SOLUTION: A metallurgist has one alloy containing 43% aluminum and another containing 63% aluminum. How many pounds of each alloy must he use to make 43 pounds of a third alloy containing 49

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Question 1197807: A metallurgist has one alloy containing 43% aluminum and another containing 63% aluminum. How many pounds of each alloy must he use to make 43 pounds of a third alloy containing 49% aluminum?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A type of mixture problem asked very frequently. There are millions of these answered.

43% and 63%;
want mixture 43 pounds of 49%;

L=43
H=63
T=49
M=43, pounds of the mixture
y, unknown pounds of the H% alloy (63% alloy)

Hy%2BL%28M-y%29=TM
-
Hy%2BLM-Ly=TM
%28H-L%29y=TM-LM
highlight%28y=M%28%28T-L%29%2F%28H-L%29%29%29
and you can find M-y.

Substitute your given values.


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How much pounds the 63%:
43%28%2849-43%29%2F%2863-43%29%29
43%286%2F20%29
43%283%2F10%29
12.9------pounds of the 63% Al alloy