document.write( "Question 410648: How many gallons of a solution of 20% acid must be blended with a solution of 5% acid to obtain 40 gallons of a solution of 12% acid? \n" ); document.write( "

Algebra.Com's Answer #288721 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi
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document.write( "to obtain 40 gallons 
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document.write( "Let x and (40gal-x) represent the amount of the 20% and 5% solutions respectively
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document.write( "Question states***
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document.write( " .20x + .05(40gal-x) = .12*40gal
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document.write( "Solving for x
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document.write( "     .15x = .07*40gal
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document.write( "        x = .07*40gal/.15
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document.write( "        x = 18 2/3 gal\r
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document.write( "CHECKING our Answer***
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document.write( " .20*18.667 + .05*21.333 = 4.8 = .12*40 \n" );
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