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(#M40030160) ELITMUS Numerical Ability question Keep an EYE

A cow standing on a bridge, 5m away from middle of the bridge. A train was coming towards bridge from the end nearest to cow. Seeing this cow ran towards train and managed to escape when train was 2m away from bridge. If it had run in opposite direction(i.e. away from train) it would have baaen hit by train 2m before the end of bridge. What is length of bridge in meters assuming speed of train is 4 times of cow.

Asked In Elitmus (10 years ago)

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Read Solution (6) :

sarab  (10 years ago)

speed of train= 4 * speed of cow
let bridge length be 2*l
half length l,
let train be t meters away from bridge.
first case 1:
distance traveled by cow l-5
distance traveled by train t-2
relation between distances
(t-2)=4(l-5)

second case:
distance by train: t+2l-2
distance by cow: 5+l-2 =>l+3
relation b/w distance :
t+2l-2=4*(l+3)
substituting t-2=4(l-5) from first case relation
we get
4(l-5)+2l=4*(l+3)
we get l=16 2l=32
length of bridge= 32 meters... hence proved... ;-)

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ROHIT RAJ  (10 years ago)

train=> T--------|--/________________.________________/
4x 2m c-5m-M .2m
|

T- Train
/ - bridge start and end
M - middle of the bridge
c - cow

let cow's speed = x
∴ train's speed = 4x
let bridge length = 2L
half bridge length = L
again, let train be at some distance from the bridge = d

now, opposite direction (Miss case)
train travels d-2, and
cow travels L-5
∴ d-2 = 4*(L-5)
=> d = 4L-20+2
=> d = 4L-18 -------------(i)

again
same direction (Hit case)
train travels d+2L-2 and
cow travels L+5-2 = L+3
∴ d+2L-2 = 4*(L+3)
=> d = 4L+12-2L+2
=> d = 2L+14 ------------(ii)

∴ from i & ii
4L-18 = 2L+14
=> 2L = 32

hence length of bridge is 32 metres..
Thankyou.. :)

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Mahesh  (10 years ago)

Let the length of the bridge be 2B metres, the reason for assuming 2B is, as we are talking about the middle of the bridge, it will be easy for us .
Let the distance that the train has to cover to enter onto the bridge be T metres.

In the first case the train is just two metres away from entering onto the bridge, so the train covered a total distance of (T-2) metres
In the mean time, the cow covered came just out of the bridge, as it is 5 metres away from the middle of the bridge it covered half of the bridge less 5 metres, so the distance covered by the cow is (B-2) metres.

In teh second case, the train reached the bridge and then covered the length of the bridge less two metres, so the total distance covered by the train is (T+2B-2) metres
In the mean time the cow which is 5 metres away from the bridge covered that 5 metres but fell short of 2 metres in covering the remaining half of the bridge, so the cow covered (5+B-2)= (B+3) metres

First Second
Train (T-2) (T+2B-2)
Cow (B-5) (B+3)

From first case to second case the train covered the distance equal to the length of the bridge more whereas the cow covered 8 metres more.

As it is given the speed of train is four times that of the cow, in the time that the cow covers 8 metres, the train coveres 32 metres

So, the length of the bridge is 32 metres.

Answer is 32 metres.

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deepak kr  (10 years ago)

train!---------y----------!--2m---(bridge starts here)||==============cow at (x/2 -5)=====(x/2)==============|==2m==||(Bridge ends here)

when cow was moving opp to the direction of the train
y/4s=(x/2-5)/s

hence y=2x-20

consider when cow is moving in the direction of the train.

y+x/4s=x/2+3)/s

2x-20 +x/ 4s = (x/2 +3)/s

x=32

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MANISH KUMAR  (10 years ago)

ans:22 meters.
let speed of cow x. so train's is 4x
let remaining distance covered by cow is d. so train covers distance 4d in same time.
so bridge half length= (5+d)...... since cow just escaped
now if cow runs in opposite direction and get hit at time t :-
time=speed/distance
so, equating cow's & train's time respectively
x/(5meter+bridge half length-2meter)=4x/(4d+2meter remaining+2*bridge half length-2meter)
=>x/(5+5+d-2)=4x/(4d+2+10+2d-2)
=>d=6
so half bridge length=(5+d)=(5+6)=11.
so bridge is 22 meters

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Prasoon  (10 years ago)

The answer is 32 meters.

Lets assume the length of bridge is x meters. Now the cow is standing 5m away from middle of the bridge and the train is coming towards bridge from the end nearest to cow and the cow runs towards train and manages to escape when train is 2m away from bridge. If we assume the distance covered by train before reaching bridge is y meters and the speed of the cow is a m/s then we get following eq.

(x/2 - 5)/a = (y-2)/4a say this as eq. 1 (given that the speed of the train is 4 times the speed of cow)

Now consider another case when cow runs away from train and get hit by train 2m before the end of bridge, we get following eq.

(5+(x/2 -2))/a = (y+(x-2))/4a say this as eq. 2

Now on solving these two eq. you will get value of x as 32 meters. That is the length of bridge. 'a' will cancel out. Considering the distance travelled by train before reaching bridge was the key point. Rest is simple. We have just equalised the time.

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