Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 7)
7.
Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
Answer: Option
Explanation:
Let the length of each train be x metres.
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
| = |  |
10 x | 5 |  m/sec |
| 18 |
| = | |
25 | m/sec |
| 9 |
 |
2x | = | 25 |
| 36 | 9 |
2x = 100
x = 50.
Discussion:
232 comments Page 15 of 24.
Semanth said:
1 decade ago
If faster train passes the slower train in 36 seconds then slower train passes the faster train in how many seconds?
Tintu said:
1 decade ago
Cant we do like:
relatv speed=10km/hr
Time=36 sec=.6hr
Then 2x distance is given by 10*.6 hours=6hrs
So x=3hrs
relatv speed=10km/hr
Time=36 sec=.6hr
Then 2x distance is given by 10*.6 hours=6hrs
So x=3hrs
D. R. Chaudhary said:
1 decade ago
Two trains are on parallel lanes then the one has to cross only 100 m or say x. So the correct answer is 100.
Siree said:
6 years ago
s1-s2=(L1+L2)/time.
46-36=2(L1+L2)/36,
10 * 5/18=2(L1+L2)/36,
50=2(L1+L2)/2,
50*2=2(L+1L2).
L1+L2=100/2 = 50.
46-36=2(L1+L2)/36,
10 * 5/18=2(L1+L2)/36,
50=2(L1+L2)/2,
50*2=2(L+1L2).
L1+L2=100/2 = 50.
Mahesan said:
1 decade ago
The 2nd train has 36 sec, 1st train speed has 2 times of the slower train, so it is calculated as above ?
Faraz said:
8 years ago
Isn't it a case of relative speed. If it is can't we solve this problem by that method? please tell me.
Nelson said:
6 years ago
Why have you subtracted the speeds to get the relative speeds instead of adding them? Please explain.
Suntth said:
1 decade ago
How the relative velocity is V2-V1? Since tthe trains are in same direction, velocity shall be V1+V2
Pankaj said:
7 years ago
Why are we calculating relative speed?
I used 36= 2x/(46+36). Using this getting 410m. Am I right?
I used 36= 2x/(46+36). Using this getting 410m. Am I right?
Prati said:
1 decade ago
I think we can do like:
Relative speed is (46-36);
Length = (10*5/18)*36;
=100both
=50 each.
Relative speed is (46-36);
Length = (10*5/18)*36;
=100both
=50 each.
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