The mixture problem involves mixing different concentrations of input solutions in order to obtain a desired concentration of an output solution.
Think in terms of:
For example, you may have one gallon of a 25% solution of sugar in water. We do not care about the water. It is not important. We do care about the sugar. Sugar is important. We have 25% of a gallon of sugar, which is one quart of sugar. We have one quart of sugar. Period. End of story. You can hold it in your hands. You can visualize it. You can see it. One quart of sugar.
Let's say that you want to combine a 10% solution of sugar in water and a 30% solution of sugar in water to yield a 25% solution of sugar in water.
| Define | x = number of gallons of the 10% solution |
| y = number of gallons of the 30% solution |
| Therefore | the amount of STUFF you got is: | (0.10)·x + (0.30)·y |
| And | the amount of STUFF you want is: | (0.25)·(x + y) |
If you mix together the STUFF you got to become the STUFF you want, then:
| (0.10)·x + (0.30)·y = (0.25)·(x + y) | (Equation 1) |
Pause a bit to savor the moment, to let the meaning of Equation 1 set in.
If we multiply both sides of Equation 1 by one hundred, we obtain:
| 10·x + 30·y = 25·x + 25·y | (Equation 2) |
Rearranging Equation 2, we obtain:
| 5·y = 15·x | (Equation 3) |
Simplifying Equation 3, we obtain:
| y = 3·x | (Equation 4) |
Note that we have not said anything about the total volume (e.g., in gallons) of our final solution. Maybe we might not need to. Maybe all we are seeking is the ratio of one input solution to another. In this case, for every gallon of the 10% solution (represented by x) we will need 3 gallons of the 30% solution (represented by y), because y is three times as large as x (Equation 4).
If, however, we want to have a final volume of 12 gallons, then we can write:
| x + y = 12 | (Equation 5) |
Substituting 3·x for y from Equation 4 into Equation 5, we obtain:
| x + 3·x = 12 | (Equation 6) |
Or:
| 4·x = 12 | (Equation 7) |
Or:
| x = 3 | (Equation 8) |
If x = 3 and y = 3·x, then:
| y = 9 | (Equation 9) |
Therefore, if you mix 3 gallons of a 10% solution of sugar in water with 9 gallons of a 30% solution of sugar in water, then you will get 12 gallons of a 25% solution of sugar in water.
Now, focusing on the important STUFF, which is the sugar and not the water, we will get 0.3 gallons of pure sugar from the first solution (3 gallons of 10% solution), plus 2.7 gallons of pure sugar from the second solution (9 gallons of 30% solution), which add up to 3.0 gallons of pure sugar. The total volume of liquid is 12 gallons (3 + 9), which means that we have 3 gallons or pure sugar in 12 gallons of liquid, or a 25% solution of sugar in water.