Two types of mixture problems often appear on the GMAT. Because these types of problems fall into two very specific categories, we can simply learn the correct strategy for each, and then apply it on test day.
The first type of problem will have a mixture of two components and will ask you to alter the make-up of the mixture by adding or subtracting one of the components. For example, a problem could tell you that a 50-ounce mixture of sugar and water is made up of 40% sugar and 60% water, and then ask how much sugar you need to add so that the mixture is 60% sugar and 40% water. Always remember: the key to these problems is the component that does not change. In this case, we are adding sugar, while water remains constant. Therefore, we will focus on the water for most of the problem. First determine how much water is in the mixture; 60% of 50 is 30 ounces. Because we are not adding or subtracting any water, we will still have 30 ounces of it after we add more sugar. However, we now want that 30 ounces to represent 40% of the total, rather than 60%. 30 is 40% of 75 ounces (note: if you struggled to find that, you could set up the algebraic equation 30 = .4x, and solve for x which gives you 75). Since the increase in total volume is only made up of additional sugar, and we went from 50 total ounces to 75 total ounces, we must add 25 ounces of sugar, which would be the answer to the question.
The second type of mixture problem will involve combining two mixtures. For example, we could be told that mixture X is 10% acid and that mixture Y is 30% acid. We could then be asked, if, when mixture X and mixture Y are combined, our new mixture is 15% acid, what percent of the new mixture is mixture X? To solve this problem, set up an equation using the percents you are given. Let’s use ‘X’ to represent the total amount of mixture X, and ‘Y’ to represent the total amount of mixture Y. The amount of acid in mixture X is then .1X and the amount of acid in mixture Y is .3Y. If we add .1X and .3Y, we end up with our total amount of acid, which is 15% of the total mixture. Therefore, we can say .1X + .3Y = .15(X+Y). We solve this equation in the following way:
.1X + .3Y = .15(X+Y)
.1X + .3Y = .15X + .15Y
.15Y = .05X
15Y = 5x
3Y = X
X/Y = 3/1
This means that for every 3 parts of mixture X we have 1 part of mixture Y. The ratio of mixture X to the total is 3 to 4, so mixture X makes up 75% of the total.