SOLUTION: how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 103174This question is from textbook Elementary and intermediate algebra
: how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid? This question is from textbook Elementary and intermediate algebra

Found 2 solutions by checkley75, stanbon:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.05x+.10*30=.08(30+x)
.05x+3=2.4+.08x
.05x-.08x=2.4-3
-.03x=-.6
x=-.6/-.03
x=20 gallons of 5% acid will be needed.
proof
.05*20+3=.08*50
1+3=4
4=4

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
how many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?
---------------
10% liquid DATA:
Amt = 30 gals ; amt of active ingredient = 0.10*30 = 3 gallons
-----------------
5% liquid DATA:
Amt = x gals ; amt of active = 0.05x gallons
-----------------
Mixture DATA:
Amt = 30+x gallons ; amt of active = 0.08(30+x) = 2.4 + 0.08x gallons
---------------
EQUATION:
active + active = active
3 + 0.05x = 2.4 + 0.08x
0.6 = 0.03x
x = 0.6/0.03 = 20 gallons ( amt of 5% acid solution that must be added)
===============
Cheers,
Stan H.