document.write( "Question 200265: Hi!\r
\n" ); document.write( "\n" ); document.write( "A biologist has two brine solutions, one containing 1% salt and another containing 4% salt. How many milliliters of each solution should he mix to obtain 1 L of a solution that contains 2.8% salt?\r
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Algebra.Com's Answer #150635 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
Since the 2 solutions must add to 1 liter, then they must add to 1,000 ml\r
\n" ); document.write( "\n" ); document.write( "Let the amount of 1% salt solution in the 2.8% salt solution be x.
\n" ); document.write( "Then the amount of 4% solution in the 2.8% salt solution is 1,000 – x\r
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\n" ); document.write( "\n" ); document.write( "Therefore, we get: .01(x) + .04(1,000 – x) = .028(1,000)\r
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\n" ); document.write( "\n" ); document.write( "This equation becomes: .01x + 40 - .04x = 28\r
\n" ); document.write( "\n" ); document.write( "-.03x + 40 = 28
\n" ); document.write( "-.03x = - 12
\n" ); document.write( "\"x++=++%28-12%29%2F%28-.03%29\" = 400\r
\n" ); document.write( "\n" ); document.write( "Therefore, he needs to mix 400 ML of the 1% salt solution, and 600 (1,000 – 400) ML of the 4% salt solution to get 1,000 ML, or 1 Liter of 2.8% salt solution.
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