document.write( "Question 896843: A 100% concentration is to be mixed with a mixture having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentration will be needed? (Round answers to nearest whole gallon.)\r
\n" ); document.write( "\n" ); document.write( "I've tried setting up the problem and this is what I've done-
\n" ); document.write( "100%con.+40%con=55 gal w/ 75% con. I'm really confused and I mainly need help setting up the problem and the best way to go about solving this. \n" ); document.write( "
Algebra.Com's Answer #543794 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
100% concentration is to be mixed with a mixture having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentration will be needed? (Round answers to nearest whole gallon.)
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\n" ); document.write( "You can do it this way
\n" ); document.write( "Let x = amt of 100% concentration required
\n" ); document.write( "the total will be 55 gal, therefore:
\n" ); document.write( "(55-x) = amt of 40% mixture required
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\n" ); document.write( "100x + 40(55-x) = 75(55)
\n" ); document.write( "100x + 2200 - 40x = 4125
\n" ); document.write( "100x - 40x = 4125 - 2200
\n" ); document.write( "60x = 1925
\n" ); document.write( "x = 1925/60
\n" ); document.write( "x = 32.0 gal of 100% stuff required
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\n" ); document.write( "You can check this for yourself, find the amt of 40% stuff
\n" ); document.write( "55 - 32 = 23 gal
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\n" ); document.write( "See if this comes out right
\n" ); document.write( "100(32) + 40(23) = 75(55) a slight error because we rounded down to 32 gal
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\n" ); document.write( "Hopefully, this unconfused you a little?
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