document.write( "Question 1195288: A container contains 40L of wine which is 80% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 70%? \n" ); document.write( "
Algebra.Com's Answer #827735 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "The other tutor has the correct answer of 5 liters.\r
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\n" ); document.write( "\n" ); document.write( "Here's how to check it:
\n" ); document.write( "If 5 liters are poured out, then 80% of that is pure alcohol
\n" ); document.write( "80% of 5 = 0.80*5 = 4
\n" ); document.write( "So 4 liters of pure alcohol is poured out and 1 liter is water.\r
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\n" ); document.write( "\n" ); document.write( "The 40 L of wine drops to 40-5 = 35 L after the wine is poured, but then we replace those five liters with water to get back to 35+5 = 40 L again.\r
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\n" ); document.write( "\n" ); document.write( "The amount of pure alcohol is not replaced. We go from 0.80*40 = 32 L to 32-4 = 28 L\r
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\n" ); document.write( "\n" ); document.write( "Then notice that 28/40 = 0.70 = 70%\r
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\n" ); document.write( "\n" ); document.write( "Therefore, the answer has been confirmed.
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