document.write( "Question 953330: 2. 15 gallons of a 20% acid solution is to be mixed with a 12% acid solution to yield an 18% acid solution. How many gallons of the 12% acid solution will be needed?
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Algebra.Com's Answer #582296 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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5 gallons of a 20% acid solution is to be mixed with a 12% acid solution to yield an 18% acid solution.
\n" ); document.write( " How many gallons of the 12% acid solution will be needed?
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\n" ); document.write( "let = amt of 12% acid required
\n" ); document.write( "then
\n" ); document.write( "(x+5) = resulting amt of the 18% solution
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\n" ); document.write( "A mixture equation in decimal form
\n" ); document.write( ".20(5) + .12x = .18(x+5)
\n" ); document.write( "1.0 + .12x = .18x + .90
\n" ); document.write( "1.0 - .90 = .18x - .12x
\n" ); document.write( ".1 = .06x
\n" ); document.write( "x - .1/.06
\n" ); document.write( "x = 1 $\"2%2F3\"$ gal of 12% solution
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\n" ); document.write( "Check this on your calc:
\n" ); document.write( "enter .2(5) + .12(1.6667)
\n" ); document.write( "then
\n" ); document.write( "enter .18(1.6667+5) \n" ); document.write( "

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