SOLUTION: One acetic acid solution is 70% water and another is 30% water. How many liters of each solution should be mixed to produce 20 liters of a solution that is 60% water? Solve using s

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Question 1204800: One acetic acid solution is 70% water and another is 30% water. How many liters of each solution should be mixed to produce 20 liters of a solution that is 60% water? Solve using system of equations
Found 2 solutions by josgarithmetic, mananth:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Viewpoint is usually solute. Your question is only about solvent. So for focusing on the percents water:

v of the 70%water
20-v of the 30% water
Want 60% water


70v%2B30%2820-v%29=60%2A20
-
70v-30v=60%2A20-30%2A20
%2870-30%29v=20%2860-30%29
highlight_green%28v=20%28%2860-30%29%2F%2870-30%29%29%29
.
.

You could use a system of equations if you really want, but obviously something like v+x=20, whatever you want to choose for variables...


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SYSTEM of two equations
v of the 70%
x of the 30%
v%2Bx=20

The "pure water", 70v%2B30x=60%2A20
or simplifying,
7v%2B3x=60%2A2
7v%2B3x=120

and a system can be system%287v%2B3x=120%2Cv%2Bx=20%29.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
One acetic acid solution is 70% water and another is 30% water. How many liters of each solution should be mixed to produce 20 liters of a solution that is 60% water? Solve using system of equations
One acetic acid solution is 70% water let the quantity be x liters
another is 30% water Let quantity be y liters
We need 20 liters of acid in final mix
x+y=20-----------------------(1)
water percentage equation. 60% water in 20 liters= 12 liters
0.7x+0.3y= 12-----------------------(2)
multiply by 10
7x+3y= 120
3x+3y=60 (1) multiplied by 3. subtract
4x=60
x=15
Therefore y=5
Check
15+5=20
0.7*15+5*0.3= 12
15 liters of the 70% water acid and 5 liters of the 30% water acid are added