SOLUTION: one alloy containing 41% aluminum and the other containing 70% aluminum. How many pounds of each alloy must he use to make 50 pounds of a third alloy containing 61%

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Question 901586: one alloy containing 41% aluminum and the other containing 70% aluminum. How many pounds of each alloy must he use to make 50 pounds of a third alloy containing 61%
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887) (Show Source):
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one alloy containing 41% aluminum and the other containing 70% aluminum. How many pounds of each alloy must he use to make 50 pounds of a third alloy containing 61%
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Equation:
alum + alum = alum
0.41x + 0.70(50-x) = 0.61*50
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41x + 70*50 - 70x = 61*50
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-29x = -9*50
x = (9/29)50
x = 15.52 lbs (amt of 41% alloy needed)
50-x = 34.48 lbs (amt of 70% alloy needed)
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Cheers,
Stan H.
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Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
Aluminum
percent ---------------- quantity
Type I 41.00% ---------------- x pounds
Type II 70.00% ------ 50 - x pounds
Mixture 61.00% ---------------- 50
Total 50 pounds
41.00% x + 70.00% ( 50 - x ) = 61.00% * 50
41 x + 70 ( 50 - x ) = 3050
41 x + 3500 - 70 x = 3050
41 x - 70 x = 3050 - -3500
-29 x = -450
/ -29
x = 15.52 pounds 41.00% Type I
34.48 pounds 70.00% Type II

m.ananth@hotmail.ca