document.write( "Question 1094741: How many gallons of a 90% antifreeze solution must be mixed with 80 gallons of 30% antifreeze to get a mixture that is 80% antifreeze? \n" ); document.write( "

Algebra.Com's Answer #709313 by greenestamps(13200) $\"\"$ $\"About$

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First, here is an easy way to solve mixture problems like this, if you can understand it....

\n" ); document.write( "(1) Look to see how far (or close) the percentage of the mixture is to the percentages of the two ingredients:
\n" ); document.write( " $\"90-80+=+10\"$
\n" ); document.write( " $\"80-30+=+50\"$

\n" ); document.write( "The percentage of the mixture, 80%, is \"5 times as close\" to 90% as it is to 30%. That means there must be 5 times as much of the 90% ingredient as the 30% ingredient. Since there are 80 gallons of the 30% antifreeze, the number of gallons of 90% antifreeze needed is 5*80 = 400.

\n" ); document.write( "If you want to use the slow traditional algebraic solution method...

\n" ); document.write( "let x = liters of 90% antifreeze
\n" ); document.write( "80 = liters of 30% antifreeze

\n" ); document.write( "The total mixture is (80+x) liters; the amount of antifreeze is 30% of the 80, plus 90% of the x. You want the antifreeze to be 80% of the total mixture, so
\n" ); document.write( " $\".30%2880%29%2B.90%28x%29+=+.80%2880%2Bx%29\"$
\n" ); document.write( " $\"24%2B.9x+=+64%2B.8x\"$
\n" ); document.write( " $\".1x+=+40\"$
\n" ); document.write( " $\"x+=+400\"$

\n" ); document.write( "You need to use 400 gallons of the 90% antifreeze. \n" ); document.write( "

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