 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
A man buys sheep, goats, and hogs, to the number of 100, for 100 crowns; the sheep cost him 1/2 a crown a-piece; the goats cost him 1-1/3 crowns a-piece; and the hogs cost him 3-1/2 crowns a-piece. |
La,
You have two linear equations here, and three variables so you cannot solve fully over the rationals. But you can solve down to a line in three-dimensional (sheep,goat,hog) space (most easily given in parametric form, say (a-bh, c-dh, h)). Any point on this line is a solution - if you allow the sale of fractional and negative animals!
If on the other hand we assume actual animals were sold, not bought, and even that he got at least one of each, you are looking for a point with positive integer coordinates on that line. If we had (these are not the values from your problem!)
g = 95 - (17/3)h s = 155 - (19/2)h
we would know that h was at least 1 (hogs positive) and at most 16 (sheep positive). The goat-positive condition is implied by the sheep-positive condition.
Finally, if (17/3) h and (19/2) h are integers, h is divisible by 6. So h=6 and h=12 give valid answers.
Good Hunting!
-RD
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.