Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 9)
9.
Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
Answer: Option
Explanation:
Relative speed = (60+ 90) km/hr
| = |  |
150 x | 5 |  m/sec |
| 18 |
| = | |
125 | m/sec. |
| 3 |
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
| Required time = | |
2000 x | 3 | sec = 48 sec. |
| 125 |
Discussion:
107 comments Page 5 of 11.
Dipti said:
1 decade ago
Relative speed = (60+ 90) km/hr 150 is ok
= 150 x 5/18 m/sec
but how this 5/18 came
could you please discuss on this
= 150 x 5/18 m/sec
but how this 5/18 came
could you please discuss on this
Sabarish said:
1 decade ago
@Mohd. Nafees
why you multiply 2000 with 3/125. why it is not multiply by 125/3????
apply this formula in this step:
length of the train= speed*time taken
here speed of the train is (125/3)in m/sec and length of the train is 2000m.
time is required so assumed as x ;
now apply above formula :
2000=(125/3)*time
2000*3=125*time
6000=125*time
finally we get time=(6000/125)=48sec.
have a nice day!
why you multiply 2000 with 3/125. why it is not multiply by 125/3????
apply this formula in this step:
length of the train= speed*time taken
here speed of the train is (125/3)in m/sec and length of the train is 2000m.
time is required so assumed as x ;
now apply above formula :
2000=(125/3)*time
2000*3=125*time
6000=125*time
finally we get time=(6000/125)=48sec.
have a nice day!
Sabarish said:
1 decade ago
@Dipti
1km=1000m
1hr=60min=3600sec[one min contains 60sec then 60min having 3600sec]
here we want to convert speed(km/hr) into m/sec
so , [(1000/3600)=(5/18)]
have a nice day!
1km=1000m
1hr=60min=3600sec[one min contains 60sec then 60min having 3600sec]
here we want to convert speed(km/hr) into m/sec
so , [(1000/3600)=(5/18)]
have a nice day!
Rajkumar said:
1 decade ago
You all use this formula and do the sum.
The sum which is in opposite direction.
So, formula for time=(L1+L2)/(S1+S2)sec
The sum which is in opposite direction.
So, formula for time=(L1+L2)/(S1+S2)sec
Amul said:
1 decade ago
Wow super explanation. Im workouting by doing these derivations. All usefull problems with formulas. Thanks.
Sara said:
1 decade ago
Please explain this
(1.10+0.9)/(60+90)*5/18
=48
(1.10+0.9)/(60+90)*5/18
=48
Prateek said:
1 decade ago
What is the time taken for faster train to cross the slower train?
Harish Raj said:
1 decade ago
@Prateek
The time taken for faster train to cross the slower train is as equal as the time taken for slower train to cross the faster train.
The time taken for faster train to cross the slower train is as equal as the time taken for slower train to cross the faster train.
Vaasudevarao said:
1 decade ago
Formula (a+b)/(x+y) =1.10+.90/60+90
=2km/150km/h
=2000m/150*5/18
=2000*3/125
=48 sec
=2km/150km/h
=2000m/150*5/18
=2000*3/125
=48 sec
Aarti said:
1 decade ago
Why are you multiplying 150*5/8?
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