document.write( "Question 24050: Angela needs 20 quarts of 50% antifreeze in her radiator. She plans to obtain this by mixing some pure antifreeze (100%) with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use? \n" ); document.write( "

Algebra.Com's Answer #12760 by Earlsdon(6294) $\"\"$ $\"About$

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You can set up the appropriate equation for this situation from the problem description.
\n" ); document.write( "Let x = the required number of quarts of pure (100%) antifreze, then (20-x) = the required number of quarts of 40% antifreeze solution which, when added together, will give 20 quarts of 50% antifreeze solution. Change the percentages to their decimal equivalents. The equation is:
\n" ); document.write( " $\"x+%2B+%2820-x%29%280.4%29+=+20%280.5%29\"$ Simplify and solve for x.
\n" ); document.write( " $\"x+-+0.4x+%2B+8+=+10\"$
\n" ); document.write( " $\"0.6x+%2B+8+=+10\"$ Subtract 8 from both sides of the equation.
\n" ); document.write( " $\"0.6x+=+2\"$ Divide both sides by 0.6
\n" ); document.write( " $\"x+=+2%2F0.6\"$ = 3 1/3
\n" ); document.write( "x = 3 1/3 quarts of pure antifreeze.
\n" ); document.write( "20-x = 16 2/3 quarts of 40% antifreeze.
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