SOLUTION: A metallurgist needs to make 12.4 lbs. 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 o

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Question 637707: A metallurgist needs to make 12.4 lbs. 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) About Me  (Show Source):
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A metallurgist needs to make 12.4 lbs. 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?
:
Common sense says that it would be half 50% alloy and half 40% alloy
But here is a the mixture equation
:
Let x = amt of 60% gold alloy required
the total is to be 12.4 lbs, therefore:
(12.4-x) = amt of 40% alloy required
:
Decimal equiv equation
.60x + .40(12.4-x) = .50(12.4)
.60x + 4.96 - .40x = 6.2
.60x - .40x = 6.2-4.96
.20x = 1.24
x = 1.24/.2
x = 6.2 lbs of the 60% alloy
then
12.4 - 6.2 = 6.2 lbs of 40% alloy
:
:
Check this
.6(6.2) + .4(6.2) = .5(12.4)
3.72 + 2.48 = 6.2
6.2 = 6.2