# SOLUTION: A deal has two kinds of mets; one costs 90 cents a pound, the other 60 cents a pound. He wishes to make 50 pounds of a mixture that will cost 72 cents a pound. How many pounds of e

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A deal has two kinds of mets; one costs 90 cents a pound, the other 60 cents a pound. He wishes to make 50 pounds of a mixture that will cost 72 cents a pound. How many pounds of e      Log On

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 Click here to see ALL problems on Mixture Word Problems Question 130568: A deal has two kinds of mets; one costs 90 cents a pound, the other 60 cents a pound. He wishes to make 50 pounds of a mixture that will cost 72 cents a pound. How many pounds of each kind should he use?Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!Let x=amount of 90 cent meat needed Then 50-x=amount of 60 cent meat needed Now we know that the value of the meat before it's mixed ((0.90x+0.60(50-x)) has to equal the value of the meat after it's mixed (50*0.72). So our equation to solve is: 0.90x+0.60(50-x)=50*0.72 get rid of parens 0.90x+30-0.60x=36 subtract 30 from each side 0.90x+30-30-0.60x=36-30 collect like terms 0.30x=6 divide both sides by 0.30 x=20 lb---------------------------------amount of 90 cent meat needed 50-x=50-20=30 lb----------------------amount of 60 cent meat needed CK 20*0.90+30*0.60=50*0.72 18+18=36 36=36 Hope this helps---ptaylor