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A metallurgist has one alloy containing 36% aluminum and another containing 68% aluminum. How many pounds of each alloy must he use to make 47 pounds of a third

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metallurgist has one alloy containing 36% aluminum and another containing 68% aluminum. How many pounds of each alloy must he use to make 47 pounds of a third alloy containing 46% aluminum?
I am giving the answer for each one of the alloys because there is another constraint missing.
The equation will be: x*0.36 + y*0.68 = 47*0.46
Total aluminum required 21.62 pounds.
Separated we will need 21.62/0.36 of the first alloy =60.05 pounds
or 21.62/0.68 = 31.79 pounds of the second alloy.
To combine the two alloys, we need something else in the data.