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 4 of 8.
Ravi tomar said:
6 years ago
220m=0.220km : 6 sec = 1/600 hr.
0.220/ (50+x) = 1/600
0.220x600 = (50+x) => 132 = 50+x
x = 132-50=82,
x=82 km/hr.
0.220/ (50+x) = 1/600
0.220x600 = (50+x) => 132 = 50+x
x = 132-50=82,
x=82 km/hr.
Muthulakshmi said:
1 decade ago
Hai Mr.Ramesh
You are Correct.Thanks for explaining the problem clearly
You are Correct.Thanks for explaining the problem clearly
MAHI said:
6 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,
Swati hande said:
5 years ago
Thanks all for explaining.
Kajal said:
5 years ago
Why have we assumed x in km/he why not in m/s?
Please, someone, explain it.
Please, someone, explain it.
Shubham said:
5 years ago
Thanks @Mahi.
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
Rakshikasaravanan said:
5 years ago
First train speed 50sec.
Second train speed?
S = l1 + l2/t1 + t2.
Relative speed = x + 50s.
X + 50s = 108+112/6+t2.
X + 50s =792/6.
6x + 300=792.
6x = 792-300.
6x = 492.
X = 492/6.
X = 82.
Second train speed?
S = l1 + l2/t1 + t2.
Relative speed = x + 50s.
X + 50s = 108+112/6+t2.
X + 50s =792/6.
6x + 300=792.
6x = 792-300.
6x = 492.
X = 492/6.
X = 82.
Gargi Banerjee said:
1 decade ago
Please give an appropriate answer with perticular reason, this is not clear to me.
Deepa Singh said:
1 decade ago
Please solve this part clearly 220/(5x+250)/18=6
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