# Newton's puzzle

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Hey people, how are you doing?

I heard this couple days ago and tried to figure it out myself but I seem not to be able to:(

It goes like this:

|=|A cow and a goat eat all the grass of a pasture in 45 days. The (same) cow and a duck eat the grass in 60 days, and the duck and the goat east the grass in 90 days. The cow eats the same amount of grass as the duck and the goat together. But you have to take into consideration that new grass grows continuously.|=|

At first I got 45 days, then I got 41.5... days but I think neither of them is the answer.

Any suggestion?

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The real question is: "What color is the cow?"

I might have missed it, but what's the question?

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OMFG, sorry!

The question is "How many days does it take the three of them together to eat all the grass."

30 days?

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30 days?

Could you explain how you got your answer (I don't know if its correct)?

Here's how I got my 41.5 days

The cow can finish it alone in 90 days. Since it take the goat twice as more time to finish when it is with the duck, compared to the cow, then (I assume) it would take the duck 180 days to finish to finish the entire thing alone. And similarly I got 135 days for the goat (though I'm not so sure on this one).

And if you try to get the number of days that would take for the three of them together, then I tried

x/90 + x/180 + x/135 = 1

than I got that x=41.5....

But there's another way to look this.

If the cow finishes alone in 90 days, then if would finish half in 45 days. And since it takes the duck and the goat 90 days to finish it together, the three of them would finish it in 45 days:(

But the problem (I think) is that none of these takes into consideration the new grass growing:-(

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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.

Edited by Sisyphus
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I believe that's it:-)

Good work Sisyphus

Consecutive posts merged.
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• 2 weeks later...

lets say x = amount of pasture a cow can eat in 15 days

y = amount of pasture a duck can eat in 15 days

z = amount of pasture a goat can eat in 15 days

3x + 3z = 1

4x + 4y = 1

6y + 6z = 1

x = (1-3z)/3

...

(4-12z)/3 + 4y = 1

4/3 - 4z + 4y = 1

4y = 4z -1/3

...

6(4z-1/3) + 6z = 1

30z - 2 = 1

30z = 3

z = 10

i'm sorry i messed up somewhere

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• 1 month later...

c + g = 45

c + d = 60

d + g = 90

-------------

therefore...

d = 90- g

c = 45 - g

so, if c + d = 60...

(45 -g) + (90 - g) = 60 solve for g

g = 37.5

-----

c + g = 45

c + 37.5 = 45

c = 7.5

-----

d + 37.5 = 90

d = 52.5

-----

c= 7.5 days

d=52.5

g= 37.5

the values for c,d, and g are as stated....if they cow, duck or goat were eating by themselves... i shall try to work on what time it would take if they gave a group effort...i have to think about it more...probably using percentages?

Edited by sjbouscher
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|=|A cow and a goat eat all the grass of a pasture in 45 days. The (same) cow and a duck eat the grass in 60 days, and the duck and the goat east the grass in 90 days. The cow eats the same amount of grass as the duck and the goat together. But you have to take into consideration that new grass grows continuously.|=|

(Ci+Cng)(Gi+Gng)=45

(Ci+Cng)(Di+Dng)=60

(Di+Dng)(Gi+Gng)=90

(Ci+Cng)=(Di+Dng)(Gi+Gng)

lol...They can never finish the grass.

Edited by Ranjha
e
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The main issue with the last few posters trying to solve it via algebra is that at its heart, this is a calculus problem. I.e. everything is rates, not just algebraic relations.

Let $g(t)$ = amount of grass at time t, and $g_0$ is the initial amount of grass (a constant).

