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:
247 comments Page 14 of 25.
Dinesh said:
1 decade ago
As the train crosses, which means the train should start at the starting of bridge and completely leave the bridge.
So, (130+x+130) should be the distance traveled in 30 secs.
= (130+x+130)/30 = (45*5/18).
x = 115.
So, (130+x+130) should be the distance traveled in 30 secs.
= (130+x+130)/30 = (45*5/18).
x = 115.
Jitendra said:
1 decade ago
Can we try this way?
Length of train + Length of bridge = (Distance which train need to be cover in 30 sec).
(130 m + X)---(1)-X = Suppose length of bridge.
Speed of trains = 45 km/h speed given in km/h hence convert it in m/sec.
45*5/18 = 25/2 ----- (2).
From equation (1) and (2).
= 25/2 = (130+X)/30.
= 25*30 = 260+2X.
= 750-260 = 2X.
= 490/2 = X.
= 245 = X.
Hence length of bridge (X) will be 245 m.
Length of train + Length of bridge = (Distance which train need to be cover in 30 sec).
(130 m + X)---(1)-X = Suppose length of bridge.
Speed of trains = 45 km/h speed given in km/h hence convert it in m/sec.
45*5/18 = 25/2 ----- (2).
From equation (1) and (2).
= 25/2 = (130+X)/30.
= 25*30 = 260+2X.
= 750-260 = 2X.
= 490/2 = X.
= 245 = X.
Hence length of bridge (X) will be 245 m.
Pradnya said:
1 decade ago
I can't understand your answer please explain.
Shashi said:
1 decade ago
We can solve by using formula also @Anusha.
Anusha 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 = 245 m.
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 = 245 m.
Madan said:
1 decade ago
45*5/18 = 12.5.
12.5*30 = 375.
375-130 = 245.
12.5*30 = 375.
375-130 = 245.
Bala said:
1 decade ago
45*5/18 = 12.5.
12.5*30 = 375.
375-130 = 245.
12.5*30 = 375.
375-130 = 245.
Deep said:
1 decade ago
Easy method is by using time and distance formula.
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)
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)
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