document.write( "Question 688822: How many gallons of a 30% acid solution must be mixed with 20 gallons of a 19% solution to obtain a solution that is 25% acid? \n" ); document.write( "
Algebra.Com's Answer #425703 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Let x=amount of 30% solution needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure acid that exists before the mixture takes place (0.30x+0.19*20) has to equal the amount of pure acid that exists after the mixture takes place(0.25(20+x)). Sooooooo
\n" ); document.write( "0.30x+0.19*20=0.25(20+x)
\n" ); document.write( "0.30x+3.8=5+0.25x subtract 0.25x and also 3.8 from each side
\n" ); document.write( "0.05x=1.2
\n" ); document.write( "x=24 gal---amount of 30% acid solution needed
\n" ); document.write( "CK
\n" ); document.write( "0.30*24+3.8=0.25*44
\n" ); document.write( "7.2+3.8=11
\n" ); document.write( "11=11\r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor
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