We want to see how much pure water and 80% salt solution should be mixed to get 20 ounces of a 50% salt solution, we will solve this by using a system of equations. We will see that we need to use 7.5 ounces of pure water and 12.5 ounces of the 80% salt solution.
How to get the system of equations?
We should start by defining the variables we will be using, these are:
- x = ounces of pure water used
- y = ounces of the 80% solution used.
We know that we want to end with 20 ounces of the mixture, then we have:
x + y = 20
We also know that the percentage of salt in the final mixture must be 50%, which means that a 0.5 of the final mixture is salt, then we have:
0*x + 0.8*y = 0.5*20
(the first 0 is because we have no salt in the pure water, and the 0.8 is because we have 80% of salt in the 80% mixture)
So we got the two equations needed, thus the system of equations is:
x + y = 20
0*x + 0.8*y = 0.5*20
With the second one we can get the value of y:
0*x + 0.8*y = 0.5*20
0.8*y = 0.5*20 = 10
y = 10/0.8 = 12.5
This means that we must use 12.5 ounces of the 80% solution, and we need to use:
x = 20 - y = 20 - 12.5 = 7.5
7.5 ounces of pure water.
If you want to learn more about systems of equations, you can read:
https://brainly.com/question/13729904
