SOLUTION: A metallurgist has one alloy containing 25% copper and another containing 62% copper. How many pounds of each alloy must he use to make 45 pounds of a third alloy containing 34% co
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Question 516387: A metallurgist has one alloy containing 25% copper and another containing 62% copper. How many pounds of each alloy must he use to make 45 pounds of a third alloy containing 34% copper? Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! Ally A
Amount = x
Concentration = 25%=0.25
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Alloy B
Amount = 45-x pounds
Concentration = 62%=0.62
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Resultant Alloy
Amount=45
Concentration=34%=0.34
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(Amount of A * Concentration of A)+(Amount of B * Concentration of B) = (Amount of Resultant * Concentration of Resultant)
(x*0.25)+(45-x)(0.62)=(45)*0.34
0.25x+27.9-0.62x=15.3
-0.37x=15.3-27.9
-0.37x=-12.6
-0.37x/-0.37=-12.6/-0.37
x=34.05 pounds
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Ally A
Amount = 34.04 pounds
Concentration = 25%=0.25
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Alloy B
Amount = 45-x=45-34.04=10.95 pounds
Concentration = 62%=0.62
