Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 20)
20.
A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
Answer: Option
Explanation:
Let the length of the train be x metres and its speed by y m/sec.
| Then, | x | = 8  x = 8y |
| y |
| Now, | x + 264 | = y |
| 20 |
8y + 264 = 20y
y = 22.
 Speed = 22 m/sec = |
 |
22 x | 18 |  |
km/hr = 79.2 km/hr. |
| 5 |
Discussion:
46 comments Page 2 of 5.
Mahesh said:
1 decade ago
You see the question he said total is 20 sec.
In that, The train take 8sec to rich post and remaining time is used to cross the bridge.
i.e., 20-8=12.
So 264/12=22.
And 22* (18/5) =79.2km/hr.
In that, The train take 8sec to rich post and remaining time is used to cross the bridge.
i.e., 20-8=12.
So 264/12=22.
And 22* (18/5) =79.2km/hr.
Rakesh said:
1 decade ago
22meters*60(sec)=1.32km/min
1.32*60(mins)=79.2km/hr
1.32*60(mins)=79.2km/hr
Divyang said:
1 decade ago
Above in ans x and y are unknown..so that simplify very easily
x/8 = (x+264)/20
3x = 528
x = 176
speed y = 176/8
y = 176/8 *18/5
y = 79.2 km/hr
x/8 = (x+264)/20
3x = 528
x = 176
speed y = 176/8
y = 176/8 *18/5
y = 79.2 km/hr
Sumitra said:
1 decade ago
Here we don't know the speed of train, So let x be the speed of train , According to the formula t = a+b/u+v.
1st case where train crosses the platform,
let b be the length of platform.
t=300+b/x+0 => 39=300+b/x => 39x=300+b ---- eqn 1.
2nd case where train passes the pole,
t=300+0/x+0 =>18=300/x =>x=300/18 ----eqn 2.
We go value for x. So substitute in equation 1.
39*300/18=300+b.
650-300=b.
b=350----->length of platform.
1st case where train crosses the platform,
let b be the length of platform.
t=300+b/x+0 => 39=300+b/x => 39x=300+b ---- eqn 1.
2nd case where train passes the pole,
t=300+0/x+0 =>18=300/x =>x=300/18 ----eqn 2.
We go value for x. So substitute in equation 1.
39*300/18=300+b.
650-300=b.
b=350----->length of platform.
SK Lodhi said:
1 decade ago
T = 8 Sec.
T+264 = 20 Sec.
---------------
T SPEED = 264/12 m/s = 22 m/s Where 12 Sec. = 20-8(Difference).
T = length of Train.
Speed = 22*18/5 = 79.2 km/hr Answer.
If calculate length of Train then 8 Sec*22 m/s = 176 m.
T+264 = 20 Sec.
---------------
T SPEED = 264/12 m/s = 22 m/s Where 12 Sec. = 20-8(Difference).
T = length of Train.
Speed = 22*18/5 = 79.2 km/hr Answer.
If calculate length of Train then 8 Sec*22 m/s = 176 m.
Krish said:
1 decade ago
S = x/8 m/s.
S = d/t.
x/8 = (264+x)/20.
Solve it and you will get:
x = 176 m.
S = 176/8 = 22 m/s.
Now convert m/s to km/hr.
22*(18/5) = 79.2 km/hr.
That's it.
S = d/t.
x/8 = (264+x)/20.
Solve it and you will get:
x = 176 m.
S = 176/8 = 22 m/s.
Now convert m/s to km/hr.
22*(18/5) = 79.2 km/hr.
That's it.
Neelam said:
1 decade ago
How you people got 3x = 528?
Subham kumar said:
1 decade ago
A train passes telegraph post is 40 seconds morning at a rate of 36 km/hr. Then the length of the train is: 8.
Hussain said:
1 decade ago
How we got 18/5? Can any one explain me clearly?
Vamsy said:
1 decade ago
@Hussaian.
1) km/hr to m/s conversion:
km/hr = a x 5km/18hr.
2) m/s to km/hr conversion:
m/s = a x 18km/5hr.
1) km/hr to m/s conversion:
km/hr = a x 5km/18hr.
2) m/s to km/hr conversion:
m/s = a x 18km/5hr.
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