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 9 of 13.
Vaibhav said:
1 decade ago
How come 15/4 could you explain?
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.
Ashok said:
1 decade ago
How come to 15 could you explain?
Ramme said:
1 decade ago
Time taken=15/(4*75) how its comes.
Because t=d*s means (15 * 75)/4 only possible am confused.
Because t=d*s means (15 * 75)/4 only possible am confused.
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
Subhankar said:
1 decade ago
Here the speed given in metre/hr and distance given in mile then how it possible to evaluate time without converting length in same unit.
Swetha said:
1 decade ago
Hi @Arun kumar. This method is so confusing can you please tell me other method to convert into 15/4.
Vivek said:
1 decade ago
How can do it (15/4*75) hrs,
Because 15/4 unit-mils.
and
75 unit-mph.
Then how can possible (15/4*75)hrs?
Because 15/4 unit-mils.
and
75 unit-mph.
Then how can possible (15/4*75)hrs?
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.
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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