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1. 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

Solution:

\begin{aligned} Ratio of speeds \\ = \sqrt{4} : \sqrt{6 \frac{1}{4}} \\ = \sqrt{4} : \sqrt{\frac{25}{4}} \\ = 2 : \frac{5}{2} \\ = 4 : 5 \end{aligned}

Let the speeds of the two trains be 4x and 5x km/hr respectively
Then time taken by trains to meet each other

\begin{aligned} = (\frac{270}{4 x + 5 x}) hr \\ = (\frac{270}{9 x}) hr  = (\frac{30}{x}) hr \\ Time taken by slower train to travel \\ 270 km  = (\frac{270}{4 x}) hr \\ ∴ \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 \\ Hence speed of slower train \\ =  4 x \\ = 24 km/hr \end{aligned}

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