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Cow and Bridge

A cow was standing on a bridge, 5feet away from the middle of the bridge. suddenly a lightning
express with 90 miles/hr was
coming towards the bridge from nearest end of the cow.seeeing this the cow ran towards the
express and managed to escape
when the train is one feet away from the bridge. if it would have ran to opposite direction
(ie away from train) it would have been
hit the train one ft away from the end of the bridge.
Calculate the length of bridge.
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rajikrishna
1 Solution

Commented:
Hey, y'all watch this.

<ooooeeeeeeooooo>I have control of your monitor.  I am watching you.  I can see you.  You are studying for an Infosys exam.  You wish the kind experts here to help you prepare.

Freaky, huh?
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Commented:
Ok, seriously, we just answered this problem (or one very like it) a week or two ago, but I think it was deleted, because asking Infosys questions is considered equivalent to cheating on homework, which is not allowed here.  What I can do is point you in the direction of how to solve it yourself.

Lay out the problem with the starting point of the train, the cow, and the ends of the bridge.  Assign variables for the speed the cow runs, the length of the bridge, and the distance from the train to the bridge.  Once you have all that laid out, you can post here again if you're still stuck.
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Commented:
I want to know from where these guys get all their questions. Shouldn't these questions be kept under lock and key somewhere so that potential infosys employees don't cheat?
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Commented:
I think there is some information missing. Are you talking about a laden or an unladen cow? And is this a European or an African cow?
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Commented:
I want to know how I can get a job at Infosys, writing entrance exams.  Think they'd let me telecommute?
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Commented:
African cows aren't migratory.
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Author Commented:
Thank you for YOUR WONDERFUL RESPONSE!!!!!!!!!!!!!!
puzzle is a puzzle WHETHER it is Infosys or IBM it doesnt matter.

KNOWLEDGE IS MEANT TO BE SHARED AND NOT TO BE MADE FUN OF.

thank you so much

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Commented:
Hey look, Arawn, we touched a nerve.  Think we can get this moved to the lounge where we can be civil about it?

rajikrishna: We're happy to help you out with this puzzle, but you have to meet us halfway (or 5 feet either side of halfway in this case).  If you post your thoughts on the puzzle, what you've tried, and what you don't understand, then we will try to give you a nudge in the correct direction.  But don't expect us to solve your problems for you ... remember, we won't be there when you take the test, so you'll have to do it on your own.
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Commented:
I agree with rajikrishna.

Knowledge is meant to be shared, and not made fun of.

But LACK of knowledge on the other hand, where you don't know the answer so you post it here hoping we'll help you get a job that you don't deserve, IS meant to be fun of.  We'll share our knowledge if you, as snoyes said, meet us halfway and explain to us what parts you don't understand or whatnot.

We won't do your work for you, but we'll HELP you do your work.  You, in the end, are the one who needs to learn how to do the puzzle, not us.  If we give you an answer it won't help you unless you're wanting to cheat.

Give a man a fish, he'll ask for another.
Teach a man to fish, and he'll leave you alone.
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Commented:
excuse me . i'm taking a home-study math today.  can u please solved these for me:

How many badgers does it take to screw in a lightbulb?
What would be the wattage of a lightbulb that was powered by potato's?  (spud light)
x + 0 = 1  Solve for X
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Commented:
phileoca:
a) That's not an appropriate questions outside the lounge.
b) Idaho
c) Well, X is probably the antiderivative of x.  Rearranging, we get x = 1; integrate, and we get X = x + c.
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Commented:
>>How many badgers does it take to screw in a lightbulb?
None. The little suckers get their human slaves to do it.

>>What would be the wattage of a lightbulb that was powered by potato's?  (spud light)
How quckly are the potatoes running on the treadmill?

