<PRE><B>Question 58158</B>
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 
</PRE>