document.write( "Question 278492: mixed 2 liters of 80% acid solution with 6 liters of 20% acid solution. What was the percent of acid in resulting mixture? \n" ); document.write( "

Algebra.Com's Answer #202685 by ptaylor(2198) $\"\"$ $\"About$

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Let x=percent of acid in resulting mixure
\n" ); document.write( "Now we know that the percent of acid in the 80% solution (0.80*2)+the percent of acid in the 20% soluton(0.20*6)=the amount of acid in the resulting mixture (x*8). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.80*2+0.20*6=8x
\n" ); document.write( "1.6+1.2=8x
\n" ); document.write( "2.8=8x
\n" ); document.write( "x=0.35 or 35%\r
\n" ); document.write( "\n" ); document.write( "Hope this helps--ptaylor \n" ); document.write( "

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