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Commented:
thanks snoyes.
After someone else posts with those exact same answers, i'll be sure to award them the points.  ^^
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Commented:
>> African cows aren't migratory.
What if an entire flock of African swallows caried the cow?
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Commented:
>What if an entire flock of African swallows caried the cow?
Not a bad idea.  In my experience, wasps are far more interested in roast beef than they are in strawberry jam.  (No, I'm not going to explain it).
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Commented:
>>How many badgers does it take to screw in a lightbulb?
1001, 1 to hold the lightbulb and 1000 to rotate the house

>>What would be the wattage of a lightbulb that was powered by potato's?  (spud light)
approximately 1.5712 x the wattage of a lightbulb powered by carrots

>>c) Well, X is probably the antiderivative of x.  Rearranging, we get x = 1; integrate, and we get X = x + c.
so we get: X = x + (Well, X is probably the antiderivative of x.  Rearranging, we get x = 1; integrate, and we get X = x + c)
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Commented:
Hooray!  Recursion and the Endless Loops!  eh, a mediocre name for a rock band, at best.
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Commented:
>>How many badgers does it take to screw in a lightbulb?
None their nocturnal (that means they like the dark)

>>What would be the wattage of a lightbulb that was powered by potato's?  (spud light)
According to the reading I'm getting from my flux capacitor, exactly 1.21 gigawatts.

:)

And just for the record, I'm deathly afraid of bridge crossing cows.
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Commented:
phileoca:
a) That's not an appropriate questions outside the lounge.
b) Idaho
c) Well, X is probably the antiderivative of x.  Rearranging, we get x = 1; integrate, and we get X = x + c.

now gimme points
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Administrateur Systeme et ReseauxCommented:
hey guys if some one doesnt wanna answer ......dont answer but please refrain from using yr IQ guessing how the question got outta the Infosys exams or telling the person that he doesnt deserve to be an INFOSYS employee if he cannot answer that question.
r all of us (capable of answering that question) deserving to be Infosys employee.....
having a bagfull of point at EE should not make a person VAIN!!!!!!!!
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Author Commented:
Hi Guys!!

If not for the comment made by iamthecreator i wouldnt have come to this web site again,atleast
of you all "experts!!" he is the only person who understood me.
Everyobody wants to come up in life and i am not an exception.I am trying hard to pass the exam
but when i learnt that i am not capable to be an Infosys employee it really hurt me a lot and
i lost all my hope.Dont think that i blindly post the problems.After sitting with it for
hrs together and if i dont get only then i disturb you EXPERTS.If you could have told me that i shuld
tell u people how fari have done i wouldnt have been hurt at all.
Ok anyway i am a dumb girl no body can change that and one think whether i pass the
exam or not it doesnt matter ,if i want to improve my problem
solving capability can you please tell me what i should do or is it an inborn talent??

Here is my sum :
we have to find out the length of the bridge assume it to be x mts.
The cow is standing 5 ft away from x/2 and the train hit it 1 ft from the end of the bridge
so,
the cow covers (x/2 - 5 + 1) 5 [5 feet away] and 1[the train hit it 1 ft from the bridge]
at the same time the train covers x+1 ft .here only i amm stuck we should add some feet to this
(x+1) i dont know how much we should add
Similarly we have to form another set of equation where the cow ran towards the train and managed to escape.
here the cow covers (x/2+5) at the same time the train covers ????how mcuh
if we can form these two sets of equations we can find out
Thats all guys
Thank you in advance

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Commented:
Looks like you're getting there...
x is the length of the bridge, so the cow is at x/2-5 from the end of the bridge.
y is the current distance from the train to the bridge.
So the train goes y - 1 in the same time the cow goes x/2 - 5 + 1, and the cow escapes.

Alternatively, the train goes y + x - 1 in the same time the cow goes x/2 + 5 - 1, and the cow gets hit.

So you have two equations, two unknowns:
y - 1 = x/2 - 5 + 1
y + x - 1 = x/2 + 5 - 1

Rearrange, and solve for x.
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Commented:
>> if i want to improve my problem solving capability can you please tell me what i should do

Some books that would help:

Puzzles to Puzzle you - Shakuntala Devi
More puzzles - Shakuntala Devi
Quantitative Aptitude - R.S.Agarwal

Most of the puzzles appearing in Infosys tests are variants of the puzzles in these books.
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Commented:
rajikrishna

The reason that all of us "experts" have been commenting like we have is because for the last few weeks the Puzzles and Riddles section has become a dumping ground of questions from people who want to pass the Infosys exam. Some people have been devious in their attempts to disguise the fact that they were trying to use our skills to pass an exam. Others would only post questions and wait for results without providing any sort of feedback to show they have done some work. You at least have the courtesy (after a bit of prodding) to post your work and show that you have made an attempt. So when we saw your question, we all naturally assumed that you were just another person who wanted us to do all the work for you while you go and apply for a job for which you might not be qualified. Sorry about any hurt feelings or misunderstandings.

