Only a single rail track exists between station A and B on a railway

Question

Only a single rail track exists between station A and B on a railway line. One hour after the north bound super fast train N leaves station A for Station B, a south bound passenger train S reaches station A from station B. The speed of the super fast 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 super fast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the super fast train so that passenger train S reaches exactly at the scheduled time at station A on that day.

(A) 1 : 3
(B) 1 : 4
(C) 1 : 5
(D) 1 : 6
(E) 1 : 7

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(A) 1 : 3 
(B) 1 : 4 
(C) 1 : 5 
(D) 1 : 6
(E) 1 : 7

Are You Up For the Challenge: 700 Level Questions

DevangiK · Accepted Answer

So, the current ratio between the super fast train N and Passenger train S = 1:4
Time taken by train S = 4/5*60 = 48 mins
Time taken by train N =1/5*60 = 12 mins

On a certain day, when N leaves 20 mins behind the schedule, it doubles its speed,
Now it will take 12/2 = 6 mins to cover the distance

This means that it is 14 mins behind schedule (20*6=14 mins)
So, Train S will be late by 14 mins as well = 48-14=34 mins

Now, the ratio changes to 6:34 = 1:6
Thus, the correct ans is Option D.

Hope it helps !

BhishmaNaidu99 · Answer

IMO its 1 : 5

considering the trains leave with no waiting once one train reaches opposite station,

total time take for super fast and passenger = 1hr = 60 min

\frac{D}{s} + \frac{D}{p} = 60  min. give s=4p ,  \frac{D}{4p} + \frac{D}{p}= 60  ,  \frac{5D}{4p} = 60  --1 ,

During the delay, the trains increase speeds to be on right time, so total time taken here is 60-20=40 min

gives super fast increases speed by twice. let passengers new speed is xp

D/2s + D/xp = 40 min , D/8p + D/xp = 40 -- 2

diving equation 1 and 2 ,

\frac{ 5}{4p} / \frac{1}{8p} + \frac{1}{ xp}  = \frac{3}{2}

p/xp = 17 /24

the questions the ratio os xp / 2s = 3/17 = 1 : 5 APPROX

nick1816 · Answer

Ratio of speed of N to that of S= 4:1

Average speed of both trains=  \frac{2*4x*x}{(4x+x)}= 1.6x

Time to cover the distance on a particular day is reduced from 1 hour to 40 minutes.

Time is reduced to 2/3 times of regular time taken by both train; hence, average speed is increased to 3/2 times of regular average speed in order to maintain the schedule.

Average speed on that particular day=  1.6x*(\frac{3}{2})= 2.4x

Speed of N train on that particular day= 4x*2=8x

2.4x= \frac{2*8x*v}{(8x+v)}

v=\frac{24x}{17}

\frac{v}{8x}=\frac{24x}{17*8x}= \frac{1}{5.67 }  ≈  \frac{1}{6}

D

effatara · Answer

The question stem should have mentioned that the passenger train S departs as soon as the super-fast train N reaches Station B because any delay affects the answer. For example, a delay of 10 minutes (logical under the circumstances because N would have to get out of the way for S to get in position since it is a single track) would cause the ratio to be exactly 1:5.

Bentsi · Answer

DevangiK 
Thank you for your explanation.
One little question, how 6:34 = 1:6?

DevangiK · Answer

Hi ! So, 6:34 = 1:5.66667 which is approximately equal to 1:6. I hope it clearer now, we don't have a aproximate symbol so I used equal to. But, thank you for the question.

tyildirim92 · Answer

The question stem should be clarified further. Please specify whether more than one train can use the track at the same time. If yes, the answer is 1/5, if no, then 1/6

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Only a single rail track exists between station A and B on a railway

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Only a single rail track exists between station A and B on a railway

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