SOLUTION: How many ounces of pure water must be added to 80oz. of a 35% salt solution to make a 14% salt solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How many ounces of pure water must be added to 80oz. of a 35% salt solution to make a 14% salt solution?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 94305: How many ounces of pure water must be added to 80oz. of a 35% salt solution to make a 14% salt solution?
Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How many ounces of pure water must be added to 80oz. of a 35% salt solution to make a 14% salt solution?
-----------
35% solution DATA;
Amt = 80 oz ; amt of salt = 0.35*80 = 28 oz
-------------
Pure water DATA;
Amt = x oz ; amt of salt = 0
------------------
14% solution DATA:
Amt = 80+x oz ; amt of salt = 0.14(80+x) = 11.2 + 0.14x oz
==================
EQUATION:
salt + salt = salt in mixture
28 + 0 = 11.2 + 0.14x
0.14x = 16.8
x = 120 oz (amount of pure water that must be added)
===========
Cheers,
Stan H.

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
oz*%=oz*%
80*.35=x*.14
divide both sides by .14 %2880%2A.35%29%2F.14=200oz
200-80=120oz to be added
Ed Jones