How many gallons of 20% alcohol solution and 50% alcohol solution must be mixed to get 9 gallons of 30% alcohol solution

The problem asks you to find how many gallons of 20% solution and 50% solution, respectively, are needed to create a 12 gallon solution that is 30% alcohol.

To solve this, set up the following equation using the given values: that you have some quantity of a 20% solution, a quantity of a 50% solution, and that you need to have 12 gallons of a 30% solution. We will use a variable, x, together with the "goal" quantity of 12 gallons to solve this problem with only one unknown, even though we really have two unknowns (the quantity of 20% solution and that of the 50% solution).

Remember that when writing out percentages, a 20% solution is also a 0.2 solution, and a 50% solution is a 0.5 solution.

Let x = an unknown quantity of the 20% (or 0.2) solution. We therefore have 0.2x, since we have x gallons of the 20% solution. (Remember that "of" is a key word that denotes multiplication).

Therefore, 12 - x is the other unknown and is multiplied by the 0.5 (50%) solution, since we have this quantity of the 50% solution. Really, we could say that we have 0.5y (if we want to introduce another variable; however, it is much simpler to put the other quantity in terms of x, which is easy to do, since we know that the two unknown quantities must amount to 12 gallons). Now, we must multiply our second unknown, 12 - x, by 0.5. This gives us 0.5 (12 - x).

Our two unknown quantities, 0.2x and 0.5 (12 - x) add up to 12 gallons of a 30% solution. Putting all the words together to form an equation we get:

0.2x + 0.5 (12 - x) = 0.3 (12)

Now we expand the equation (call it distributing out the parentheses).

0.2x + 6 + 0.5x = 3.6

Combine like terms:

-0.3x + 6 = 3.6

Solve for x:

-0.3x = - 2.4

x = 8

Plug x = 8 back into the equation to check that this is possible.

x = 8, and 12 - x = 4

0.2 (8) + 0.5 (12 - (8)) = 0.3 (12)

0.2 (8) + 0.5 (4) = 0.3 (12)

3.6 = 3.6

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