document.write( "Question 1158615: how many mL of sterile water will need to be mixed with 60 mL of a 60% solution to form a 40 % solution? what will the total volume of the 40% solution ? \n" ); document.write( "
Algebra.Com's Answer #781565 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Algebraically....

\n" ); document.write( "x mL of 0% solution, plus 60 mL of 60% solution, equals (x+60) mL of 40% solution:

\n" ); document.write( "\"0%28x%29%2B.60%2860%29+=+.40%28x%2B60%29\"

\n" ); document.write( "Solve using relatively simple algebra.

\n" ); document.write( "And here is an informal way to solve this kind of mixture problem quickly and easily.

\n" ); document.write( "You are starting with 60% solution and adding 0% solution, stopping when you get to 40%.
\n" ); document.write( "40 is one-third of the way from 60 to 0.
\n" ); document.write( "Therefore, 1/3 of the mixture is what you are adding.
\n" ); document.write( "That means the ratio of 60% solution to 0% solution is 2:1.
\n" ); document.write( "Since the amount of 60% solution is 60 mL, the amount of 0% solution required is half of that, which is 30mL.

\n" ); document.write( "Another variation of the informal approach is to reason that the target 40% is \"twice as close\" to 60% as it is to 0%; and therefore the mixture needs to have twice as much of the 60% solution as it has of the 0% solution. So the 60 mL of the 60% solution is twice as much as the amount you need of the 0% solution, leading again to the answer that you need 30 mL of the 0% solution.

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