Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 25)
25.
A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:
Answer: Option
Explanation:
Let the speed of the second train be x km/hr.
| Relative speed | = (x + 50) km/hr | |||||||
|
||||||||
|
![]](/_files/images/aptitude/1-sym-cbracket-h1.gif)
Distance covered = (108 + 112) = 220 m.
 |
220 | = 6 | ||
|
250 + 5x = 660
x = 82 km/hr.
Discussion:
77 comments Page 5 of 8.
Rutika said:
10 years ago
S = D/T.
So, D = 108+112 = 220 m.
= 220/1000 km/hr.
T = 6 sec = 6/3600 = 1/600 hr.
S = 50 km/hr.
We assume second train speed 'x'.
Relative speed = 50 km/hr+x km/hr.
S = D/T.
= 50+x = 220/1000*600.
= 50+x = 22*6 = 50+x = 132.
x = 82 km/hr.
So, D = 108+112 = 220 m.
= 220/1000 km/hr.
T = 6 sec = 6/3600 = 1/600 hr.
S = 50 km/hr.
We assume second train speed 'x'.
Relative speed = 50 km/hr+x km/hr.
S = D/T.
= 50+x = 220/1000*600.
= 50+x = 22*6 = 50+x = 132.
x = 82 km/hr.
Moin said:
9 years ago
But can't do in direct formula of a + b/u + v = t? Then finding us there.
Rabvah said:
9 years ago
50*5/18 = 13.889 m/sec.
108 + 112/13.889 + x = 22.778.
Since it has to be in km/hr.
23.778*18/5 = 82 km/hr.
108 + 112/13.889 + x = 22.778.
Since it has to be in km/hr.
23.778*18/5 = 82 km/hr.
Priyanka said:
9 years ago
Total distance = 112 + 120 = 220 m.
Time = 6 sec.
Relative speed = total distance / time,
=> 220/6 m/s
Speed in km/h,
=> 220/6 * 18/5 = 132.
Speed of second train = relative speed - speed of 1st train,
=> 132- 50 = 82 km/h.
Time = 6 sec.
Relative speed = total distance / time,
=> 220/6 m/s
Speed in km/h,
=> 220/6 * 18/5 = 132.
Speed of second train = relative speed - speed of 1st train,
=> 132- 50 = 82 km/h.
Azam Khan said:
9 years ago
How Comes 5 in answer? I can't Understand please explain this.
Ravi said:
9 years ago
Simple solution method:.
Length of train 1st is = 108m.
Speed of train 1st is = 50km/h => 50 * 5/18 == 125/9 m/sec.
Length of trian - 2nd is = 112m.
Now, let the speed of 2nd tarin is = x m/sec.
Relative speed =x + (125/9).
Relative distance = 108 + 112 = 220 m.
Time = 6 sec.
Now, s = d/t.
=> x + 125/9 = 220/6 (therefore 220/6= 110/3).
=> x = (110/3) - (125/9).
=> x = 205/9 m/sec (205/9) * (18/5).
x = 82 km/h answer.
Length of train 1st is = 108m.
Speed of train 1st is = 50km/h => 50 * 5/18 == 125/9 m/sec.
Length of trian - 2nd is = 112m.
Now, let the speed of 2nd tarin is = x m/sec.
Relative speed =x + (125/9).
Relative distance = 108 + 112 = 220 m.
Time = 6 sec.
Now, s = d/t.
=> x + 125/9 = 220/6 (therefore 220/6= 110/3).
=> x = (110/3) - (125/9).
=> x = 205/9 m/sec (205/9) * (18/5).
x = 82 km/h answer.
Apoorv said:
9 years ago
Sir, if the train is moving in an opposite direction the total distance one has to cover is the larger length. So why have we added the two?
Amitesh said:
9 years ago
The solution is not clear for it. Can anybody solve it easily?
Amitesh said:
9 years ago
220/6 = 250 + 5x/18,
By solving it we get,
3960 = 1500 + 30x,
2460 = 3x,
So, x = 82.
By solving it we get,
3960 = 1500 + 30x,
2460 = 3x,
So, x = 82.
Soumitra Das said:
9 years ago
D = 108 + 112 = 120
Converting the speed of 1st Train = 50x 5/18 m/s
Let the speed of 2nd Train= V m/s
We get,
220/ (50x 5/18) + V = 6 sec
=> 220= 6x(250+18V)/18,
=> 220=(250+18V)/3,
=>660= 250+18V,
=>18V=390,
=> V= 390/18 m/s.
Converting the speed of the second Train we Get
=> (390/18) x (18/5)
=> 78 Kmph.
This is the right answer
Converting the speed of 1st Train = 50x 5/18 m/s
Let the speed of 2nd Train= V m/s
We get,
220/ (50x 5/18) + V = 6 sec
=> 220= 6x(250+18V)/18,
=> 220=(250+18V)/3,
=>660= 250+18V,
=>18V=390,
=> V= 390/18 m/s.
Converting the speed of the second Train we Get
=> (390/18) x (18/5)
=> 78 Kmph.
This is the right answer
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