I presented the problem for Part 1 here. The answer is as follows.
The merchant would need to buy 1 boar, 9 sows, and 90 piglets. As there are three variables and only two equations, trial and error must be employed. My solution is as follows.
X=Number of Boars
Y=Number of Sows
Z=Number of Piglets
[Eq 1] X+Y+Z = 100
[Eq 2] 10X+5Y+0.5Z = 100
Solving Equation 1 for Z and substituting into Equation 2 and simplifying you get the following equation.
[Eq3] 1.9X + 0.9Y = 10
This is where trial and error comes in and you get X=1, Y=9. Back substituting into Equation 1, you get that Z=90.
Hope you had fun!
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