document.write( "Question 61082This question is from textbook Intermediate Algebra
\n" ); document.write( ": E-chem Testing has a solution that is 80% base and another that is 30% base. A technician needs 150L of a solution that is 62% base. The 150L will be prepared by mixing the two solutions on hand. How much of each should be used? \n" ); document.write( "

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Suppose that 'x' liters of 80% base solution is taken.\r
\n" ); document.write( "\n" ); document.write( "Then quantity of 30% base solution = (150 - x)liters [as the total is 150L)\r
\n" ); document.write( "\n" ); document.write( "Strength of 150L solution = 62%\r
\n" ); document.write( "\n" ); document.write( "The equation becomes,
\n" ); document.write( "x*0.8 + (150-x)*0.3 = 150*0.62
\n" ); document.write( "==> 0.8x + 45 - 0.3x = 93
\n" ); document.write( "==> 0.5x + 45 = 93
\n" ); document.write( "==> 0.5x = 93 - 45 [adding -45 to both the sides]
\n" ); document.write( "==> 0.5x = 48
\n" ); document.write( "==> 0.5x/0.5 = 48/0.5
\n" ); document.write( "==> x = 96\r
\n" ); document.write( "\n" ); document.write( "so 150-x = 150-96
\n" ); document.write( " = 54\r
\n" ); document.write( "\n" ); document.write( "Thus 96L of 80% base is to be added to 54L of 30% base to get 150L of 62% base. \n" ); document.write( "

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