SOLUTION: A metallurgist needs to make 12.4 lb. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of e
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Question 232695: A metallurgist needs to make 12.4 lb. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of each should he use? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A metallurgist needs to make 12.4 lb. of an alloy containing 50% gold.
He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold.
How much of each should he use?
:
Since 50 is half way between 60 and 40 it would be half and half, 6.2, 6.2
But here is the method for any mixture
:
Let x = amt of 60% gold required
then since the resulting amt is 12.4
(12.4-x) = amt of 40% gold
:
Write an "amt of gold" equation, using the decimal equiv of %
.6x + .4(12.4-x) = .5(12.4)
:
.6x + 4.96 - .4x = 6.2
.6x - .4x = 6.2 - 4.96
.2x = 1.24
x =
x = 6.2 lb of 60% gold
