SOLUTION: A metallurgist has one alloy containing 20% copper and another containing 70% copper. How many pounds of alloy must he use to make 50 pounds of the a third alloy containing 50% cop

Question 58158: A metallurgist has one alloy containing 20% copper and another containing 70% copper. How many pounds of alloy must he use to make 50 pounds of the a third alloy containing 50% copper?
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Found 2 solutions by ankor@dixie-net.com, Nate:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A metallurgist has one alloy containing 20% copper and another containing 70% copper. How many pounds of alloy must he use to make 50 pounds of the a third alloy containing 50% copper?
:
This is a typical mixture problem, if you learn this method you can handle most
of these problems
;
Let x = amt of the 20% alloy, Since the resulting amt is 50 pounds we can say:
(50-x) = amt of the 70% alloy:
:
.20(x) + .70(50-x) = .50(50)
.2x + 35 - .7x = 25
.2x - .7x = 25 - 35
-.5x = -10
x = -10/-.5
x = + 20 lb of the 20% alloy
:
The 70% alloy = 50 - 20 = 30 lb
:
Check:
.2(20) + .7(30) = .5(50)
4 + 21 = 25

Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
x = amount of 20% copper
y = amount of 70% copper
*Remember: x = 50 - y
(0.2x + 0.7y)/50 = 0.5
0.2x + 0.7y = 25
0.2(50 - y) + 0.7y = 25
10 - 0.2y + 0.7y = 25
0.5y = 15
y = 30
x = 50 - y = 50 - 30 = 20
30 pounds of 70% copper
20 pounds of 20% copper
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