<PRE><B>Question 122501</B>
an alloy of tin is 15% tin weighs 20 pounds. a second alloy is 10% tin. how much of the second alloy must be added to the first alloy to get a 12% mixture.
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Let x = no. of pounds of 10% alloy that must be added
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The resulting 12% mixture would weigh (x+20) lbs
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A typical mixture equation
.15(20) + .10x = .12(x+20)
:
 3 + .1x = .12x + 2.4
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.1x - .12x = 2.4 - 3
:
-.02x = - .6
x = {{{(-.6)/(-.02)}}}
x = +30 lbs of 10 % tin required
:
:
Check solution, we know the final amt will weigh 20+30 = 50
.15(20) + .10(30) = .12(50)
 3 + 3 = 6; confirms out solution</PRE>