SOLUTION: A lab needs to make 100 gallons of an 18% acid solution by mixing a 12% acid solution with a 20% solution. How many gallons of each solution are needed?

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A lab needs to make 100 gallons of an 18% acid solution by mixing a 12% acid solution with a 20% solution. How many gallons of each solution are needed?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1023988: A lab needs to make 100 gallons of an 18% acid solution by mixing a 12% acid
solution with a 20% solution. How many gallons of each solution are needed?
Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A lab needs to make 100 gallons of an 18% acid solution by mixing a 12% acid
solution with a 20% solution. How many gallons of each solution are needed?
-----------------------
Quantity:: t + w = 100 gallons
Acid Eq:: 0.12t + 0.20w = 0.18*100
----------------------------------------
Modify for elimination::
12t + 12w = 12*100
12t + 20w = 18*100
-------------
Subtract and solve for "w"::
8w = 6*100
-----
w = 75 gallons (amt. of 20% solution needed)
------------
Solve for "t"::
t = 100 - 75 = 25 gallons (amt. of 12% solution needed)
-----------------
Cheers,
Stan H.
---------------

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
v gallons of the 20% acid and 100-v gallons of the 12% acid. Concentration wanted is 18%.

%2812%28100-v%29%2B20v%29%2F100=18


--
some steps....
12%28100-v%29%2B20v=18%2A100
6%28100-v%29%2B10v=9%2A100
3%28100-v%29%2B5v=9%2A50
300-3v%2B5v=9%2A50
keep going until solved.....