document.write( "Question 5478: how much of a 25% acid solution should be mixed with a 50% acid solution to obtain 500 gallons of a 34%acid solution?
\n" ); document.write( "thanks for the help! \n" ); document.write( "
Algebra.Com's Answer #2844 by Abbey(339)\"\" \"About 
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let the 25% acid solution quantity (volume) = x
\n" ); document.write( "Let y= volume of 50% acid solution.
\n" ); document.write( "because the total volume can equal only 500 gallons, we can say that 500 gallons - the quantity of 25% solution = the quantity of 50%, right? So:
\n" ); document.write( "500-x=y
\n" ); document.write( "To figure the solution we can multiply the percentage by the volume to have a \"standard\" to work with:
\n" ); document.write( "Let 25*x=25% standard
\n" ); document.write( "Let 50*y = 50% standard
\n" ); document.write( "Let 34*(500)=the final solution \r
\n" ); document.write( "\n" ); document.write( "This means we can build an equation that says, mix the amount of the 25% with an amount of the 50% and get 500 gallons of 34%:
\n" ); document.write( "25x+50y=34(500)
\n" ); document.write( "and now we can substitute a form of x in for our y value (because we know that 500-x=y) and solve for x:
\n" ); document.write( "25x+50y=34(500)
\n" ); document.write( "25x+50(500-x)=34(500)
\n" ); document.write( "25x+25000-50x=17000
\n" ); document.write( "-25x+25000=17000
\n" ); document.write( "25000=17000+25x
\n" ); document.write( "8000=25x
\n" ); document.write( "320=x\r
\n" ); document.write( "\n" ); document.write( "So we need 320 gallons of the 25% solution and 180 gallons of the 50% solution.\r
\n" ); document.write( "\n" ); document.write( "Check the results:
\n" ); document.write( "320 gallons *25% acid = 80 gallons of acid
\n" ); document.write( "180 gallons *50% acid = 90 gallons of acid
\n" ); document.write( "80+90=170
\n" ); document.write( "170 gallons/500 gallons = 34%
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