Cheers,

Arawn
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Commented:
An off-topic comment, forgive me.

>> r all of us (capable of answering that question) deserving to be Infosys employee.....
Whether we are deserving or not, I do not know. But with the amount of recruitment that Infosys does (10,000 people a year), we can be sure that most people capable of answering that question would get through the tests. Until recently, written tests were the sole criterion for recruiting people at Infosys. No interviews!

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Author Commented:
ok guys cool it
no more heart feelings

u are all my brothers and sisters  and hence i give you the full freedom to comment on me
i believe whatever you tell me it will be for my benefit only.
i am not upset with you guys anymore

ok see you
have a nice day
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Commented:
rajikrishna:

So, have you solved for x yet?  What value do you get?
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Author Commented:
just came from office

i solved for x and got  8  for x which is OBVIOUSLY NOT CORRECT SEEMS TO BE ILLOGICAL
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Commented:
Right, we forgot to account for the fact that the train and the cow move at different speeds.  The train moves at a rate of 90mph, or 7920 ft/min, which we'll call T, and the cow goes at an unknown rate, which we'll call C.

So our equations should have been:
1/T * (y-1) = 1/C * (x/2 - 5 + 1)
1/T * (y + x - 1) = 1/C * (x/2 + 5 - 1)

Now we have two equations and three unknowns, which means there are many possibilities.  Does the original problem state how fast the cow runs?
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Commented:
Snoyes, I believe there is a small flaw in your first equation, since the cow escaped it got of the tracks at the end of the bridge and did not cover the extra foot to reach the train, so:

1/T * (y-1) = 1/C * (x/2 - 5)
1/T * (y + x - 1) = 1/C * (x/2 + 5 - 1)

We know T and can eliminate either y or C, hence the solution will depend on the remaining unknown, so unless somebody can tell us how fast the average cow (african or not) can run towards an approaching train, or whether we can estimate y by a cow's hearing ability some info is missing from the question.

rajikrishna: glad to hear your are not hurt and working on the solution with us.

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Author Commented:
Thank you all for ur help
I think thers some flaw in the sum or some data missing.
Rather than the answer i was much bothered abt the approach to such kind of sums.Anyway i am trying to solve this.

Thank you once again.
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Commented:
Idefiks, I think you're right about the equation. But also consider that the cow needs room to get off the bridge.  If the cow is standing at the bridge entrance when the train is one foot from the entrance, it's still going to get creamed.  So it needs to go a little farther than just (x/2 - 5).  Train tracks are what, 4 feet wide?  So it's got to go at least two feet perpendicular to the track.  Plus we have to consider the length of the cow, and its acceleration, and how long it takes to turn one direction or the other, and where the railings for the bridge end so the cow can move sideways off the track, and...
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Commented:
just a quick side questions, what is infosys and why do they have these ridiculous entrance exams?
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Commented:
Big tech company in India, I think.  I guess they are looking for people with good math and problem-solving skills.
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Commented:
thanks.
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Commented:
I'm not sure if this really helps, as it doesn't take into consideration the velocity of the train, nor the velocity of the cow (in either direction).  I'm just hoping that this different way of looking at things may help one of you solve the problem.

