How many gallons of distilled water must be mixed with 50 gallons of a 30% alchohol solution to get a 25% solution?

A 30% solution (I assume we are talking about volume/volume percent here) is 30 gallons alcohol + 70 gallons water OR

[30 gallons alcohol/(30 gal alcohol+70gal water)]*100. 50 gallons of a 30% solution would contain 15 gal alcohol + 50 gal solution (or 15 gal alcohol + 35 gal water).
[(15 gal alcohol)/(15 gal alc+35 gal water+ x gal water)]*100.
Solve for x gal water. Check my thinking. Check my work.

To solve this problem, we can use the concept of the mixture equation. Let's denote the amount of distilled water to be added as 'x' gallons.

Step 1: Write the equation representing the amount of alcohol in the mixture:
0.30 * 50 = 0.25 * (50 + x)

Step 2: Solve the equation for 'x':
15 = 12.5 + 0.25x
0.25x = 15 - 12.5
0.25x = 2.5

Step 3: Divide both sides of the equation by 0.25:
x = 2.5 / 0.25
x = 10

Therefore, 10 gallons of distilled water must be mixed with 50 gallons of a 30% alcohol solution to obtain a 25% alcohol solution.

To determine the number of gallons of distilled water needed to achieve a 25% alcohol solution, we can follow a simple equation:

Let "x" represent the number of gallons of distilled water.

Quantity of alcohol in the 30% alcohol solution = Quantity of alcohol in the final 25% alcohol solution.

Using this information, we can set up the equation as follows:

0.30 * 50 = 0.25 * (50 + x)

Now, let's solve for "x":

15 = 0.25 * (50 + x)

Divide both sides of the equation by 0.25:

15 / 0.25 = 50 + x

60 = 50 + x

Subtract 50 from both sides of the equation:

60 - 50 = x

10 = x

Therefore, you will need to mix 10 gallons of distilled water with the 50 gallons of the 30% alcohol solution to obtain a final solution with a 25% alcohol concentration.