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: $\displaystyle 0.30 \times 30 \,=\,9$ gallons of water.

Then $\displaystyle x$ gallons of water is added.

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

But we know that the final mixture will be $\displaystyle x+30$ gallons which is 40% water.

. . It contains: $\displaystyle 0.40(x+30)$ gallons of water. .[2]

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

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

Got it?

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