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 6 of 8.
Ashok said:
1 decade ago
220/250+5x/18 = 6.
step 1: 18*220 = 3960.
step 2: 250+5x * 6 = 1500+30x
step 3: 3960 = 1500+30x.
step 4: 3960-1500 = 30x
step 5: 2460 = 30x.
step 6: x=2460/30.
step 7: x=82.
step 1: 18*220 = 3960.
step 2: 250+5x * 6 = 1500+30x
step 3: 3960 = 1500+30x.
step 4: 3960-1500 = 30x
step 5: 2460 = 30x.
step 6: x=2460/30.
step 7: x=82.
Roji said:
1 decade ago
Why should we convert to m/sec as in above problem (x+50)*5/18.
But in answer the required ans in km/hr again why we are taking 5/18.
Can anyone explain me please?
But in answer the required ans in km/hr again why we are taking 5/18.
Can anyone explain me please?
Roji said:
1 decade ago
Hey shyam sharma if I do in your method what you said the problem formula not getting answer yar.
t = 220/ (v+50)*(5/18).
t = 6.
v = 82 got directly.
And hence no need 2 convert again 2 m/sec why?
t = 220/ (v+50)*(5/18).
t = 6.
v = 82 got directly.
And hence no need 2 convert again 2 m/sec why?
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.
Ram said:
1 decade ago
Easiest method is,
220*18 = 6(250+5x).
Multiply by 6(250+5x) you get 1500+30x.
3960 = 1500+30x.
Subtract the value of 3960-1500 = 2460.
Then divide by this form,
2460/30 = 82 kmph
220*18 = 6(250+5x).
Multiply by 6(250+5x) you get 1500+30x.
3960 = 1500+30x.
Subtract the value of 3960-1500 = 2460.
Then divide by this form,
2460/30 = 82 kmph
Arunkumar said:
1 decade ago
108+112=220.
220/6 is the relative speed per sec.
(220/6)*(18/5)=132 is the relative speed per hour.
Speed of the 2nd train is 132-50 = 82kmph.
220/6 is the relative speed per sec.
(220/6)*(18/5)=132 is the relative speed per hour.
Speed of the 2nd train is 132-50 = 82kmph.
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)
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.
@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
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.
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