document.write( "Question 1100733: You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 70% alcohol mixture. How much of each mixture will you need to add to obtain the desired solution? \n" ); document.write( "

Algebra.Com's Answer #715265 by ikleyn(52786) $\"\"$ $\"About$

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\n" ); document.write( "You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 70% alcohol mixture.
\n" ); document.write( "How much of each mixture will you need to add to obtain the desired solution?
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document.write( "Let X = \"How much of the 30% mixture to mix\", in mL.\r\n" );
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document.write( "Then the volume of the 70% mixture to mix is (600-X) mL.\r\n" );
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document.write( "X mL of the 30% mixture contribute 0.3*X mL of the pure alcohol to the new mixture.\r\n" );
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document.write( "(600-X) mL of the 70% mixture contribute 0.7*(600-X) mL of the pure alcohol to the new mixture.\r\n" );
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document.write( "In all, two input mixtures provide  0.3*X + 0.7*(600-X)  mL of the pure alcohol in the output mixture.\r\n" );
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document.write( "The output volume  is 600 mL,  and it contains 0.3*X + 0.7*(600-X)  mL of the pure alcohol, as we find out above.\r\n" );
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document.write( "Hence, the concentration of the new mixture is   as the fraction.\r\n" );
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document.write( "According to the condition, it must be  45% or 0.45. It gives you an equation\r\n" );
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document.write( " = 0.45.\r\n" );
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document.write( "It is you major/basic equation, so called \"concentration\" equation.\r\n" );
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document.write( "As soon you got it (and understand it), the setup part is completed.\r\n" );
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document.write( "To solve the equation, first multiply both sides by 600. You will get\r\n" );
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document.write( "0.3*X + 0.7*(600-X) = 0.45*600.\r\n" );
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document.write( "Simplify and solve it:\r\n" );
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document.write( "0.3X + 0.7*600 - 0.7X = 0.45*600  ====>  -0.4X = 0.45*600-0.7*600 = -0.25*600  ====>  X =  = 375.\r\n" );
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document.write( "Thus you need 375 mL of the 30% mixture and  600-375 = 225 mL of the 70% mixture.\r\n" );
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document.write( "Check.   = 0.45  ! correct concentration !\r\n" );
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document.write( "Answer.  You need 375 mL of the 30% mixture and  225 mL of the 70% mixture.\r\n" );
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\n" ); document.write( "There is entire bunch of introductory lessons covering various types of mixture problems\r
\n" ); document.write( "\n" ); document.write( "    - Mixture problems\r
\n" ); document.write( "\n" ); document.write( "    - More Mixture problems\r
\n" ); document.write( "\n" ); document.write( "    - Solving typical word problems on mixtures for solutions \r
\n" ); document.write( "\n" ); document.write( "    - Word problems on mixtures for antifreeze solutions \r
\n" ); document.write( "\n" ); document.write( "    - Word problems on mixtures for alloys \r
\n" ); document.write( "\n" ); document.write( "    - Typical word problems on mixtures from the archive\r
\n" ); document.write( "\n" ); document.write( "in this site.\r
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\n" ); document.write( "\n" ); document.write( "Read them and become an expert in solution mixture word problems.\r
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\n" ); document.write( "\n" ); document.write( "Also, you have this free of charge online textbook in ALGEBRA-I in this site\r
\n" ); document.write( "\n" ); document.write( "    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.\r
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\n" ); document.write( "\n" ); document.write( "The referred lessons are the part of this textbook in the section \"Word problems\" under the topic \"Mixture problems\".\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Save the link to this online textbook together with its description\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Free of charge online textbook in ALGEBRA-I
\n" ); document.write( "https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson\r
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\n" ); document.write( "\n" ); document.write( "to your archive and use it when it is needed.\r
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