Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 7)
7.
Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
Answer: Option
Explanation:
Let the length of each train be x metres.
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
| = |  |
10 x | 5 |  m/sec |
| 18 |
| = | |
25 | m/sec |
| 9 |
 |
2x | = | 25 |
| 36 | 9 |
2x = 100
x = 50.
Discussion:
232 comments Page 12 of 24.
Kousalya said:
10 years ago
My question is why we are including ten in the calculation its not given in the data right?
Prasun jana said:
10 years ago
Fast train 46*(5/18)*36 = 460.
2nd train 36*(5/18)*36 = 360.
Difference = 460-360 = 100.
2 train = 100/2 = 50 (ans).
2nd train 36*(5/18)*36 = 360.
Difference = 460-360 = 100.
2 train = 100/2 = 50 (ans).
Malli said:
10 years ago
Friends he said it takes 36 sec to cross a train then it means the difference in distance b/n them is x (which is length of train).
So the answer is:
x/(Relative speed) = 36;
x = 100;
Not 50 its wrong.
So the answer is:
x/(Relative speed) = 36;
x = 100;
Not 50 its wrong.
Ponjit said:
10 years ago
Since we are taking relative speed that means the slower train is stationary. And thus the faster train needs to travel only the distance of the slower train. Since both the trains start (assuming) at the same time and with the engines at the same line. The answer comes to be 100 m.
Viswanath said:
10 years ago
As two trains are in parallel (same direction). So we can subtract the speeds of other trains. Relative speed is (25/9) m/sec.
NOTE: Time in contact of 2 trains at a time MEANS the time at which faster train engine meet the last part of slower train TO faster train last part meet the engine of slower train.
Both trains relative speed = Distance of 2 trains/(Time in contact of 2 trains at a time).
(25/9) m/sec = 2x/36.
NOTE: Time in contact of 2 trains at a time MEANS the time at which faster train engine meet the last part of slower train TO faster train last part meet the engine of slower train.
Both trains relative speed = Distance of 2 trains/(Time in contact of 2 trains at a time).
(25/9) m/sec = 2x/36.
Patani said:
10 years ago
2x/36=25/9 why use 36?
SAM said:
10 years ago
Why should we take relative speed is (46-36) in this case they are parallel?
Bharathi said:
10 years ago
I can't understand can anyone explain clearly?
Why we take 2x inspite 2 trains are equal length but they reach different distance only then how it possible?
Why we take 2x inspite 2 trains are equal length but they reach different distance only then how it possible?
Vichu119 said:
10 years ago
Ya if t trains are running from same starting point above answer is wrong because length of train is 100 meter. If the faster train's front starts from slower train's back then the length of train is 50 mt.
Ghanshyam said:
10 years ago
There is some data is not given properly that the faster train start from ending point of slower train that's why the answer is 100 meters and if the trains start from same point at that time answer is 100 meters.
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