# Analyzing a 700+ Data Sufficiency Question

As a full-time GMAT geek, the question of what exactly constitutes a difficult GMAT question occupies more of my time than I’d like to admit. This curiosity was only reinforced two weeks ago, when I took my annual official GMAT. Much to my relief, I was able to again score a perfect 51 on the Quant (and, for anyone who’s curious, I have the score report on the desk at my office), so I’m pretty sure the questions I saw were representative of the type of reasoning the GMAT test-makers expect. Let me first start off by saying the following:

I saw zero questions testing combinatorics, zero questions testing probability, and zero questions testing work and rate. This was not surprising. Despite the fear that these topics strike in most test-takers’ hearts, they’re some of the least-frequently-tested topics and, as such, do not merit the same level of attention as other topics on the exam. You might be thinking that the absence of these topics would make the test decidedly easier, but, at least in this tutor’s humble opinion, that’s not the case at all. The GMAT test-makers want to test your abstract and deductive reasoning skills, and those Advanced Word Problems, though difficult, don’t test these skills in the same way that seemingly simple topics might. What topics, you ask?

– Number Properties

– Exponents

– Inequalities

– Absolute Values

Especially in Data Sufficiency. Now I’m not saying that the test focused exclusively on these topics, nor am I saying that these are the only topics tested in Data Sufficiency, but if you want to get a high score on the test, you will HAVE to master these topics. I’ve had students get 47 – 49 on the GMAT despite having substantial difficulty with combinatorics, but all of these students knew the aforementioned topics inside-out, and they understood how these topics were tested on the exam.

To understand how the GMAT will tweak simple number theory concepts, let’s look at the following question step-by-step:

If *m* and *n* are positive integers, is *m* + *n *odd?

*1) **m = p ^{2} *+

*4p*+

*4*

*2) **n *=* p ^{2} *+

*6p*+

*9*

First, let’s understand what the prompt is *really* asking. This is a skill that many people haven’t fully-honed for the GMAT, and, without it, they often find themselves in difficult spots. The question is asking whether the sum of these two values is odd. What has to be true for a sum to be odd? The numbers have to be different with respect to odd/even. One must be even and one must be odd. So that’s our question: Are *m* and *n* different with respect to even-ness and odd-ness. In number theory terminology, we can view this question as asking: Do *m *and *n *have different polarities?

Now that we know what we’re trying to determine, let’s dive into the statements:

The statements don’t look fun. We have quadratics, which can often be frustrating. So, before anything else, let’s see what answer choices we can definitely eliminate. We’re looking for the relationship between *m *and *n*. Since each individual statement addresses only one of those variables, we know that neither statement alone is sufficient. The answer must be C or E.

Now let’s look at what we can extract from each statement. Statement 1 gives us an annoying-looking quadratic. Hmm.. when we see a quadratic, what should be going through our minds? Factoring works. But before we go through that process, let’s think about special products. *p ^{2} *+

*4p*+

*4*is actually one of our special products and can be written in the form: (

*p*+ 2)

^{2 }

So what we really know from statement 1 is that *m *= (*p *+ 2)^{2 }

So how does this help us? Since we’re dealing with an odd/even question, let’s interpret the above information through the lens of the properties of odds and evens. If two numbers are equal, then they have the same polarity. Therefore, *m* and (*p *+ 2)^{2} are both odd or both even. Let’s now work with (*p *+ 2)^{2 }. The exponent implies that (*p *+ 2)^{2} is really a product. If the base of this exponential expression is odd, then the product will be odd, and if the base is even, the product is even. So the exponent doesn’t even matter! If (*p *+ 2)^{2 }is odd/even, then the same will hold true for (*p *+ 2). So, we can interpret this information to tell us that the polarity of *m* is the same as the polarity of (*p + 2*) — in other words, if *m *is odd, then *p + *2 will be odd, and if *m *is even, then *p *+ 2 will be even. Continuing with this approach, we also know that if you add 2 to an even number, the result is even, and if you add 2 to an odd number, the result is odd. Therefore, *p* will have the same polarity as *p *+ *2*. Okay, so let’s put it all together. What we now know is that *m* has the same polarity as *p *+ 2, which has the same polarity as *p*. Therefore,** m and p have the same polarity**.

Now let’s look at what we can take away from Statement 2. We again have a quadratic, and this quadratic can also be factored into one of our special squares: (*p *+ 3)^{2}

So we know that *n *= (*p *+ 3)^{2}

We can now use the same reasoning we used for Statement 1. Do you see the similarities? As we deduced in Statement 1, we can deduce that *n *has the same polarity as (*p *+ 3). We can further deduce that *p *and (*p *+ 3) have DIFFERENT polarities. We know this because adding an odd to any number results in a sum whose polarity will be the opposite of the original value. For example, take 2 (even) and add 3 (odd). The result is 5 (odd), which has a polarity different from that of 2 (even). Take 5 (odd) and add 3 (odd). The result is 8 (even), which has a different polarity from the original value 5. So, we know that *n *has the same polarity as *p *+ 3, and that *p *+ 3 has a different polarity from *p*. Therefore, *n *and *p *have DIFFERENT POLARITIES.

Whew. That’s a lot. Now let’s combine the statements. We know that *m *and *p *have the same polarity and that *n *and *p* have different polarities. Therefore, we know that ** n and m have different polarities. **In other words, if

*n*is even, then

*m*is odd, and if

*n*is odd, then

*m*is even. When you add an odd and an even, the result is always odd. So, we have now

*proved*that

*m*+

*n*is odd. The answer is C.

If you thought that this involved a lot of steps, you’d be correct. I’m in no way disputing that it requires a very strict and methodical form of thought and analysis. But that’s what the GMAT is testing! The test-makers care about your ability to synthesize and analyze information, and one key demonstration of that ability is using deductive reasoning to analyze abstract situations. Most people’s initial inclination would be to plug in numbers here. I’m not saying that won’t work, but plugging in numbers can be tedious, and, importantly, plugging-in numbers is prone to error. You have to try every possible situation, and if you miss something, you’ll miss the question. Getting a 700+ on the GMAT isn’t about shortcuts or memorizing rules. It’s about being efficient with your thought process. It might be difficult now, but the more you practice the proper way of thinking, the better off you’ll be come test day.

[…] Analyzing a 700+ Data Sufficiency Question Getting a 700+ on the GMAT isn’t about shortcuts or memorizing rules. It’s about being efficient […]

I had a different approach to this question.

1. m = p^2 + 4p + 4

– 4p + 4 is always even

– p^2 is either odd or even

-> m is either odd or even

2. n = p^2 + 6p + 9

– 6p + 9 is always odd

– p^2 is either odd or even

-> n is either odd or even

m + n = 2.p^2 + 12p + 13

– 2.p^2 + 12p is always even

-> m + n is always odd

-> C is the answer

I too had a different approach although am not sure if this always works or if it just happened to be correct in this situation.

In order for m and n to add to be odd, one would have to be even the other odd (logic explained in original answer).

The difference between m (p^2+4p+4) and n (p^2+6p+9)

is 2p+5…For any integer value of p (1,2,3,etc…) that value of 2p+5 between M and N will always be odd.

There is no way that the difference between 2 odd numbers or 2 even numbers will be odd, so we must be dealing with one odd and one even.

Once we know that one is odd and one is even, C is our answer.

Yes, John! That approach definitely works and has a lot of merit in this situation. To see how my approach can be a bit more useful, check out the OG 12th edition, P. 288, # 172.