document.write( "Question 804046: Chemist has 10% solution and 60%solution; how many liters of each solution does the chemist need to make 200 liters of a solution that is 50%acid? \n" ); document.write( "

Algebra.Com's Answer #550577 by AlgebraLady88(44) $\"\"$ $\"About$

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In this question, we can see that we have two types of solutions here: the 10 % solution and the 60% solution and that altogether, we need 200 liters of a solution that is 50 % acid.\r
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\n" ); document.write( "\n" ); document.write( "x stands for the 10 % solution and y stands for the 60 % solution. We need 200 liters of the solution. Hence:\r
\n" ); document.write( "\n" ); document.write( "x + y = 200 (a)\r
\n" ); document.write( "\n" ); document.write( "The next equation would be what is needed to make 200 liters that is 50% acid.
\n" ); document.write( "Hence:\r
\n" ); document.write( "\n" ); document.write( "0.10x+0.60y = 0.50 * 200
\n" ); document.write( "0.10x+0.60y = 100
\n" ); document.write( "Multiply by 10 to get rid of decimals: x+ 6y = 1000 (b)\r
\n" ); document.write( "\n" ); document.write( "Multiply (a) by -1 and then we can add equations (a) and (b)\r
\n" ); document.write( "\n" ); document.write( "-x-y = -200 (a)
\n" ); document.write( "x+6y = 1000 (b)\r
\n" ); document.write( "\n" ); document.write( "5y = 800
\n" ); document.write( " y = 160\r
\n" ); document.write( "\n" ); document.write( "This will give us x+y = 200
\n" ); document.write( " x = 200- 160
\n" ); document.write( " x = 40\r
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