# percent mixture problem

• Sep 20th 2011, 03:37 PM
rainbowx
percent mixture problem
"How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze?"

I am having trouble figuring out how to set up the problem.
If I do .6(30+x) = .7(30) + 1x the answer is -7.5 gallons.
How do I set up this equation?
• Sep 20th 2011, 03:59 PM
Soroban
Re: percent mixture problem
Hello, rainbowx!

Quote:

How much water should be added to 30 gallons of a solution that is 70% antifreeze
in order to get a mixture that is 60% antifreeze?

Since water is added to the solution, we must think in terms of water.

The original solution is 30 gallons which is 30% water.
. . It contains: $0.30 \times 30 \,=\,9$ gallons of water.

Then $x$ gallons of water is added.

The final mixture contains: $x + 9$ gallons of water. .[1]

But we know that the final mixture will be $x+30$ gallons which is 40% water.
. . It contains: $0.40(x+30)$ gallons of water. .[2]

We just described the final amount of water in two ways.

There is our equation! . . . $x+9 \;=\;0.4(x+30)$

Got it?

• Sep 20th 2011, 05:57 PM
bjhopper
Re: percent mixture problem
another way to solve this.
The conc line starts at 0% AF goes to 60 and then to 70%. Therefor 60 =parts 70% and 10 =parts water so answer for 30 gals 70% is 5 gals water