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GMATPrep Quant Question, Part Two: What is this?

by Stacey Koprince on June 7th, 2012
15 comments
Stacey is a GMAT Instructor living in Montreal. Click here to read more articles from Manhattan GMAT and to learn more about Manhattan GMAT's classes.
Posted in
  • Arithmetic
  • Fractions/Ratios
  • GMAT Math
  • Integer Properties

Recently, I gave you a GMATPrep® question and started out by asking “What is this thing, anyway?” I’ve got another one (along similar lines) for you this week (also a GMATPrep problem).

By the way, I love this problem. Yes, I know I’m a complete dork. But it does such an amazing job of disguising what’s going on, and it looks deceptively simple, but then it’s hard to figure out an efficient way to tackle it. There’s so much to learn on this one – that’s why I love it.

Try it out (2 minutes!):

“Are x and y both positive?

“(1) 2x – 2y = 1

“(2) {x/y}>1”

It can’t be that hard, right? It’s just asking whether they’re positive, and the equation and inequality look pretty simple, and… well, let’s see how we do.

This is a theory question, first of all. How do we know that? Because they’re asking whether something is true, that thing is a characteristic (in this case, positive), and the information they give us is clearly not enough to determine a single value for x and y. Therefore, those statements are actually disguising other characteristics that can help us to tell whether these variables are always positive.

Second, this seems to be testing us on number properties theory. The first clue is that word “positive.” Are there others? Well, sort of but there’s something a little off. That second statement, the inequality, looks a little different than I’d normally expect for a pure positive / negative theory problem. In a pure problem, they’d tell me that x/y was greater than zero (or less than zero, or something related to zero). But this one says 1, not 0, so I’m going to need to figure out the significance of that.

(Note: at the beginning of the previous paragraph, I said that the question “seems to be” testing NP theory. I don’t want to mislead you with my language: it is testing that, but it may also be testing something else. I haven’t figured that out yet. I’m just intrigued by the 1 in that inequality. :) )

Okay, so the question itself is pretty straightforward. I do need to make sure I remember that they’re asking about both of the variables, so I write that down on my scrap paper.

Next, do I start with statement 1 or statement 2? If statement 2 said {x/y}>0, I would definitely start with that, because I already know what it means (from previous study). I decide to start with that one this time, too.

“(2) {x/y}>1”

At the least, I know that x and y have the same sign, because “greater than 1” is also telling me that it’s positive, and when you divide one number by another, they have to have the same sign in order to result in a positive number. I’ve just figured out one really important thing actually: if any future information does tell me that either x or y is positive (or negative), then I know the same is true of the other variable. I only actually need to determine the sign of one of them now.

Can I tell whether they’re both positive or both negative? Nope, not just via statement 2; either outcome is possible. Great; I can eliminate answer choices B and D.

Now I have a choice. I know there’s more to learn about statement 2 because I haven’t yet figured out why they put 1 here instead of 0. But I also know that this statement is insufficient, and I haven’t looked at statement 1 yet. If it turns out that statement 1 is sufficient, I don’t have to worry about why they put 1 here for statement 2. So I decide not to keep working on statement 2 right now, but I’m going to keep this question in the back of my mind in case I need to come back to it: What is the significance of saying >1 instead of >0?

“(1) 2x – 2y = 1”

Hmm. Just looking at that, I don’t really know what the significance is. I am a little annoyed by the fact that I don’t have “pure” variables, so I decide to divide by 2:

x – y = 1/2

And then I don’t like the subtraction sign so much:

x = y + 1/2

Okay, I can translate that into normal-person-think: x is larger than y and it’s specifically ½ larger. The two variables, then, are not both integers, and possibly neither one is.

Does that tell me anything about whether they’re positive or negative? Nope. I can think of numbers where they’re both negative, x is positive and y is negative, or they’re both positive. Okay, so statement 1 is not sufficient and now I am going to have to go back and figure out what’s going on with statement 2. Cross off answer A.

“(2) {x/y}>1”

(1) x = y + 1/2 (note: this is my manipulation of statement 1, not the original)

Okay, so I know now that x and y have the same sign, and that x is larger by ½. I can try some numbers now to try to understand what’s going on, or I can think theoretically. We’ll start here by trying numbers and use that to show what is actually going on theoretically. If you can then learn the theory, you won’t have to try numbers next time!

