SOLUTION: AN alloy containing 40% gold and an alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold. How much of each alloy is needed?

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Question 513472: AN alloy containing 40% gold and an alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold. How much of each alloy is needed?
Answer by Maths68(1474) About Me  (Show Source):
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Alloy A
Amount = x lb
Concentration =40% =0.4
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Alloy B
Amount = 50-x lb
Concentration =70% = 0.70
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Final Alloy
Amount =50 lb
Concentration =60%=0.6
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[Amount Alloy A * Concentration A] + [Amount Alloy B * Concentration of B] = Amount of Final Alloy * Concentration of Final Alloy
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(x)(0.4)+(50-x)(0.7)=(50)(0.6)
0.4x+35-0.7x=30
0.4x-0.7x=30-35
-0.3x=-5
-0.3x/-0.3=-5/-0.3
x=16.66
===========================
Alloy A
Amount = 16.66 lb
Concentration =40% =0.4
================================
Alloy B
Amount = 50-16.66=33.33 lb
Concentration =70% = 0.70
==============================
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A 16.66 pounds alloy containing 40% gold and a 33.33 pounds alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold