document.write( "Question 384774: The chemistry lab at the University of Hardwoods keeps two acidic solutions on hand. One is 20% acid and the other is 35% acid. How much 20% acid solution and how much 35% acid solution should be used to prepare 25 liters of a 26% acid solution? \n" ); document.write( "

Algebra.Com's Answer #272377 by stanbon(75887) $\"\"$ $\"About$

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One is 20% acid and the other is 35% acid. How much 20% acid solution and how much 35% acid solution should be used to prepare 25 liters of a 26% acid solution?
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\n" ); document.write( "Quantity Eq: x + y = 25 liters
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\n" ); document.write( "Acid Eq::0.20x + 0.35y = 0.26*25
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\n" ); document.write( "Multiply the Quantity Eq. by 20
\n" ); document.write( "Multiply the Acid Eq. by 100
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\n" ); document.write( "20x + 20y = 20*25
\n" ); document.write( "20x + 35y = 26*25
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\n" ); document.write( "Subtract and solve for \"y\":
\n" ); document.write( "15y = 6*25
\n" ); document.write( "y = 2*5 = 10 liters (amt. of 35% solution needed in the mixture)
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\n" ); document.write( "Siince x+y = 25, x = 15 liters (amt. of 20% solution needed in the mixture)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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