document.write( "Question 327704: how many ounces of an alloy containing 30% gold on the world market must be mixed with an alloy containing 5% gold to obtain 25 ounces of an alloy containing 20% gold? \n" ); document.write( "

Algebra.Com's Answer #234777 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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how many ounces of an alloy containing 30% gold on the world market must be mixed
\n" ); document.write( " with an alloy containing 5% gold to obtain 25 ounces of an alloy containing 20% gold?
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\n" ); document.write( "Let x = amt of 30% alloy
\n" ); document.write( "The resulting total is to be 25 oz, therefore:
\n" ); document.write( "(25-x) = amt of 5% alloy
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\n" ); document.write( "A typical mixture equation:
\n" ); document.write( ".30x + .05(25-x) = .20(25)
\n" ); document.write( ".30x + 1.25 - .05x = 5
\n" ); document.write( ".30x - .05x = 5 - 1.25
\n" ); document.write( ".25x = 3.75
\n" ); document.write( "x = $\"3.75%2F.25\"$
\n" ); document.write( "x = 15 oz of 30% alloy required
\n" ); document.write( "then
\n" ); document.write( "25-15 = 10 oz of 5% alloy
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\n" ); document.write( "Check solution
\n" ); document.write( ".30(15) + .05(10) = .20(25)
\n" ); document.write( "4.5 + .5 = 5
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