A 10% acid solution is to be mixed with a 50% acid solution in order to get 120 ounces of 20% acid solution. How many ounces of the 10% solution and 50% solution should be mixed?

To solve this problem, we can use the concept of the amount of acid in a solution. Let's denote the amount of the 10% acid solution as x ounces and the amount of the 50% acid solution as y ounces.

We can use the formula for the amount of acid in a solution:

Amount of acid = (acid concentration / 100) * volume of solution

Given that we want to end up with 120 ounces of a 20% acid solution, we can set up the following equation:

0.10x + 0.50y = 0.20 * 120

Now, let's solve this equation to find the values of x and y.

0.10x + 0.50y = 24 (because 0.20 * 120 equals 24)

To make it easier to work with, let's multiply the entire equation by 10 to remove the decimal:

1x + 5y = 240

Now, we have a system of two equations:

1x + 5y = 240 (Equation 1)
0.10x + 0.50y = 24 (Equation 2)

To solve this system, we can use either elimination or substitution method. Let's use the elimination method here:

Multiplying Equation 1 by 0.10, we get:

0.10x + 0.50y = 24

Multiplying Equation 2 by 10, we get:

1x + 5y = 240

Now, subtract Equation 1 from Equation 2 to eliminate the x variable:

(1x + 5y) - (0.10x + 0.50y) = 240 - 24

0.90x + 4.50y = 216

Simplifying the equation, we have:

0.90x + 4.50y = 216 (Equation 3)

Now, we have a new system of equations:

0.10x + 0.50y = 24 (Equation 2)
0.90x + 4.50y = 216 (Equation 3)

We can solve this system by elimination again:

Multiplying Equation 2 by 9, we get:

0.90x + 4.50y = 216

Now, subtract Equation 3 from Equation 2 to eliminate the x variable:

(0.90x + 4.50y) - (0.90x + 4.50y) = 216 - 216

0 = 0

Since the equation simplifies to 0 = 0, it means that the two equations are actually the same line. This indicates that there are infinitely many solutions to this system of equations.

In the context of this problem, it means that there are infinitely many combinations of the 10% and 50% acid solutions that can be mixed to obtain 120 ounces of a 20% acid solution.

Therefore, we cannot determine a unique solution to this problem.