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2 cows in 4 weeks can eat all the grass on 2 acres and all the grass that grows on these 2 acres during the 4 weeks. 3 cows in 2 weeks can eat all the grass on 2 acres and all the grass that grows on these 2 acres during the two weeks. How many cows can eat in 6 weeks all the grass on 6 acres and all the grass that grows on these 6 acres during the 6 weeks? Assume the grass has a constant growth rate and is distributed evenly.


I feel like I should use a 3 by 3 system to solve this problem but I only got to:

$x$= amount of grass 1 cow eats in a week

$y$= amount of grass that grows on an acre per week.

What should I do next?

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    $\begingroup$ You also need z = initial amount of grass per acre $\endgroup$ – WW1 Dec 13 '16 at 22:50
  • $\begingroup$ @WW1 That's what I'm missing! Thanks! :) $\endgroup$ – Yuna Kun Dec 13 '16 at 22:56
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You don't need a 3 by 3 system.

let a be the amount of grass that can grow in 1 acre in one week

Case 1:the amount of grass consumed by the cows is $$ 2+4\times 2\times a $$ Then the rate at which one cow can eat is simply $$ \frac{(2+4\times 2\times a)}{2\times 4} $$ Case 2: the amount of grass consumed by the cows is $$ 2+2\times 2\times a $$ And so the rate at which one cow can eat is $$ \frac{(2+2\times 2\times a)}{3\times 2} $$ Now we set the expressions from case 1 and case 2 equal: $$ \frac{(2+4\times 2\times a)}{2\times 4}=\frac{(2+2\times 2\times a)}{3\times 2} $$ We get a = 0.25 acres/week

So the rate at which one cow can eat is 0.5 acres/week (after substituting it back into one of the two expressions)

Case 3: the amount of grass consumed by the cows is $$ 6+6\times 6\times a=6+6\times 6\times0.25=15 $$ So the rate at which one cow can consume grass is $$ \frac{15}{6\times (number\;of\;cows)} $$ Since we know one cow can eat at a rate of 0.5 acres/week $$ \frac{15}{6\times (number\;of\;cows)}=0.5 $$ $$ number\;of\;cows=5 $$

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