Latest CAT Aptitude Question SOLUTION: a cow was standing on a bridge,5m away from middle of the bridge.a train was coming towards the bridge from the end nearest to the cow.seeing this the cow ran towards the train and

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(#M40010358) CAT Logical Reasoning question Keep an EYE

a cow was standing on a bridge,5m away from middle of the bridge.a train was coming towards the bridge from the end nearest to the cow.seeing this the cow ran towards the train and managed to escape when the train was 2m away frm the bridge.if it had run in the opposite direction (away frm train) it would have been hit by the train 2m before the end of the bridge.what is the length of bridge in meters assuming speed of train is 4times that of the cow...

Asked In CAT deepika (11 years ago)

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Deepak  (11 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 (22) | No (4)  

jiten amin  (11 years ago)

in the second scenario cow is at the distance of 5m away from half of the bridge(opposite side of the train) & it is hit by train 2m away from bridge's end

so total distance covered by cow before hit by train in second scenario is (if bridge's lenth is B then) (B/2)-(5+2) & of by train is B-2

in both case time is equal & velocity is diff. if we take velocity of cow as X then of train is 4X as given

& t1=t2 (time)
so ((B/2)-7/X)=(B-2)/4x
by solving this we will get B=26m

you can telly the ans

Is this solution Helpfull? Yes (3) | No (15)  

hari priya   (10 years ago)

Deepak . i need explanation about what is m here?

Is this solution Helpfull? Yes (3) | No (3)  

BIKASH YADAV  (10 years ago)

Let the train travel distance x when the cow just manages to escape the bridge of total length d .So
x/(d/2 - 5) = 4 ;
x = 2d - 20 ;
Now when the cow moves away from the train, the train moves (x+2+d-2) = x+d;
the cow moves (d/2+5-2) = (d/2+3); thus
(x+d)/(d/2+3) = 4
x + d = 2d + 12 ;
2d - 20 + d = 2d + 12;
d = 32.

Is this solution Helpfull? Yes (3) | No (2)  

siddharth  (10 years ago)

a__.__________.____c._______b ctrain

let speed cow x train 4x given
cow is at c so when moved towards train time taken by cow
d/2-5/x=d-10/x=t
same time t take train travel distace so distance travel by trainn till cow escape is d-10/x*4x(train speed)=2d-20=distance train travel
now if cow moves oppp

so distace travel =d/2+5-2
time taken =d/2+5-2/x
distance by train = first distance that we calculated

2d-20+2+d-2=3d-20
time taken by train is 3d-20/4x
equate both
3d-20/4x =d+6/2x
d=32 ans

Is this solution Helpfull? Yes (1) | No (3)  

Soniya  (10 years ago)

here m is the distance covered by train from some unkown point to 2m away from the bridge with in the time period when cow crooses the bridge

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