How to use extreme way of thinking to tackle impossible GMAT questions 

Consider the following non- GMAT example:

Two identical containers are each half filled with identical amounts of wine. One container is filled with red wine and the other with white wine. Using a third container, a certain amount of red wine is transferred from the red wine container to the white wine container, and mixed. Using the same container, exactly the same amount of the white wine and the red mix is transferred back to the red wine container. Is there more red wine in the white wine container, or white wine in the red wine container?

Most of my students said that there is more red wine in the white wine container. Their explanation was that when the red wine was transferred, the third container contained only red wine, but on the way back it contained a mix of red and white. Their intuition when encountering this situation for the first time led them to a certain conclusion. When asked to prove their answer, most students started to plug in numbers just to find out it is not as simple as they thought. After experimenting with numbers for a few minutes, almost all students returned to their initial conclusion.

A good method that can be used to solve such a question is to use extreme situations. Since the size of the container used to transfer the wine was not defined in the question, it is possible to assume that any size we choose has to give the same result. Let’s begin by assuming it has a volume of zero. Hence, no red wine was transferred to the white container and no wine was returned. This makes the amounts equal. (No red in the white and no white in the red). We can also assume the size of the third container was larger than each of the other two containers, so all the red wine was transferred to the white. After mixing, the amounts of the red and the white are equal and they remain so when transferring the same amount back (since the concentration remains 50%). Again, the amounts are equal.

By looking at extreme situations (0 and 100%) it was possible to get to the right answer. Try this method on questions where a definition can fit many situations and plugging numbers does not help.


Another example:

In which of the following ranges is the sum of all the numbers in the following set: 1/101, 1/102, 1/103, 1/104….1/200? 
     a. 0-1/3
     b. 1/3 -1/2
     c. 1/2 -1
     d. 1-2
     e. 2-10

In this question, the method of using extreme situations can be employed to find the range. In the set there are 100 numbers. If all the integers in the set were equal to the largest number in the set (1/101), the sum of the set was 100/101, which is a little less than 1. If all the numbers in the set were equal to the least number in the set (1/200), the sum of the set was 100/200 or 1/2. This way, the range is found. The sum of the set has to be less than 1 and more than 1/2. Hence, the best answer is 1/2-1.

Another example: 
 


A point inside a rectangle (O) is linked to the vertices using lines AO, BO, CO and DO. What is the value of [(AO)² + (BO)²] – [(CO)² + (DO)²]?


At first glance, this seems to be a no–solution question. Then, using the Pythagorean Theorem, it is possible to express AO, BO, CO and DO in similar terms. This takes a few minutes to complete. 
 



However, since point O can be any point inside the rectangle, you can choose it so it will be easy for you to solve.
You can choose point O to be the meeting point of the diagonals, such that AO = BO = CO = DO.
Now, [(AO)²+(BO)²]=[(CO)²+(DO)²] and [(AO)²+(BO)²]–[(CO)²+(DO)²]=0