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 20 of 24.
Umer said:
7 years ago
How is the speed of the train mentioned as the length we needed?
Joydeep Mondal said:
7 years ago
From where 18/5 came?
Mohammad yasir said:
7 years ago
If the first train crosses slower train by relative speed 10kmph. It means the whole length of this train crosses slower train in 36 seconds.
Which literally means the same train crosses the slower train twice when we treat it a point.
Which literally means the same train crosses the slower train twice when we treat it a point.
SHUBHaM SaKHaRE said:
7 years ago
Let the length of both trains be 'x'.
Distance covered by slower train = 36*(36*5/18),
Hence total distance covered by faster train= 2x+36*(36*5/18),
Hence we get;
46*5/18=(2x+36*(36*5/18))/36,
By solving this on calc we get;
X=50.
Distance covered by slower train = 36*(36*5/18),
Hence total distance covered by faster train= 2x+36*(36*5/18),
Hence we get;
46*5/18=(2x+36*(36*5/18))/36,
By solving this on calc we get;
X=50.
Gurpreet said:
7 years ago
36sc=a+b/u-v.
36 =x+x/46-36kmh,
36s= 2x/10kmh,
36s=2x/10*5/18msec,
36s=2x/25/9msec,
36s=2x*9/25,
36=18x/25,
36*25/18=x,
2*25=x,
50=x.
36 =x+x/46-36kmh,
36s= 2x/10kmh,
36s=2x/10*5/18msec,
36s=2x/25/9msec,
36s=2x*9/25,
36=18x/25,
36*25/18=x,
2*25=x,
50=x.
Pankaj said:
7 years ago
Why are we calculating relative speed?
I used 36= 2x/(46+36). Using this getting 410m. Am I right?
I used 36= 2x/(46+36). Using this getting 410m. Am I right?
Shashi said:
7 years ago
Why the distance covered is taken as 2x?
Sarang said:
7 years ago
Distance = speed * time,
D = 46-36= 10* time,
D = 10 5/18* 36,
D= 10 * 5 * 2 = 100,
The length both train is 100/2 = 50.
D = 46-36= 10* time,
D = 10 5/18* 36,
D= 10 * 5 * 2 = 100,
The length both train is 100/2 = 50.
Aminul Islam said:
7 years ago
Let, the length of each train be x metres.
In 36 seconds The faster train passes the slower one. so total distance covered by two train is 2x metres sharp 36 seconds.
According to formula 9 given above,
t = (a+b)/(u-v).
or, 36 = (x+x)/(46-36),
or, 36 = 2x/(46-36)*5/18 (to change km/hr to m/s,multiflied by 5/18).
x = (36/2)*(25/9),
= 50.
In 36 seconds The faster train passes the slower one. so total distance covered by two train is 2x metres sharp 36 seconds.
According to formula 9 given above,
t = (a+b)/(u-v).
or, 36 = (x+x)/(46-36),
or, 36 = 2x/(46-36)*5/18 (to change km/hr to m/s,multiflied by 5/18).
x = (36/2)*(25/9),
= 50.
Ritik said:
7 years ago
We should consider the extreme case, where starting point of faster train coincides with the finishing point of slower train.
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