document.write( "Question 1007738: How many gallons of 5% acid solution should be mixed with 20gallons of a 10% acid solution to obtain 8% acid solution?
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Algebra.Com's Answer #623586 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "We are mixing a weaker acid solution with a stronger acid\r\n" );
document.write( "solution to obtain a medium-strength acid solution.\r\n" );
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document.write( "Let the number of gallons of the weaker \r\n" );
document.write( "5% solution be x\r\n" );
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document.write( "Then the number of gallons of the final medium-strength\r\n" );
document.write( "solution will be x + 20\r\n" );
document.write( "                                     \r\n" );
document.write( "                                      Number of  \r\n" );
document.write( "                         Percent       Gallons\r\n" );
document.write( "  Type       Number       acid         of PURE\r\n" );
document.write( "   of          of       expressed       ACID\r\n" );
document.write( "Solution     gallons   as a decimal   contained\r\n" );
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document.write( "Weaker          x         0.05          0.05x\r\n" );
document.write( "Stronger       20         0.10        0.10(20)\r\n" );
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document.write( "Mixture       x+20        0.08      0.08(x+20) \r\n" );
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document.write( "The equation comes from the column on the right\r\n" );
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document.write( "  \r\n" );
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document.write( "        0.05x + 0.10(20) = 0.08(x+20)\r\n" );
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document.write( "Clear of decimals by multiplying through by 100\r\n" );
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document.write( "             5x + 10(20) = 8(x+20)\r\n" );
document.write( "                5x + 200 = 8x+160\r\n" );
document.write( "                     -3x = -40 \r\n" );
document.write( "                       x = 40/3\r\n" );
document.write( "                       x = 13 1/3 gallons of 5% acid solution.\r\n" );
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document.write( "Edwin

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