document.write( "Question 925865: How many ounces of a base metal (no silver) must be alloyed with 100 ounces of 39% silver alloy to obtain an alloy that is 15% silver? \n" ); document.write( "
Algebra.Com's Answer #561814 by TimothyLamb(4379)\"\" \"About 
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base metal % in 39% silver allow = (1 - 0.39)
\n" ); document.write( "base metal % in 15% silver allow = (1 - 0.15)
\n" ); document.write( "---
\n" ); document.write( "(1 - 0.39)*100 + 1.0*x = (1 - 0.15)*(100 + x)
\n" ); document.write( "61 + x = (1 - 0.15)*(100 + x)
\n" ); document.write( "61 + x = 0.85*(100 + x)
\n" ); document.write( "61 + x = 85 + 0.85x
\n" ); document.write( "0.15x = 24
\n" ); document.write( "x = 24/0.15
\n" ); document.write( "---
\n" ); document.write( "answer:
\n" ); document.write( "x = 160 ounces of base metal
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