document.write( "Question 701981: How man milliliters of 100% acid must be added to 50 milliliters of a 40% acid solution to obtain a 50% acid solution? \r
\n" ); document.write( "\n" ); document.write( "I had set up and figured the problem like this:\r
\n" ); document.write( "\n" ); document.write( "1(x)+.4(50)=.5(50+x)
\n" ); document.write( "20x=25+.5x
\n" ); document.write( "19.25x=25
\n" ); document.write( "x=50/39 mL\r
\n" ); document.write( "\n" ); document.write( "But I have a gut feeling that x does not equal 50/39.
\n" ); document.write( "What did I do wrong?
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\n" ); document.write( "

Algebra.Com's Answer #432679 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
How much pure acid will be in the solution? Letting x equal how much of the 100% material to add,
\n" ); document.write( "Pure acid in the resulting solution is 1*x+50*(0.40).\r
\n" ); document.write( "\n" ); document.write( "How much solution must result when the two parts are added and mixed?
\n" ); document.write( "The amount of resulting solution is x+50 milliliters.\r
\n" ); document.write( "\n" ); document.write( "You want a result of 50% acid, or as fraction, 0.5 as acid.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28x%2A1%2B50%2A%280.04%29%29%2F%28x%2B50%29=0.50\"
\n" ); document.write( "Solve for x.\r
\n" ); document.write( "\n" ); document.write( "Skipping a step,...
\n" ); document.write( "\"x%2B20=0.5%2Ax%2B25\"
\n" ); document.write( "Continue....Finish...
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