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If x and y are positive, is 4x > 3y? (1) x > y - x (2) x/y < 1

Given: \(x>0\) and \(y>0\). Question: is \(4x>3y\)?

(1) \(x>y-x\) --> \(2x>y\) --> \(4x>2y\), not sufficient to say whether \(4x\) is more than \(3y\).

(2) \(\frac{x}{y}<1\) --> as \(y>0\) we can cross multiply --> \(x<y\) --> \(4x<4y\), not sufficient to say whether \(4x\) is more than \(3y\).

(1)+(2) We know that \(4x\) is more than \(2y\) but less than \(4y\) --> \(2y<4x<4y\), but from this we can not find the relationship between \(4x\) and \(3y\) (we have \(2y---3y---4y\), \(4x\) is somewhere between endpoints but we can not say whether it's between \(2y\) and \(3y\), equal to \(3y\) or between \(3y\) and \(4y\)). Not sufficient.

If x and y are positive, is 4x > 3y? (1) x > y - x (2) x/y < 1

from the stem: \(x/y>3/4\)? 1) \(2x>y\) \(x/y > 1/2\) not sufficient as \(x/y\) could be greater or smaller than \(3/4\) 2) \(x/y < 1\) not sufficient for the same reason both: \(1/2 < x/y < 1\) still not sufficient for the same reason

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14 Dec 2010, 19:48

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ajit257 wrote:

if x and y are positive is 4x > 3y ?

(1) x > y - x (2) x/y < 1

Can someone provide good pointers into handling such question. I also have a doubt about the ans. thanks

Since x and y are positive, you can multiply/divide the inequality by them. Now, another thing is that it is hard to wrap your head around a question if it has more than one variables. So try to get the variables together so that you can consider them a single variable. is 4x > 3y ? I can change this to: is x/y > 3/4?

Stmnt 1: x > y - x 2x > y x/y > 1/2 So we know that x/y is greater than 1/2 but we do not know whether it is greater than 3/4. Not sufficient.

Stmnt 2: x/y < 1 So we know that x/y is less than 1 but we still do not know whether x/y is greater than or less than 3/4. Not sufficient.

Using both statements, we know x/y is greater than 1/2 and less than 1 but we still do not know whether it is greater than or less than 3/4. Not sufficient.

Re: if x and y are positive is 4x > 3y [#permalink]

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17 Dec 2010, 09:11

Karishma, Thanks! I do have a small doubt though...when should you solve such questions using algebraic approach and when should you take the intuitive approach. I always get caught up in this dilemma. _________________

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17 Dec 2010, 19:36

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ajit257 wrote:

Karishma, Thanks! I do have a small doubt though...when should you solve such questions using algebraic approach and when should you take the intuitive approach. I always get caught up in this dilemma.

First of all, one method is not better than the other. They both have their pros and cons. I like logical approach for most questions since it helps in finding the answer very quickly. On the other hand, using the algebraic approach, you don't need to think very much. Just make equations and solve. Also the probability that you will overlook something is less using algebraic approach. Ideally, you need to be comfortable with both. Different questions demand a different approach but during the exam, pick the one which comes first to mind and with which you are comfortable. For inequality questions, I prefer to use algebra since using logic too takes time.. It isn't as intuitive as in the case of arithmetic... Numbers between 0 and 1 behave in a different manner and those between 1 and infinity behave in a different manner. Similarly negative number follow their own rules too.... So if I can see simple straight forward algebra, I use it for inequalities. Else you have to use logic, plug in number etc.... _________________

Re: if x and y are positive is 4x > 3y [#permalink]

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28 Dec 2010, 09:53

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animeshsen wrote:

Hi, Can anyone plz explain the above question using graphical method.

Thanks Animesh

Look at the diagrams below.

Attachment:

Ques5.jpg [ 43.99 KiB | Viewed 5213 times ]

The first one shows the region asked in the question stem: is 4x > 3y? or is 4x - 3y > 0? (The Green region in the first diagram. If you have a doubt on how the regions were drawn, let me know. I will guide you to the related blog.) The second and third show the statements 1 and 2 respectively 1. x > y - x or 2x - y > 0 (shown by the Red region) 2. x/y < 1 or x - y < 0 (shown by the Blue region) Since Red and Blue regions have points that are not in Green region, a point satisfying 2x - y > 0 or x - y < 0 doesn't necessarily satisfy 4x - 3y > 0. So each alone is not sufficient. Look at the last diagram now. Using both 1 and 2 together, we get the points that satisfy both inequalities: 2x - y > 0 and x - y < 0. (Region where Red and Blue overlap) Even here, some points will lie in Green region, some will not. Hence, even together, the statements are not enough. Answer (E).

I intend to take this question in detail on the Veritas Blog. If interested, go to http://www.veritasprep.com/blog/ and search for "Quarter Wit Quarter Wisdom" in the upcoming days. _________________

Re: if x and y are positive is 4x > 3y [#permalink]

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29 Dec 2010, 00:33

Thanks Karishma, The explanation is really nice and what exactly I wanted.... Can you please tell me more about solving inequalities using graphs(some links..etc)

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29 Dec 2010, 20:18

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animeshsen wrote:

Thanks Karishma, The explanation is really nice and what exactly I wanted.... Can you please tell me more about solving inequalities using graphs(some links..etc)

In addition, try solving questions using graphs and if you get stuck, pm me... Also, I think shrouded1 here uses graphs frequently too so you can check some of his/her solutions as well.. _________________

Re: if x and y are positive is 4x > 3y [#permalink]

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29 Dec 2010, 20:25

I'm finding that the best way to get better at these problems is to see heaps and heaps of problems

I still get caught up with silly mistakes - I've really gotta start getting into some sample questions to start getting a feel for all of the different types of questions, and develop a logical system to approach all questions. _________________

Re: if x and y are positive is 4x > 3y ? (1) x > y - x (2) [#permalink]

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21 May 2014, 20:24

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Re: If x and y are positive, is 4x > 3y? [#permalink]

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10 Feb 2016, 22:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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