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There's a hall with 100 seats. I want to fill up these seats with men, women and children; they're going to purchase seat positions. The cost per seat for men is 5 USD, women 1 USD, and children 0.05 USD. I want to get 100 USD from selling all 100 seats. How many men, women and children would be needed to fill up the 100 seats and land me 100 USD? Condition: there has to be at least one of each (man, woman and child).

I've formulated 2 equations ($m$, $w$ and $c$ denote the number of men, women, and children respectively):

$$5m + w + 0.05c = 100$$ $$m + w + c = 100$$

The first one is for the price (price/person * number of persons) giving me the 100 USD. The second equation is just the total of the persons which is obviously 100, because there are 100 seats that I need to fill. That's two equations and three variables! What do I do?!

Now, I actually know the answer to this, but I got this answer by just plugging in random numbers and guessing until I got it. I'm not too into math and stuff, but shouldn't a problem like this be possible to be solved with equations to get a definite answer, or is there some...thing...that makes this problem only solvable through iterations? Maybe there's more than one answer and the actual mathematical answer is some sort of "range", or another equation.

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  • $\begingroup$ Can you sell tickets to $2.5$ children? Can you sell tickets to $-3$ women? That should give you some more constraints. $\endgroup$ – JimmyK4542 Jul 25 '14 at 1:25
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    $\begingroup$ A variant of a $2000$ year old classic that originated in China, the Hundred Fowls Problem. $\endgroup$ – André Nicolas Jul 25 '14 at 1:45
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If there is a solution, you find it by solving the set of equations you have: subtract the second from the first, giving $4m-\frac{19}{20}c = 0$, so that $m = \frac{19}{80}c$. Substitute this for $m$ in the second equation, giving $\frac{19}{80}c + w + c = 100$, or $w = 100 - \frac{99}{80}c$. But all numbers must be integers, so set $c=80$, giving $m=19$, $c=80$, $w = 1$. Note that it could have turned out that under different conditions, for example, $w<0$, so that there were no solutions in positive integers.

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It is clear that $c$ must be a positive integral multiple of $20$, and the condition that $m,n,c \ge 1$ and $m+w+c=100$ implies that it is sufficient to iterate over four values of $c$ ($20,40,60,80$). We are then solving a classical system of linear equations in $m$ and $n$ with the condition that $m$ and $n$ are positive integers.

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