Last printed 9/20/2007 1:40:00 PM

Mixture Problem – Explanation
There is a difference between percent and amount. Percent is a rate. Amount is a quantity. If I told you I drove 60 mph for 3 hours and asked how far I had driven, you would not respond 60 mph (a rate) but rather you would state the quantity180 miles. It is important that you make the distinction between rate and amount, particularly throughout the process of solving a mixture problem. When analyzing mixture problems it is essential to convert to and discuss quantities (amounts) not rates. When analyzing a mixture problem, as in many other problems, a winning strategy is to find some quantity (amount) in the problem which can be expressed in two different ways thus arriving at an equation which can be solved. Consider the following mixture problem. Problem: What quantity (amount) of a 60% acid solution must be mixed with a 30% acid solution to produce 300 mL of a 50% acid solution? Analysis: In this analysis (an integral part of the solution process) I will state the steps in logical order. Beneath each step, in small blue indented type, I will state the reason/logic/explanation for that step. (1) Let x be the amount of the 60% solution to be added.
Begin by assigning a variable to the quantity (not the rate) of material you are to determine.

(2) The amount of the final solution is 300 mL.
The problem statement dictates that the final quantity be 300 mL.

(3) The amount of acid in the final solution is (0.5)(300)
The problem statement dictates that the final solution be half acid (a 50% solution). Observe that (0.5)(300) is one way of describing the amount of acid in the final solution. Our strategy is to find another way of describing the amount of acid in the final solution. Those two ways of describing the amount of acid in the final solution will produce an equation. That equation is the