document.write( "Question 640207: I am in college algebra. My problem is:\r
\n" ); document.write( "\n" ); document.write( "A mechanic finds the car with a 20 quart radiator has a mixture containing 30% antifreeze in it. How much of this mixture would he have to drain out and replace with pure antifreeze to get a 50% antifreeze mixture?
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Algebra.Com's Answer #403106 by KMST(5328) $\"\"$ $\"About$

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Let $\"x\"$ be the number of quarts drained and replaced.
\n" ); document.write( "After draining $\"x\"$ quarts, you have $\"20-x\"$ quarts of 30% solution in the radiator. That solution is 30% antifreeze and 70% water (100%-30%=70%).
\n" ); document.write( "The amount of water in the radiator at that point is
\n" ); document.write( " $\"0.70%2A%2820-x%29\"$ quarts.
\n" ); document.write( "After you add $\"x\"$ quarts of antifreeze, you end up with $\"20\"$ quarts of solution, that is supposed to be 50% antifreeze and 50% water.
\n" ); document.write( "The amount of water in the radiator at that point is supposed to be 50% of 20 quarts, or
\n" ); document.write( " $\"0.50%2A20\"$ quarts = $\"10\"$ quarts.
\n" ); document.write( "Since you did not add any water, all that water is exactly the same amount you had after the draining step, or
\n" ); document.write( " $\"0.70%2A%2820-x%29\"$ quarts.
\n" ); document.write( "So $\"10=0.70%2A%2820-x%29\"$
\n" ); document.write( " $\"10=0.70%2A%2820-x%29\"$ --> $\"10%2F0.70=0.70%2A%2820-x%29%2F0.70\"$ --> $\"10%2F0.70=20-x\"$ --> $\"10%2F0.70%2Bx-10%2F0.70=20-x%2Bx-10%2F0.70\"$ --> $\"x=20-10%2F0.70\"$ --> $\"highlight%28x=5.7%29\"$ quarts (rounding)
\n" ); document.write( "NOTE: In $\"x=20-10%2F0.70\"$ or x=20-10/0.70, according to order of operations rules (conventions) the division is done first. Most calculators know that, and will do it that way if you enter \"20-10/0.7\" before pressing \"=\". \n" ); document.write( "

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