Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 16)
16.
A train travelling at a speed of 75 mph enters a tunnel 3
miles long. The train is 
mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
miles long. The train is mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?Answer: Option
Explanation:
| Total distance covered |
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 Time taken |
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| = 3 min. |
Discussion:
129 comments Page 4 of 13.
Sudheer said:
1 decade ago
Given data is
Distance is (7/2 ) + (1/4 ) = (15/4)
Speed is 75mph
Formula for this is speed=(length/time) i.e.,Time=(length/speed)
Time=(15/4)/75=(15/(4*75))=3 mins
Distance is (7/2 ) + (1/4 ) = (15/4)
Speed is 75mph
Formula for this is speed=(length/time) i.e.,Time=(length/speed)
Time=(15/4)/75=(15/(4*75))=3 mins
Tarun said:
1 decade ago
@Nidhi, Shro.
If the tunnel lengh is given in KMPH, then we need to convert the speed also in KMPH, here tunnel length in mph, so we will keep the given speed mph as it is.
If the tunnel lengh is given in KMPH, then we need to convert the speed also in KMPH, here tunnel length in mph, so we will keep the given speed mph as it is.
Vinit said:
1 decade ago
In question it is asked that "How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?" so 1/4 will be added twice i.e.
Total distance would be 7/2 + 1/4 + 1/4
And Finally answer is 3.2 mins
Total distance would be 7/2 + 1/4 + 1/4
And Finally answer is 3.2 mins
Jaani said:
1 decade ago
@vinit: Nope! Look, we need to calculate time taken by the train to cover the "inside" of the tunnel :
1)When "Front emerges".. no part of the train is 'inside' the tunnel.. therefore distance covered till nw = 0
2) when the front emerges out.. it has covered the whole distance of the tunnel = 7/2
3) rest of the train is still inside the tunnel (as only the front has emerged out).. so now the remaining train comes out.. Hence 7/2 + 1/4
1)When "Front emerges".. no part of the train is 'inside' the tunnel.. therefore distance covered till nw = 0
2) when the front emerges out.. it has covered the whole distance of the tunnel = 7/2
3) rest of the train is still inside the tunnel (as only the front has emerged out).. so now the remaining train comes out.. Hence 7/2 + 1/4
Umakant said:
1 decade ago
Answer is 3.2, because Train length=1/4 mile,Tunnel=3.5 i.e 7/2 mile, When train enter in to the tunnel Time need to start till last rear end come out from the tunnel.
that's why 7/2+1/4+1/4=16/4= 4 miles.
Time stop[ end 7/2 tnl start ](Time start)
1/4 Srt<-------End
Start<---------End.
this above figure shows "3 min",but we need to count till rare end of train.
I hope u understand.
that's why 7/2+1/4+1/4=16/4= 4 miles.
Time stop[ end 7/2 tnl start ](Time start)
1/4 Srt<-------End
Start<---------End.
this above figure shows "3 min",but we need to count till rare end of train.
I hope u understand.
Karthiga said:
1 decade ago
I can't understand how the total distance calculated.
Aneesh said:
1 decade ago
Hey guys anyone clearly say how does the value 7/2 came?
GSC said:
1 decade ago
Its 7/2 + 1/4 + 1/4 miles as the distance of the train has to be counted twice (in the question it says, the rear has to come out). That makes it 3.2 mins.
Aman said:
1 decade ago
Total distance = 4 miles.
Time = 3.2 mins.
Time = 3.2 mins.
Akshay said:
1 decade ago
Everyone saying that the length of the train has to be added twice is right because really we are asked about the moment REAR EMERGES, not THE FRONT.
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