Hungry Oxen
In 4 weeks, 12 oxen consume 3 1/3 acres of pasture land. In 9 weeks, 21 oxen consume 10 acres of pasture land. Assuming the grass grows at a uniform rate, how many oxen are required to consume 24 acres in a period of 18 weeks?
posted by Zaux at
9:35 PM
11 Comments:
The figures don't look right to me.
In 4 weeks, 12 oxen consume 3 1/3 acres. So, in the same 4 weeks, 36 oxen would consume 10 acres.
Yet the second statement says that in 9 weeks (more than twice as long) only 21 oxen are needed to consume 10 acres.
This implies that in the extra 5 weeks of grass growing, 15 fewer oxen can be supported!
This doesn't seem right.
Also, wouldn't you expect that there would be some grass in a pasture before you put a herd into it? I know this is a mathematical puzzle, totally unconnected to the real world, but, if there is no grass in a pasture, and the herd eats faster than the grass grows, you're going to end up with a dead herd pretty soon!
My point is that if there is grass in the pasture to start with, then the question arises: how much grass? This obviously has some bearing on how long it would take a herd to eat all the grass in a pasture, i.e. the initial amount plus what grows in the time allowed.
36 oxen required.
Bit fiddly to type up now. But is only fiddling about with the equation: NCW = A(S+GW), where N = number of oxen, C = rate at which one ox consumes/week, A is area, S is the initial amount of grass/acre, G is the rate at which grass grows/acre and W is the number of weeks.
Using the initial information, can e.g. find C and S in terms of G.
... it turns out that C = (10/9)G and S =12G. The rest is trivial.
Hi Chris ...36 is correct.
Well done, Chris, that's exactly right.
I must be spending too much time in the sun here in Oz. In my objection to the question above I overlooked the obvious fact that, of course, fewer oxen would take longer to eat the grass. I had this the wrong way round.
Just to be a little bit pedantic, the question assumes the same initial amount of grass per acre in each case without any mention at all of initial amounts. Maybe that was intentional, just to confuse slow thinkers like myself.
Hi Wiz. LOL. Take a look back to the 16th October 2009, feeding Frenzy". A very similar problem. Someone calling himself "The Wizard of Oz" solved that ;)
Zauz. Thank you for all your hard work. You've posted some very good problems with a nice range of difficulty. I haven't seen most of them before.
Zaux, sorry about the spelling - a typo, not profound ignorance.
Hi Wiz. I've just re-read your comment. I've always assumed a code with most of these questions, it boils down to only making simplifying assumptions (where possible).
... I also assume that the poster has taken great care when setting a problem and that not providing information is intentional.
I'll admit that I often initially think that a question has been set badly, but usually it turns out not to be the case.
At least Ragknot did't have a pop at us for overgrazing.
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