SOLUTION: How much of an alloy that is 10% copper should be mixed with 600 ounces of an alloy that is 90% copper in order to get an alloy that is 40% copper?
Question 688891: How much of an alloy that is 10% copper should be mixed with 600 ounces of an alloy that is 90% copper in order to get an alloy that is 40% copper? Found 2 solutions by checkley79, solver91311:Answer by checkley79(3341) (Show Source):
You can put this solution on YOUR website! .10X+.90*600=.40(X+600)
.10X+540=.40X+240
.10X-.40X=240-540
-.30X=-300
X=-300/-.30
X=1000 OZ. OF 10% COPPER IS USED.
PROOF:
.10*1000+540=.40(1000+600)
100+540=.40*1600
640=640
600 ounces of 90% copper is 540 ounces of pure copper. ounces of 10% copper is ounces of pure copper. ounces of 40% copper is ounces of pure copper, so:
Solve for
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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