SOLUTION: A metallurgist has one alloy containing 28% aluminum and another containing 64% aluminum. How many pounds of each alloy must he use to make 49 pounds of a third alloy containing 4

Question 1021994: A metallurgist has one alloy containing 28% aluminum and another containing 64% aluminum. How many pounds of each alloy must he use to make 49 pounds of a third alloy containing 45% aluminum? (Round to two decimal places if necessary.)
In the equation it has it equaled out as .28x+.64y=22.05
How do they get the 22.05?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Amount of the alloys: x+y=49

Amount of Aluminum in the blended alloy: 28x+64y=45*49

The two equations should make sense. If not, that is the explanation and help you need. The aluminum accounting shows percent of the 49 pound alloy blend. The percents can be expressed as decimal fractions.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

A metallurgist has one alloy containing 28% aluminum and another containing 64% aluminum. How many pounds of each alloy must he use to make 49 pounds of a third alloy containing 45% aluminum? (Round to two decimal places if necessary.)
In the equation it has it equaled out as .28x+.64y=22.05
How do they get the 22.05?

.28x + .64y = .45(49) --------> .28x + .64y = 22.05. As you can see, .45(49) = 22.05.
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