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.
Neeraja said:
5 years ago
Actually, whenever a train crosses platform we are considering the length of train+length of the platform, similarly, in this case, we consider the length of tunnel+length of train i.e7/2+1/4.
Shyam said:
3 years ago
The speed of the train = 75mph.
=75 * 1/60.
Train speed = 5/4seconds.
The total distance is (7/2 + 1/4),
= 15/4mile.
Formula of time = distance /speed.
= 15/4/5/4.
= 15/4 * 4/5.
= 3seconds.
=75 * 1/60.
Train speed = 5/4seconds.
The total distance is (7/2 + 1/4),
= 15/4mile.
Formula of time = distance /speed.
= 15/4/5/4.
= 15/4 * 4/5.
= 3seconds.
(8)
GURAVAREDDY said:
7 years ago
Given data is D1=3(1/2), D2=1/4 S=75m/s.
respective length is (7/2+1/4)=15/4,
S=D/T.
75=(15/4)/T,
T=(15/4)/(1/75),
=15/4*75,
=1/20.
SO ITS IN SEC.
CONVERT INTO HRS.
= (1/20)*60.
= 3 MINS.
respective length is (7/2+1/4)=15/4,
S=D/T.
75=(15/4)/T,
T=(15/4)/(1/75),
=15/4*75,
=1/20.
SO ITS IN SEC.
CONVERT INTO HRS.
= (1/20)*60.
= 3 MINS.
Gaurav said:
1 decade ago
I think the total distance = 7/2 + 1/4 + 1/4 = 4 miles as the summation of length of the tunnel, length of train when it enters and length of train when it emerges out should be taken.
Shubham said:
4 years ago
Ans must be 3.2 min.
We have to take the length of train twice.
7/2 + 1/4 + 1/4 = 4 miles.
Then,
Time = 4/75.
=0.05333 hr.
In seconds 0.05333*60= 3.198 which is 3.2 min.
We have to take the length of train twice.
7/2 + 1/4 + 1/4 = 4 miles.
Then,
Time = 4/75.
=0.05333 hr.
In seconds 0.05333*60= 3.198 which is 3.2 min.
(23)
Aditya shukla said:
1 decade ago
Because train to pass through the tunnel from the moment the front enters to the moment the rear emerges. So we should add double the length of train then answer should be 3.2 min.
Veda said:
5 years ago
@Palla Nagesh.
7/2+1/4 = 14 + 1/4.
Take LCM for 2 & 4 then final lcm is 4.
Because we write 4 at denominator then 2*2=4 2*7=14 4*1=4
And finally, we get the ans 15/2.
7/2+1/4 = 14 + 1/4.
Take LCM for 2 & 4 then final lcm is 4.
Because we write 4 at denominator then 2*2=4 2*7=14 4*1=4
And finally, we get the ans 15/2.
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.
Praveen said:
7 years ago
So, if I change the speed to 90mph, then according to the way this is solved, the time would increase. Now that does not sound right to me. Can anyone explain this, please ?
Sravya Sree said:
2 years ago
@Atikraja.
It is a mixed fraction.
First, multiply 2(denominator)*3(first term numerator) = 6 then add 1 which is 7 and include the denominator as it is;
i.e 3*2+1=7/2.
It is a mixed fraction.
First, multiply 2(denominator)*3(first term numerator) = 6 then add 1 which is 7 and include the denominator as it is;
i.e 3*2+1=7/2.
(2)
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