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Suppose you have a grassy field, and cows eat grass at a constant rate. Keep in mind, the grass keeps growing continuously. 48 cows can clear all the grass off the field in 90 days. 120 cows can clear all the grass off the field in 30 days. How many cows would be needed to clear all of the grass in 16 days? Round up to the nearest whole cow.

Mathematics
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48 cows can clear all the grass off the field in 90 days. 120 cows can clear all the grass off the field in 30 days. ??
yeah...
no contradiction. the grass keeps growing. \[initial + (grassgrowthrate * days) = (numcows * eatingrate) \] for simplification x + y * z = i * j in the context of this problem x + 30y = 120j x + 90y = 48j

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Other answers:

what is initial?
the amount of grass that the field had initially
yeah i figured out that.
so you set them equal to each other because you assumed they ate all the grass
I didn't assume, they said that they "cleared the field"
errr, that was given
so what exactly are you equating , what is your left side and right side
total grass grown = total grass eaten (over the days )
the initial grass plus the growth of the grass is equal to the grass that the cows ate in other words, when you subtract the total grass minus the grass eaten, it should equal zero, thus my equation.
ok but thats 2 equations in 3 unknowns
the grass is growing while they are eating
grass growth must be some function of days
i started this way, 1 cow eats an ammount of grass each day is m the grass grows by the amount n each day and if the initial is a then we get a+90n=48*90m a+30n=120*30 m
well you can subtract equation 1 from equation 2, that eliminates x the initial
by eliminating the initial i will find the relation between m and n
how did you get grass growth rate is 90 and 30 ?
nevermind
that was days
Yeah sritama I was actually just about to correct my equations to add the days on both sides.
yeah,i took the grass growth rate n/each day ... so it must b 90n and 30n
oh ok gingerkid 101
ginger can you redo your equation
i like your equation, its very logical :)
x + 30y = 120 * 30 * j x + 90y = 48 * 90 * j where x is in grass, y is in grass per day j is in grass per cow per day
yeah thats a little odd, grass per cow per day
how does that come out in units, grass/ (cow/day) or
probably it means the amount of grass taken by each cow per day
grass/(cow * day) or grass/cow/day it's all the same.
in other words, if you have a certain number of cows, you can multiply it by that number to get 30 cows * 1 grass/cowday = 30 grass/day is resulting from having 30 cows.
anyway the equations fully played out are... x + 30y = 3600j x + 90y = 4320j that's what I'm at right now.
ok now its linear algebra, one moment
x + 30y - 3600j=0 x + 90y- 4320j=0
I get x = 3240 j y = 12j j = j
actually we know we want 16 , so we have another equation
x + 30y = 3600j x + 90y = 4320j x + 16 y = 16 i * j
However, we also know that 8 additional cows are needed to balance the original proportion, which can possibly give us a 3rd equation? 8j - 60y = 0 or else the proportion would be even. therefore, we now have a 3 part system 8j - 60y = 0 x + 30y - 3600j=0 x + 90y- 4320j=0
how did you get 8 additional cows?
yeah,thats my question too
8 additional to 1/3 of 120. maybe I'm just talking out of my retricenow haha
I don't even know. meh, I never had to do anything like this and I took up to calc 3 haha
well this is a solid start
i got 12j = y , so 12 cows it takes to eat the grass grown in 1 day

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