SOLUTION: I've got another one I'd like help with please.
A metallurgist has one alloy containing 21% aluminum and another containing 67% aluminum. How many pounds of each alloy must he us
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A metallurgist has one alloy containing 21% aluminum and another containing 67% aluminum. How many pounds of each alloy must he us
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Question 599684: I've got another one I'd like help with please.
A metallurgist has one alloy containing 21% aluminum and another containing 67% aluminum. How many pounds of each alloy must he use to make 57 pounds of a third alloy containing 23% aluminum? the answer should be rounded to the nearest hundredth. if this could be explained how to set it up with the answer, so hopefully I can help my step daughter with her remaining questions. Thank you Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A metallurgist has one alloy containing 21% aluminum and another containing 67% aluminum.
How many pounds of each alloy must he use to make 57 pounds of a third alloy containing 23% aluminum?
:
Let x = amt of 67% Alum
the result is to be 57 lb, therefore
(57-x) = amt of 21% Alum
:
A typical mixture equation using decimal equiv
.67x + .21(57-x) = .23(57)
.67x + 11.97 - .21x = 13.11
.67x - .21x = 13.11 - 11.97
.46x = 1.14
x = 1.14/.46 =
x = 2.48 lb of the 67% alloy required
then
57 - 2.48 = 54.52 lb of the 21% alloy
