document.write( "Question 12332: A 100% concentrate is to be mixed with having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentrate will be needed? \n" ); document.write( "
Algebra.Com's Answer #6271 by longjonsilver(2297)\"\" \"About 
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we have a volume of liquid1 added to a volume of liquid2 to produce a mixture.\r
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\n" ); document.write( "\n" ); document.write( "Let x = volume of 100% concentrate
\n" ); document.write( "Let y = volume of 40% concentrate\r
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\n" ); document.write( "\n" ); document.write( "so we have
\n" ); document.write( "(x gallons of 100%) + (y gallons of 40%) = (55 gallons of 75%)\r
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\n" ); document.write( "\n" ); document.write( "converting this into Algebra gives:
\n" ); document.write( "(x*1) + (y*0.40) = (55*0.75) --> or in fact, keep the percentages as percentages, it doesn't matter.\r
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\n" ); document.write( "\n" ); document.write( "x + 0.4y = 41.25\r
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\n" ); document.write( "\n" ); document.write( "OK...so we have 1 equation with 2 unknowns, which we therefore cannot solve. However, we also know that x+y=55. So, y=55-x, hence:\r
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\n" ); document.write( "\n" ); document.write( "x + 0.4(55-x) = 41.25
\n" ); document.write( "x + 22 - 0.4x = 41.25
\n" ); document.write( "0.6x = 19.25\r
\n" ); document.write( "\n" ); document.write( "--> x = 32.08333 gallons
\n" ); document.write( "--> x = 32.08 gallons (to 4 significant figures)\r
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\n" ); document.write( "\n" ); document.write( "jon.
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