SOLUTION: A mettallurgist needs to make 12.4lb of an alloy containing 50% gold. He is going to melt and combine on emetal that is 60% gold with another metal that is 40% gold. how much of ea

Question 448243: A mettallurgist needs to make 12.4lb of an alloy containing 50% gold. He is going to melt and combine on emetal that is 60% gold with another metal that is 40% gold. how much of each should he use?
Found 2 solutions by stanbon, dmhargrove:
Answer by stanbon(75887) (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?
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Equation:
gold + gold = gold
0.60x + 0.40(12.4-x) = 0.50*12.4
----
Multiply thru by 100 to get:
60x + 40*12.4 - 40x = 50*12.4
------
20x = 10*12.4
---
x = 6.2 lb (amt. of 60% alloy needed)
---
12.4-x = 6.2 lb (amt. of 40% alloy needed)
===========================================
Cheers,
Stan H.
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Answer by dmhargrove(1) (Show Source): You can put this solution on YOUR website!
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?
-------------------------------
Equation:
gold + gold = gold
0.60x + 0.40(12.4-x) = 0.50*12.4
----
Multiply thru by 100 to get:
60x + 40*12.4 - 40x = 50*12.4
------
20x = 10*12.4
---
x = 6.2 lb (amt. of 60% alloy needed)
---
12.4-x = 6.2 lb (amt. of 40% alloy needed)
===========================================

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