So let’s start with a positive pair because, hey, positive numbers are easier. Let’s try y = 2 and x = 2.5. (remember that x has to be the larger of the two). Hmm… 2.5 / 2 equals… something larger than 1. This always has to be true because 2/2 = 1, so (something larger than 2)/2 has to be larger than 1.

Okay, so in this case, the answer to our original question (are x and y both positive) is yes. But what about negative numbers – could those work too? Let’s test a pair.

Let’s try basically the same numbers: y = -2 and x = -2.5. What do we do first?

We stop. Be really careful when you flip the numbers. The x must be larger, and -2.5 is not larger than -2. We need to try something like y = -2.5 and x = -2. Hmm… -2 / -2.5 = 2 / 2.5 = 2 / (5/2) = 4/5. That’s less than 1, so this pair isn’t a valid pair to test – I’m required to pick something that makes statement 2 true (x/y > 1). But is there another pair of negative numbers that will work, a pair that does make statement 2 true?

Before, we said that 2/2 always equals 1, and that therefore (something larger than 2) / 2 would always be larger than 1. Guess what? We can reverse that, too. In this case, 2.5 / 2.5 = 1. (something smaller than 2.5) / 2.5 = something smaller than 1. Not sure? It’s annoying having a fraction on the denominator here, isn’t it? Okay, let’s try a different pair: y = -2 and x = -1.5. -2 / -2 = 2/2 = 1. -1.5 / -2 = 1.5 / 2 = less than 1. Easier to see with that example, isn’t it?

So, the only numbers that will make both statements true are positive numbers – therefore, we can definitely say that x and y are both positive using the two statements.

The correct answer is C.

Here are the fraction rules that we just deduced (assume that we know that zero is not a possibility, as statement 2 told us in this problem):

x/x = 1 (of course!)

If the numerator and denominator have the same sign, then turn both top and bottom to positive. Then:

(1) (something larger than x with the same sign) / x = greater than 1

(2) (something smaller than x with the same sign) / x = less than 1 (and greater than zero because they have the same sign!)

What about statement 2? In future, when you see {x/y}>1 , you’ll know that this is testing you on both number properties and fraction theory. This statement tells you that the two variables have the same sign. If (after turning the numerator and denominator to the positive), you’re also told that the numerator is the larger number, then x and y are both positive. The converse is also true: if we’re told the numerator is the smaller number, then x and y are both negative.

Not sure why? Test it out in the same way that we tested out the other scenario above and convince yourself. And if you forget in the moment, you can always test out a number or two to figure out the rule.

Key Takeaways:

(1) The GMAT test writers are experts at writing deceptively simple-looking and really quite hard problems. Many of these problems are testing math theory – and you can often handle these by picking numbers and testing things out. This takes time, though, so dive in right away, and lots of practice – you’ll get faster as you practice more. Note that sometimes you’ll still just have to let go.

(2) If you haven’t already, create a “When I see X, I’ll think / do Y” file or set of flashcards. Things to add:

* All quotes copyright and courtesy of the Graduate Management Admissions Council. Usage of this material does not imply endorsement by GMAC.

If you liked this article, let Stacey Koprince know by clicking Like.

RELATED ARTICLES

15 comments

  • ngufo06 on June 6th, 2012 at 11:26 pm

    Thanks for wonderful insight to how to tackle such questions.
    Though I did this wrong...wanted to add.

    From statement 2 x/y>1 we can find
    if y>0 x>y
    if y<0 x0 x>y ---0---y---x
    if y<0 x0 x>y ---0---y---x
    so C

    Reply to this comment
    • Stacey Koprince on June 7th, 2012 at 6:08 am

      Thanks! :)

      I'm not following your explanation entirely - first, there are two "x0" comments in the middle - I'm guessing there was supposed to be an inequality symbol or something like that in the middle?

      Also, in the final two statements, you start by saying "if y y" (though x would be <y if y < 0, according to statement 2), then show "---0---y---x" - is that a number line showing the placement of the numbers? You put y in between 0 and x there, but you started the statement with y < 0, so y shouldn't be placed between 0 and x. (Or is your point that y is simultaneously 0, which obviously isn't possible, so these pieces aren't valid and x and y must be positive?)

      Did something get messed up or lost in the formatting when you posted?

