Question 67717: Here is the problem. A parts person has to spend exactly $100 and must buy 100 parts. Good parts cost 5$, regular parts cost 1$, and cheap parts cost 25 cents. He has to buy 100 parts for $100 and have some of each part. I tried the trial and error method and got an answer, but wanted to get an anwer algebraically, (sp?)
I tried: 5x + y + .25z = $100
x + y + z = 100 and substituting variables but could not get it to work. I appreciate your help.
George
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A parts person has to spend exactly $100 and must buy 100 parts. Good parts cost 5$, regular parts cost 1$, and cheap parts cost 25 cents. He has to buy 100 parts for $100 and have some of each part.
:
Let the three types of parts = x, y, z
Three unknowns, two equations calls for some assumptions
:
We know that the $5 part (x) has to be 4 or less,
lets assume x = 3 good parts at $5 ea
:
3 + y + z = 100
y + z = 97
and:
3(5) + y + .25z = 100
15 + y + .25z = 100
y + .25z = 85
:
See if these two equations will give an integer answer for z
y + z = 97
y + .25z = 85
-------------- subtract
.75z = 12
z = 12/.75
z = 16 ea cheap parts
Find y from the equation y + z = 97
y + 16 = 97
y = 97 - 16
y = 81 ea regular parts
:
we have 3 good, 81 regular and 16 cheap parts
:
Check in the cost equation:
3(5) + 81 + .25(16) =
15 + 81 + 4 = $100
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