SOLUTION: A metallurgist wants to make 150 pounds of an alloy that is 55% aluminum. If he mixes an alloy that is 20% aluminum with an alloy that is 70% aluminum, how much of each should he u

Question 677743: A metallurgist wants to make 150 pounds of an alloy that is 55% aluminum. If he mixes an alloy that is 20% aluminum with an alloy that is 70% aluminum, how much of each should he use?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 70% aluminum
Then 150-x=amount of 20% aluminum
Now we know that the amount of pure aluminum that exists before the mixture takes place(0.70x+0.20(150-x)) has to equal the amount of pure aluminum that exists after the mixture takes place(0.55*150).
Sooooooo
0.70x+0.20(150-x)=0.55*150
0.70x+30-0.20x=82.5
0.50x=52.5
x=105 lb-------------------amount of 70% aluminum needed
150-x=150-105=45 lb---------amount of 20% aluminum needed
CK
0.7*105+0.2*45=0.55*150
73.5+9=82.5
82.5=82.5
Hope this helps---ptaylor

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