The SBC GMAT Files
Mixture Questions
Pavel, Staff, Test Prep New York/Test Prep San Francisco
Ratios, proportions, and percent calculations are one of the most frequently tested math concepts on the GMAT. But GMAT questions rarely ask to find a ratio of two explicitly stated numbers or to find a percentage of a given number. Instead, you can often come across real life problems that can be solved using these concepts, and an example of such can be seen in mixture problems.
Even though mixture problems can seem daunting, they only require you to apply your knowledge of percentages, ratios and proportions. One of the most important skills you need to approach these problems is the ability to convert between absolute amounts, proportions, and percentages. Let’s work together through a relatively easy example to see how it works.
300 ml of a solution or vinegar and water contains 33.3% of vinegar. If 100 ml of water are added to the solution, what percent of the new solution will be vinegar?
Let’s start by calculating the absolute amount of vinegar in the initial solution. 300 ml results in 100ml of the initial solution being calculated as vinegar. After 100 ml of water had been added, the total volume of the new solution became 300+100 = 400 ml. Since no vinegar was added, there is still 100 ml of vinegar in this solution. Now we need to convert our absolute amounts back to percentages, or more precisely, we need to calculate what percentage of 400 is 100. If you have a natural feel for numbers, percentages, and fractions, you should be able to say right away that 100 is 25% of 400, but if you are not able to do these fast mental conversions, you can still calculate it manually. You can also work through this via a chart, similar to how you’d work through a ratio question:
Ratio  Multiply  Whole  
Vinegar  1/3  X 300  100  
Water  2/3  X 300  200  + 100 
Total  1  X 300  300 ml  + 100 
New total 400 ml
100 (vinegar)/400 (total) = ¼ = 25%
Another closely related subtype of such problems deals with costs instead of concentrations. These often require the ability to convert word questions into algebraic equations. Let’s go through another example:
How many liters of paint, which costs $4 per liter, need to be mixed with 20 liters of paint, which costs $8 per liter, to create a mixture of paint that costs $5 per liter?

10

25

60

90

100
To answer this question we will need to set up two expressions for the final cost of the mixture. As these two expressions will represent the same amount (the final cost), we will be able to designate them as equal, and form an equation.
Let’s let x stand for the number of liters of $4 paint that is required to prepare the mixture. Then we can say that the final volume of mixture will be x + 20 liters. As this mixture will cost $5 per liter the total cost (number of liters Ã— price per liter) will be 5(x+20)””this is our first expression.
The second way to express the final cost is to express it in terms of the cost of the cheaper paint plus the cost of the more expensive paint that will be used. We will use x liters of $4 paint, which will cost 4x, and 20 liters of $8 paint, which will cost 20 Ã— 8 = 160 dollars. So the total cost of the final mixture will be 4x+160””this is our second expression.
Now, all that we need is to set these two expressions for the final cost equal and solve for x:
5(x+20) = 4x + 160
5x + 100 = 4x + 160
x = 60
The correct answer is C.
Another way of answering this question is to use the answer choices themselves. Called Backsolving, or Answer Substitution, this may streamline your efforts.
How many liters of paint, which costs $4 per liter, need to be mixed with 20 liters of paint, which costs $8 per liter, to create a mixture of paint that costs $5 per liter?

10

25

60

90
 100
Starting with (C ), since it’s usually the middle value, plug it right into the questions where it identifies that this is the answer it’s seeking:
Does (60) liters of paint, which costs $4 per liter, need to be mixed with 20 liters of pain,t which costs $8 per liter, to create a mixture of paint that costs $5 per liter?
Then do the math:
(60) (4) + (20) (8) = (80)(5) ?
240 + 160 = 400? The answer is yes, so the answer is, indeed, C.
If you had a number that did not work, you’d go to a larger or smaller answer choicenumber depending on what direction you needed to go. For example, if we had started with 25, choice B, instead of 60, this would be the equation:
Does (25) liters of paint, which costs $4 per liter, need to be mixed with 20 liters of paint, that costs $8 per liter, to create a mixture of paint that costs $5 per liter?
(25) (4) + (20) (8) = (80)(5) ?
100 + 160 = 400?
The answer is no, and in fact it’s much less than what we’re looking for,
therefore we need a number larger than 25.
Use common sense when you’re plugging in or using this substitution method – – sometimes you can tell when certain answer choices are way too big or small, so think about always using the middle number, even if you’ve eliminated 1 or 2 answers. With plugging in, you want to make sure you pick exceptions, like 0, 1, fractions or decimals, or really large numbers.
On the GMAT, especially if you are doing very well, you may see a harder question that asks you something like this:
If paint which costs $4 per liter were mixed with 20 liters of paint which costs $8 per liter to create a mixture of paint that costs $5 per liter, what percent of the new mixture would be constituted by the cheaper paint?
This sounds almost unreal if you cannot break the question into smaller steps. So how would you proceed if you did get a question like this? First, determine what you need to get your answer. To calculate a percentage you need the part and the whole. The part in this case is the volume of the cheaper paint; the whole is the total volume of the final solution. So you proceed just as we did in the previous question to calculate that you will need 60 liters of the cheap paint, and add 20 liters of the expensive paint to get the final total volume of 80 liters. The last step is to determine what percentage of 80 is 60. You can spot that it’s ¾ or 75%, or if you cannot spot this, you can calculate it using the general percentage formula: or in our case
Feared by many, mixture problems are not as complex as they appear. If you can convert between fractions, decimals, ratios, and percentages, if you can divide complex problems into easier steps, and if you know how to form linear equations from word problems””you will do just fine. If, however, you lack any of these skills, then simply invest some time to obtain them!
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