document.write( "Question 575062: What quantity of a 50% acid solution must be mixed with a 20% solution to produce 150 mL of a 40% solution? \n" ); document.write( "

Algebra.Com's Answer #369443 by KMST(5328) $\"\"$ $\"About$

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Mixture problems are solved by setting up two equations: one for the total mixture amount balance, and another for the amount balance for the compound of interest.
\n" ); document.write( "In algebra, little attention is paid to the science, or the units, and any volume changes due to mixing are ignored.
\n" ); document.write( "Chemists are truly a dying breed
\n" ); document.write( "http://www.chemicalprocessing.com/articles/2011/cartoon_caption_62.html
\n" ); document.write( "So I'll pretend I'm not a chemist.
\n" ); document.write( "Let x be the volume of 50% solution to be used.
\n" ); document.write( "Let y be the volume of 20% solution to be used.
\n" ); document.write( "Total solution balance
\n" ); document.write( " $\"x%2By=150\"$ (all quantities are in mL, and volumes are additive)
\n" ); document.write( "Amount of acid balance:
\n" ); document.write( "Let's assume that the percentages are in v/v (volume to volume basis), so that a 50% solution contains 50 mL of acid (whatever is meant by that) in a total volume of 100 mL.
\n" ); document.write( "The amount (in mL) of acid in x mL of 50% solution is (0.50)x, and the amount (in mL) of acid in y mL of 20% solution is (0.20)y.
\n" ); document.write( "The total amount of acid in x mL of 50% solution plus y mL of 20% solution is:
\n" ); document.write( " $\"0.5x%2B0.2y\"$
\n" ); document.write( "We want that to be the amount of acid in 150 mL of 40% solution, which is (0.40)150, and
\n" ); document.write( " $\"0.40%2A150=60\"$ , so
\n" ); document.write( " $\"0.5x%2B0.2y=60\"$
\n" ); document.write( "We need to solve the system of equations
\n" ); document.write( " $\"system%28x%2By=150%2C0.5x%2B0.2y=60%29\"$
\n" ); document.write( "There are several ways to solve the system.
\n" ); document.write( "USING ELIMINATION:
\n" ); document.write( "You could multiply the second equation times 2 to get
\n" ); document.write( " $\"x%2B0.4y=120\"$ and then subtract it from the first equation to get
\n" ); document.write( " $\"x%2By-%28x%2B0.4y%29=150-120\"$ --> $\"x%2By-x-0.4y=30\"$ --> $\"y-0.4y=30\"$ --> $\"y%281-0.4%29=30\"$ --> $\"0.6y=30\"$ --> $\"0.6y%2F0.6=30%2F0.6\"$ --> $\"y=50\"$
\n" ); document.write( "And then, substituting the value found for y into $\"x%2By=150\"$ , you would get
\n" ); document.write( " $\"x%2B50=150\"$ , which you could solve for x to find $\"x=100\"$ .
\n" ); document.write( "You need to mix $\"highlight%28100%29\"$ mL of 50% solution with $\"highlight%2850%29\"$ mL of 20% solution to get 150 mL of 40% solution. \n" ); document.write( "

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