Let's start off with naming some variables:

X = Length of Bridge
A = Starting point of Train
B = Side of Bridge nearest to the Train
C = Side of Bridge furthest from the Train
D = Cow's starting location
E = Center of the Bridge
X = Length of Bridge
Y = Distance from A to B

Now let's define things in terms of these points, starting with the cow running toward the train and escaping by a single foot's margin:

The cow will travel from (D to B) in the same time that the train travels from (A to B - 1).  NOTE:  Assume that B - 1 means one foot short of point B.
The cow's travel can also be defined as:     X/2 - 5
The train's travel can also be defined as:     Y - 1

Now let's look at them with the cow running away from the train and being hit only a foot from the safety of the far side of the bridge:

The cow will travel from (D to C - 1) in the same time that the train travels from (A to C - 1).
The cow's travel can also be defined as:     X/2 + 4
The train's travel can also be defined as:     (Y - 1) + X

This seems like a good start, but I'm lost from here.  Hopefully someone can take this and run with it...good luck everyone.
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Commented:
The train travels the length of the bridge in the same time that it takes the cow to run 9 feet.
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Author Commented:
I feel that the train speed is irrelevant
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Commented:
Length of bridge = 810/(speed of cow)
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Commented:
Hey raji,

Not able to get into infosys is not the end of the world. I too had tried and failed a few years back. So, take it easy.....
Or to feel better, maybe you can do what I did - marry someone who works in Infy...:-)).......but seriously, my wife does work in Infy!!

As for the riddle I dont have the habit of solving such types using equations as I spend too much time to form the equations. My GUESS solution is - Length of the bridge is 80mts.
Train moving @ 40m/s, cow @ 4m/sec, cow covers 36m in 9 sec. to escape. Converted all to meters as Iam more comfy using that.
I might be wrong, better to chk this out.

As for people who are taing up Infy exams, you better try seeing sites like this one - <<http://members.rediff.com/sudarshanbr/info.htm>>
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Commented:
Snoyes: The cow we are dealing with here starred in one of the X-files episodes, once it reaches the end of the bridge, before being killed by the train, it can teleporte itself sideways ... now you see it, now you don't ... Why can it not do this when it is in the middle of the bridge you ask? Because the huge metal railings of the bridge guide the energy field created by the cow when teleporting and form a magnetic field which the cow can not cross in its energised state. The risk of being absorbed by the magnetic field and remain trapped in the railing forever is just too big ...

LemmeC: You came up with the same solution I did, the cow's speed also needs to be expressed in mph.

Rajikrishna: I feel the train speed IS rellevant, imagine this: the train is extremely slow and therefor the cow can stare at it for a while - as cows sometimes do - and easily stroll of at either side. Or, it is a high speed magnetic train chasing at you at 500 Kph, the cow looks at either side of the bridge and before it can say BOOOOO it is transformed into corned beef ...
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Author Commented:
bcus the speed didnt come in the equations which i framed i told like that but
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Commented:
I would also think the speed of the cow is irrelevant, as long as it is CONSTANT (no acceleration or deceleration).  Of course, this assumes that our super-cow can not only teleport to safety at the end of the bridge, but also that she can go from 0 mph to her final speed instantaneously. Seriously, there must be some bovine enhancements at work here.
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Commented:
Oh -- but the speed of the train WOULD be relevant insofar as it would help determine the distance traveled by the train from Point A (1 foot away from where the cow leaps to safety) to Point B (where the cow meets her doom 1 foot from the far end of the bridge).
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Commented:
Shoot the cow where it stands, and let the train bring you the meat. (What's left anyways)
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Commented:
ok I am not an american, WHAT ON EARTH IS Infosys  ???
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Commented:
david, i'm an american, i still didnt know what infosys is. look up, i asked. its a company in india.
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Commented:
ok... here's the problem
the speed of the cow does not matter.
it is standing 5 feet from the middle of the bridge. (this gives you an Idea to how long the bridge is, (and we aren't talking about the golden gate)
The cow runs toward the train.
the cow makes it to safety by doing this
would have to assume at this point that the cow was standing 5 feet from the center of the bridge toward the train.
had the cow been running away from the train, and made it, it would have had to have been standing on the five feet from the middle away from the train.
do you understand?

All logic problems, which is what this boils down to. are confusing at first. but once you figure out the wording, and mark down the facts, the solution is easier to reach.

I add one last comment which may solve your problem,  (but you still have to work for it.

Because the cow was standing five feet away from the middle towards the train, it escaped it's doom. had it run the other direction, it would have been hit 1 foot before the end of the bridge.

