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Probability & Statistics/curious solution to a maths' problem

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Hi Clyde,

i have a mathematics textbook called  "Mathematical Quickies 270 stimulating problems with solutions" and the question # 105 is called the "Farmer's Dilemma". it reads thus:

A farmer must buy 100 head of animals with $100. Calves cost $10 each, lambs cost $3 each, and pigs cost 50cents each. IF the farmer buys at least one of each kind of animal, how many of each kind does he buy?

SOLUTION GIVEN IN THE TEXT BOOOK:

The average cost per animal is $1. Each calf's cost differs from the average by +$9, each lamb's cost by +$2 and each pig's cost by -$1/2(half a dollar). Hence for each calf he must have 18 pigs and for each lamb he must have 4 pigs. Therefore, since 5(1+18) + (1+4) = 100, he must buy 5 calves, 1 lamb and 94 pigs.

My question is, how did the solver come about the 5(1+18) + (1+4) = 100?

Thanks

Answer
The solution makes use of the idea of the average. Because you have 100 animals and 100 dollars, the average animal in the purchase must be exactly 1 dollar.

Because calves cost $9 over the average, if you bought a calf and didn't make up for it in some way, you'd be over budget. In order to balance out the $9 overage, you need 18 pigs.

So if you bought calves, you would have to buy 18 pigs to compensate, meaning it's a sort of "package" of 19 animals that is $1 per animal (which is the target).

The same package can be made from 1 lamb and 4 pigs, which is a package of five animals.


Finally, we have to decide how to split up our 100 animal total into packs of 5 or 19.

If you use the number 19 only once, that leaves 81 left over (not divisible by 5 -- so this doesn't work). Likewise if you use 19 twice, three times, or four times.

You can't use 19 zero times (you must buy one calf) and you can't use it more than five times (that puts you over 100), so you have to use it five times, leaving 100 - 195 = 5 left over. That's exactly enough for one pack of sheep/pigs.

That's how they arrived at the answer:

-- five "packs" of calf/pigs (1 calf, 18 pigs each)
-- one pack of sheep/pigs (1 sheep, 4 pigs each)

That gives you a total of 100 animals, $100, 5 calves, 1 sheep, 94 pigs.

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Clyde Oliver

Expertise

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

Experience

I am a PhD educated mathematician working in research at a major university.

Organizations
AMS

Publications
Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.

Education/Credentials
BA mathematics & physics, PhD mathematics from a top 20 US school.

Awards and Honors
Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

Past/Present Clients
In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.

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