document.write( "Question 1202656: How much 9% saline solution should Kent mix with 70 cubic centimeters (cc) of a 17% saline solution to produce a 13% saline solution? \n" ); document.write( "

Algebra.Com's Answer #837623 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "We mix x of the weaker solution with 70 of the stronger to get y of the medium\r\n" );
document.write( "strength solution:\r\n" );
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document.write( "                  amt.| %  |total \r\n" );
document.write( "                   in |as  |amt of\r\n" );
document.write( "                  ccs |dec.|saline\r\n" );
document.write( "weaker          |  x  |0.09|0.09x\r\n" );
document.write( "stronger        | 70  |0.17|0.17(70)\r\n" );
document.write( "medium strength |  y  |0.13|0.13y\r\n" );
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document.write( "The equations comes from the first and last columns:\r\n" );
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document.write( "Multiply the 2nd equation through by 100 to clear decimals:\r\n" );
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document.write( "Substitute x+70 for y in the 2nd equation:\r\n" );
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document.write( "9x+1190 = 13(x+70)\r\n" );
document.write( "9x+1190 = 13x+910\r\n" );
document.write( "    280 = 4x\r\n" );
document.write( "     70 = x\r\n" );
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document.write( "So as it turns out, we use 70 ccs of each and get 140 ccs. of\r\n" );
document.write( "the medium strength.  We could have told that because 13% just \r\n" );
document.write( "happens to be exactly half-way between 9% and 17%.  But that's \r\n" );
document.write( "just a coincidence.  In another problem that won't be the case. \r\n" );
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document.write( "Edwin

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