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.
Dharamjeet Singh said:
1 decade ago
How can the final answer come in Km/hr ??
Harisha said:
1 decade ago
Here the ans is in km/hr. so frm starting onwards we sould calculate
the problem without converting into m/s.
Let speed of d second train= xkm/hr
Distance covered=(108+112)=220m=0.22km
Time=6sec= 6/3600hr =1/600hr
So
1/600hr= 0.22km/(50+x)
x=82km/hr
the problem without converting into m/s.
Let speed of d second train= xkm/hr
Distance covered=(108+112)=220m=0.22km
Time=6sec= 6/3600hr =1/600hr
So
1/600hr= 0.22km/(50+x)
x=82km/hr
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
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
Ankit said:
1 decade ago
Nice explanation Sanjay.
Karthika said:
1 decade ago
Hey the final answer is in m/s its not in km/h can some one clarify my doubt.
Darshan Kumar said:
1 decade ago
@Karthika , see when we take the series 220/5x+250/18=6 than the 18 will go up side and new series came 220*18/5x+250=6 , so here the formula 18/5 km/h is applicable .
I think ur doubt is clear
I think ur doubt is clear
Thiyam surjakumar singh said:
1 decade ago
In the equation all the units meter, second get cancelled so the remaining is x only, whose units is km/hr as we have choosen at beginning of the problem. Hence the result is in km/hr.
@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
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.
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