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 11 of 11.
Hsu Hsu said:
6 years ago
Why do you sum the relation speed of the train in the opposite direction?
If direction are same, will you sum?
Please, someone, explain to me, I need help!
If direction are same, will you sum?
Please, someone, explain to me, I need help!
Srinivas said:
6 years ago
Simply, the solution is;
2 km =2000.
=150 * 5/18 is 41.66.
So 2000/41.66 = 48.
2 km =2000.
=150 * 5/18 is 41.66.
So 2000/41.66 = 48.
Crack said:
6 years ago
I know why 5/18 is calculate. But in this particular math they didn't mention anything related to meter. Only answer is given in second, so we should only convert hr to seconds right? why we converted the km also? train length and speed both are in km, also in answer there's no km unit, only second. So please explain this part.
Gou said:
6 years ago
How to solve (2000*3÷125) ? please explain in breif.
Rony said:
5 years ago
Why not solve in km?
Speed & distance should be in km, right?
Speed & distance should be in km, right?
Nanda said:
1 month ago
Please explain the answer.
Subhash said:
3 days ago
T = D/S.
D = 1100m + 900m = 2,000.
S(relative) = (60+90) * 5/18 = 41.667,
T = 2000/41.667 =48.
D = 1100m + 900m = 2,000.
S(relative) = (60+90) * 5/18 = 41.667,
T = 2000/41.667 =48.
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