Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 29)
29.
A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is
Answer: Option
Explanation:
Let the length of the first train be x metres.
| Then, the length of the second train is |  |
x |  |
metres. |
| 2 |
| Relative speed = (48 + 42) kmph = | |
90 x | 5 | |
m/sec = 25 m/sec. |
| 18 |
 |
[x + (x/2)] | = 12 or | 3x | = 300 or x = 200. |
| 25 | 2 |
Length of first train = 200 m.
Let the length of platform be y metres.
| Speed of the first train = | |
48 x | 5 | |
m/sec = | 40 | m/sec. |
| 18 | 3 |
| (200 + y) x |
3 | = 45 |
| 40 |
600 + 3y = 1800
y = 400 m.
Discussion:
48 comments Page 1 of 5.
Ruko said:
4 months ago
I think 600m is the correct option.
Royal said:
6 months ago
Can anyone explain simply?
Raj said:
4 years ago
Guys,
Why can't we assume length of 1st train 2L metres and length of 2nd train L metres?
Please explain me.
Why can't we assume length of 1st train 2L metres and length of 2nd train L metres?
Please explain me.
(3)
Naveen said:
5 years ago
Why we didn't use the length of the second train?
(1)
Monika said:
5 years ago
Not getting this clearly, please explain in detail.
N.Rohith said:
5 years ago
We can do this problem in another way.
Length of train travelling at 48 km/h = 'X' metres;
Let speed of the train travelling at 48 km/h be 'u'.
Length of train travelling at 42 km/h = 'X/2' meters;
Let speed of the train travelling at 42 km/h be 'v'.
Time taken by both trains to cross each other = (X+X/2)/(u+v) {FROM FORMULA 8 IN IMPORTANT FORMULAS}.
u=48 km/h = 13.33 m/sec {CONVERTED TO metres/sec}
v=42 km/h = 11.66 m/sec {CONVERTED TO metres/sec}.
i.e, (X+X/2)/(13.33+11.66) = 12 sec.
i.e on solving , (3X)/(49.98) =12 sec.
X=199.2 metres.
Let platform length = 'Y' metres.
Time taken by a train of length 199.2 m to cross-platform of length Y m = time taken by train to cover (199.2+Y) metres. {FROM FORMULA 5 IN IMPORTANT FORMULAS}.
i.e, 45 sec = (199.2+Y)/(48 km/h).
i.e, 45 sec =(199.2+Y)/(13.33 m/sec).
From this Y=400.65 metres ; approx = 400 metres.
Length of train travelling at 48 km/h = 'X' metres;
Let speed of the train travelling at 48 km/h be 'u'.
Length of train travelling at 42 km/h = 'X/2' meters;
Let speed of the train travelling at 42 km/h be 'v'.
Time taken by both trains to cross each other = (X+X/2)/(u+v) {FROM FORMULA 8 IN IMPORTANT FORMULAS}.
u=48 km/h = 13.33 m/sec {CONVERTED TO metres/sec}
v=42 km/h = 11.66 m/sec {CONVERTED TO metres/sec}.
i.e, (X+X/2)/(13.33+11.66) = 12 sec.
i.e on solving , (3X)/(49.98) =12 sec.
X=199.2 metres.
Let platform length = 'Y' metres.
Time taken by a train of length 199.2 m to cross-platform of length Y m = time taken by train to cover (199.2+Y) metres. {FROM FORMULA 5 IN IMPORTANT FORMULAS}.
i.e, 45 sec = (199.2+Y)/(48 km/h).
i.e, 45 sec =(199.2+Y)/(13.33 m/sec).
From this Y=400.65 metres ; approx = 400 metres.
(2)
G.One said:
5 years ago
@Ranjith.
In Qn. clearly mentioned as another train length is half of the first train's length than the length of another train would be (x/2) or 0.5x rather than 2x.
In Qn. clearly mentioned as another train length is half of the first train's length than the length of another train would be (x/2) or 0.5x rather than 2x.
Ranjith said:
5 years ago
Well I have a little doubt in here, why can't we take the length of another train as 2x and length of this train as X so the things won't change but I can't get the ans, someone make me clear about this.
Sunny Talukdar said:
6 years ago
@Sai.
RS=(48+42)=90kmh
Then 90*5/18=25m/s.
So, First Train + 2nd Train Total Length = 25m/s * 12second = 300M.
ATQ,
First Train = 200M.
2Nd Train = 100M.
Now,
First Train Speed in Ms = 48 * 5/18 = 40/3.
So, First train+platform Total Length = 40/3 * 45 = 600M.
So, platform length is =600M-200M = 400M.
@Reka.
We use first train Speed because it's moving towards the platform and more important For ATQ we use First train speed we can't use both the speed because 2nd train is moving opposite.
RS=(48+42)=90kmh
Then 90*5/18=25m/s.
So, First Train + 2nd Train Total Length = 25m/s * 12second = 300M.
ATQ,
First Train = 200M.
2Nd Train = 100M.
Now,
First Train Speed in Ms = 48 * 5/18 = 40/3.
So, First train+platform Total Length = 40/3 * 45 = 600M.
So, platform length is =600M-200M = 400M.
@Reka.
We use first train Speed because it's moving towards the platform and more important For ATQ we use First train speed we can't use both the speed because 2nd train is moving opposite.
(7)
Sayli said:
7 years ago
@Riya.
In the question it is mentioned that first train completely passes the second train and also passe the platform that's why we have to consider the first trains speed.
For that first we have to calculate the first train.
In the question it is mentioned that first train completely passes the second train and also passe the platform that's why we have to consider the first trains speed.
For that first we have to calculate the first train.
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