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 | |||||||
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![]](/_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 5 of 8.
Srinivas8126 said:
1 decade ago
Hi Sundar.
What you said is right we have taken x in terms on km/hr, considering above explanation while converting relative speed we use (x + 50) km/hr both speeds converted into meters/seconds in next step. So final answer will be in m/s how it will be in km/hr?.
What you said is right we have taken x in terms on km/hr, considering above explanation while converting relative speed we use (x + 50) km/hr both speeds converted into meters/seconds in next step. So final answer will be in m/s how it will be in km/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.
Rajgowtham said:
1 decade ago
How the final answer comes in km per hour? First we are converting into metre per second so we need to convert it into km per hour.
Purushottam said:
2 months ago
It's (220/6*18/5) - 50 = 82km/h.
Dharamjeet Singh said:
1 decade ago
How can the final answer come in Km/hr ??
Rajata said:
1 decade ago
Total distance = 108+112 = 220 m.
Time = 6s.
Relative speed = distance/time = 220/6 m/s = 110/3 m/s.
= (110/3)x(18/5) km/hr = 132 km/hr.
=> 50 + speed of second train = 132 km/hr.
=> Speed of second train = 132-50 = 82 km/hr.
Time = 6s.
Relative speed = distance/time = 220/6 m/s = 110/3 m/s.
= (110/3)x(18/5) km/hr = 132 km/hr.
=> 50 + speed of second train = 132 km/hr.
=> Speed of second train = 132-50 = 82 km/hr.
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.
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