Wait.
If 70/30 is the same ratio 24/60, then it doesn’t matter which pairing we use to solve for x, right?
I came up with 18.75 cows.
Busy weekend, sorry I didn’t get back to this eariler.
18.75 isn’t the right answer.
Here is a hint. The amount of grass eaten = starting amount + amount grown.
But if the initial height is the same and the grass grows at the same rate, that’s a constant and we don’t need to figure that in, right?
I think you still have to factor in the growth rate.
For instance, if you had a billion cows, the effect of the growth rate would be negligible. They’d eat all the grass before it had a chance to grow even a millimeter.
If you had only one cow, it’s possible that the growth rate would outpace the cow’s consumption rate, and the grass would never be wholly eaten.
So, I think the way to solve this problem is to use the two data points given to figure out the growth rate of the grass, and then factor that in when figuring out the solution.
If we assume that a cow eats a unit of grass per day we can write:
N t = I + G t
where N is the number of cows, t is the time in days, I is the initial units of grass available, and G is the number of grass units that grow per day. This assumes that the grass continues to grow along at the same rate until poof it is all consumed.
Apply that equation for each scenario and solve the system of equations to get I=1600 and G=3.333…
Apply it once again to find the number of cows is
20
That’s the correct number of cows.
How I did it, with the equations I used.
x = grass consumption rate in amount/day
y = grass growth rate in amount/day
z = starting amount of grass
C = number of cows for 96 days
amount of grass eaten = starting amount + amount grown
70 (24x) = z + 24y
z = 1680x - 24y
30 (60x) = z + 60y
z = 1800x - 60y
1680x - 24y = 1800x - 60y
(10/3)x = y
C (96x) = z + 96y
C (96x) = (1800x-60y) + 96y
C (96x) = (1800x-60((10/3)x)) + 96((10/3)x))
factor out the x
C = (1800 - 60(10/3) + 96(10/3)) / 96
20 cows