Let $\dot{C}$ = amount of grass the cow consumes per unit time

Let $\dot{G}$ = amount of grass the goat consumes per unit time

Let $\dot{D}$ = amount of grass the duck consumes per unit time

Let $\dot{r}$ = amount of grass that grows per unit time

The equations of the statements given in the problem are:

$\frac{dg(t)}{dt} = -\dot{C} - \dot{G} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 45 days, g=0

The other two are similar:

$\frac{dg(t)}{dt} = -\dot{C} - \dot{D} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 60 days, g=0

$\frac{dg(t)}{dt} = -\dot{G} - \dot{D} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 90 days, g=0

Finally we are given:

$\dot{C} = \dot{G} + \dot{D}$

If you assume all the rates are constant -- and that isn't completely straightforward b/c while I can buy that the food consumption is probably a constant, I suspect that grass growth rate isn't. If all the grass is dead (g=0) then no new grass can grow. That would mean that growth rate would be a function of both time and amount of grass ($\dot{r} = \dot{r}(g,t)$). But, if you do assume that they are all constant, you actually will end up with 6 equations (though 3 of them are trivial), 2 from each of the boundary conditions of the differential equations. Combine that with the rate relationship that the cow consumes at the same rate as the goat and duck combined and you have enough equations to solve for the 4 unknowns.

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ranjha...look again...the cow can eat it all by itself in 7.5 days. the grass won't grow back in time...and even it it could, the constant trampling by the animal would turn the grassfield into a dirt field...

consider my previous answers for the time it takes each animal to do it by itself:

cow= 7.5 days

duck= 52.5 days

goat= 37.5 days

---------------

the cow can eat the grass 7x faster than the duck:

52.5 / 7.5 = 7

the cow can eat 5x faster than than the goat:

37.5 / 7.5 = 5

the duck's time and the goat's time is subtracted from the cow's time since they are helping the cow out:

cow's time - duck's time -goat's time

cow - 1/5 cow - 1/7 cow

7.5 - (1/5)7.5 - (1/7)7.5 = total time

7.5 - 1.5 - 1.1 = total time

4.9 days total

Edited by sjbouscher
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ranjha...look again...the cow can eat it all by itself in 7.5 days.

No, this cannot be right at all. How can the cow eat all the grass in 7.5 days when it takes both the cow AND the goat to finish in 45 days? The goat is making more grass? I don't think so. Any animal by itself MUST take a longer amount of time to finish than when any two of them team up.

Again, the problem is not algebraic, it is calculus based as its heart. The numbers are rates, you can't just add them directly up, you have to multiply the rate by the period of time to get total amounts.

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ranjha...look again...the cow can eat it all by itself in 7.5 days.

Just to add to what Bignose has already said, look at the problem statement. The answer to the question of how long it will take the cow by itself to eat the grass is right there. "The duck and the goat eat the grass in 90 days. The cow eats the same amount of grass as the duck and the goat together." In other words, the cow will eat the grass in 90 days, not 7.5.

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Bignose and DH, please refer to my post on september 10th...the math confirming each animal's time to do it by themselves is all there. those numbers are indeed correct. please take time to read it over.

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Your numbers are incorrect, sjbouscher. Read Bignose's post. Read the correct answer in post #6. Read the problem. It says right up front that the cow eats the same amount as the duck and goat combined and that the duck and goat eat all the grass in 90 days.

You set the problem up incorrect in post #10 and arrived at a nonsense answer. What is the right way to set up the problem?

The underlying assumption is that the grass grows at some unspecified linear rate, call this rate r. If the initial height of the grass is h, the height h_0 (the subscript 0 denotes no critters are eating the grass) at time t is $h_0(t)=h+rt$. That is assuming no critter is munching on it, of course.

This is not a particularly realistic assumption (grass stops growing at some height), but it becomes a lot closer to reality when critters keep the grass trimmed. Suppose some critters are munching on the grass -- the cow and goat, for instance. These critters are going to eat fixed amounts of grass per day. Denote the amount eaten by the cow as c and the goat as g. Now the height as a function of time is $h_{c,g}(t) = h+(r-(c+g))t$. The cow and goat eat all the grass in 45 days. In other words, the height is zero at t=45. In math, $h+45(r-(c+g)) = 0$ or $45(c+g-r)=h$. Continuing in this regard,

$45(c+g-r)=h$

$60(c+d-r)=h$

$90(d+g-r)=h$

$c=d+g$

That final equation reflects the given information that the cow eats as much as the duck and goat combined.

One thing that makes this problem interesting is that the problem is uniquely solvable even though there are only four equations and five unknowns (c, d, g, h, and r).

