels is
(a)16 km/hr.
(b)24 km/hr.
(c)25 km/hr.
(d)30 km/hr.
Answer
Answer (as printed): B
Explanation
Ratio of speeds $=\sqrt{4}: \sqrt{6 \frac{1}{4}}=\sqrt{4}: \sqrt{\frac{25}{4}}=2: \frac{5}{2}=4: 5$. Let the speeds of the two trains be $4 x$ and $5 x \mathrm{km} / \mathrm{hr}$ respectively. Then, time taken by trains to meet each other $$=\left(\frac{270}{4 x+5 x}\right) \mathrm{hr}=\left(\frac{270}{9 x}\right) \mathrm{hr}=\left(\frac{30}{x}\right) \mathrm{hr} .$$ Time taken by slower train to travel $270 \mathrm{km}=\left(\frac{270}{4 x}\right) \mathrm{hr}$. $$\begin{aligned} & \therefore \frac{270}{4 x}=\frac{30}{x}+6 \frac{1}{4} \Rightarrow \frac{270}{4 x}-\frac{30}{x}=\frac{25}{4} \Rightarrow \frac{150}{4 x}=\frac{25}{4} \\ & \Rightarrow 100 x=600 \Rightarrow x=6 . \end{aligned}$$ Hence, speed of slower train $=4 x=24 \mathrm{km} / \mathrm{hr}$.