document.write( "Question 1166872: Forty liters of a 60% salt solution are reduced to a 45% solution. How much solution must be drained off and replaced with distilled water so that the resulting solution contains only 45% solution. \n" ); document.write( "

Algebra.Com's Answer #791441 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "We can overlook the fact that the people who write these problems don't know the science that says a 60% salt solution is not possible. The problem is intended to give the student practice solving mixture problems.

\n" ); document.write( "For a formal algebraic solution, we are mixing x liters of a 0% solution (distilled water) and (40-x) liters of a 60% solution to get 40 liters of a 45% solution:
\n" ); document.write( " $\"0%28x%29%2B.60%2840-x%29+=+.45%2840%29\"$

\n" ); document.write( "Solve using basic algebra....

\n" ); document.write( "If a formal algebraic solution is not required, here is a quick and easy way to solve any 2-part mixture problem like this.

\n" ); document.write( "(1) You are starting with a 60% solution and adding a 0% solution, so you are heading towards a 0% solution; you are stopping when you get to a 45% solution.
\n" ); document.write( "(2) 45% is 1/4 of the way from 60% to 0%. (Picture the numbers on a number line if it helps....)
\n" ); document.write( "(3) That means 1/4 of the 40 liter total is the 0% solution you are adding.

\n" ); document.write( "ANSWER: Drain off 1/4 of the 40 liters, or 10 liters, and replace it with distilled water.

\n" ); document.write( "CHECK:
\n" ); document.write( ".60(30) + 0(10) = 18
\n" ); document.write( ".45(40) = 18

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