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I encountered this problem about Newton's problem of cows and fields:

In a field, 17 cows can finish the whole grass in the field for 30 days. 19 cows can finish in 24 days. If a group of cows eat the grass for 6 days, then 4 cows are sold, the remaining cows can finish the grass in 2 days. What is the initial number of cows before 4 of them are sold?

I could do this problem using algebraic method but when I consulted another source, a different approach is used and I could not understand that approach.

Here is my way:
First we find the amount of grass that grows each day: $(17\times30-19\times24)\div(30-24)=9$ (the amount of grass is equivalent to the amount of grass that 9 cows could eat in a day).

Then we find the initial amount of grass before any cow eats: $(17\times30-9\times30)=240$ (the amount of grass is equivalent to the amount of grass that 240 cows could eat in a day).

Then we let the initial number of cows before 4 of them are sold be $x$. Then $(x-9)\times6+(x-4-9)\times2=240$. So $x=40$.

Another approach that confuses me is $(240+8\times9+1\times2\times4)\div28=40$. The explanation is: Suppose the 4 cows are not sold, then we have $240+8\times9+2\times4=320$ amount of grass (where we use the same unit as above). Then we have $320\div(6+2)=40$ cows.

My question is: when the cows are not sold, then we won't need 2 more days to finish the grass, why should we add $2\times4$? Basically I don't understand the reason for adding up $240+8\times9+2\times4$ and then finally the sum is divided by 8?

Thanks very much.

user71346
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1 Answers1

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We are computing the total grass that the whole herd would eat if the four cows had not been sold. There was $240$ grass at the start, $8 \times 9$ is the grass that grew while the herd was eating, and $2 \times 4$ is the grass that the cows that were sold would have eaten in the last two days. This gives the grass that the whole herd would have eaten in $8$ days.

Ross Millikan
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  • Yes, I understand that. But what troubles me is: if the 4 cows are not sold, then the whole herd will probably finish the grass in less than 8 days. Then what is the point of adding $2\times4$? in those 2 days, the grass is eaten by cows less than the sold ones, if we add another 4 cows to eat the grass, then they will finish in less than 2 days, right? I hope my question is not confusing :) – user71346 Oct 12 '13 at 04:11
  • Maybe I should put it this way: If we assume the cows are not sold, then they will need less than 2 days to finish the grass, but why are we adding $2\times4$? We assume each cow eats a fixed amount of grass. – user71346 Oct 12 '13 at 04:25
  • It is two days of four cows not eating grass-the two days after they have been sold. Yes, the herd would have finished quicker than 2 days, but we want the total grass that would have been eaten by the whole herd in the 8 days. – Ross Millikan Oct 12 '13 at 04:34
  • hmm.. tbh, I'm still in doubt.. sorry. should the total grass that would have been eaten by the whole herd in the 8 days be the initial grass plus the grass that grows during the 8 days i.e. $240+8\times9$? the total grass in 8 days does not change regardless of how many cows eat it, right? – user71346 Oct 12 '13 at 04:46
  • That is correct. But we are imagining we didn't sell the cows and calculating how much grass the whole herd would eat in 8 days. So we need the actual grass plus imaginary grass for the four imaginary cows for two days. – Ross Millikan Oct 12 '13 at 04:58