How many gallons of a 12% indicator solution must be mixed with a 20% indicator solution to get 10 gal of a 14% solution?
.12 x + .2(10-x) = .14(10)
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To find out how many gallons of a 12% indicator solution must be mixed with a 20% indicator solution to get 10 gallons of a 14% solution, we can set up a simple equation based on the principles of mixing solutions.
Let's say x represents the number of gallons of the 12% solution to be mixed.
So, the amount of the 20% indicator solution needed would be (10 - x) gallons, as the total volume is 10 gallons.
To determine the concentration of the final solution, we calculate the weighted average of the two concentrations using the formula:
(concentration of solution 1 * volume of solution 1 + concentration of solution 2 * volume of solution 2) / total volume of the mixture.
Using the given information, we can set up the equation:
(0.12 * x + 0.20 * (10 - x)) / 10 = 0.14.
Simplifying the equation, we get:
0.12x + 2 - 0.20x = 1.4.
Combining like terms:
-0.08x = -0.6.
Dividing both sides by -0.08:
x = -0.6 / -0.08.
x = 7.5.
Therefore, you would need to mix 7.5 gallons of the 12% indicator solution with 2.5 gallons of the 20% indicator solution to obtain 10 gallons of a 14% indicator solution.