document.write( "Question 1204463: Pure acid is to be added to a 10% acid solution to obtain 45L of a 40% acid solution.
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Algebra.Com's Answer #840742 by ikleyn(52786) $\"\"$ $\"About$

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\n" ); document.write( "Pure acid is to be added to a 10% acid solution to obtain 45L of a 40% acid solution.
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\n" ); document.write( "\n" ); document.write( " There is an elegant method to solve this problem (and many other similar problems)\r
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document.write( "45 liters of the 40% acid contain  0.4*45 = 18 liters of the pure acid and  45 - 12 = 27 liters of water.\r\n" );
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document.write( "When we add pure acid to the 10% acid mixture, we do not add water.  \r\n" );
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document.write( "Hence, 27 liters of water was in 10% acid solution initially.\r\n" );
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document.write( "Again: the initial 10% acid mixture contained 27 liters of water, and water was 9 \r\n" );
document.write( "of 10 parts of the volume of this 10% mixture.\r\n" );
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document.write( "Hence, the pure acid in this 10% acid mixture was 3 liters , while the water was 27 liters.\r\n" );
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document.write( "To get 18 liters of the pure acid in the final 40% mixture, 18-3 = 15 liters of the pure acid should be added\r\n" );
document.write( "to the initial 10% mixture.\r\n" );
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document.write( "ANSWER.  15 liters of the pure acid and 3+27 = 30 liters of the 10% mixture shoud be used.\r\n" );
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\n" ); document.write( "\n" ); document.write( "To see many other similar solved problems, look into the lessons\r
\n" ); document.write( "\n" ); document.write( "    - Special type mixture problems on DILUTION adding water\r
\n" ); document.write( "\n" ); document.write( "    - Increasing concentration of an acid solution by adding pure acid \r
\n" ); document.write( "\n" ); document.write( "    - Increasing concentration of alcohol solution by adding pure alcohol \r
\n" ); document.write( "\n" ); document.write( "    - How many kilograms of sand must be added to a mixture of sand and cement \r
\n" ); document.write( "\n" ); document.write( "    - How much water must be evaporated \r
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