document.write( "Question 1191884: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 55%. They want to use all three solutions to obtain a mixture of 216 liters containing 30% acid, using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used? \n" ); document.write( "

Algebra.Com's Answer #823734 by ikleyn(52776) $\"\"$ $\"About$

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\n" ); document.write( "A chemist has three different acid solutions. The first acid solution contains 20% acid,
\n" ); document.write( "the second contains 35% and the third contains 55%.
\n" ); document.write( "They want to use all three solutions to obtain a mixture of 216 liters containing 30% acid,
\n" ); document.write( "using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
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document.write( "35% solution:  x liters;\r\n" );
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document.write( "55% solution:  3x liters;\r\n" );
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document.write( "20% acid:      216 - x - 3x = 216 - 4x liters.\r\n" );
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document.write( "The pure acid in ingredients is the same as the pure acid in the mixture\r\n" );
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document.write( "    0.35x + 0.55*(3x) + 0.2*(216-4x) = 0.3*216.\r\n" );
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document.write( "From this equation\r\n" );
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document.write( "    x =  = 18.\r\n" );
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document.write( "ANSWER.  18 liters of the 35% solution; 3*18 = 54 liters of the 55% solution and (216-4*18) = 144 liters of the 2-% solution.\r\n" );
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