document.write( "Question 100503: How many liters of 80% alcohol must be mixed with 40 liters of 70% alcohol to produce a mixture that is 75% alcohol? \n" ); document.write( "

Algebra.Com's Answer #73172 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of 80% alcohol needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure alcohol in the 80% solution (0.80x) plus the amount of pure alcohol in the 70% solution ((0.70)(40)) has to equal the amount of pure alcohol in the final mixture (0.75)(40+x)). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.80x+0.70*40=0.75(40+x) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.80x+28=30+0.75x subtract 28 and also 0.75x from both sides\r
\n" ); document.write( "\n" ); document.write( "0.80x-0.75x+28-28=30-28+0.75x-0.75x collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.05x=2 divide both sides by 0.05\r
\n" ); document.write( "\n" ); document.write( "x=40----------------------------number of gal of 80% alcohol needed\r
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\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "0.80*40+0.70*40=0.75*80
\n" ); document.write( "32+28=60
\n" ); document.write( "60=60\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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