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 2 of 8.
Vijayalakshmi said:
1 decade ago
Solution for 220
__________ =6 is
(250+5x/18)
220*18
_______ =6 => 220*18=6(250+5x)
250+5x
=> 220*3=(250+5x)
=>660=250+5x
=> 660-250=5x
=>410=5x
=>x=82
__________ =6 is
(250+5x/18)
220*18
_______ =6 => 220*18=6(250+5x)
250+5x
=> 220*3=(250+5x)
=>660=250+5x
=> 660-250=5x
=>410=5x
=>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
Lakshmi said:
1 decade ago
Distance = 108+112 = 220m.
Time = 6s.
Effective speed = x+y (+ because in opp direction).
=> let x = 50 kmph in terms of m/s = 50*(5/18) = 250/18.
eff.sp = 250/18+y (since v need to find y).
So distance = speed*time=>220 = (250/18+y)6.
220 = (250+18y)/18*6.
On simplification,
we get, 410/18.
Converting m/s to kmph, (410/18)*(18/5) = 82kmph.
Time = 6s.
Effective speed = x+y (+ because in opp direction).
=> let x = 50 kmph in terms of m/s = 50*(5/18) = 250/18.
eff.sp = 250/18+y (since v need to find y).
So distance = speed*time=>220 = (250/18+y)6.
220 = (250+18y)/18*6.
On simplification,
we get, 410/18.
Converting m/s to kmph, (410/18)*(18/5) = 82kmph.
Ambika said:
1 decade ago
Length of the trains=108+112=220m=(220/1000)=(11/50)km
Let the speed of second train be xkm/hr
therefore,relative speed=(50+x)km/hr(since it is opposite direction the speeds should be added)
time=6sec=(6/3600)=(1/600)hr
then solve using the formula time=(distance/speed)
i.e (11/50)/(50+x)=(1/600)
after solving x will be 82km/hr
Let the speed of second train be xkm/hr
therefore,relative speed=(50+x)km/hr(since it is opposite direction the speeds should be added)
time=6sec=(6/3600)=(1/600)hr
then solve using the formula time=(distance/speed)
i.e (11/50)/(50+x)=(1/600)
after solving x will be 82km/hr
Vishal Kumar said:
1 decade ago
But Raj, When we calculate the speed of second Train, then we also convert the time in hour. But in the above question time write it as it is, without any conversion of time.
If we want result in Km/hr then we also need to change the time in hour because both the trains takes 6 seconds to cross each other.
If we want result in Km/hr then we also need to change the time in hour because both the trains takes 6 seconds to cross each other.
Ranjith said:
1 decade ago
Total distance of the 2 trains are 220 (108+112) :first don't change anything. Just change the speed 50 km/hr into m/sec.
If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.
(125/9) +x=220/6.
We get x=410/18 m/sec.
Now convert to km/hr we get x= (410/18) (18/5).
X=82.
If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.
(125/9) +x=220/6.
We get x=410/18 m/sec.
Now convert to km/hr we get x= (410/18) (18/5).
X=82.
Ranjith said:
1 decade ago
Total distance of the 2 trains are 220 (108+112) :first don't change anything. Just change the speed 50 km/hr into m/sec.
If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.
(125/9) +x=220/6.
We get x=410/18 m/sec.
Now convert to km/hr we get x= (410/18) (18/5).
X=82.
If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.
(125/9) +x=220/6.
We get x=410/18 m/sec.
Now convert to km/hr we get x= (410/18) (18/5).
X=82.
(1)
Sanjay sharma said:
1 decade ago
I want to explain clearly for those who have confusio b/w km/h & m/s.
Let take v M/s is the speed of the second train.
Thus formula is
T=220/(v+50*5/18)
after solv. it
we get
v=410/18 in M/s
convert v into km/hr
thus
v=410*18/(18*5)km/hr
then
v=82 km/hr.
I think its very clear now
Let take v M/s is the speed of the second train.
Thus formula is
T=220/(v+50*5/18)
after solv. it
we get
v=410/18 in M/s
convert v into km/hr
thus
v=410*18/(18*5)km/hr
then
v=82 km/hr.
I think its very clear now
@shok Kumar said:
1 decade ago
Here we assumed speed of second train as 'x' km/hr which is in km/hr,okay
then in problem it is converted in to m/sec as 'x*(5%18)' okay and
finally we got answer in terms of 'x' whose units are km/hr as
x=82 km/hr,
which can be in m/sec as
x*(5%18) = 82*(5%18) m/sec
then in problem it is converted in to m/sec as 'x*(5%18)' okay and
finally we got answer in terms of 'x' whose units are km/hr as
x=82 km/hr,
which can be in m/sec as
x*(5%18) = 82*(5%18) m/sec
MAHI said:
5 years ago
Here, time is in seconds & distance is in meters that's why we have to convert the speed into m/sec. Initially assume that X is in km/hr. SO we have to convert the speed in m/sec as all parameters is in meter & seconds. After solving where get the value of x in km/hr,
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