document.write( "Question 132499This question is from textbook Holt Algebra 1
\n" ); document.write( ": How much pure water must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% anitfreeze solution? \n" ); document.write( "
Algebra.Com's Answer #96856 by stanbon(75887)\"\" \"About 
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How much pure water must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% antifreeze solution?
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\n" ); document.write( "40% solution DATA:
\n" ); document.write( "Amt. = 2 liters ; amt. of active ingredient = 0.4*2 = 0.8 liters
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\n" ); document.write( "Pure water DATA:
\n" ); document.write( "Amt. = x liters ; amt of active ingredient = 0*x = 0 liters
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\n" ); document.write( "Mixture DATA:
\n" ); document.write( "Amt. = (2+x) liters ; amt of active = 0.25(2+x) = 0.5+0.25x liters
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\n" ); document.write( "EQUATION:
\n" ); document.write( "active + active = active
\n" ); document.write( "0.8 + 0 = 0.5+0.25x
\n" ); document.write( "0.3 = 0.25x
\n" ); document.write( "x = 1.2 liters ( amt. of water needed for the mixture )
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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