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:
252 comments Page 4 of 26.
WDL said:
4 years ago
Speed = 45*(5/18)=25/2 m/s.
Displacement = speed * time(here time is 30sec).
X + 130 = 25/2 * 30.
2(X + 130) =750.
2X + 260 = 750.
X = (750-260)/2= 245m.
Displacement = speed * time(here time is 30sec).
X + 130 = 25/2 * 30.
2(X + 130) =750.
2X + 260 = 750.
X = (750-260)/2= 245m.
(12)
Prakash said:
4 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.
(9)
Deepak Sahoo said:
4 years ago
Thanks for explaining the answer @Fathima.
(1)
Sid said:
5 years ago
Why it is +x, it also be -x, right?
(1)
Sonik Salam said:
5 years ago
260+2x=750.
2x=750-260,
2x=490,
And x = 490/2 = 245m.
2x=750-260,
2x=490,
And x = 490/2 = 245m.
(1)
Arup Kumar subudhi said:
5 years ago
Train Speed is 45 km/sec = (45*5)/18 = 25/2 m/s.
= (25*15)/(2*15) m/s,
= 375/30 m/s.
375 m in 30 sec, so bridge length is = 375 - 130 = 245m.
= (25*15)/(2*15) m/s,
= 375/30 m/s.
375 m in 30 sec, so bridge length is = 375 - 130 = 245m.
(2)
Sanilam Guha said:
5 years ago
(0.130+x) = (45 * 30)/3600.
If assume x is bridge length;
OR, 0.130 + x = 0.375.
OR, X = 0.375 - 0.130.
OR, X = 0.245 KM OR 245 M.
Am I right?
If assume x is bridge length;
OR, 0.130 + x = 0.375.
OR, X = 0.375 - 0.130.
OR, X = 0.245 KM OR 245 M.
Am I right?
(2)
DURAI said:
6 years ago
Well said, thanks @Coolburn.
(1)
Priyanka Priyadarshini Samal said:
6 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)
Kaushik vinay said:
6 years ago
45*18/5 = 25/3.
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