1' |              X                   |               5' |             Y             |
now from this, you can determine the speed of the cow, and the length of the bridge it is on.
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Commented:
note, I didn't catch that it stated the cow was running toward the train in the question, just a logical conclusion based on the fact the cow lived.
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Commented:
So... the time it takes for the cow to travel HALF the length of the bridge less her 5-foot head start can be Tx. The time it takes for the cow to travel HALF the length of the bridge PLUS the 5 feet but LESS the fateful 1 foot at the far end of the bridge can be Ty. The difference between Ty and Tx is the amount of time required for the train to travel exactly the length of the bridge.

Just my 2 cents. I still can't work it out... Too distracted...
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Commented:
I really hate to disappoint everybody but this is nothing new.
These were the 2 equations I posted a week ago:

1) 1/T * (y-1) = 1/C * (x/2 - 5)
2) 1/T * (y + x - 1) = 1/C * (x/2 + 5 - 1)

The 'new' equation would be:
9ft/C = x/T

However if we do 2) - 1) we get the exact same thing and a linear combination of equations is NOT a new equation ...
It is just another way of writing down the same thing.

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Commented:
Shattuc,

you wrote:
The cow runs toward the train.
the cow makes it to safety by doing this
would have to assume at this point that the cow was standing 5 feet from the center of the bridge toward the train.
had the cow been running away from the train, and made it, it would have had to have been standing on the five feet from the middle away from the train.
do you understand?

No, I don't understand because that is not correct. If the cow is standing 5 ft away from the middle, away from the train, and ran away from the train then the cow would also need to run a distance of x/2-5 to get to safety. The train on the other hand would have to cover the distance of the bridge extra to catch it so the situation is not the same.
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Commented:
Lets finish this off:
Let A and B be the ends of the bridge, and the train is approaching from the B side at 90mph = 132 feet per sec (fps).
Let width of bridge be x feet. Cow stands x/2 - 5 feet from B.
Let the train be a distance d from point B
Let cow's speed be y fps.

cow covers distance x/2-5 feet in
t1 = (x/2 -5)/y
At that point of time the train is 1 foot from the cow i.e 1 foot from B.
t1 = (d-1)/132
Therefore  (x/2 -5)/y = (d-1)/132
This is the first equation

In the second case, when the cow runs AWAY from the train, it covers x-1-(x/2-5) feet before being hit, = x/2+4 feet.
t2 = (x/2+4)/y seconds
train covers d+x-1 feet in the same time
t2 = (d+x-1)/132
Therefore (x/2+4)/y = (d+x-1)/132
This is the second equation

We have TWO equations & THREE unknowns. Hence there is no single solution for the bridge width.

simplifying equation 1 we get
(x-10)/2y = (d-1)/132

equation2:  (x+8)/2y = (d-1)/132 + x/132

subtracting eqn 1 from eqn 2 we get

18/2y = x/132
or
9/y = x/132
or
x = 1188/y

If there is really a solution I would like to know.

There is no lower limit to y. But we can assume the cow would run like hell. How fast do you think a cow can run?
Anyway, how did the cow manage to calculate that if it ran TOWARDS the train it would escape?
Maybe the COW should join INFOSYS.
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Commented:
b=Bridge Length in feet
c = top speed of the cow in feet/second

The train is travelling at 132'/s
The cow escapes with just 1/132 seconds to go.
To escape the cow has travelled 1/2b-5'
If the cow runs away from he train by the time the train is 1' away from the bridge the cow has travelled 1/2b-5' towards the end of the bridge after starting 1/2b+5' from the end so he (the cow) only has 10' to go before reaching the end of the bridge. The train has just distance b to go before hitting the cow.

So the time it takes the cow to run 10' is the time it takes the train to cover the length of the bridge.

Say the cow travels at c feet per second then it will take 10/c secs to run 10'
The train is travelling at 132 feet/sec so it would travel 132*(10/c) feet in the same time - this is also the time it take to travel the length of the bridge.

