## GMAT Studying: Correct Answers Can Be a Bridge to Success.

For about a year, I always used the same method to solve the following GMAT problem:

How many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 5 percent sugar solution?

“This is simple percentages,” I would say. “Just start by taking 3% of 50 liters, which is 3 over 100 times 50, which comes out to 1.5 liters sugar…”

But one day, teaching this same quantitative problem, a student’s hand shot straight up. “Yes, James?” I said. (That wasn’t his real name, by the way, but it will do.)

“Eli, who cares about the sugar?”

I paused. “Well, the sugar will help us figure out the solution.”

“But you don’t need it!” James explained. “I’ve been a chemical engineer for years, so I do this problem all the time. The sugar is a constant. The amount of sugar doesn’t change, and that amount is always equal to the concentration times the volume. So just do CV = CV; 50 times 3 is equal to the final volume times 5!”

I paused, impressed, and amazed—and have taught his timesaving shortcut ever since.

However, there is a bigger lesson here than simple mixture problems.  I had approached that problem uncritically.  I “knew” how to find the right answer, so I never gave it a second thought.  I spent far more time prepping the combination and probability problems given their complexities and hidden challenges.  As a result of my complacency, I made extra work for myself.

## GMAT Permutations: Putting Things in Order

I want you to think back to when you were in grammar school (strange how often the test for business school has us thinking about grammar school).  Specifically, I want you to think about gym class and every time you needed to pick teams.  For purposes of today’s discussion, let’s say we are picking a baseball team from the students in class.  Now I realize that for some of you this was a traumatic experience, and GMAT teachers are no exception to that – we are not particularly notable for our athletic prowess.  However, this scenario can help you understand a difficult question type on the GMAT – permutation problems.

A permutation problem will ask you to determine the number of ordered subsets of a certain size that exist in a group.  If we move away from GMAT speak, this means permutations will occur when you must perform two actions.  First, you will be creating groups.  In our example of a baseball team in a gym class, the class may have 30 students, but you need to select nine.  Second, you will put the selected entities in an order.  On our baseball team, you can think of this as the batting order.  If A is first, B is second, and C is third you have a different outcome than if C is first, B is second, and A is third.  However, A, B, and C may not even be on the team – you will have lots of other outcomes in which they are left out.

Thus, we need to both select nine players to be on the team AND put those players in an order.  On GMAT test day, when we must perform both of these actions, we will have a permutations problem.  The formula to determine the number of permutations is n!/(n – k)!, where n is the number of total entities, in our example 30, and k is the number of entities we want to select, in our example nine.

See how you do on this next problem, and ask questions if you have them…

Problem:

Kim has four trophies which she wishes to display in a cabinet with five shelves, with only one trophy to a shelf.  How many different ways are there to arrange the trophies?

(A) 4

(B) 5

(C) 20

(D) 24

(E) 120

Solution:

This is an arrangement question. Often you can simply count the different possible arrangements, making sure to do so systematically, but it is usually easier to remember that the number of ways of arranging x objects is x !

There are too many arrangements here to count by simply writing them out, so you will have to use factorials to count the arrangements.

In this case you have 4 objects, but 5 spaces. So you have five possibilities for each shelf-it could contain one of the four trophies, or it could be empty. So the number of possible arrangements is 5! = 120. The answer is (E)