document.write( "Question 63246: How many grams of salt must be added to 50 grams of a 20% solution in order to increase the content to 25%? \n" ); document.write( "

Algebra.Com's Answer #44013 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of salt to be added\r
\n" ); document.write( "\n" ); document.write( "We know that the amount of salt added (x) plus the amount of salt in the original 50 gms (50)(.20) must equal the amount of salt in the final solution (50+x)(.25). Thus, our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "x+(50)(.20)=(50+x)(.25) simplifying, we get:
\n" ); document.write( "x+10=12.5+.25x or
\n" ); document.write( ".75x=2.5
\n" ); document.write( "x=3.34 gms of salt\r
\n" ); document.write( "\n" ); document.write( "A slightly easier way to solve this problem would be to realize that the amount of the \"otherstuff\" in the solution does not change. Thus, the amount of \"otherstuff\" in the original 50 gms (50)(.80) equals the amount of \"otherstuff\" in the final solution (50+x)(.75). So our equation would be:
\n" ); document.write( "(50)(.80)=(50+x)(.75) simplifying, we get:
\n" ); document.write( "40=37.5+.75x or
\n" ); document.write( ".75x=2.5
\n" ); document.write( "x=3.34 gms of salt \r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );