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Question:

Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day?

Explanation:

Let the speed of E be 2x km/hr.

∴ Speed of N = 4x km/hr and speed of S = x km/hr

According to the problem,

Time taken by N to cover AB + Time taken by S to cover BA = 1 hour

$∴ \frac{d}{4 x} + \frac{d}{x} = 1$

∴ 5d = 4x or d = $\frac{4 x}{5}$

∵ To maintain the schedule (gap of 20 min), N doubles its speed and speed of S becomes y.

$∴ \frac{\frac{4 x}{5}}{8 x} + \frac{\frac{4 x}{5}}{y} = \frac{2}{3}$

$∴ \frac{1}{10} + \frac{4 x}{5 y} = \frac{2}{3}$

$∴ \frac{4 x}{5 y} = \frac{17}{30}$

$∴ \frac{4 x}{y} = \frac{17}{6}$

$∴ \frac{8 x}{y} = \frac{17}{3} o r \frac{y}{8 x} = \frac{3}{17} \approx \frac{1}{6}$

$∴ \frac{S p e e d o f p a s s e n g e r t r a i n}{S p e e d o f s u p e r f a s t t r a i n} \approx \frac{1}{6}$

Hence, option (d).

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