document.write( "Question 269725: how much of an alloy that is 10% copper should be mixed with 500 ounces of an alloy that is 70% copper in order to get an alloy that is 20% copper? \n" ); document.write( "

Algebra.Com's Answer #197636 by josmiceli(19441) $\"\"$ $\"About$

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In words:
\n" ); document.write( "(ounces of copper in final mixture)/(ounces of final mixture)= 20%
\n" ); document.write( "Let $\"x\"$ = ounces of 10% copper alloy needed
\n" ); document.write( " $\"%28.1x+%2B+.7%2A500%29%2F%28x+%2B+500%29+=+.2\"$
\n" ); document.write( " $\".1x+%2B+350+=+.2%2A%28x+%2B+500%29\"$
\n" ); document.write( " $\".1x+%2B+350+=+.2x+%2B+100\"$
\n" ); document.write( " $\".1x+=+250\"$
\n" ); document.write( " $\"x+=+2500\"$
\n" ); document.write( "2500 ounces of 10% copper alloy are needed
\n" ); document.write( "check:
\n" ); document.write( " $\"%28.1x+%2B+.7%2A500%29%2F%28x+%2B+500%29+=+.2\"$
\n" ); document.write( " $\"%28.1%2A2500+%2B+.7%2A500%29%2F%282500+%2B+500%29+=+.2\"$
\n" ); document.write( " $\"%28250+%2B+350%29%2F%282500+%2B+500%29+=+.2\"$
\n" ); document.write( " $\"600%2F3000+=+.2\"$
\n" ); document.write( " $\"600+=+600\"$
\n" ); document.write( "OK \n" ); document.write( "

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