SOLUTION: Solve algebraically. How many liters of a 20% salt solution must be added to 60 liters of 40% solution to obtain a solution that is 35% salt?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Solve algebraically. How many liters of a 20% salt solution must be added to 60 liters of 40% solution to obtain a solution that is 35% salt?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1048300: Solve algebraically. How many liters of a 20% salt solution must be added to 60 liters of 40% solution to obtain a solution that is 35% salt?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = liters of the 20% solution needed
+.2x+ = liters of salt in the 20% solution
-------------------
+%28+.2x+%2B+.4%2A60+%29+%2F+%28+x+%2B+60+%29+=+.35+
+%28+.2x+%2B+24+%29+%2F+%28+x+%2B+60+%29+=+.35+
+.2x+%2B+24+=+.35%2A%28+x+%2B+60+%29+
+.2x+%2B+24+=+.35x+%2B+21+
+.15x+=+3+
+x+=+20+
20 liters of the 20% solution are needed
--------------
check:
+%28+.2x+%2B+.4%2A60+%29+%2F+%28+x+%2B+60+%29+=+.35+
+%28+.2%2A20+%2B+.4%2A60+%29+%2F+%28+20+%2B+60+%29+=+.35+
+%28+4+%2B+24+%29+%2F+80+=+.35+
+28+=+.35%2A80+
+28+=+28+
OK