document.write( "Question 969630: How much water should we add to 20 kg of seawater to change its concentration from 3% to 2%?
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Algebra.Com's Answer #592459 by Boreal(15235)\"\" \"About 
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x=amount of water in kg. (0 salt)\r
\n" ); document.write( "\n" ); document.write( "start with 20 kg at .03 (3%) (salt)=.60 \r
\n" ); document.write( "\n" ); document.write( "add x kg of water with 0 salt.\r
\n" ); document.write( "\n" ); document.write( "will have (20 + x) kg of water with 2 % salt.
\n" ); document.write( "Distribute. 02* (20+x)=0.4 +0.02x \r
\n" ); document.write( "\n" ); document.write( "This will equal .60, the desired concentration.\r
\n" ); document.write( "\n" ); document.write( "So, 0.4 +0.02x=.60 \r
\n" ); document.write( "\n" ); document.write( "Multiply the whole equation by 100. Not necessary, but will be easier to work with\r
\n" ); document.write( "\n" ); document.write( "40 + 2 x=60
\n" ); document.write( "Subtract 40 from both sides\r
\n" ); document.write( "\n" ); document.write( "2x=20
\n" ); document.write( "Divide by 2 on both sides.\r
\n" ); document.write( "\n" ); document.write( "x=10 liters.\r
\n" ); document.write( "\n" ); document.write( "Check. We are decreasing the concentration 1/3.\r
\n" ); document.write( "\n" ); document.write( "We add 10 liters of water to 20, and the final amount, 30, has 1/3 less salt, and 1/3 of it was without salt. \r
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