# SOLUTION: A metallurgist needs to make 12.4 lbs. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much o

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 Question 637707: A metallurgist needs to make 12.4 lbs. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of each should he use?Answer by ankor@dixie-net.com(15651)   (Show Source): You can put this solution on YOUR website!A metallurgist needs to make 12.4 lbs. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of each should he use? : Common sense says that it would be half 50% alloy and half 40% alloy But here is a the mixture equation : Let x = amt of 60% gold alloy required the total is to be 12.4 lbs, therefore: (12.4-x) = amt of 40% alloy required : Decimal equiv equation .60x + .40(12.4-x) = .50(12.4) .60x + 4.96 - .40x = 6.2 .60x - .40x = 6.2-4.96 .20x = 1.24 x = 1.24/.2 x = 6.2 lbs of the 60% alloy then 12.4 - 6.2 = 6.2 lbs of 40% alloy : : Check this .6(6.2) + .4(6.2) = .5(12.4) 3.72 + 2.48 = 6.2 6.2 = 6.2