document.write( "Question 252630: how many liters of water must be added to 10L of 50% acid solution to produce 80% acid solution? \n" ); document.write( "
Algebra.Com's Answer #184662 by drk(1908)\"\" \"About 
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This is a mixture problem. The thing to remember is that water is 0% everything. Here is the table:
\n" ); document.write( "liquid . . . . % . . . . Liter . . . . %Liter
\n" ); document.write( "acid . . . . . 50 . . . . 10 . . . . . 500
\n" ); document.write( "water . . . . 0 . . . . . L . . . . . 0
\n" ); document.write( "mixture . . . 80 . . . . 10 + L . . . 800 + 80L
\n" ); document.write( "The liters must be added to get the total amount of liquid.
\n" ); document.write( "Now,
\n" ); document.write( "500 + 0 = 800 + 80L
\n" ); document.write( "This is not possible. You can't add water and make it stronger. If it were 80% and you wanted a mixture of 50%, then we are good. Suppose that is what you meant. The new table looks like this:
\n" ); document.write( "liquid . . . . % . . . . Liter . . . . %Liter
\n" ); document.write( "acid . . . . . 80 . . . . 10 . . . . . 800
\n" ); document.write( "water . . . . 0 . . . . . L . . . . . 0
\n" ); document.write( "mixture . . . 50 . . . . 10 + L . . . 500 + 50L
\n" ); document.write( "Now,
\n" ); document.write( "800 + 0 = 500 + 50L
\n" ); document.write( "300 = 50L
\n" ); document.write( "L = 6. You would need 6 liters of water to dilute the mixture from 80 to 50%.
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