document.write( "Question 689761: How much pure acid must be added to 25 gallons of 25% solution to obtain a 50% solution?\r
\n" ); document.write( "\n" ); document.write( "What I tried: \r
\n" ); document.write( "\n" ); document.write( "25(100x + 25) = 50(100x)
\n" ); document.write( "2500x + 625 = 5000x
\n" ); document.write( "625 = 2500x
\n" ); document.write( "x = 1/4 (.25)\r
\n" ); document.write( "\n" ); document.write( "But something doesn't seem quite right! Thank you in advance!
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Algebra.Com's Answer #426042 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Stop guessing.\r
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\n" ); document.write( "\n" ); document.write( "There are gallons of pure acid in gallons of pure acid. There are gallons of pure acid in 25 gallons of 25% solution. And there are gallons of pure acid in 25 plus gallons of 50% solution. Putting it all together:\r
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\n" ); document.write( "\n" ); document.write( "Solve for \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "Egw to Beta kai to Sigma
\n" ); document.write( "My calculator said it, I believe it, that settles it
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\"The

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