document.write( "Question 248129: how many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
\n" ); document.write( "How many kg of pure salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt solution?
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Algebra.Com's Answer #181094 by ankor@dixie-net.com(22740)\"\" \"About 
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how many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
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\n" ); document.write( "Let x = amt of water
\n" ); document.write( "a simple \"amt of acid\" equation (the amt of acid does not change only the percent)
\n" ); document.write( ".40(8) = .10(x+8)
\n" ); document.write( "3.2 = .10x + .8
\n" ); document.write( "3.2 - .8 = .1x
\n" ); document.write( "2.4 = .1x
\n" ); document.write( "x = \"2.4%2F.1\"
\n" ); document.write( "x = 24 liters of pure water required
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\n" ); document.write( "Check for equal acid:
\n" ); document.write( ".4(8) = .10(24+8)
\n" ); document.write( "3.2 = .1(32)
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\n" ); document.write( "How many kg of pure salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt solution?
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\n" ); document.write( "Let x = amt of pure salt
\n" ); document.write( ".10(20) + x = .25(x+20)
\n" ); document.write( "2 + x = .25x + 5
\n" ); document.write( "x - .25x = 5 - 2
\n" ); document.write( ".75x = 3
\n" ); document.write( "x = \"3%2F.75\"
\n" ); document.write( "x = 4 kg of salt to be added
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\n" ); document.write( ":
\n" ); document.write( "Check
\n" ); document.write( ".10(20) + 4 = .25(4+20)
\n" ); document.write( "2 + 4 = .25(24)
\n" ); document.write( "6 = 6
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