SOLUTION: What quantity of a 80% acid solution must be mixed with a 40% solution to produce 600 mL of a 50% solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: What quantity of a 80% acid solution must be mixed with a 40% solution to produce 600 mL of a 50% solution?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 878683: What quantity of a 80% acid solution must be mixed with a 40% solution to produce 600 mL of a 50% solution?
Found 2 solutions by josgarithmetic, harpazo:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Typical two-part mixture percents concentration problem. Learn how to solve them in this lesson:
Mixing two concentrations for known amount of mixture

H = 80
L = 40
T = 50
M = 600
u = volume of 40%
v = volume of 80%

%28Lu%2BHv%29%2FM=T and u%2Bv=M

Answer by harpazo(655) About Me  (Show Source):
You can put this solution on YOUR website!

Let x be what we are looking for.
0.80x + 0.40x = 0.50(600)
1.20x = 300
1.20x = 300
x = 250
Now that we know x, multiply 0.80 by 250 to find the quantity of an 80% acid solution.
So, 250 * 0.80 = 200
The answer is 200.
If we want to know the quantity for the 40% acid solution, multiply 250 by 0.40.
So, 250 * 0.40 = 100.
Here is the Prove
Let x = 250 in the equation above.
0.80(250) + 0.40(250) = 300
200 + 100 = 300
300 = 300
Did you follow? The most important thing to master when dealing with word problems is setting up the right equation leading to the answer. This comes with practice.