  • Shreeraj on June 7th, 2012 at 6:18 am

    Great post!
    As you said,practicing such problems is necessary to develop speed.Which are the topics to mainly practice for these kind of sums? I assume inequalities and algebraic equations will give the right exposure for these kind of sums. Also,when picking numbers to test,is there any standard set to be followed( I usually test for +ve nos,-ve nos, 0 and fractions)

    Reply to this comment
    • Stacey Koprince on June 8th, 2012 at 7:12 am

      I think the trickiest problems are the ones that look like their algebra or inequalities, but they actually turn out to be Number Properties in disguise - dealing with fractions, divisibility, prime, odds, evens, positive, negative, and so on.

      Re: testing cases: 0, 1, 2, 3, -1, -2, fractions between 0 and 1, fractions between 0 and -1.

      You don't want to test all of those cases all of the time, but you want to think about which ones might make the difference on one particular problem. The above range hits 0, integers, positives and negatives, primes, odds and evens, and fractions.

  • niksworth on June 7th, 2012 at 6:25 am

    Can we not solve this algebraically?

    It is easy to discern that either of the statements are not sufficient by themselves.

    Combining them,
    x=y+1/2-------(1)
    x/y>1-----------(2)

    Substituting (1) in (2)
    (y+1/2)/y>1
    or, 1+1/(2y)>1
    or, 1/(2y)>0

    This is only possible when y>0. Thus combining the two statements we can conclude that y>0, and if y>0, this implies x>0.

    Thus both statements are together sufficient. (C)

    P.S. Algebraically, I thought it was pretty straightforward. Did I miss something?

    Reply to this comment
    • karthikchandru on June 7th, 2012 at 3:04 pm

      Yup. I found it straightforward too, algebraically.

  • dhonu121 on June 7th, 2012 at 8:50 am

    I really like the way you boil to each stuff to very basics..However, the questions that you take up are not really the very difficult ones that we see on GMAT..
    As a suggestion, why don't you sometimes pick you really hard problems and illustrate them the same way as you do here.

    Thanks.

    Reply to this comment
    • Stacey Koprince on June 8th, 2012 at 7:15 am

      Go look through my article archive. I've written nearly 150 articles for BTG now and many of them cover VERY difficult problems.

      I purposely select a range of difficulty levels in order to accommodate a wide variety of students / student levels.

    • dhonu121 on June 8th, 2012 at 7:36 am

      Stacey,
      Can you please direct me to your archive ?

    • Stacey Koprince on June 8th, 2012 at 7:51 am

      All MGMAT content on BTG is archived by date here:
      http://www.beatthegmat.com/mba/manhattan-gmat

      There isn't a page that shows only my stuff. But there's a lot of other great stuff in there to check out, too, including Ron's videos!

  • Karthik on June 7th, 2012 at 2:08 pm

    Nice article and it was really easy understand the concept

    Reply to this comment
  • rs on June 10th, 2012 at 7:56 am

    i dont quite get this.....

    when x&y is positive ....x/y>1
    when x&y are negative x/y<1

    so isnt b sufficient here.

    Reply to this comment
    • Stacey Koprince on June 10th, 2012 at 8:10 am

      I'm going back up trying to find where you got the two lines that you typed, but I don't see them, so I'm not following your question 100%. Are you asserting that when x and y are negative, then x/y < 1 in general? That's not necessarily true - the only thing you know when x and y are both negative is that x/y is greater than zero (because dividing a negative by a negative produces a positive). That positive number could be less than 1 but it could also be greater than 1.

      Also, possibly this is your source of confusion: on a yes/no question, an "always yes" answer is sufficient, but so is an "always no" answer. If a question says "is Alice 13 years old" and a statement tells me she's 20, then that statement is sufficient because the answer "definitely no" is a sufficient ansewr.

  • rs on June 13th, 2012 at 8:31 pm

    I have another question :
     when you say x/y>1;
    dosent it mean x>y
    OR -X > -Y
    did we need A to determine x>y,
    in other words why is the answer not b

    Reply to this comment
    • Stacey Koprince on June 14th, 2012 at 8:36 am

      when you multiply or divide an inequality by a negative number, the sign flips. If you multiply or divide an inequality by an unknown (for which you don't know the sign), then two outcomes are possible:
      x/y > 1 --> x > y IF y is positive
      x/y > 1 --> x < y IF x is negative

      Try all of this out with real numbers to see how it works - then it's easier to remember the "abstract" forms with variables.

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