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Bignose and DH, please refer to my post on september 10th...the math confirming each animal's time to do it by themselves is all there. those numbers are indeed correct. please take time to read it over.

That math is incorrect. Or rather, you're using the wrong math. Think about what those numbers represent, and if the equations and the eventual answer make sense. Think of a simpler example: you and I work at the same rate, and together we can accomplish a given task in 2 days. The way you have it set up (you + me = 2), either of us could finish that task in 1 day. Obviously that doesn't make sense. It would take one of us twice as long to finish as both of us, not half as long.

Check out my first post in this thread (#6), where I explain the answer in words.

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Your numbers are incorrect, sjbouscher. Read Bignose's post. Read the correct answer in post #6. Read the problem. It says right up front that the cow eats the same amount as the duck and goat combined and that the duck and goat eat all the grass in 90 days.

You set the problem up incorrect in post #10 and arrived at a nonsense answer. What is the right way to set up the problem?

The underlying assumption is that the grass grows at some unspecified linear rate, call this rate r. If the initial height of the grass is h, the height h_0 (the subscript 0 denotes no critters are eating the grass) at time t is $h_0(t)=h+rt$. That is assuming no critter is munching on it, of course.

This is not a particularly realistic assumption (grass stops growing at some height), but it becomes a lot closer to reality when critters keep the grass trimmed. Suppose some critters are munching on the grass -- the cow and goat, for instance. These critters are going to eat fixed amounts of grass per day. Denote the amount eaten by the cow as c and the goat as g. Now the height as a function of time is $h_{c,g}(t) = h+(r-(c+g))t$. The cow and goat eat all the grass in 45 days. In other words, the height is zero at t=45. In math, $h+45(r-(c+g)) = 0$ or $45(c+g-r)=h$. Continuing in this regard,

$45(c+g-r)=h$

$60(c+d-r)=h$

$90(d+g-r)=h$

$c=d+g$

That final equation reflects the given information that the cow eats as much as the duck and goat combined.

One thing that makes this problem interesting is that the problem is uniquely solvable even though there are only four equations and five unknowns (c, d, g, h, and r).

That would be because it is effectively a problem in 4 unknowns:

h/c, h/g, h/d, and h/r.

=Uncool-

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• 2 months later...

Ok, so what was the answer? BTW read my signature line for those trying to make sense of what I wrote.

Post #6.

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• 4 weeks later...

I may be reading to much into this question, but wouldn't the rate of growth change? And I'm not talking about how fast an individual blade of grass's growth rate changes, but rather the percent of growth of the field.

You have the entire field that the cow and goat are starting to eat. We know it will take 45 days. So if the growth rate is X%, well X% of the % of grass left at the end of the first day is very different than the X% of the % of the grass left at the end day 44.

Yes, the grass grows at a constant rate, but the amount of increase of total grass changes day to day.

Sisyphus states that the cow can eat half the grass + half of the total growth in 45 days. Well, in the next 45 days the cow has to eat the other half of the grass + the initial growth + the new growth over the next 45 days. That means the cow cannot eat the rest of the grass in the next 45 days.

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• 2 months later...

Hmmm...

The part where the cow tramples over the grass and kills it is not necessary to take into consideration. What is important though is the fact that the grass grows continuously at a constant rate. The word "Continuously" means it grows immediately. So, it is impossible to finish up all the grass. But the question states that the animals could finish it up. So we need to find out the rate at which each animal eats.

c+G take 45 days. So they finish up 1/45 per day. If the grass does not grow, it will eat together at a rate of 1/45/day+g/day. c+d eat at a rate of 1/60+g/day and G+d eat at a rate of 1/90+g/day. The cow's rate is 1/90+g/day. So G's rate is 1/90/day. so d's rate is g/day. That means c's rate is 1/60/day. so g= 1/180/day of the field.so d's rate is 1/180/day . In total, their rate to finish it up would be, 1/36/day. So they will take 36 days to finish up the job of eating it.

Merged post follows:

Consecutive posts merged
The main issue with the last few posters trying to solve it via algebra is that at its heart, this is a calculus problem. I.e. everything is rates, not just algebraic relations.

Let $g(t)$ = amount of grass at time t, and $g_0$ is the initial amount of grass (a constant).

