Aptitude - Problems on Trains - Discussion

Length of the train l1 = 130m;
Speed s = 45kmph;
time t = 30 sec ;
length of the bridge l2 = ?

Here total distance D= (length of the train + length of the bridge)
so d=l1+l2;
we know the formula of distance (d)=speed(s)*time(t);
but the speed in kmph we have to convert into mph so speed (s) = 45*(5/18)=25/2;
then d=s*t;
then substitute all the values in it.
(l1+l2)=(25/2)*30;
we know the l1 value l1=130m;
(130+l2)=(25*15)
(130+l2)=375
then l2 =375-130
length of the bridge is l2=245;

Speed=Distance/Time.

45*5/18=Distance/30,
375= Distance,
Length= 375-130=245.

How do you take 5/18?

1 km --- 1000 metre & 1 hour --- 3600sec whenever you divide 1km/1 hour then 1000 metre/3600 sec equals to 5/18 m/s.

Distance = speed*time,
D= 45* 5/18*30,
D= 5*5/2* 30,
D= 5 * 5 * 15 = 375,
D= 375- 130= 245.

@Sarang you are correct this is the easiest way to solve this problem.

As
Distance= time * speed.

D=45*5/18=25/2(speed)*30(time),
Actual distance=25 * 15=375.
Then bridge distance=Actual distance (375)- train Distance (130)=245.

45*5/18 = 12.5,
12.5*30 = 375,
375-130 = 245.

Good explanation, Thanks @Nursat Fathima.

Distance= Time * speed.

1km=1000m.
1hr=3600sec.
Time=d/t.

=1000/3600=5/18.
There distance: 5/18*45=25/2.
Length=t/d*s.
130+x=25/2*30.
130+x=375.
X = 375-130 = 245.

Shouldn't we are considering from the time the train's front enters the bridge till the tail is completely off the bridge as 30 seconds?

Then it would be 375- (130+130) =115m.