document.write( "Question 1020060: How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 90% copper in order to get an outlier 30% copper \n" ); document.write( "

Algebra.Com's Answer #636045 by LinnW(1048) $\"\"$ $\"About$

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$\"%280.90%28400%29%2B0.10x%29%2F%28400%2Bx%29+=+30%2F100\"$
\n" ); document.write( "Since we want 30% copper, we want 30 parts pure copper to 100 parts of the whole.
\n" ); document.write( "This is the right side of the above equation.
\n" ); document.write( "The numerator on the left hand side represents the total amount of pure
\n" ); document.write( "copper, that is 90% of the 400 ounces plus 10% of the alloy.
\n" ); document.write( "The denominator is the total amount of material in the new mixture.
\n" ); document.write( "Now we just need to solve.
\n" ); document.write( "Beginning with the cross products, we have
\n" ); document.write( "30(400+x) = 100(0.90(400)+0.10x)
\n" ); document.write( "12000 + 30x = 90(400) + 10x
\n" ); document.write( "12000 + 30x = 36000 + 10x
\n" ); document.write( "add -10x to each side
\n" ); document.write( "12000 + 20x = 36000
\n" ); document.write( "add -12000 to each side
\n" ); document.write( "20x = 24000
\n" ); document.write( "divide each side by 20
\n" ); document.write( "x = 1200
\n" ); document.write( "To verify, substitute 1200 for x in $\"%280.90%28400%29%2B0.10x%29%2F%28400%2Bx%29+=+30%2F100\"$
\n" ); document.write( " $\"%280.90%28400%29%2B0.10%2A%281200%29%29%2F%28400%2B1200%29+=+30%2F100\"$
\n" ); document.write( " $\"%28360%2B120%29%2F%28400%2B1200%29%29=+30%2F100\"$
\n" ); document.write( " $\"480%2F1600+=+30%2F100\"$
\n" ); document.write( "This checks out, so we need 1200 ounces of the alloy. \n" ); document.write( "

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