Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 12)
12.
A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
Answer: Option
Explanation:
| Speed = |  |
72 x | 5 |  m/sec |
= 20 m/sec. |
| 18 |
Time = 26 sec.
Let the length of the train be x metres.
| Then, | x + 250 | = 20 |
| 26 |
x + 250 = 520
x = 270.
Discussion:
39 comments Page 3 of 4.
Rakesh said:
1 decade ago
Distance = speed * Time.
So Distance = X+250.
Where x= Length of train.
Speed = 72*5/18 m/s (Conversion from km/s to m/s).
i.e., 20 m/s.
And time = 26 seconds.
Therefore X+250 = 20 * 26.
X = 520-250.
X = 270 m.
So Distance = X+250.
Where x= Length of train.
Speed = 72*5/18 m/s (Conversion from km/s to m/s).
i.e., 20 m/s.
And time = 26 seconds.
Therefore X+250 = 20 * 26.
X = 520-250.
X = 270 m.
Sandip said:
1 decade ago
At 72 km/hr train goes 520 mt.
Length of the plat form = 250.
So length of the train = 520-250 = 270.
Length of the plat form = 250.
So length of the train = 520-250 = 270.
Saurabh said:
1 decade ago
Speed in meters = 72 * (5/18) = 20.
Length = speed * time.
250 + length of platform = 20*26.
Length of platform = 520-250 = 270.
Length = speed * time.
250 + length of platform = 20*26.
Length of platform = 520-250 = 270.
Benhur Dsilva said:
1 decade ago
First convert 72km/hr in m/sec
So, 72/1*1000/3600
=72*5/18
=20m/sec.
D = s * t
= 20*26
= 520.
Its a distance of Train + platform; this is because full length of train pass full length of platform.
So,
520-250=270.
So, 72/1*1000/3600
=72*5/18
=20m/sec.
D = s * t
= 20*26
= 520.
Its a distance of Train + platform; this is because full length of train pass full length of platform.
So,
520-250=270.
Manovasanth said:
1 decade ago
@Sonia.
Here relative speed in sense where two forces come in opposite directions and in turn of collide force.
In above said question the trans come in same direction so it is added if it in opposite direction it is subtracted.
Here relative speed in sense where two forces come in opposite directions and in turn of collide force.
In above said question the trans come in same direction so it is added if it in opposite direction it is subtracted.
Naresh said:
1 decade ago
Here they given train speed=72kmph now convert into m/s that is 72*5/18=20m/s.
And time 26sec given. This time taken by the train to cross the total platform.
So now we know train speed, and time. Speed=distance/time.
Distance=speed*time=20*26=520m.
So this is total distance of platform and train. i.e (platform+train length).
Platform+Train length=520.
250+Train length=520.
Train length=520-250=270m.
And time 26sec given. This time taken by the train to cross the total platform.
So now we know train speed, and time. Speed=distance/time.
Distance=speed*time=20*26=520m.
So this is total distance of platform and train. i.e (platform+train length).
Platform+Train length=520.
250+Train length=520.
Train length=520-250=270m.
Kunal kanchan srivastava said:
1 decade ago
Station is crossed when tail of train pass, so we can't take length of train be x it should be 2x and length of the train must be 135m.
Sujith.m said:
1 decade ago
(x+250)/26=20 Can anybody explain how to come this?
Raj said:
1 decade ago
@Savi
here l=x*250 or l=x+250 ?
here l=x*250 or l=x+250 ?
Akber said:
1 decade ago
@Sania.
Here platform is stationary that is the reason we are adding.
Here platform is stationary that is the reason we are adding.
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