Newton's puzzle

[JordanRHughes]

I got 32 days. I bet you that is correct.....

[Chriton]

You are all wrong, Grass dose not grow at a constant rate, it depends on rainfall, no rain no growth, if there is no rainfall in the 90 days the grass will not grow,so the equation will not work unless you know the rainfall and exact rate of the grass growth.

[JN.]

I think this problem seems to be about maths, but if we look closer, we'll see that this is just to make people laugh. I think that the most important here is that grass grows continuously, so it's impossible for the animals to eat them all. I think that's how it should be interpreted but, of course, I may be wrong.

[D H]

Just because grass grows continuously does not mean it can tolerate the onslaught of an excessive number of herbivores. Overgrazing is a real problem. This is a simplified version of an overgrazing problem.

[raj bhramar]

Please let me know this is correct or not...!

 

Say

Empty pasture is full of grass in p days.

Then rate of grass re-growth is 1/p pasture per day.

Cow eats full pasture in c days, so in 1 day it eats (1/c -1/p) = (p – c)/pc pasture

Goat eats full pasture in g days, so in 1 day it eats (1/g - 1/p) = (p - g)/pg pasture.

Duck eats full pasture in d days, so in 1 day it eats (1/d - 1/p) = (p - d)/pd pasture.

Therefore

Cow + goat eat {(p – c)/pc + (p - g)/pg} pasture in 1day.

It is given that cow + goat eat Full + re-grown grass in 45 days. So in 1 day they eat 1/45 pasture.

[Note 1/45 pasture they apparently eat in 1 day = what they have actually eaten – the grass re-grown]

So (p – c)/pc + (p - g)/pg = 1/45 ………………………………………………………………………………………………….. [1]

Similarly (p – c)/pc + (p - d)/pd = 1/60 ….…………………………………………………………………………………….. [2]

Similarly (p – g)/pg + (p - d)/pd = 1/90 ….…………………………………………………………………………………….. [3]

Also it is given that the amount of grass cow eats equals the amount of grass goat and duck eat together.

Therefore 1/c = 1/g + 1/d …………………………………………………………………………………………………………… [4]

Which can be rewritten as (1/c – 1/p) = (1/g – 1/p) + (1/d – 1/p) +1/p

OR (p – c)/pc = (p - g)/pg + (p - d)/pd + 1/p……………………………………………………………………………….… [5]

Hence from [3] & [5],

(p – c)/pc = 1/90 + 1/p or (1/c – 1/p) = 1/90 + 1/p or (1/c – 2/p) = 1/90

OR (p - 2c)/pc = 1/90 or 90p – 180c = pc or 90p – pc = 180c

OR p(90-c) = 180c …………………………………………………………………………………………………………………………[6]

From equation [6] we may assume p = 180 ……………………………………………………………..……………….. [7]

And then 90 – c = c; so c = 45 …………………………………………………………………………………………………….. [8]

Putting the values of c and p in the equations [1] & [2], we get g = 90 and d = 180 ………………..…. [9]

From equation [9] it may be observed that a duck eats the same amount of grass as is re-grown, making its contribution as nil when all three eat the grass together.

Therefore when all three eat the grass together ONLY COW & GOAT CONTRIBUTE IN FINISHING THE GRASS.

And it is already given as 45 days, or may be calculated from the equation [1].

 

 

[D H]

Please let me know this is correct or not...!

 

...

 

Therefore when all three eat the grass together ONLY COW & GOAT CONTRIBUTE IN FINISHING THE GRASS.

And it is already given as 45 days, or may be calculated from the equation [1].

 

Incorrect. See post #6 for the correct answer.

[raj bhramar]

Thank you D H,

I think it could be more interesting if we consider the rate of grass re-growth as per square unit. It does not seem logically correct that grass regrows irrespective of the space /area available for its growth.

Whatsoever my answer in my last post was wrong, reposted answer here is the right one.

 

Incorrect. See post #6 for the correct answer.

Say

Empty pasture is full of grass in p days. Then rate of grass re-growth is 1/p pasture per day.

Cow eats full pasture in c days, so in 1 day it eats 1/c grass.

Goat eats full pasture in g days, so in 1 day it eats 1/g grass.

Duck eats full pasture in d days, so in 1 day it eats 1/d grass.

Therefore

Cow + goat eat (1/c + 1/g) pasture in 1day.

