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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Discussion:
129 comments Page 3 of 13.
Siddhant said:
7 years ago
@All.
Assume that there's a platform of length 7/2 miles and train of length 1/4 miles.
So now we just have to calculate the time taken by train to pass the platform.
Hence total distance = 15/4 miles.
No need to add 1/4 twice.
Hope you will understand.
Assume that there's a platform of length 7/2 miles and train of length 1/4 miles.
So now we just have to calculate the time taken by train to pass the platform.
Hence total distance = 15/4 miles.
No need to add 1/4 twice.
Hope you will understand.
Akash sanghvi said:
1 decade ago
Ya as train enters time starts so distance 1/4 and as train leave from tunnel distance covered is more 1/4 since it has covered xtra 1/4 distance after coming out of tunnel.
---------front[ (tunnel) ]rear------------
1/4 3/2 1/4
---------front[ (tunnel) ]rear------------
1/4 3/2 1/4
Sunny Singla said:
8 years ago
Aas per my understanding it's 3.2.
Total distance cover by train in minutes : 75/60 = 1.25,
Total Distance need to cover : 3.5+.25(enter)+.25(exit) = 4 miles,
In one minute it cover : 1.25 m.
Minutes required to travel 4 miles = 4/1.25 = 3.2.
Total distance cover by train in minutes : 75/60 = 1.25,
Total Distance need to cover : 3.5+.25(enter)+.25(exit) = 4 miles,
In one minute it cover : 1.25 m.
Minutes required to travel 4 miles = 4/1.25 = 3.2.
Ishant said:
1 decade ago
According to me answer will be c)3.2min
Because the lenth of train has to be taken twice bec. intially train of length 1/4 will enter and then it wil leave the tunnel
Therefore distance travelled=7/2+1/4+1/4=4
Please check it out
Because the lenth of train has to be taken twice bec. intially train of length 1/4 will enter and then it wil leave the tunnel
Therefore distance travelled=7/2+1/4+1/4=4
Please check it out
Raju Ahmed said:
9 years ago
Total distance covered = 7/2(tunnel) + 1/4(length of train) + 1/4(length of train since the rear part is to be considered emerging) = 4 miles.
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is [C].
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is [C].
Arunkumar said:
1 decade ago
Hi,
Resolving the fractions here is 7/2 + 1/4 is equal to 15/4, how
Since (x/y+a/b) is (x*b/y*b) + (y*a/y*b) = (x*b + y*a)/y*b
Therefore
(7/2 + 1/4 )=> (7*4/2*4) + (2*1/2*4)=> (28/8 + 2/8)=> (30/8)=> 15/4.
Resolving the fractions here is 7/2 + 1/4 is equal to 15/4, how
Since (x/y+a/b) is (x*b/y*b) + (y*a/y*b) = (x*b + y*a)/y*b
Therefore
(7/2 + 1/4 )=> (7*4/2*4) + (2*1/2*4)=> (28/8 + 2/8)=> (30/8)=> 15/4.
Madhur said:
1 decade ago
It is 3 min. For sure as the train enters into the tunnel distance inside the tunnel counts and not the back part of train.
So when rear emerges, only 1/4 is added to 7/2 and not 1/4+1/4. So 3 min. Is absolutely correct.
So when rear emerges, only 1/4 is added to 7/2 and not 1/4+1/4. So 3 min. Is absolutely correct.
S.D.Brite said:
8 years ago
@Ashu.
They ask only for the time to cross the tunnel...
The train is driven by the engine, which is in the front part of the train.
;So we need to calculate the distance as;
7/2 mile + 1/4 mile = 3.75 mile.
They ask only for the time to cross the tunnel...
The train is driven by the engine, which is in the front part of the train.
;So we need to calculate the distance as;
7/2 mile + 1/4 mile = 3.75 mile.
Arbino said:
4 years ago
@All.
In question it is clearly mentioned that the rear part of the train got emerge in the tunnel, so we have to calculate the only the length of the train which is 1/4mile.
Correct me, if I am wrong.
In question it is clearly mentioned that the rear part of the train got emerge in the tunnel, so we have to calculate the only the length of the train which is 1/4mile.
Correct me, if I am wrong.
(1)
Harsh said:
1 decade ago
Can't we just do it this way
3 1/2 =3.50 and 1/4 = .25
So 3.50+0.25 = 3.75
(3.75/75 as t = dist/speed) t = 3.75/75 = 0.05
Now 0.05 is in hr so 0.05*60 = 3
This method is better then above.
3 1/2 =3.50 and 1/4 = .25
So 3.50+0.25 = 3.75
(3.75/75 as t = dist/speed) t = 3.75/75 = 0.05
Now 0.05 is in hr so 0.05*60 = 3
This method is better then above.
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