SOLUTION: A metallurgist needs exactly 600 g of an alloy containing 52% copper.He has two alloys, one containing 80% copper and the other 20% copper.How much of each alloy should he use?
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-> SOLUTION: A metallurgist needs exactly 600 g of an alloy containing 52% copper.He has two alloys, one containing 80% copper and the other 20% copper.How much of each alloy should he use?
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Question 973743: A metallurgist needs exactly 600 g of an alloy containing 52% copper.He has two alloys, one containing 80% copper and the other 20% copper.How much of each alloy should he use?
You can put this solution on YOUR website! A metallurgist needs exactly 600 g of an alloy containing 52% copper.He has two alloys, one containing 80% copper and the other 20% copper.How much of each alloy should he use?
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let x=amt of 80%copper alloy to use
let(600-x)=amt of 20%copper alloy to use
..
80%x+20%(600-x)=52%*600
.80x+120-.20x=312
.60x=192
x=320
600-x=280
amt of 80%copper alloy to use=320 g
amt of 20%copper alloy to use=280 g
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