document.write( "Question 731016: We have 20% alcohol solution and 50% acid solution, how many pints must be added from each to obtain 8 pints of 30%?\r
\n" ); document.write( "\n" ); document.write( "-Because it mixes alcohol and acid solutions, I'm not sure if the same equation for mixtures works for this problem.
\n" ); document.write( "

Algebra.Com's Answer #446956 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Bad question. No specification for the ratio of the alcohol to the acid in the 8 pints mixture, and you cannot expect to INCREASE the amount of alcohol from 20% to 30%. If you want to accept this as an open-ended question, then you could pick any ratio between the alcohol and the acid that is possible. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "One way to approach is to make a total percent concentration rational equation and give one of the conditions that the volumes of each quantity of the available solutions must sum to 8 pints. The thought is like this:\r
\n" ); document.write( "\n" ); document.write( "x for the alcohol volume, y for the acid volume,
\n" ); document.write( "\"%2820x%2B50y%29%2F8=p\" and \"x%2By=8\". You would not care what the ratio is between x and y; only that you care that the value of p is 30, for 30%. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "According to that, you are back to a typical two component mixture problem without regard to the identity of the pure material in the available solutions or the resulting solution.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%2820x%2B50y%29%2F8=30\" and \"x%2By=8\"
\n" ); document.write( "Solve for x and y.
\n" ); document.write( "
\n" );