document.write( "Question 913624: Solution A is 10 percent salt, by weight, and solution B is 20 percent salt, by weight. how many grams of solution B must be added to 200 grams of solution A to make a new solution that is percent salt, by weight? \n" ); document.write( "

Algebra.Com's Answer #554622 by josgarithmetic(39618) $\"\"$ $\"About$

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You omitted or do not yet know the target percent of the mixture, so call this T percent salt, by weight.\r
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\n" ); document.write( "\n" ); document.write( "100(salt/mixture)=T percent\r
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\n" ); document.write( "\n" ); document.write( " $\"highlight%28100%2A%28200%2A%280.10%29%2Bv%2A%280.20%29%29%2F%28v%2B200%29=T%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "The number, T, is expected to be given, but you want to solve for v, assigned as the unknown MASS IN GRAMS of solution B to add.\r
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\n" ); document.write( "\n" ); document.write( "Study this lesson: http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev \n" ); document.write( "

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