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 24 of 24.
Yoo said:
2 months ago
The Length of the train be x.
(u-v) = 46 - 36 = 10km/hr = 25/9,
T = (a+b)/(u-v).
T is the time difference of 36 sec.
(u-v)relative speed train moves in the same direction.
By Substitute all values;
36 = (x+x)/(25/9),
36 * 25/9 = 2x,
x = 50.
(u-v) = 46 - 36 = 10km/hr = 25/9,
T = (a+b)/(u-v).
T is the time difference of 36 sec.
(u-v)relative speed train moves in the same direction.
By Substitute all values;
36 = (x+x)/(25/9),
36 * 25/9 = 2x,
x = 50.
(2)
Pavani said:
1 week ago
The Length of L1 = length of L2.
L1 = L2.
So, length=2L.
Speed of the train t1 = 46km/hr.
Speed of the train t2 = 36km/hr.
The relative speed = t1 - t2 = 46 - 36 = 10km/hr.
convert to meters a = 10x5/18.
Then time = 36sec.
Now using formula:
2L = 10x5/18 x36.
2L = 100,
L = 50m.
L1 = L2.
So, length=2L.
Speed of the train t1 = 46km/hr.
Speed of the train t2 = 36km/hr.
The relative speed = t1 - t2 = 46 - 36 = 10km/hr.
convert to meters a = 10x5/18.
Then time = 36sec.
Now using formula:
2L = 10x5/18 x36.
2L = 100,
L = 50m.
(7)
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