 

But in 1 day 1/p grass re-grows.

 

So cow + goat finish (1/c + 1/g) – 1/p grass in 1 day.

It is given that cow + goat eat Full + re-grown grass in 45 days. So in 1 day they finish 1/45 pasture.

 

1/c + 1/g – 1/p = 1/45 ……………………………………………………………………………………………………….. [1]

Similarly 1/c + 1/d - 1/p = 1/60 ….………………………………………………………………………………..……. [2]

Similarly 1/g + 1/d - 1/p = 1/90 ….……………….………………………………………………….………..……….. [3]

Also it is given that the amount of grass cow eats equals the amount of grass goat and duck eat together.

So 1/c = 1/g + 1/d ………………….………………………………………………………………………………………… [4]

From equations [3] & [4] 1/c – 1/p = 1/90 .………………………………………………………………….….… [5]

Hence from [1] & [5],

1/g + 1/90 = 1/45 or 1/g = 1/45 + 1/90 or 1/g = 1/90 ……………………………………………………….. [6]

Similarly from [2] & [5], 1/d = 1/ 180 ..……………………………………………………………………………… [7]

From equation [4], [6], & [7], 1/c = 1/60 …………..………………..…………………………..……………….. [8]

And from [5] & [8], 1/p = 1/180 ……………………………………………………………………………………….. [9]

Hence when all three eat the grass together, they finish 1/60 + 1/90 +1/180 – 1/180 = 1/36 grass in 1 day.

 

So all three cow + goat + duck together finish the grass in 36 days.

 

I used 'spoiler'. But somehow it did not work....?

Edited March 26, 2012 by raj bhramar

[imatfaal]

/snipped

 

I used 'spoiler'. But somehow it did not work....?

 

[spoiler] your answer here [/spoiler]

[Pradeep Sridharan]

Remember the "does this answer make sense" test. The cow and goat finish in 45 days. Therefore all three must take less time.

 

The animals eat at a constant rate and the grass grows at a constant rate, so the amount of grass in any scenario decreases at a constant rate. For example, with just the cow, after 45 days half the initial grass is left. The cow must have therefore consumed half the initial grass + whatever grass grew in 45 days.

 

Now, the cow and goat finish in 45 days, half the time as the cow alone. But as stated above, the cow has eaten half of his 90 day total, which is half the initial grass + the growth of grass in 45 days. What's left for the goat to have eaten is the other half of the initial grass. So the goat can eat the initial grass by itself in 90 days. The goat needs the duck's help to eat initial grass plus 90 day's growth in 90 days, so it follows that the duck eats grass exactly as fast as it grows.

 

The cow and the duck can eat everything in 60 days. Which means (since the duck cancels out the growth) that the cow can eat the initial grass in 60 days. The goat needs 90 days, so the cow can eat 1.5 times as fast as the goat. And, since goat + duck = cow, the duck eats 0.5 times as fast as the goat.

 

So:

The cow eats 1/60th (or 3/180) initial grass per day.

The goat eats 1/90th (or 2/180) initial grass per day.

The duck eats 1/180th initial grass per day.

The grass grows at 1/180th initial grass per day.

 

So all three eating together plus the grass growing means a net decrease of 5/180 inititial grass per day, or 1/36th per day. They finish in 36 days.

 

Sorry to bring back an old thread to life.

 

I was trying to solve this and was fascinated by this solution.

 

Though I do not to understand one point.

 

Th First assumption that a cow would have eaten half the initial Grass in half the time it takes to finish the full pasture, does not seem to fit an equation with rates.

One obvious issue with that assumption is If the Cow can finish Half the pasture + growth from a full pasture in 45 days, It should be able to handle Half the pasture+ growth from this half pasture in less than 45 days.

Am I missing something here?

[derek w]

The rate at which the grass grows is a compound interest problem.

 

fg=pg(1 + i)^n

 

where:-

fg= future grass

pg = present grass

i = fixed interest rate/n

n = no. of days

 

If e = the amount of grass a cow eats in 1 day

and dog + goat is = to cow

 

then the grass is eaten at a rate of 2e by the cow dog goat combination

 

therefore fg=pg(1 + i - 2e)^n

Edited May 17, 2012 by derek w

[derek w]

2e is in wrong place should be fg=(pg(1+i)-2e)^n