Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 3)
3.
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Answer: Option
Explanation:
| Speed = |  |
45 x | 5 |  m/sec |
= | |
25 | m/sec. |
| 18 | 2 |
Time = 30 sec.
Let the length of bridge be x metres.
| Then, | 130 + x | = | 25 |
| 30 | 2 |
2(130 + x) = 750
x = 245 m.
Video Explanation: https://youtu.be/M_d8WufJWKc
Discussion:
245 comments Page 4 of 25.
Jagabandhu patra said:
1 decade ago
(45000/3600*30)-130 = 245.
(1)
Vishali said:
1 decade ago
Hi friends I did in a different way:
Let the length of train b = 130.
Let the length of bridge b = x.
Therefore distance = x+130.
Time = 30 sec.
S = 45 km/h = 45*5/18 = 25/2 m/s.
So, in this case distance = 375 m.
So, d = x+130 substituting the value of d (distance).
We get 375-130 = x.
x = 245m.
Let the length of train b = 130.
Let the length of bridge b = x.
Therefore distance = x+130.
Time = 30 sec.
S = 45 km/h = 45*5/18 = 25/2 m/s.
So, in this case distance = 375 m.
So, d = x+130 substituting the value of d (distance).
We get 375-130 = x.
x = 245m.
(1)
Sohail Khan said:
1 decade ago
We know that:
Time = Distance/Speed.
Suppose length of the bridge is X mtr.
Length of the train is 130 mtr.
So total distance = 130+X.
30 = (130+X)/45*(5/18).
Then the value of X is 245 mtr.
Time = Distance/Speed.
Suppose length of the bridge is X mtr.
Length of the train is 130 mtr.
So total distance = 130+X.
30 = (130+X)/45*(5/18).
Then the value of X is 245 mtr.
(1)
Ramkumar gadi said:
1 decade ago
Time taken by the train to cross object which is in motion in relative motion we have formulas:
T = (a+b)/(U+V) and T = (a+b)/(U-V).
Let be b = x.
So here we have U = Speed of train; a = Length of train = 130m.
V = Speed of object; B = Length of object.
V = 0; b = x; T = 30s.
S = 45 km/h = 45*5/18 = 25/2 m/s.
30 = (130+x)/(25/2+0).
= 30+25/2 = 130+x.
x = 245m.
T = (a+b)/(U+V) and T = (a+b)/(U-V).
Let be b = x.
So here we have U = Speed of train; a = Length of train = 130m.
V = Speed of object; B = Length of object.
V = 0; b = x; T = 30s.
S = 45 km/h = 45*5/18 = 25/2 m/s.
30 = (130+x)/(25/2+0).
= 30+25/2 = 130+x.
x = 245m.
(1)
Sivakumar.s said:
1 decade ago
Train 130 meters long. Travel 45 km/hr.
To convert m/s = 25/2.
So 2 second train will move 25 meter. So 30 second means.
2*15 = 30 second.
25*15 = 375.
So train travelling 375 meters to reduce the train meters.
375-130 = 245.
So the bridge meter is 245 meters.
To convert m/s = 25/2.
So 2 second train will move 25 meter. So 30 second means.
2*15 = 30 second.
25*15 = 375.
So train travelling 375 meters to reduce the train meters.
375-130 = 245.
So the bridge meter is 245 meters.
(1)
ARYAN said:
7 years ago
Length of the train = 130m.
speed =45km/h so, 45/3.6=12.5.
12.5*time=12.5*30=375m.
length of the bridge = [375-130] = 245m ANS.
speed =45km/h so, 45/3.6=12.5.
12.5*time=12.5*30=375m.
length of the bridge = [375-130] = 245m ANS.
(1)
Banda said:
7 years ago
Hi, Good explanation. Thanks all.
(1)
Ahmad said:
7 years ago
(130 + x)/30 = 25/2.
What is meant by this?
What is meant by this?
(1)
Priyanka Priyadarshini Samal said:
5 years ago
We have to calculate speed.
Speed = 45*5/18=25/2 m/s.
Time is given t= 30 sec.
Let's find the length of the bridge,
Let the length of the bridge is X.
And whenever a train crosses the bridge then the total distance covered by the train is the length of the bridge +it's own length (train's length),
According to the motion's 1st law distance or displacement = speed *time.
X+130 =25/2*30.
2(X+130)=750.
2X+260= 750.
X= (750-260)/2= 245m.
Speed = 45*5/18=25/2 m/s.
Time is given t= 30 sec.
Let's find the length of the bridge,
Let the length of the bridge is X.
And whenever a train crosses the bridge then the total distance covered by the train is the length of the bridge +it's own length (train's length),
According to the motion's 1st law distance or displacement = speed *time.
X+130 =25/2*30.
2(X+130)=750.
2X+260= 750.
X= (750-260)/2= 245m.
(1)
DURAI said:
5 years ago
Well said, thanks @Coolburn.
(1)
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