Avoiding Traps in GMAT Quant Questions

Question

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A common mantra at Veritas Prep is that the GMAT is a test of how you think, not of what you know. This shouldn’t be interpreted to mean that you can go into the exam without knowing anything and expect to get a good score. Rather, it means that how you apply concepts is crucial in this exam. You need to have a strong base, like the foundation of a house, but the difficulty is in using the information you have to solve the problem in front of you.

As can be expected, different quantitative questions will pertain to different mathematical notions. However some more advanced questions will begin to blur the lines (#BlurredLines) between multiple concepts. A question can ask you to solve an equation using variables from a given shape, incorporating geometry, algebra and even arithmetic concepts in one fell swoop. It’s important to note that all these seemingly disparate topics you’re studying while preparing for the GMAT can be combined into one question. These questions tend to be more difficult, but mostly because they require more steps, and therefore more opportunities to make mistakes.

The mathematical concepts don’t have to be any harder on these questions; the simple fact of merging them into a Frankenstein’s monster question can make the problem harder than the sum of its parts. (The question wants you to use your BRAINS). Add to this the time pressure of having to solve such questions in roughly two minutes, and you can imagine how longer questions combining various elements can frustrate even the most experienced student.

Let’s review a question and examine the various pitfalls we can fall into:

If you select two cards from a pile of cards numbered 1 to 10, what is the probability that the sum of the numbers is less than the average of the pile?

(A) 1/100

(B) 2/45

(C) 2/25

(D) 4/45

(E) 1/10

The first hurdle here is interpreting the question. To paraphrase, if I were to choose two random cards, would their sum be less than a certain other number. This is essentially a probability question, as evidenced by the answer choices as fractions. However there are a couple of elements to keep in mind. The first task is to determine the average of the pile.

Given 10 numbers, we could simply sum them up and divide by 10, but it’s probably much faster to recognize that the mean of an evenly spaced set is equal to the median of the set. A set with 10 numbers has a median that’s the average of the 5th and 6th elements (Not the Bruce Willis movie). Conveniently, the 5th element is 5 and the 6th element is 6, yielding an average of 5.5. Since we’re dealing with integers, we must now determine the number of possibilities that give a sum of 5 or less.

The options are limited enough that we can just reason out the choices. A good strategy is just to assume that the first card is a 1, and figure out what numbers work for the second number. If we pick 1, the next smallest card is 2. Thus the possibility (1,2) works. Similarly, we can see that (1,3) and (1,4) will work. (1,5) is too big, so we can stop there as any other option would only be bigger than this benchmark. It’s worth noting that the question is set up so that there’s no repetition, thus the option (1,1) cannot be considered. If the first card picked is a 1, there are three options that will keep the average below 5.5 (like a Russian judge at the Winter Olympics).

Next, supposing that the first card were a 2, there would be the separate option of (2,1). Since the order matters, (2,1) is not the same as the aforementioned (1,2). This is another valid choice. (2,2) is eliminated because of duplication, leaving us only with (2,3) that will also work if the first card is a 2. Since (2,4) is too big, we don’t need to examine any further. That’s two more options to add to our running tally.

Continuing, if the first card were a 3, then (3,1) and (3,2) would work. (3,3) is above the average, and it is a duplicate, so it can be eliminated for either reason. That gives us two more options for our running tally. The final option is to start with a 4, giving (4,1). Anything bigger is above the average. Similarly, anything starting with 5, 6, 7, 8, 9 or 10 will be above the average. Only eight options work out of all the possibilities.

The question is almost over, but there is one final trap we need to avoid before locking in our answer. The stimulus purported 10 different cards to select. If we were to compile all the possibilities, a natural total to think of would be 100 (10×10). However, since there is no replacement, we’re first selecting from 10 choices, and then from 9 choices. Exactly as a permutation of two selections out of 10, this gives us a total of 90 possible choices. If there are eight options that satisfy the conditions out of 90 choices, then the correct answer must be 8/90, which simplifies to 4/45. Answer choice D.

Examining the answer choices, we can see some of the more obvious traps. Compiling eight options out of 100 choices would give us the erroneous 2/25 fraction in answer choice C. Overlooking the lack of replacement would give us 10 total choices (the same eight plus (1,1) and (2,2) out of 100 possibilities, or answer choice E. The exam is designed to ask tricky questions, which means that the answer choices will often be answers you can get if you make a single calculation error or unfounded assumption. Be vigilant until the end of the question, as you don’t want to spend a full two minutes on a complicated question just to falter at the finish line. Questions can have many aspects to consider and many steps to execute, but by continuously thinking in a logical manner, you can solve any GMAT question. Remember that even the longest journey begins with a single step.

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Ron Awad  is a GMAT instructor for Veritas Prep based in  Montreal , bringing you  weekly advice  for success on your exam.  After graduating from McGill and receiving his MBA from Concordia, Ron started teaching GMAT prep and his Veritas Prep students have given him rave reviews ever since.

teja8341 · Answer

Hi  Bunuel

I approached the question in a different way and still got the same answer.
While the explanation considers Permutations, I considered Combinations of numbers.

Selecting 2 numbers from 10 = 10C2 = 45
Picking two numbers such that the sum is less than 5.5 = 
(1,2) or (2,1)
(1,3) or (3,1)
(1,4) or (4,1)
(2,3) or (3,2)

Total 4 ways.
So, the probability is 4/45.

But in the explanation, it is mentioned that " Since the order matters "
Is there a flaw in my approach?
Please help me understand this.

SuyashA · Answer

Bunuel ,

I also solved the problem in this way. Can you please help us understand?

Thanks.

Bunuel · Answer

You can find discussion of this question here: https://gmatclub.com/forum/if-you-selec ... 96042.html Hope it helps.

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