What mathematics sometimes provides for us is a templet to think with. Essentially the unnecessary details are stripped away when we consider something mathematically, and what we see are the bare innards of the problem itself. This problem is one which exemplifies this concept, and brings to light just how basic math can make situations sometimes hilariously straightforward.
Far away in the lush green fields of Switzerland, lies a field of grass with some quantity of grass on it. Of course, there are cows, and 60 of them take 30 days to finish off all the grass. On the same exact field with the same grass, somewhere in another Swiss city, a set of 30 cows take 80 days to ravage the field of all its grass. How can this be? Well grass grows… We know the grass does grow at a uniform rate, and that the cows are all the same, and each consume the same amount of grass every day. Now how many days would it take for 20 cows to finish eating this same field of grass, and what’s the maximum number of cows could live on the field happily, sustaining themselves for an infinite amount of time (until of course they reproduce and overwhelm the supply).
I do implore you to try this question out yourself, because it really is quite simple. A kid with no mathematical experience past the fourth grade has the tools to solve this one, which makes it all the more interesting. The solution lies beneath this lovely picture of a healthy Swiss cow.
Notice that the question is as ambiguous as possible regarding the grass itself, and so to simplify things, let us assume that each cow eats one “unit” of grass every day. We know that there was initially the same number of grass units in each case, which we can take as Y. We also know that the rate of the grass growing in each case is the same, which we will take as X. What do we have now:
Situation 1: 60 cows finish the grass in 30 days
Situation 2: 30 cows finish the grass in 80 days
Therefore the total number of grass units eaten in each case was:
Situation 1: 60 x 30 = 1800 units totally eaten
Situation 2: 30 x 80 = 2400 units totally eaten
The total number of units eaten in each case is simply the units of grass which we had at the beginning, which was Y, which is the same for both, plus the extra grass which was grown in that time, which is simply the rate of growing x the time period. So now let’s put this into equations:
1800 = Y + 30X
2400 = Y + 80X
These are basic simultaneous equations, which we know how to solve easily. If you subtract the first equation from the second, you get:
50X = 600
X = 12 units per day
Therefore, upon substituting X back into one of the original equations:
Y = 1800 – 30 x 12 = 1440 units
Now we know the starting units of grass on the field is 1440, and that the grass grows at a rate of 12 units/day. To answer to the first question, we will just take the total number of units eaten as 20X, which makes sense, because it is the number of cows x the number of days. This equals the initial grass, 1440, plus the extra growth, which is just 12X:
20X = 1440 + 12X
X = 1440/8 = 180 days
We come to our answer, which is that 20 cows would take 180 days to finish off the field of grass. As for the number of cows which could sustainably live on this field: that would be the rate of growing, as each day the growing would replenish the needs of the cows, so 12 cows could live on this field forever. On and on they will live, killing off the oldest cow every time a new calf is born, in order to sustain their renewable structure of living. Maybe humans should learn a little something from this basic problem, and realize that we are really not going to be able to keep exploiting the Earth’s resources without inevitably disastrous consequences…