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Tue 26 Sep, 2006 07:28 pm
Help!!!
Jody paid for her $44 bill at the hardware store using a combination of $10, $5, and $1 bills. If she paid with 13 bills in all, how many of each bill did she use?
Well you've got there two basic equations...
10x + 5y + z = 44
x+y+z = 13
It can't be directly solved without a 3rd equation.
So let's look at the minimal number of bills solution,
x=4, y=0, z = 4
That's only 8 bills. While keeping z=4, the most bills we can possible use is:
x=0, y=8, z=4
Which is only 12 bills. Therefore z needs to be > 4. Since z can only be increased in increments of 5 it is a safe bet that z=9.
This leaves a remainder of 35.
x=3, y=1 makes 35.
9+3+1=13. Hooray, the problem is solved.
Now maybe you're teacher will find this and give you an F for cheating...haha