Let $\dot{C}$ = amount of grass the cow consumes per unit time

Let $\dot{G}$ = amount of grass the goat consumes per unit time

Let $\dot{D}$ = amount of grass the duck consumes per unit time

Let $\dot{r}$ = amount of grass that grows per unit time

The equations of the statements given in the problem are:

$\frac{dg(t)}{dt} = -\dot{C} - \dot{G} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 45 days, g=0

The other two are similar:

$\frac{dg(t)}{dt} = -\dot{C} - \dot{D} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 60 days, g=0

$\frac{dg(t)}{dt} = -\dot{G} - \dot{D} + \dot{r}$

with boundary conditions that

@ t = 0 days, g=$g_0$

@ t = 90 days, g=0

Finally we are given:

$\dot{C} = \dot{G} + \dot{D}$

If you assume all the rates are constant -- and that isn't completely straightforward b/c while I can buy that the food consumption is probably a constant, I suspect that grass growth rate isn't. If all the grass is dead (g=0) then no new grass can grow. That would mean that growth rate would be a function of both time and amount of grass ($\dot{r} = \dot{r}(g,t)$). But, if you do assume that they are all constant, you actually will end up with 6 equations (though 3 of them are trivial), 2 from each of the boundary conditions of the differential equations. Combine that with the rate relationship that the cow consumes at the same rate as the goat and duck combined and you have enough equations to solve for the 4 unknowns.

You do not need to go that far. Take a look at my method; it is pretty simple to convey. You don't need calculus; just pure velocity rates (and some fairly simple simultaneous equations.)

Merged post follows:

Consecutive posts merged

The key to it is to convert the rates into fractions to make them easier to compare as they have the same base. There is no need to use percentages of the field. Besides; this question ought to be in a Maths forum don't you think?

Merged post follows:

Consecutive posts merged
lets say x = amount of pasture a cow can eat in 15 days

y = amount of pasture a duck can eat in 15 days

z = amount of pasture a goat can eat in 15 days

3x + 3z = 1

4x + 4y = 1

6y + 6z = 1

x = (1-3z)/3

...

(4-12z)/3 + 4y = 1

4/3 - 4z + 4y = 1

4y = 4z -1/3

...

6(4z-1/3) + 6z = 1

30z - 2 = 1

30z = 3

z = 10

i'm sorry i messed up somewhere

You forgot the fact that the grass continues to grow forever.

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• 1 month later...

Gotta' to run this by you to check your humor!

A hog ate a peck of beans before it started

Ran a mile befor it f--ted

Give me the total distance before it s--t?

Edited by rigney
Consecutive posts merged.
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• 1 month later...
That math is incorrect. Or rather, you're using the wrong math. Think about what those numbers represent, and if the equations and the eventual answer make sense. Think of a simpler example: you and I work at the same rate, and together we can accomplish a given task in 2 days. The way you have it set up (you + me = 2), either of us could finish that task in 1 day. Obviously that doesn't make sense. It would take one of us twice as long to finish as both of us, not half as long.

Check out my first post in this thread (#6), where I explain the answer in words.

I think in this case, the c, g, etc. should represent the amount of grass the cow, goat, etc. eat per day, so its a rate, rather than a simple quantity...

so you can set up a matrix of the original equations as (sorry it looks horrible, I did try in this forum's latex thing but it didn't work...)

[ 1 0 1 -1 ] [c] = [1/45]

[ 1 1 0 -1 ] [d] = [1/60]

[ 0 1 1 -1 ] [g] = [1/90]

[ 1 -1 -1 0 ] [r] = [ 0 ]

And you can invert it to get

[ 1 1 -2 -1 ] [1/45] = [c]

[ 0 1 -1 -1 ] [1/60] = [d]

[ 1 0 -1 -1 ] [1/90] = [g]

[ 1 1 -3 -2 ] [ 0 ] = [r]

Which gives

[c] = 1/60

[d] = 1/180

[g] = 1/90

[r] = 1/180

which is exactly the same as what Sisyphus got using logic. (Since it's pretty much the same argument just in algebraic form...)

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It said the grass grows continuously there for there is always going to be grass

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