Latest Dell Aptitude Question SOLUTION: Q. 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 passen

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(#M40018050) DELL HR Interview question Rail track Keep an EYE

Q. 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 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.

Option
a) 1 : 3
b) 1 : 4
c) 1 : 5
d) 1 : 6

Asked In Dell Govind (11 years ago)

Solved Aman Read Solution (2)

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Aman  (11 years ago)

4. 1 : 6

As per condition , initial ratio of the speed of passenger train to that of the super fast train is 1:4.
Initially, passenger train takes 4/5 or 48 mins hr to travel from B to A and superfast train train takes 1/5 hr or 12 mins to reach stn A from Stn B.
when 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, it covers distance in 6 mins and remains 14 mins behind schedule.
to maintain schedule, passenger train has to cover the distance in 48-14 = 34 mins.

so final ratio of speeds will be 6:34 or 1:6 (approx.)

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V.Yuvaraj  (5 years ago)

initial ration is 1:4 because, E/2 divided by 2E we get, the same.
now, we know dist/speed=time.Form an eq. with this formula:-
so from the ratio 4x is the speed of superfast train and x is the speed of pass.train
D/4X (Time taken by superfast train) + D/X (Time taken by pass.train) = 60min
solving we get,
D/X =48, means that Time taken by pass.train is 48 mins.
12mins for superfast train.

For 20 mins later schedule,
superfast will double its speed, so it takes on 12/2 = 6mins.
Bal.14mins remains behind schedule.
48-14=34mins, means pass.train covering distance.
Final ratio=6:34 = 1:6 (round off 5.6)
Thank you. All the best!

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