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.

- anonymous

- schrodinger

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- experimentX

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.
??

- anonymous

yeah...

- anonymous

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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## More answers

- perl

what is initial?

- anonymous

the amount of grass that the field had initially

- experimentX

yeah i figured out that.

- perl

so you set them equal to each other because you assumed they ate all the grass

- anonymous

I didn't assume, they said that they "cleared the field"

- perl

errr, that was given

- perl

so what exactly are you equating , what is your left side and right side

- perl

total grass grown = total grass eaten (over the days )

- anonymous

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.

- perl

ok but thats 2 equations in 3 unknowns

- perl

the grass is growing while they are eating

- experimentX

grass growth must be some function of days

- anonymous

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

- perl

well you can subtract equation 1 from equation 2, that eliminates x the initial

- anonymous

by eliminating the initial i will find the relation between m and n

- perl

how did you get grass growth rate is 90 and 30 ?

- perl

nevermind

- perl

that was days

- anonymous

Yeah sritama I was actually just about to correct my equations to add the days on both sides.

- anonymous

yeah,i took the grass growth rate n/each day ... so it must b 90n and 30n

- anonymous

oh ok gingerkid 101

- perl

ginger can you redo your equation

- perl

i like your equation, its very logical :)

- anonymous

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

- perl

yeah thats a little odd, grass per cow per day

- perl

how does that come out in units, grass/ (cow/day) or

- anonymous

probably it means the amount of grass taken by each cow per day

- anonymous

grass/(cow * day) or grass/cow/day
it's all the same.

- anonymous

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.

- anonymous

anyway the equations fully played out are...
x + 30y = 3600j
x + 90y = 4320j
that's what I'm at right now.

- perl

ok now its linear algebra, one moment

- perl

x + 30y - 3600j=0
x + 90y- 4320j=0

- perl

I get x = 3240 j
y = 12j
j = j

- perl

actually we know we want 16 , so we have another equation

- perl

x + 30y = 3600j
x + 90y = 4320j
x + 16 y = 16 i * j

- anonymous

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

- perl

how did you get 8 additional cows?

- anonymous

yeah,thats my question too

- anonymous

8 additional to 1/3 of 120. maybe I'm just talking out of my retricenow haha

- anonymous

I don't even know. meh, I never had to do anything like this and I took up to calc 3 haha

- perl

well this is a solid start

- perl

i got 12j = y , so 12 cows it takes to eat the grass grown in 1 day

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