document.write( "Question 1184009: How many quarts of pure antifreeze must be added to 5 quarts of a 30% antifreeze solution to obtain a 50% antifreeze solution \n" ); document.write( "
Algebra.Com's Answer #814537 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "A classic formal algebraic solution would go something like this....

\n" ); document.write( "You are mixing 5 quarts that is 30% antifreeze and x quarts that are 100% antifreeze to get (5+x) quarts that is 50% antifreeze:

\n" ); document.write( "\".30%285%29%2B1.00%28x%29=.50%285%2Bx%29\"
\n" ); document.write( "\"1.5%2Bx=2.5%2B.5x\"
\n" ); document.write( "\".5x=1\"
\n" ); document.write( "\"x=1%2F.5=2\"

\n" ); document.write( "ANSWER: 2 quarts

\n" ); document.write( "An informal method that will give you the answer within 15 seconds if you understand it....

\n" ); document.write( "(1) Look at the three percentages 30, 50, and 100 on a number line and observe/calculate that 50 is 2/7 of the way from 30 to 100. (30 to 100 is a difference of 70; 30 to 50 is a difference of 20; 20/70 = 2/7.)
\n" ); document.write( "(2)That means 2/7 of the mixture is the 100% antifreeze that you are adding.
\n" ); document.write( "(3) So the 5 quarts you started with is 5/7 of the mixture; that means 2/7 of the mixture is 2 quarts.

\n" ); document.write( "ANSWER: 2 quarts

\n" ); document.write( "
\n" ); document.write( "
\n" );