So that's the answer:
The length of the bridge is 1320/c feet
Where c is the top speed of the cow.
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Commented:
oops. Spotted a mistake:
should have said:

b=Bridge Length in feet
c = top speed of the cow in feet/second

The train is travelling at 132'/s
The cow escapes with just 1/132 seconds to go.
To escape the cow has travelled 1/2b-5'
If the cow runs away from he train by the time the train is 1' away from the bridge the cow has travelled 1/2b-5' towards the end of the bridge after starting 1/2b+5' from the end so he (the cow) only has 10' to go before reaching the end of the bridge and 9' to go before being hit by the train. The train has just distance b to go before hitting the cow.

So the time it takes the cow to run 9' is the time it takes the train to cover the length of the bridge.

Say the cow travels at c feet per second then it will take 9/c secs to run 9'
The train is travelling at 132 feet/sec so it would travel 132*(9/c) feet in the same time - this is also the time it take to travel the length of the bridge.

So that's the answer:
The length of the bridge is 1188/c feet
Where c is the top speed of the cow.

PS: doraiswamy had the correct answer all along you should give him the points.
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Commented:
Others have expressed the same answer earlier in different units.
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Commented:
Assuming a spherical cow of uniform density and constant speed....

In one case, our instant of interest is when the train is 1' before reaching (the near end of ) the bridge.  In the other case, our instant of interest is when the train is 1' before reaching the FAR end of the bridge.  "Obviously" the time delta between these two instants is the time it takes the train to cover the length of the bridge.

In the first case, the cow travels (x/2)-5 feet.  In the second case, it travels (x/2)+5-1 feet.  "Obviously" the distance delta between these two cases is 9'.  (An unwary problem-solver might come up with 10'...)

So we have that the train covers the length of the bridge in the same time as the cow would cover 9'.  IF we had the speed of the cow, we could work that out to an exact number; since we don't, we end up with an expression in terms of the speed of the cow.

Note that this only works if (x/2)-5 >= 0.  i.e., if the bridge is less than 10' wide, the cow travels less than 9' and so the solution is no longer along the same line as if the bridge is at least 10' wide.  In the extreme case where the bridge is 0' wide (by definition it absolutely cannot be *negative*...), the cow can travel at any speed, and it both escapes AND is hit by the train -- obviously some sort of quantum singularity (see Schroedinger's Cat).

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Commented:
Simultaenously geting hit AND escaping? It sounds as though this may require redefining what it means to get hit by a train. Perhaps the impact of the train is precisely what sets the cow free from a universe where a bridge can be 0' wide?
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Commented:
And so the cow joined Infosys,
doraiswamy got the points,
and we should all be hit by a train to be set free from a universe where this thread exists.
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Commented:
I realize that I got here late and this topic is closed, but I think the answer is that the bridge is 12 feet long.

Initial assumptions:

B=length of bridge
(B/2)-5 = distance cow runs and lives
(B/2)+5 = distance cow runs and dies

In both scenarios, the cow starts running when it sees the train at the same location.  The distance the train travels to the end of the bridge, then, is constant and can be ignored.  That leaves us with the cow dying at B-1; therefore the train covers B-1 feet in the same time that the cow covers (B/2)-5 feet, or:
B-1=(B/2)+5
2B-2=B+10
B=12

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Commented:
Ok all my dear Eiensteins and galileos,
ppl r equating distances of cow and train.
some r putting up 2 eqns. but 3 vars , so infinte solns.

chk this:

c = cow's speed
x = length of bridge
y = distance train is at from bridge end

(x/2 - 5)/c = (y-1)/90 ----------------------(1)
(x/2 + 5 - 1)/c = (y + x - 1)/90 --------------(2)

solving thse 2 u get,
cx = 810 -------------(3)

ok,

now consider,

x/90 = (x/2 + 5 - 1)/c

plug in cx = 810

u get x = 1612/90 = bridge length

bbye
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Commented:
>>(x/2 + 5 - 1)/c = (y + x - 1)/90 --------------(2)
>>now consider,
>>x/90 = (x/2 + 5 - 1)/c

combining both you get (y + x - 1)/90=x/90 or y=1

sorry Einstein, this riddle remains unsolved ...
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