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(#M40152322) ELITMUS Numerical Ability question Time and distance Keep an EYE

a cow was standing on a bridge 5m away from the middle of the bridge. he saw a train coming from the end nearest to him. the cow ran towards the train and managed to escape when train was 2m away from the bridge. if it had ran away from the train,it would have been hit 2m before the end of the bridge. what is the length of the bridge assuming speed of train is 4 times speed of cow.
OPTIONS- 32,36,40,cannot be determined

Asked In Elitmus Souradeep Guhathakurta (8 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Abhishek nkumar  (8 years ago)

|-------------Center------Cow----|-------------|Train|




When cow runs towards train then
Distance covered by cow=(x/2-5)
let speed of cow is c and it takes t1 time
then x/2-5=c*t1---(1)
distance covered by train is
m-2=4c*t1---(2)
When cow runs far away to train then
let time for collision is t2 then
distance covered=x/2+5-2 ---(3)
So x/2+3=c*t2
by train m+x-2=4c*t2 ----(4)
by subtracting eqn 1 ,3 and 2,4
c(t1-t2)=-8
4c(t1-t2)=-x
by dividing both

x=32

Is this solution Helpfull? Yes (6) | No   

Devendra Marghade  (8 years ago)

32 m.

If train starts from the distance 'y' m. from the one end(say left) of the bridge and length of the bridge is 'x' m. and if speed of the train =t and cow=c, given t=4c
First condition:
Distance covered by the train=y-2 and by cow=(x/2)-5
As (y-2)/t =[(x/2)-5]/c or 2x-y=18 ----(i)

Second condition:
Distance covered by the train(towards right) =y+x-2 and by cow=5+(x/2)-2
So (y+x-2)/t = [5+(x/2)-2]/c or -x +y=14 ----(ii)
On solving (i) & (ii), x=32

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Anubhaw  (8 years ago)

Let train is "td" distance away from bridge.
It is given
If cow speed is "c"
Then train speed is 4c. (ratio are given we also assume it as k and 4k )
Position of cow according to question are
Left end_____________d/2____5m(cow)________rightend_________________train
So the time taken by cow to escape from train=(d/2-5)/c
And in the same time train travel some distance let assume it "TD"
So
TD/4C=(d/2-5)/c
TD=(2d-20)
But when cow escape train was 2 m away from bridge
So total distance of train is (2d-18)

Now pick second
The time taken by cow to run towards leftend = time taken by train to travel (total distance of train from bridge and d-2 distance)
(d/2-2 + 5)/c=(2d-18 + d-2)/4c
3d-20=2d-12
d=32m(ans)

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Rishu Shrivastawa  (8 years ago)

speed of cow =v,speed of train=4v,distance from bridge train was when cow saw it=l, length of bridge=x
in time t
4vt=l-2, vt=x/2-5
dividing both we get 2x-l=18......(1)
now for second case
in time t1
5+x/2-2=vt1,l+x-2=4vt1
dividing both
l-x=14......(2)
adding (1) & (2)
x=32m

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pradeep singh rawat  (8 years ago)

32 is the ans..

Is this solution Helpfull? Yes (1) | No   

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