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Posted - 01/31/2012 : 21:15:49 For some unknown reason, we want to spend exactly $100 on 100 animals at the Animal Rescue Society. Dogs cost $15 each, cats are $1, and mice are four for a dollar. If we have to buy at least one of each kind of animal, how many of each can we purchase? Enter the number of cats first, then dogs, and finally mice all separated by commas.
3 L A T E S T R E P L I E S (Newest First)
Posted - 02/04/2012 : 16:21:50
quote:Originally posted by the_hill1962
Now, the reason that I stated all the above is because you only have 2 conditions: I know this is a round about way of doing the problem but since you only have 2 equations...
There is a third "unspoken" condition - which is d, c & m must be integers. That condition makes the solution set finite.
These are also called diophantine equations.
Posted - 02/01/2012 : 13:30:10 Thanks, I got to the 2 equatipns but got gung up on solving that system. Didnt think about your way
Posted - 02/01/2012 : 13:17:58 Are you familiar with needing a system of 3 independent equations when you have 3 unknowns? That is the general rule. Here is an example: Suppose you have 3 numbers that add up to 10. This would make just one equation x + y + z = 10. There are an infinite number of solutions (such as 2, 4, 4 and 1, 2, 7 and 3, 6, 1 etc...) so you need more "conditions" (or equations). Let's add the condition that the second number is twice the first. As you see, there are still an infinite number of solutions (such as 2, 4, 4 and 1, 2, 7 and 3, 6, 1 etc...). A third condition narrows it down to just one solution. A third condition might have been something like "and the third number is equal to the second number" The solution has to be 2,2,4. Now, the reason that I stated all the above is because you only have 2 conditions: 1) 100 pets is like saying d + c + m =100 2) Cost is $100 is like saying 15d + 1c + .25m = 100 You have no more information. The fact that dogs cost $15 each, cats $1 and mice $0.25 is not information that can be written in an equation that adds to the system. So, you need to use a technique of solving literally for a variable. In doing this problem, I happened to choose using d+c+m=100 to solve for c to get c=100-m-d. I then substituted that into 15d + 1c + 0.25m = 100 to get the equation 15d + 1(100-m-d) + 0.25m = 100 and that is 15d + 100 - m - d + 0.25m = 100 and that is 14d=0.75m which is 14d = 3/4(m). This means that m=56/3(d). So, picking out a number of dogs to be a multiple of 3 works. Say you have 3 dogs. That would be 56/3(s)=56 mice. That solution will work. I leave it to you to figure out how many cats there are to make $100. I know this is a round about way of doing the problem but since you only have 2 equations...