document.write( "Question 665032: pure acid is to be added to a 10%acid solution to obtain 90l of 89% solution . what amounts of each should be used? \n" ); document.write( "

Algebra.Com's Answer #413616 by vleith(2983) $\"\"$ $\"About$

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Let the amount of pure acid be given by x. Then if the total volume is 90 liters, the amount of 10% acid is given by (90-x). Plug that in and solve
\n" ); document.write( "100%x + 10%(90-x) = 89%(90)
\n" ); document.write( " $\"x+%2B+0.1%2890-x%29+=+0.89+%2A+90\"$
\n" ); document.write( " $\"x+%2B+9+-+0.1x+=+80.1\"$
\n" ); document.write( " $\"0.9x+=+71.1\"$
\n" ); document.write( " $\"x+=+79\"$ so there must be 11 liters of 10% to start with.
\n" ); document.write( "And remember .... add acid to water, just like you oughta (and not the other way around) \n" ); document.write( "

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